ANALISA STRUKTUR LANJUTAN.pdf

Prof DrIngJohannes Tarigan Semester B

Analisa Struktur Lanjutan 1

DEPARTEMEN TEKNIK SIPIL

FT USU

ANALISA STRUKTUR

LANJUTAN

PROF DR-ING JOHANNES TARIGAN



Bab 1 Pendahuluan

11 Pengenalan Torsi Dalam analisa struktur selain Momen Gaya Lintang dan Normal maka Torsi akan menjadi salah satu yang menentukan dalam disain struktur bangunan Dalam buku ini akan dibahas khusus hanya Torsi saja Torsi pasti akan terjadi pada konstruksi portal tiga dimensi atau pada konstruksi grid Untuk itu dalam buku ini akan dipaparkan bagaimana perletakan Torsi Bidang torsi sudut puntir akibat torsi dan tegangan torsi Tegangan torsi secara umum dibagi 3 yakni sbb

1 Tampang tebal seperti tampang Lingkaran persegi segitiga

2 Tampang tipis terbuka seperti profil I WF canal dll

3 Tampang tipis tertutup seperti tampang hollow box dll Dalam buku ini akan dibahas tentang tegangan torsi untuk ketiga jenis tampang ini 12 Sistem Koordinat Dalam perhitungan di buku ini system koordinat searah sumbu batang secara umum dinamakan sumbu Z sedangkan kearah lainnya adalah sumbu X dan Y Sedangkan untuk perpindahan (displacement) searah sumbu X adalah u dan searah sumbu Y Z adalah v w

X

Y Z

u

v w υ ψ

φ



Untuk perputaran sudut dengan sumbu putar x y dan z adalah φ ψ dan υ Khusus torsi putaran sudut yang diakibatkan Torsi disebut juga sudut puntir (twist) υ 13 Perletakan Torsi

Pada jenis perletakan tanpa torsi dikenal dengan rol lihat gambar 1 dimana 0=ΔY yang berarti pada perletakan tidak diperbolehkan bergerak kearah sb y sedangkan kesumbu x boleh

Y

0

Gambar 1 perletakan rol

Z

Kemudian perletakan selanjutnya adalah sendi yang dapat dilihat digambar 2 dimana 0=ΔX dan 0=ΔZ yang berarti pada perletakan tidak diperbolehkan bergerak ke

sumbu x dan sb y

Y

0

Gambar 2 perletakan sendi

Z

Y

Gambar 3 perletakan jepit

Z

Pada gambar 3 perletakan jepit berlaku 0=ΔX 0=ΔZ dan 0=ϕ yang berarti pada perletakan tidak diperbolehkan bergerak kearah sb x dan sb y demikian juga perputaran sudut pada perletakan sama dengan nol

Khusus pada torsi maka diadakan simbol perletakan seperti pada gambar 4 yang mana pada perletakan jeit torsi ataupun sudut puntir 0=υ dan gambar 5 adalah perletakan yang bebas Torsi



Y

Gambar 4 perletakan jepit pada torsi

Z

Y

Gambar 5 Perletakan bebas pada torsi

Z

I3 Penggambaran bidang Torsi

Momen torsi terpusat Mt dapat dibuat dengan simbol seperti pada gambar 6 yakni Momen Torsi dengan dua tanda panah dapat dibuat dengan seperti 1 tanda panah dengan rotasi 90 derajat dengan menambah 1 garis ditengah tanda panah tersebut Sedangkan untuk momen Torsi terbagi rata mt dapat dibuat seperti gambar 7

Mt

Mt

L

Gambar 6 Torsi terpusat Gambar 7 Torsi terbagi rata

Mt

Dalam penggambaran bidang torsi dapat dilakukan sama seperti menggambarkan gaya lintang seperti pada gambar 8 a b dan c

`

- -

+

MT MT

MT

L frac12 L frac12 L

a b

c



`

Penggambaran tanda bidang momen sama seperti menutup dan membuka skrup Kalau arah Momen Torsi kearah menutup maka digambarkan negatif dan kalau kearah membuka maka digambar positif

I4 Analogi antara Torsi dengan Normal

Pada Tabel 1 dapat dilihat analogi antara Torsi dan Normal seperti pada Hukum Hooke yang mana regangan adalah

EFN

=ε

dimana regangan sangat tergantung kepada Normal Sedangkan regangan geser adalah

T

T

GJM

=Γ

dan regangan geser sangat tergantung kepada Torsi

mT

A B

Gambar 8 a Bidang Torsi Terpusat pada overhang b Bidang Torsi Terpusat pada balok diatas 2 perletakan dan c Bidang Torsi terbagi rata pada balok

+

-



Normal Torsi

Elemen dengan Gaya

N n N+dN Mt mt Mt +dMt

Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν

Deformasi pada batang dw = cdzEFN

+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν

Tabel 1 Analogi antara Normal dan Torsi

dz dz

dz dzdw



BAB II TORSI PADA TAMPANG BULAT

II1 Umum

Suatu tampang bulat jika mengalami Torsi permukaan tampang tidak berubah bentuk seperti gambar dibawah Tidak berubah bentuk dalam arti kata bahwa tampang mempunyai luas dan bentuk yang sama baik sebelum dan sesudah terjadi Torsi Ada istilah yang digunakan dalam Torsi yakni tidak terjadi Warping (perubahan bentuk pada tampang)

II2 Menghitung Ј Inertia Polar Tegangan geser τ dan sudut puntir ν akibat

Momen Torsi Untuk menghitung Inertia Polar Ј (centroidal polar momen of Inertia) dapat dilihat ilustrasi dari gambar II1 dimana ada Momen Torsi bekerja sebesar Mt Maka berlaku persamaan sbb Dimana dv = τ dA dv = Tegangan Torsi seluas da dA = Luas τ = tegangan geser ρ jarak dari pusat lingkaran ke titik tertentu r radius Gambar II1 penampang bulat mengalami Torsi

dA

r

ρ

MT = intρ dv Mt

Gambar II1 Torsi pada tampang bulat



Maka MT = intρ τ dA

Untuk mendapat hubungan regangan geser ( Γ ) dan sudut puntir (ν) maka dapat

dilihat digambar II 2 Maka berlaku persamaan

ΓmaxL = r ν Γ r = ρ Γmax Γ = Γmax Gambar II 2 hubungan antara sudut puntir υ dengan regangan geser Г Dari Hukum Hooke berlaku persamaan

τ = G Γ = G Γmax rρ

MT = int ρG maxΓ rρ dA

MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=

Ј adalah Inertia Polar untuk tampang bulat Ј= int ρ 2 dA = intx2 dA + inty2 dA= Ix + Iy Ј = Ix + Iy Ј =

L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r

Hanya untuk tampang bulat (lingkaran)

1 4 πr4 + 1

4 πr4

υ



Ј = Sedangkan tegangan geser didapat

Γ max = sedangkan τmax = τmax = G Γmax

= G maka

dari persamaan diatas didapat sudut puntir (angle of twist)

ν = Kesimpulan - Tampang lingkaran Ј = frac12 πr4

τmax =

ν =

Gambar tegangan geser akibat momen torsi dapat dilihat di gambar II3

1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT

Gambar II 3 adiagram tegangan pada tampang bulat b Trayektori tegangan dan c Tampang yang tidak mengalami warping

τmax = MT

Ј r

a b c



Inertia Torsi tampang Ring )(32

44 dDJT minus=π

Contoh soal

Ditanya

a Tentukan Bidang Torsi pada AB

b Tentukan sudut puntir pada titik B jika batang AB adalah tampang bulat

dengan r = 15 cm dan E=26000 MNm2 υ=02

c Hitunglah lendutan pada titik C

d Hitunglah tegangan yang terjadi akibat torsi pada batang AB kemudian

hitunglah tegangan maximum akibat Torsi dan Gaya Lintang

Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2

Maka =υ 0006968 radial π

υ2

36000696800

= = 03990

c

bull Lendutan pada titik C jika pada B di jepit EI

PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm

bull Lendutan pada titik c jika dilepas EI

PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm

bull Lendutan akibat adanya Torsi

υδ tgL 13 = = 1500006968=10452 cm

Dengan demikian lendutan dititik c adalah 321 δδδδ ++= = 17792 cm



e Tegangan Torsi rJ

M tZYZX == ττ = 15

25794813000000 =56617 Ncm2

217566 cmNZX =τ

Akibat Gaya Lintang ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N

Maka tegangan geser ( )angLtorsi ZYZYZY int)( τττ plusmn= =603915 2cmN

217566 cmNZY =τ

274537cmN

ZY =τ



BAB III

Persamaan pada Torsi

III1 Fungsi Torsi

Fungsi torsi gunanya adalah untuk dapat menghitung tegangan pada tampang tebal

tampang tipis tertutup dan terbuka Untuk menentukan fungsi torsi digunakan

persamaan keseimbangan pada elemen tiga dimensi seperti gambar III1 τzx = τxz

τzy = τyz

τyx = τxy

Dari gambar III1 diseberangnya secara keseluruhan terdapat pada sumbu x ada

tegangan σx + δσx τxz+δτxz dan τxy+δτxy demikian pada seberang arah sumbu y ada σy

+ δσy τyz+δτyz dan τyx+δτyx dan pada arah sumbu z ada σz + δσz τzy+δτzy dan τzx + δτzy

Dengan merubah tegangan menjadi gaya dimana gaya adalah tegangan dikali luas

maka dengan membuat persamaan keseimbangan gaya kearah X Y dan Z maka

diperoleh sbb

ΣX = 0

( ) ( ) 0)( =minusminus++minus++minus+ zyxXzxxyzy XYYXYXZXZXZXXXX δδδδδτδττδδτδττδδσδσσ

Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz

Gambar III1keseimbangan tegangan pada sebuah elemen

δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0

( ) ( ) 0)( =minusminus++minus++minus+ zyxYyzyxzx XYXYXYZYZYZYYYY δδδδδτδττδδτδττδδσδσσ

0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0

( ) ( ) 0)( =minusminus++minus++minus+ zyxXzyzxyx XZXZXZYZYZYZZZZ δδδδδτδττδδτδττδδσδσσ

0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ

Berdasarkan teori St Venannt bahwa jika ada torsi maka yang tegangan yang bekerja

adalah τzx dan τzy saja Maka σx = σy = σz = τxy = 0 τzx = τxz τzy = τyz τyx = τxy dan

X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip( b )

0=+X

ZY

Y

ZY

δδτ

δδτ helliphelliphelliphelliphelliphelliphelliphelliphellip( c )

Dengan persamaan cauchy

yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=

Memenuhi persamaan c



Dari hukum Hooke diketahui

[ ])(1ZYXX Ex

u σσυσδδε +minus==

[ ])(1XZYY Ey

v σσυσδδε +minus==

[ ])(1YXZZ Ex

w σσυσδδε +minus==

Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ

Lihat gambar dibawah

Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ

maka w (xyz) = Γ ϕ(xy) dimana ϕ(xy) = fungsi Warping

maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ dimana )()( zyzyu ϑminus= dan w (xyz) = Γ ϕ(xy)

Γ = νrsquo (regangan sama dengan turunan pertama sudut puntir)

⎥⎦⎤

⎢⎣⎡ minus=⎥⎦

⎤⎢⎣⎡ +minus= y

xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ

δδδτ dimana v = x ν(z) dan w = Γϕ(xy) = νrsquo ϕ(xy)

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ

δϕϑδδϕϑϑτ dimana

xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2

Persamaan (1) dan persamaan (2) dikurangkan



⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+

fungsi torsi disebut fungsi Torsi

III 2 Soap Film Analogi

Berdasarkan teori Prandrsquol bahwa persamaan membransoap film analogi (persamaan

kulit sabun) Dimana jika ada gaya p(xy) seperti gambar dibawah maka akan terjadi

perpindahan sebesar w

H H w

P (xy)

Z

X

Y

dy

dx



Dari teori Prandrsquol maka persamaan perpindahan adalah HP

yw

xw

minus=+ 2

2

2

2

δδ

δδ

Penyelesaian persamaan ini dapat diselesaikan dengan deret forrier

III3 Fungsi torsi pada Tampang empat persegi

Fungsi perpindahan untuk tampang membran persegi dengan ukuran 2s x 2t menurut

[Thimosenko and Goodier1986] adalah

nn

n Ytxnbw

2cos

531

πsuminfin

=

=

dimana nb adalah bilangan konstanta sedangkan nY adalah fungsi y

nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus=minus sum

Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π

minus = ( ) 2)1(14 minusminusminus n

nnbHP

π

Penyelesaian umum adalah

2)1(33

2

)1(162

cosh2

sinh minusminus++= n

nn bHn

PttynB

tynAY

πππ

Jika penampang simetri maka A = 0

2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ

Untuk ( ) 0=plusmn= synY maka 2)1(33

2

)1(162

cosh0 minusminus+= n

nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π

Maka dengan dapatnya Yn maka persamaan lendutan membran pada tampang persegi

akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minusinfin

=sum

Dihubungkan dengan persamaan Torsi diatas maka menyelesaikan persamaan torsi

digunakanlah teori membran berdasarkan Prandrsquol Maka hubungan antara w dengan

fungsi torsi φ dapat dikatakan setara seperti ditunjukan ditabel dibawah

Apabila fungsi Oslash diketahui maka akan dapat dicari tegangan dimana tegangan geser

sama dengan turunan pertama dari fungsi Torsi

Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas

Oslash = 0 (sisi luas)

ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=

Dengan menggunakan rumus perpindahan (displacement pada membran pada teori

soap film analogi maka didapat hubungan antara Γminus=minus GHP 2 maka fungsi torsi

pada tampang persegi akan didapat

txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ



III 4 Hubungan antara Momen Torsi dan Fungsi Torsi

dFxyM ZXZYT )(int minus= ττ

dydxxy

dydxyx

dFxy

yx

M T )( intintintintint minusminus=minusminus=δδφ

δδφ

δδφ

δδφ

Integral Partial

intint minus= dxxxdxy

Xr

Xl

1 φφδδφ

dydxdyxdydxxy

intintintintint minus= φφδδφ pada xr dan xl (sisi luar) maka 0=φ

dydxdydxxy

intintintint minus= φδδφ

analog dydxdydxxx


maka intint= dydxM T 2 φ = int dA2 φ

X

XR XL

Y

z y

x

τzx

τzy X

Y



Hubungan momen torsi dengan fungsi Torsi adalah sbb

Mt = 2 int Oslash dA

Dari persamaan tersebut diatas menyatakan bahwa Momen Torsi dua kali volume

membran

III5 Pemakaian fungsi pada tampang lingkaran

Pada tampang lingkaran berlaku VM t 2= dimana intΦ= ρπρ dV 2

dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ= int

20 rM t ΠΦ= dan dari sini 20 r

M t

Π=Φ

Dengan demikian fungsi torsi pada tampang lingkaran menjadi

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4

Torsi pada tampang Tebal

4 1 Menentukan Inertia Torsi

Jika sesuatu tampang prismatik mengalami Torsi maka akan terjadi perubahan bentuk

pada penampangnya atau disebut juga warping Hal itu dapat dilihat digambar IV1

Dengan bantuan teori soap film analogi maka Inergia torsi tampang sembarang dapat

diturunkan dari rumus

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart

Kemudian GJM T=Γ analog dengan

EIM

=χ dimana χ = kelengkungan

Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2

dengan bantuan penyelesaian memakai teori Prandrsquol maka

2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ

Gambar IV1 Tampang prismatik yang mengalami warping pada saat ada Torsi



42 Tegangan torsi pada tampang lingkaran

Fungsi torsi pada tampang lingkaran menjadi

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ

43 Tegangan torsi pada tampang persegi

Jika pada tampang bulat tegangan torsi linier maka pada tampang persegi tegangan

torsi berbentuk parabola

Tegangan torsi ⎟⎟⎠

⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t



Dengan methode soap film analogi maka τzy dan τzx dapat dihitung dari persamaan

torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart

Jika fungsi diketahui maka tegangan torsi akan dapat dicari dengan

xZY partpart

=φτ dan

yZX partpart

minus=φτ

Fungsi Torsi pada tampang persegi [Thimosenko and Goodier1986]

txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

Jika x = plusmn s dan y = 0 maka didapat

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ



Jika sgtt maka deret yang kanan bisa diabaikan sehingga diperolehrsquo

tGΓ= 2maxτ Jika s=t

sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==

τmax Gambar IV3 Tampang persegi mengalami tegangan akibat Torsi

b

a

τ b

a b c



Secara umum khusus untuk tampang persegi maka Inersia torsi

J = α a b3

Jika ab ge 2 maka J dihitung dengan rumus )63001(3

3

ababJ minus=

τmax = γ MT dimana γ = 2abβ

Sedangkan τ b = χ τmax dimana τb adalah tegangan pada sisi terpendek

Dimana α β dan χ dapat dilihat pada tabel IV2 berdasrakan perbandingan a dan b

ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743

Tabel IV2 Koefisien untuk mencari J τmax τ b pada tampang persegi

a

b



44 Tampang Elips

Pada tampang berdasarkan [] elips fungsi torsi ⎟⎟⎠

⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ

Dengan demikian tegangan yang terjadi

yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=

Untuk tampang elip didapat sbb

TMab2max2

πτ =

22

33

babaJ+

= π

4 5 Tampang Segitiga

Pada tampang segitiga maka fungsi torsi adalah

( ) ( )[ ]227222

2122

21 3 axyxyxG a minusminusminus+minus= νφ

b

a

x

y

2max2abM T

πτ =



0=partpart

=yzxφτ sedangkan yxG

x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal

Suatu tampang persegi dengan tampang seperti dibawah dimana luas tampang sama

seperti contoh soal pada tampang bulat

Ditanya

bull Tentukan sudut puntir pada titik B jika batang AB adalah tampang persegi

dimana ab=05 dan E=26000 MNm2 υ=02

bull Hitunglah lendutan pada titik C

bull Hitunglah tegangan yang terjadi akibat torsi pada batang AB kemudian


Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm

ALingkaran= Лr2= 314(15)2=70650 cm2

APersegi = ab = 2bb= 2 b2

Dengan demikian didapat b= 1879 cm dan a3758 cm

J= α a b3 dari tabel dengan 2=ba maka α = 0229 maka J = 0229375818793

Maka Jpersegi = 5709165 cm4

Jlingkaran = frac12 Л r4 = 7948125 cm4

Dengan ab=2 maka Jlingkaran= 139 x Jpersegi

G = 108333333 Ncm2

Besar sudut puntir υ = 6557091331083333

2003000000 =00097

bull b Lendutan pada titik C jika pada B di jepit EI

PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4

Sedangkan tampang lingkaran Io = 39740625 cm4

=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm




c Tegangan Torsi TZX Mγττ ==max

b= 1879 cm dan a3758 cm

Dari tabel jika ab = 2 β = 406 makaγ = 2abβ = 305 x 10-4

Maka TZX Mγττ ==max = 305 x 10-43000000= 91799 Ncm2

maxχττ =b dimana χ =0796 maka bτ = 73072 Ncm2

Akibat gaya LintangbIQS

=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N

Torsi + Gaya Lintang τmax= 91799+4248 = 96047 2cmN

73072

91799



Tampang Lingkaran Tampang persegi

2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ

Akibat Torsi dan Lintang

=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ



V Torsi pada tampang tipis terbuka

V1 Tampang tipis

Pada tampang tipis fungsi dapat dilihat seperti digambarkan dibawah fungsi torsi

seperti sebuah setengah silinder`

Pada tampang tipis berlaku

0=partpart

yφ dimana 0)(0 =minus

partpart

== yx

Gzxϕντ

cxx

=partpartφ dimana )( x

yGcxzy +

partpart

==ϕντ

Pada gambar dibawah fungsi warping adalah yx=ϕ

Dengan demikian xGxZY 2νφτ =

partpart

minus=

maka fugsi torsi didapat

[ ]cxG minusminus= 2νφ

Pada sisi luar x=2b

plusmn maka 0=φ maka didapat c=2

2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i



Fungsi torsi akan menjadi ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusminus=

22

2bxGνφ

Pada x=0 maka ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ

Dengan demikian didapat bahwa Inertia Torsi pada tampang tipis adalah

3

31 abJ =

Untuk mengitung tegangan torsi diambil adri persamaan

xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3

62 ==τ dari sini dapat diartikan abhawa tegangan

pada tampang tipis adalah linear

Pada x = 2b tegangan geser maximum 2max

3abMb

JM T

T

T ==τ

Gambar V4 Diagram tegangan geser pada tampang tipis

τmax



Dalam menghitung sudut puntir dapat dijabarkan dari Hukum Hooke dan tegangan

torsi dengan GJLM T

=ν

V2 Tampang tipis terbuka dengan bermacam bentuk

Pada struktur bangunan banyak tampang tipis yang terbuka seperti bentuk kanal INP

WF Untuk menghindari kesalahan hitung maka secara umum profil tersebut dapat

dipecah dalam berbagai bentuk seperti dibawah ini

Maka Inertia Torsi dapat dihitung dengan ii

ii tsJ sum=

31

Sedangkan untung tegangan geser iT

T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah



V3 Tegangan Warping pada Tampang I

Suatu tampang I dapat dilihat digambar dibawah dimana dimensinya dengan

ketebalan flens atas dan bawah b1 dan b3 sedangkan ketebalan stegnya b2

Inertia torsi pada tampang I adalah

3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++

Jika suatu konstruksi dengan profil dibebani dengan MT seperti pada gambar V1a

maka bidang torsinya adalah paa gambar V1b

Pada konstruksi diatas diperhatikan batang AB yang mana batang tersebut

mengalami momen torsi MT Karena profil adalah I maka perhatikan gambar V2

a1

a2

a3

b1

b2

b3

Gambar V1 Inertia torsi tampang I

A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2

Gambar V1 Bidang torsi



= +

Yang menyebabkan warping adalah flTM

Pada gambar V2 diatas maka h

MP

flT

fl =

Jika diperhatikan pada flens saja maka akan terjadi deformasi seperti gambar V3

Pf1

t1

t

b

h =

Gambar V2 Torsi pada tampang I

MT

flTM

PTM

flTM



Pada gambar V3 ada beberapa deformasi dan tegangan pada flens yakni pada

gambar V3b tegangan geser gambar V3c tegangan lentur sedangkan pada

gambar V3c dan V3d terjadi putaran sudut dan lenturan

Secara umum pada flens berlaku

fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h

τzx max = 3Q fl 2tb

σz max = Mfl b Jfl 2

W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h



dari persamaan diatas maka berlaku

2

y

flT

fl

fl

fl

flT

IE

hM

EIP

EIQy minus=minus=minus=

atau hEI

Mdx

ud

y

flT2

3

3

minus=

dengan penulisan yang lain 3

3

2 dx

udhEIM y

fl minus=

jika ν2hu fl = seperti pada gambar V3e maka

2

2νhhEI

M yflT minus=

atau 2

4

νhEI

M yflT minus= dimana fl

TM disebut Momen Sekunder Momen sekunder

hanya berperan untuk flens saja

Pada gambar V3 solah-olah hubungan antara flens dan steg terputus padahal tidak

Untuk itu maka perubahan bentuk profil I menjadi gambar V4

Maka pada gambar V4 akan berlaku

νJGM PT =

MP disebut sebagai Momen Primer akibat torsi

PT

flTT MMM +=

maka akan berlaku +minus 2

4

νhEI y νJG = TM

Persamaan tersebut adalah persamaan differential momen sekunder dan primer akibat

torsi



V3 Penyelesaian Persamaan Torsi

MT = MS + Mp

+minus 2

4

νhEI y νJG = TM dapat ditulis dengan +minus νwEC νJG = TM

dimana 4

2hIC y

w = disebut juga sebagai konstanta warping

Persamaan diatas dijadikan

minusν νwEC

GJ=

w

T

ECM

dapat ditulis dengan minusν νλ 2 =w

T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-

Gambar 1 Bidang Torsi (Momen Sekunder dan Momen Primer)

M1 Momen Sekunder M2 Momen Primer



Kondisi perletakan

Penyelesaian umum GJMzCzBzA T+++= λλν coshsinh

Mencari koefisien A B dan C dengan boundary condition Pada z =0 maka sudut puntir υ = 0

GJM

CBA T00cosh0sinh0 +++=

0 = B + C Pada z = 0 maka Momen lentur sama dengan nol dan υrsquorsquo = 0 υrsquorsquo = A λsup2 sinh λz + B λsup2 cosh λz

0 = B Dengan demikian C = 0 juga Pada z=l2 υrsquo=0

GJM

zBzA T++= λλλλν sinhcosh

GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=

Dengan didapatnya koefisien A B dan C maka persamaan sudut puntir menjadi



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν

Sehingga didapat persamaan sudut puntir adalah ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν

Akibat M1 (Tegangan Warping) M1 = - E Ιy ν Primeprime P = -EIy ν Primeprime Q = P = - E Iy ν Primeprime P = lintang

τ w = Q = -E Iy ν Primeprime S = A X = = I = frac12 Iy

Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ

primeprimeprimeminus= = ν primeprimeprimeminus

16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprimeprime

2cosh

cosh2

Lz

GJM T

λ

λλν

Diagram tegangan wτ τw Tegangan lentur (σW)

f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2

dan 2hu ν= maka νν primeprimeminus=primeprimeminus=

hEChEIM w

ff 2

M1 h



σmax pada x = 2b maka νσ primeprime

minus==

f

w

f

f

I

bh

EC

I

bM22

max

νσ primeprimeminus=4max

Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

Gambar diagram σw σmax = - ν Prime Akibat M2 τ = dimana M2 = G J ν prime maka τ = = G ν primet dimana νprime = 1 -

E b h 4

M2 t J

( G J ν prime ) t J

MT G J

Cosh λx Cosh λL2



Diagram τ Catatan Sinh x = (ex ndash e-x) frac12

Cosh x = (ex + e-x) frac12

τ = G ν primet



VI Torsi pada tampang tipis tertutup

Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r

dAm = frac12 r ds ds =

dM0 = q r = 2 q dAm

MT = Ф dM0 = Ф 2q dAm

q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =

Elemen ds d ν = τ = G Γ Γ =

dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф



Literatur 1 Boresi Arthur dkk Advanced Mechanics of Materials 1992 2 Salmon Charles G dkk Struktur Baja 1992 3 Daryl L Logan Mechanics of Materials 1991

Thimoshenko SP GoodierJN 1986 Teori Elastisitas (terjemahan Sebayang Darwin) Penerbit Airlangga Jakarta

4 Bornsheuer Vorlesungen In Baustatik Einfuehrung in die Torsion Universitas Stutgart 1980

6 Basler Konrad Torsion in Structure Springler Verlag Berlin 1966



Bab 1 Pendahuluan

11 Pengenalan Torsi Dalam analisa struktur selain Momen Gaya Lintang dan Normal maka Torsi akan menjadi salah satu yang menentukan dalam disain struktur bangunan Dalam buku ini akan dibahas khusus hanya Torsi saja Torsi pasti akan terjadi pada konstruksi portal tiga dimensi atau pada konstruksi grid Untuk itu dalam buku ini akan dipaparkan bagaimana perletakan Torsi Bidang torsi sudut puntir akibat torsi dan tegangan torsi Tegangan torsi secara umum dibagi 3 yakni sbb

1 Tampang tebal seperti tampang Lingkaran persegi segitiga

2 Tampang tipis terbuka seperti profil I WF canal dll

3 Tampang tipis tertutup seperti tampang hollow box dll Dalam buku ini akan dibahas tentang tegangan torsi untuk ketiga jenis tampang ini 12 Sistem Koordinat Dalam perhitungan di buku ini system koordinat searah sumbu batang secara umum dinamakan sumbu Z sedangkan kearah lainnya adalah sumbu X dan Y Sedangkan untuk perpindahan (displacement) searah sumbu X adalah u dan searah sumbu Y Z adalah v w

X

Y Z

u

v w υ ψ

φ





Y

0


Z


sumbu x dan sb y

Y

0


Z

Y


Z





Y


Z

Y


Z



Mt

Mt

L


Mt


`

- -

+

MT MT

MT

L frac12 L frac12 L

a b

c



`




EFN

=ε


T

T

GJM

=Γ


mT

A B


+

-



Normal Torsi

Elemen dengan Gaya


Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν


+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν


dz dz

dz dzdw




II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











Y

0


Z


sumbu x dan sb y

Y

0


Z

Y


Z





Y


Z

Y


Z



Mt

Mt

L


Mt


`

- -

+

MT MT

MT

L frac12 L frac12 L

a b

c



`




EFN

=ε


T

T

GJM

=Γ


mT

A B


+

-



Normal Torsi

Elemen dengan Gaya


Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν


+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν


dz dz

dz dzdw




II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Y


Z

Y


Z



Mt

Mt

L


Mt


`

- -

+

MT MT

MT

L frac12 L frac12 L

a b

c



`




EFN

=ε


T

T

GJM

=Γ


mT

A B


+

-



Normal Torsi

Elemen dengan Gaya


Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν


+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν


dz dz

dz dzdw




II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









`




EFN

=ε


T

T

GJM

=Γ


mT

A B


+

-



Normal Torsi

Elemen dengan Gaya


Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν


+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν


dz dz

dz dzdw




II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Normal Torsi

Elemen dengan Gaya


Persyaratan

keseimbangan n

dzdN

minus= Tmdz

dMtminus=

Deformasi pada ele-

men

w w+dw

EFN

dzdw

== ε

ν υ+dυ

T

T

GJM

dzd

=Γ=ν


+int

dw = zzEFN ε=

cdzGJM

T

T += intν

zzGJM

T

T Γ==ν


dz dz

dz dzdw




II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










II1 Umum




dA

r

ρ

MT = intρ dv Mt










MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф















MT = dAr

G intΓ2

maxρ = dA

r int 2max ρτ

= Jrmaxτ

Dimana 4

21 rJ Π=


L

r

υ ρ

Γ max Γmax =

r L

r ρ =

Γmax Γ

ρ r


1 4 πr4 + 1

4 πr4

υ





= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











= G maka



τmax =

ν =


1 2 πr4

r ν L

MT r Ј

r ν L

MT Ј = G

ν L

MT L

G Ј

MT r Ј

MT L

G Ј

τmax

τmax

τmax

MT


τmax = MT

Ј r

a b c




44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










44 dDJT minus=π

Contoh soal

Ditanya







Jawab

a

Bidang Torsi

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

d

D



b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









b sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L = 2 m = 200 cm

J = frac12 Л r4 = 7948125 cm4

)1(2 υ+

=EG =1083333 MNm2 = 1083333 x 100000010000 =108333333 Ncm2


υ2

36000696800

= = 03990

c


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=1δ 0218 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

I = frac14 Л r4= 39740625 cm4

=2δ 0516 cm


υδ tgL 13 = = 1500006968=10452 cm




e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









e Tegangan Torsi rJ


25794813000000 =56617 Ncm2

217566 cmNZX =τ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus=

2

2 134

ra

rQ

ZY πτ

AQ

aZY 34

)0max( ==τ = 21520000

34

π=37745 2cm

N


217566 cmNZY =τ

274537cmN

ZY =τ



BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









BAB III


III1 Fungsi Torsi




τzy = τyz

τyx = τxy






diperoleh sbb

ΣX = 0


Y

X

Zσz

σy σx

τxy

τxz

τzx τzy

τyx

τyz


δZ

δX

δy



0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









0=+++ Xy

YX

Z

ZX

X

X

δδτ

δδτ

δδσ

ΣY = 0


0=+++ YX

XY

Z

ZY

Y

Y

δδτ

δδτ

δδσ

ΣZ = 0


0=+++ ZX

XZ

Y

YZ

Z

Z

δδτ

δδτ

δδσ



X = Y = Z = 0

0=Z

ZX

δδτ ( a )

0=Z

ZY

δδτ


0=+X

ZY

Y

ZY

δδτ



yyx

ZX δδτ )(Φ

=

xyx

ZY δδτ )(Φ

minus=





[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










[ ])(1ZYXX Ex


[ ])(1XZYY Ey


[ ])(1YXZZ Ex


Gxy

yu XY

XYτ

δδ

δδ

=+=Γ

Gxw

zu ZX

ZXτ

δδ

δδ

=+=Γ

Gyw

zv ZY

ZYτ

δδ

δδ

=+=Γ

Khusus Torsi

0=== ZYX εεε

0=XYτ

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ

δδδτ

⎥⎦

⎤⎢⎣

⎡+=

yw

zyGZY δ

δδδτ


Yv

Xu

ν ri

ri

i i

- ui

vi

yi

xi



i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









i

ii

i

i

i

i

rr

xy

vu ϑ

==minus

)()( zxzxy ϑ=

)()( zyzyu ϑminus=

Dari 0=zw

δδ


maka

⎥⎦⎤

⎢⎣⎡ +=

xw

zuGZX δ



⎥⎦⎤



xG

xyxyGZX δ

δϕϑδ

δϕϑϑτ )(

⎥⎦

⎤⎢⎣

⎡minus= 1

yxG

yZX

δδδϕϑ

δδτ

dimana yZX δ

δτ Φ=

⎥⎦

⎤⎢⎣

⎡minus=

Φ 1

2

2

yxG

y δδδϕϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

yw

zvGZY δ


⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

yxG

yxGZY δ


xZY δδφτ minus=

⎥⎦

⎤⎢⎣

⎡+=⎥

⎦

⎤⎢⎣

⎡+=

Φy

xGy

xGx δ

δϕϑδδϕϑϑ

δδ

⎥⎦

⎤⎢⎣

⎡+=

Φyx

Gx δδ

δϕϑδδ

1

2

2




⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









⎥⎦

⎤⎢⎣

⎡+minus=

Φyx

Gy δδ

δϕϑδδ

1

2

2

⎥⎦

⎤⎢⎣

⎡+=

Φminus

yxG

x δδδϕϑ

δδ

1

2

2

-

2

2

2

2

2 ϑδδ

δφδ G

yxminus=

Φ+






H H w

P (xy)

Z

X

Y

dy

dx




yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










yw

xw

minus=+ 2

2

2

2

δδ

δδ





nn

n Ytxnbw

2cos

531

πsuminfin

=

=


nn

n Ytxn

tnb

xw

2sin

2531

ππδδ sum

infin

=

minus=

nn

n Ytxn

tnb

xw

2cos

4 2

22

5312

2 ππδδ sum

infin

=

minus=

Sedangkan

531 2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

5312

2

2cos n

nn Y

txnb

yw π

δδ sum

infin

=

=

s

s

t t



Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Sedangkan ( )txn

nHP

HP n

n 2cos14 2)1(

531

ππ

minusinfin

=


Maka diperoleh

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus +

531 2cos n

nn Y

txnb πsum

infin

=

= ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

531 2cos n

nn Y

txnb πsum

infin

=

nn

n Ytxn

tnb

2cos

4 2

22

531

ππsuminfin

=

minus = ( )txn

nHP n

n 2cos14 2)1(

531

ππ

minusinfin

=

minusminus sum

nY nY

tn

2

22

4π


nnbHP

π


2)1(33

2

)1(162

cosh2


nn bHn

PttynB

tynAY

πππ


2)1(33

2

)1(162

cosh minusminus+= n

nn bHn

PttynBY

ππ


2

)1(162


nbHnPt

tsnB

ππ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus= minus

tsntyn

bHnPtY n

nn

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π


akan menjadi

txnbw

nn 2

cos531

πsuminfin

=

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminus minus

tsntyn

bHnPt n

n

2cosh

2cosh

1)1(16 2)1(33

2

π

π

π



txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









txn

tsntyn

nHPtw n

n 2cos

2cosh

2cosh

1)1(116 2)1(

53133

2 ππ

π

π⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡


=sum






Prandrsquol Torsi

W Oslash

Syarat batas

W = 0 (sisi luas)

Syarat batas


ZXy

τδδφ

=

xw

δδ

ZYyτ

δδφ

=




txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

yw

δδ





dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











dydxxy

dydxyx

dFxy

yx


δδφ

δδφ

δδφ

Integral Partial


Xr

Xl

1 φφδδφ

dydxdyxdydxxy


dydxdydxxy


analog dydxdydxxx



X

XR XL

Y

z y

x

τzx

τzy X

Y






membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф












membran



dimana ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minusΦ=Φ

2

0 1rρ

maka ρπρρ dr

M t 2122

0⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞



M t

Π=Φ


⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ



Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Bab 4







)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


EIM


Γ=νd dz =GJM T dz

Maka dzGJML

Tint=0

ν = LGJM T

Γ=

Γ= int

G

dF

GM

J Tφ2


2

2

2

2

4

yy

dydxJ

partpart

+partpart

= intintφφ

φ






⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛minus

Π=Φ

2

2 1rr

M t ρ

42r

Mdd

t Π=

Φ=

ρρ

τ =4

21 r

M tΠ

ρ =J

M tρ dimana 4

21 rJ Π=

JrM t=maxτ





⎞⎜⎜⎝

⎛minus

partpart

= yy

GZXϕντ

⎟⎠⎞

⎜⎝⎛ +

partpart

= xx

GZYϕντ

s

s

t t




torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










torsi

)(2 2

2

2

2

zGxy

νφφminus=

partpart

+partpart


xZY partpart

=φτ dan

yZX partpart

minus=φτ


txn

tsntyn

ntG n

n 2cos

2cosh

2cosh

1)1(132 2)1(

53133

2 ππ

π

πφ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

= minusinfin

=sum

Maka didapat txn

tsntyn

nttG

x n

nzy 2

sin

2cosh

2cosh

1)1(12

32531

2)1(23

2 ππ

ππ

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=

txn

tsntyn

ntG

x n

nzy 2

sin

2cosh

2cosh

1)1(116531

2)1(22

ππ

π

πφτ sum

infin

=

minus

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡ minus

Γminus=

partpart

minus=


⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡minus⎟

⎠⎞

⎜⎝⎛ +++

Γ=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minus⎥⎦⎤

⎢⎣⎡Γ

= sumsuminfin

=

infin

=

tsnn

tG

tsnn

tGnn

2cosh

1151

31116

2cosh

11116531

2222531

22max ππππτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ

= suminfin

=

tsnn

GtGn

2cosh

11168

16531

22

2

2max πππ

πτ

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡Γ

minusΓ= suminfin

=

tsnn

GtGn

2cosh

11162531

22max ππτ





sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











sGtG Γ=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++minusΓ= 3511

23cosh9

1

2cosh

1812 2max πππτ

dydxaxn

abn

ayn

ntG

x n

nzy

2cos

2cosh

2cosh

1)1(116531

2)1(22 intint sum

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

minusminusΓ

minus=partpart

minus=infin

=

minus ππ

π

πφτ

sbn

naG

ntsG

nnzy 2

tanh1)2(641)2()2(32531

54

4

53144

3 πππ

τ sumsuminfin

=

infin

=

Γminus

Γ==


b

a

τ b

a b c




J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










J = α a b3


3

ababJ minus=




ba α β Χ

1 0141 481 1000

15 0196 433 0853

2 0229 406 0796

25 0249 388 0768

3 0263 374 0753

4 0281 355 0745

5 0291 343 0744

6 0299 335 0743

8 0307 326 0743

10 0312 320 0743

~ 0333 300 0743


a

b



44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









44 Tampang Elips


⎞⎜⎜⎝

⎛minus+minus= 12

2

2

2

by

ax

abM t

πφ


yabM

yT

zx 2πφτ minus=

partpart

=

xba

Mx

Tzy 2π

φτ =partpart

minus=


TMab2max2

πτ =

22

33

babaJ+

= π



( ) ( )[ ]227222

2122


b

a

x

y

2max2abM T

πτ =



0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









0=partpart


x azy 6( 21++=

partpart

minus= νφτ

Contoh Soal



Ditanya






Jawab

a

Bidang Torsi

sudut puntir GJ

LM t=υ

Mt = 3 t m= 300000 kgcm= 3000000 Ncm

L=2m

L1=15 m

P=2t

A

B

C

Mt=3tm

3tm

A

B

b

a



L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









L = 2 m = 200 cm








G = 108333333 Ncm2


2003000000 =00097


PL3

31

1 =δ

P= 2 ton = 2000 kg=20000 N

L1= 15 m =150 cm

E=26000 MNm2= 2600000 Ncm

I = 3

121 bh =

121 1879 37583 = 83 102 84 cm4


=1δII o 0218 cm = 0104 cm


PL3

3

2 =δ

P= 2 ton = 2000 kg=20000 N

L2= 2 m =200 cm

E=26000 MNm2= 2600000 Ncm

Io = frac14 Л r4= 39740625 cm4

=2δIIo 0516 cm = 0247 cm


υδ tgL 13 = = 150 00097=1455 cm





b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











b= 1879 cm dan a3758 cm





=τ AQ

23

max =τ =58377918

2000023 =4248 2cm

N


73072

91799




2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










2=ba

Perbanding Tampang

Lingkaran dibagi

Persegi

R=15 cm

A=70560 cm2

a=3758cm b=1879cm

A=70560 cm2

J = 7948125 cm4 J=5709165 cm4 391=

pp

o

JJ

=υ 0006968 rad =υ 0009700 rad 71800 =

ppνν

Akibat Torsi

=maxτ 217566 cmN

Akibat Torsi

=maxτ 91799 Ncm2 620=

pp

o

ττ

Akibat Gaya Lintang

AQ

34

max =τ =37745 2cmN

Akibat Gaya Lintang

AQ

23

max =τ =4248 2cmN

890=pp

o

ττ


=maxτ 603915 2cmN


=maxτ 96047 2cmN

630=pp

o

ττ




V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










V1 Tampang tipis




0=partpart


partpart

== yx

Gzxϕντ

cxx


yGcxzy +

partpart

==ϕντ



partpart

minus=



Pada sisi luar x=2b


2⎟⎠⎞

⎜⎝⎛ b

b

a x

y

I-I x

z

x

y i




⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










⎤

⎢⎢⎣

⎡⎟⎠⎞


22

2bxGνφ


⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

2

0 2bGνφ

Maka 3

2

0

3123

223222

abG

bGab

G

ab

G

dFJ =

⎟⎠⎞

⎜⎝⎛

=== intν

ν

ν

φ

ν

φ


3

31 abJ =


xGxZY 2νφτ =

partpart

minus= dimana T

T

JM

G =ν

Maka didapat xabM

xJM T

T

TZY 3




3abMb

JM T

T

T ==τ


τmax




torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










torsi dengan GJLM T

=ν






ii tsJ sum=

31


T tJM

maxmax plusmn=τ

s1

s2

s3

t1

t2

t2

dipisah

dipisah







3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф













3

1 31

nn

n

ibaJ sum

=

= = 333

322

311 3

131

31 bababa ++





a1

a2

a3

b1

b2

b3


A B Ca

GIT 2 MT

(-)

(+)

b MT

MT

L2 L2




= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









= +



MP

flT

fl =


Pf1

t1

t

b

h =


MT

flTM

PTM

flTM







fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф













fl

flT

EIM

y minus= dimana 2

yfl

II =

Pf1 = MT h

fl

b zw

xu

a

(+) Qf1

Qf1 b

Mfl c

d

e v

ufl

h



W

ν fl W = b 2

ν fl

v = u fl h2

= 2 u h




2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










2

y

flT

fl

fl

fl

flT

IE

hM

EIP


atau hEI

Mdx

ud

y

flT2

3

3

minus=


3

2 dx

udhEIM y

fl minus=


2

2νhhEI

M yflT minus=

atau 2

4

νhEI







νJGM PT =


PT

flTT MMM +=


4

νhEI y νJG = TM


torsi




MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










MT = MS + Mp

+minus 2

4


dimana 4

2hIC y



minusν νwEC

GJ=

w

T

ECM


T

ECM

Dimana wEC

GJ=2λ

2 MT

frac12L frac12L Z

Y

X

M1

M2

MT

X

+

_-





Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Kondisi perletakan



GJM




GJM


GJMLA T+=

2cosh0 λλ

Maka didapat

2cosh

1LGJ

MA T

λλminus=




⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=+=

2cosh

sinhsinhLzz

GJM

GJMzzA TT

λ

λλλ

λν


⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=

2cosh

sinhLzz

GJM T

λ

λλλ

ν



Maka didapat

2

8

4

2

yf

fY

w It

tbhEI ντ


16

2hbE

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=prime

2cosh

coshLz

GJM T

λ

λλλλ

ν

h2 4

h2 4

1 h

b

tf

b 4 x =

M1 h

h

h 4

Q S b I

h 4

b tf 2

b 4

b2 tf 8



⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛


2cosh

cosh2

Lz

GJM T

λ

λλν


f

f

f

ffw I

xM

xIM

WM

===σ

f

f

EIM

dxud

minus=2

2


hEChEIM w

ff 2

M1 h




minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










minus==

f

w

f

f

I

bh

EC

I

bM22

max


Ebh

Dimana

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

minus=primeprime

2cosh

sinhLz

GJM T

λ

λλν


E b h 4

M2 t J


MT G J

Cosh λx Cosh λL2





τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф











τ = G ν primet




Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф










Tegangan geser

q = τt dimana

t = tebal

q = shear flow

τ = Tegangan geser

dA = t ds

dM0 = dF (r) = (τ dA) (r)

= (τ t ds) r

= q ds r


dM0 = q r = 2 q dAm


q =

τ =

q = τ t

r dF = qds

ds

dAm

q τt

2 dAm r

2 dAm r

MT = 2 q Am

MT 2 Am MT

2 Am t



Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф









Inertia Bred

ν =


dν =

d ν =

ν = =

=

= Ф

ν =

maka

J Bred =

ds

v

Γ L r

Γ L r

τ G

τ Gr

MT L 2 Am t G r

Ф dv dAm Am

MT L 2 Am t G r

d Am Am

MT L 2 Am t G r

r ds 2

Am

MT 4 Am2 G

ds t

4 A2m

ds t Ф













Documents

ANALISA STRUKTUR LANJUTAN.pdf