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.
BAB I
PENDAHULUAN
1.1 Deskripsi Struktur
1.1.1 Bentuk dan Dimensi Struktur
3.00 M3.00 M 3.00 M
4.50 M
3.00 M
3.00 M
4.00 M
4.00 M
1
Panjang masing masing elemen yaitu :
1.Kolom
H1 3:=
H2 3.:=
2. Balok
A 4:=
A1 4:=
A2 4:=
3. Kemi ingan Ata!
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α 30 π180⋅:=
PanjangRafter
2 A
2⋅
cos α( ):=
PanjangRafter 4.619m=
1.1.2 Fungsi Struktur
"ungsi ge#ung a#ala$ as ama
1.1.3 Spesifkasi Material
%utu Baja & Bj'3(
%utu Baut & Bj'3)
Dari tabel 5.3 SNI 2002 didapat
fy 2100 g
cm2
:=
f! 3400 g
cm2
:=
*ekanan angin
Pang"n 1.93
g
m2:=
+enis Ata! & seng
#ata$ %c( ) 10 g
m2
:=
+a ak go #ing maksimum
s 0.6:=
%o#ulus Elastisitas
& 2 106
⋅ '
mm2
:=
%o#ulus ,ese
8 103
⋅ '
mm2
:=
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Ni-a$ Poisson
µ 0.3:=
Koe sien !emuaian
a 12 10 6−
⋅:=BAB II
PE/EN0ANAAN , /DIN,
2.1 Analisa Pembebanan
b30o
4
cos30 0.866:=
s"n30 0.5:=
4
cos30:=
4.619=
+a ak %aksimum ,o #ing s
s 0.6:=
+umla$ ,o #ing n
n
s 1+:=
n 8.698=
Di-ulatkan se$ingga :
n 9:=
+a ak Anta ,o #ing 4
*
n 1−:=
* 0.5++=
, 4:=
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a#ala$ -entang te !anjang sejaja go #ing
2.1.1 Beban Mati
Tentukan sendiri profl dari tabel profl baja light chane l
Be at P o l ,o #ing Lig$t 0$anel
%$ +.51:=
kg5m
(Baja Profl C Hal 50 dimensi 150 !5 20 3.2 tabel baja "
arena digunakan atap seng! maka "
Be at ata! 67
%c 10:=
kg5m2
(Bab 2 tabel 21 #al 12 at$ran
pembebanan "
Be at Penyam-ung
-$enyam !ng 5 0.05→:=
(bab 3 pasal 3.% at$ran
pembebanan"
arena nilai # tidak sama dengan 1 m! maka digunakan rumus "
%co %c⋅:=
%co 5.++4=
kg5m
Analisa " karena nilai $c untuk %m2 maka nilai $c terlebih dahulu dikalikandengan nilai #
*otal Be-an %ati 6#
Nilai 1.89 m & 188 ; 9 67 asumsi 188%/ %$ %co+( ) 1.05⋅:=
%/ 13.948=
kg5m
%omen #itenga$ -entang
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Kemi ingan ata! a
α 30:=
, 4:=
&'ilai bentang terpanjang pada bagian (ang mengalami tekuk)
M/ 1
8 %/⋅ cos30⋅ , 2⋅:=
M/ 24.158=
kg m
M/y 1
8 %/⋅ s"n30⋅ , 2⋅:=
M/y 13.948=
kg m
2.1.2 Beban *idup
Be-an o ang te !usat P
P 250:=
(bab 3 pasal 3.2 a&at 1 at$ran pembebanan"
Mo 1
4 P⋅ cos30⋅ , ⋅:=
Mo 216.5=
kg m
Moy 1
4 P⋅ s"n30⋅ , ⋅:=
Moy 125=
kg m
2.1.+ Beban Angin
(bab % pasal %.3 a&at 1b at$ran pembebanan
'tap se i)3 den an s$d$t *emirin an α
+30, jadi α -!5 se#in a r$m$sn&a (0.02 α)0.%""
c1 0.02 α⋅ 0.4−( ):=
c1 0.2=
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&Angin Tekan)
c2 0.4:=
&Angin *isap)
(bab % pasal %.3 a&at 1d at$ran pembebanan,
'tap se i)3 majem$* α-!5 se#in a r$m$sn&a 0.2 α ) 0.%"
Ke7e!atan Angin
a 5.56:=
m5s
&Dari Soal)
*ekanan Angin Pa
Pa a2
16:=
(bab % pasal %2 b$tir 3 at$ran pembebanan"
Pa 1.932=
kg5m2
Berdasarkan Peraturan Pembebanan ,ndonesia bab - pasal -.2 point 1 tekanan tiup minimum 2/ kg%m2
Pa 25:=
kg5m2
%a1 c1 Pa⋅:=
%a1 5=
kg5m2
%a2 c2 Pa⋅:=
%a2 10=
kg5m2
Tekanan ma0 $a $a2
%a 10:=
kg5m2
% %a *⋅:=
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% 5.++4=
kg5m
A a$ ' <
M 1
8
%⋅ , 2
⋅:=
M 11.54+=
kg m
A a$ ' y
M y 0:=
kg m
2.1.- Beban *ujan
% 1 40 0.8 α⋅−( ):=
(bab 3 pasal 3.2 a &at 2a at$ran pembebanan"
% 1 16=
kg5m2
Berdasarkan PP, bab + pasal +.2 point 2a P ma0 2 kg%m2
% 16:=
kg5m2
%r % *⋅ cos30⋅:=
%r 8=
kg5m
Mr 1
8 %r ⋅ cos30⋅ , 2⋅:=
Mr 13.856=
kg m
Mry 1
8 %r ⋅ s"n30⋅ , 2⋅:=
Mry 8=
kg m
2.1./ ombinasi Pembebanan
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(SNI 2002 #al 13, b$tir !.2.2"
Beban mati 3 &Beban 4rang atau *ujan) 3 Beban Angin
A a$ < :
M! 1 1.2 M/⋅ 1.6 Mo⋅+ 0.8 M⋅+:=
M! 1 384.62+=
kg m
M! 2 1.2 M/⋅ 0.5 Mo⋅+ 1.3 M⋅+:=
M! 2 152.251=
kg m
M! 3 1.2 M/⋅ 1.6 Mr⋅+ 0.8 M⋅+:=
M! 3 60.39+=
kg m
M! M! 1:=
M! 384.62+=
kg m
A a$ y :
M!y1 1.2 M/y⋅ 1.6 Moy⋅+ 0.8 M y⋅+:=
M!y1 216.+3+=
kg m
M!y2 1.2 M/y⋅ 0.5 Moy⋅+ 1.3 M y⋅+:=
M!y2 +9.23+=
kg m
M!y3 1.2 M/⋅ 1.6 Mry⋅+ 0.8 M⋅+:=
M!y3 51.02+=
kg m
M!y M!y1:=
M!y 216.+3+=
kg m
2.2 Desain 5ording
2.2.1 Perhitungan apasitas Penampang
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P o!e ties !enam!ang go #ing Light Channel
(Baja Profl C Hal 50
dimensi 150 !5 20 3.2 "
H 15:=
c
44.3:=
7m3
6.5:=
c
y 12.2:=
7m3
ε
/ 2:=
c
332:=
7m(
t 0.32:=
c
y 53.8:=
7m(
tf 0.32:=
c
r 5.89:=
c
A 9.56+:=
7m2
ry 2.3+:=
c
%ate ial Baja
(SNI 2002 #al , b$tir 5.1.3"
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(SNI 2002 #al 11 , tabel 5.3"
& 2000000:=
kg57m2
+enis : B+ 3(
fy 2100:=
kg57m2
( 800000:=
kg57m2
fr 0.3 fy⋅:=
= & tegangan sisa
fr 630=
kg57m2
6ek Terhadap Tekuk 7ateral
"akto Pengali %omen 0-
1.136:=
&8ntuk balok sederhana)
Nilai 7- #i!e ole$ -e #asa kan tu unan umus :
"akto Pengali %omen 0-
2,3
3.M4.M3.MM2,5M12,5
CCBAmax
maxb ≤+++
=
1.136:=
.....$nt$* balo* seder#ana (SNI 2002 #lm3/ point .3.1"
&Sumber " S', Baja *al.+9 :.+.1 point d)
12.5 Mma⋅
2.5Mma 3MA+ 4 M-⋅+ 3M+ 2.30≤:= Mma
Dimana :%A & %omen #i 15( -entang & 3532.6.l2
%B & %ma< & %omen #i 152 -entang & 15>.6.l2
%0 & %omen #i 35( -entang & 3532.6.l2
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12.5 1
8 ,
2⋅
⋅
2.5 1
8⋅ ,
2⋅ ⋅ 3 3
32⋅ ,
2⋅ ⋅+ 4 1
8⋅ ,
2⋅ ⋅+ 3 3
32⋅ ,
2⋅ ⋅+
1.1363636363636363636→:=
1.136 2.30≤
.......... 4 ;;;
?e$ingga #i!e ole$ nilai 0- & 1.13@
1.136=
7 400:=
c
&'ilai bentang terpanjang tanpa pengekang lateral)
(SNI 2002 #al 3 ,tabel .3)2"
7$ 1.+6 ry⋅ &
fy⋅:=
7$ 128.+26=
c
Dimana + yaitu konstanta !unte to si #engan umus :
3ii .hb3
1J ∑=
&Sumber " *andout Dosen)?e$ingga umus + menja#i :
1
3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=
0.35=
7m(
1 π & ⋅ ⋅ A⋅
2⋅:=
1 1.16 105×=
c
tf 3⋅
H2
12⋅
3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=
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1.51 103×=
7m@
2 4
y
⋅
⋅
2⋅:=
2 2.819 10 6−×=
c
7r ry 1
fy fr − ⋅ 1 1 2 fy fr −( ) 2⋅++⋅:=
7r 35+.899=
c
7 400:=c
L L-
(SNI 2002 #al 3 , b$tir .3.5"
35+.899 400<
....4 ;;;
M "sa, !nt! $en am$an g seg" 4
1 ,5hb61hb41
SZ 2
2
===ξ
:n t ! $rof", # ; ξ ∼ 1.09 < 1=18= )"asanya /"am )",ξ > 1=12
&Sumber " *andout Dosen)
M$ 1.5 ⋅ fy⋅:=
M$ 1.395 105×=
kg 7m
Mr fy fr −( ) ⋅:=
Mr 6.512 104×=
kg 7m
(SNI 2002 #al 3 , tabel .3)2.b"
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Mn Mr M$ Mr −( ) 7r 7−( )7r 7$−( )⋅+⋅:=
Mn 5.846 104×=
kg 7m
uat 7entur
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kg 7m
M!y 1.063 104
⋅:=
kg 7m
M!y 1.063 10 4×=kg 7m
M!y φMny≤
(SNI 2002 #al 3%, b$tir .1.2"
1.063 104× 2.+6+ 10
4×≤
....... 4 ;;;
2.2.2 Pemeriksaan ekuatan dan 7endutan
Pemeriksaan ekuatan
M! 2 104×=
kg 7m
φMn 5.262 104×=
kg 7m
M!y 1.063 104×=
kg 7m
φMny 2.+6+ 104×=
kg 7m
Mn M$:=
(SNI 2002 #al 3!, b$tir .2.3"
φM$ φMn:=
φM$y φMny:=
M!
φM$ ζ M!y
φM$y ζ+ 1.0≤
#imana
M!
φM$M!y
φM$y+ 0.+64=
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0.+64 1.0<
...... 4 ;;;
Pemeriksaan 7endutan
Len#utan Aki-at Be-an %ati Be-an ang Be-an Hujan #an Be-an Angin
Arah Sumbu x
7 400:=
c
Be-an %ati
∆ /0.05 %/⋅ cos α π
180⋅
⋅ 74⋅
384 &⋅ ⋅:=
∆ / 0.061=c
Be-an ang
∆ oP cos α π
180⋅
⋅ 73⋅
192 &⋅ ⋅:=
∆ o 0.109=
c
Be-an Hujan
∆ r 0.05 %r ⋅ cos α π
180⋅
⋅ 74⋅
384 &⋅ ⋅:=
∆ r 0.035=
c
Be-an Angin
∆ 0.05 %⋅ 74
⋅384 &⋅ ⋅:=
∆ 0.029=
c
%aka :
∆ tota, ∆ / ∆ o+ ∆r +:=
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∆ tota, 0.204=
c
Arah Sumbu
7 400:=
c
Be-an %ati
∆y/0.05 %/⋅ s"n α π
180⋅
⋅ 74⋅
384 &⋅ y⋅:=
∆y/ 0.216=
c
Be-an ang
∆yoP s"n α π
180⋅
⋅ 73⋅
192 &⋅ y⋅:=
∆yo 0.38+=
c
Be-an Hujan
∆yr 0.05 %r ⋅ s"n α π
180⋅
⋅ 74⋅
384 &⋅ y⋅:=
∆yr 0.124=
c
Be-an Angin :
∆y 0:=
%aka :
∆ytota, ∆y/ ∆yo+ ∆yr +:=
∆ytota, 0.+2+=
c
∆ ∆ tota, 2 ∆ytota, 2+:=
∆ 0.+55=
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c
end$tan I in
7 400:=
c
∆" 7240
:=
(SNI 2002 #al 15, b$tir !.%.3"
∆" 1.66+=
c
∆ ∆"≤
0.+55 1.66+≤
...... 4 ;;;
BAB III
ANALI?A PE%BEBANAN
+.1 Analisa Pembebanan
+.1.1 Beban Mati
Beban pada
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A2 4:=
m
A 4:=
m
$da)*$da 1
Rafter1( )
-gor/"ng1 %$ 0.5⋅ A1⋅:=
-gor/"ng1 15.02=
kg
$da)*$da 2
Rafter2( )
-gor/"ng2 %$ 0.5⋅ A1 A2+( )⋅:=
-gor/"ng2 30.04=
kg
$da)*$da 3
Rafter3( )
-gor/"ng3 %$ 0.5⋅ A2 A2+( )⋅:=
-gor/"ng3 30.04=
kg
$da)*$da %
Rafter4( )
-gor/"ng4 %$ 0.5⋅ A2( )⋅:=
-gor/"ng4 15.02=
kg
Beban Atap " Beban atap dijadikan beban terpusat disepanjang
ra=ter pada titik gording
$da)*$da 1
Rafter1( )
-ata$1 %co 0.5⋅ A1⋅ *⋅:=
-ata$1 6.66+=
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kg
$da)*$da 2
Rafter2( )
-ata$2 %co 0.5⋅ A1 A2+( )⋅ ⋅:=
-ata$2 13.334=
kg
$da)*$da 3
Rafter3( )
-ata$3 %co 0.5⋅ A2 A2+( )⋅ ⋅:=
-ata$3 13.334=
kg
$da)*$da %
Rafter4( )
-ata$4 %co 0.5⋅ A2( )⋅ *⋅:=
-ata$4 6.66+=
kg
Beban Mati pada Balok
A
B
A A A
A1
1 2 3 4
C
D
A2
A2
Be-an #ija#ikan -e-an t a!e4oi# atau -e-an t a!esium #an5atau -e-an segitiga
aj$r 1 dan %
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Asumsi te-al !lat lantai 18
7m
?$,at 0.10:=
(Bab 2 tabel 2.1 #al 11 4 12 at$ran pembebanan "
- eton 2400:=
kg5m3
- $,afon 40:=
kg5m2
- s$es" 21:=
kg5m3
- ! "n 24:=
kg5m2
a. Be-an Plat
-$,at1 - eton ?$,at⋅ A
2⋅:=
-$,at1 480=
kg5m
-. Be-an Pla=on ; /angka
-$r1 - $,afon A
2⋅:=
-$r1 80=
kg5m
7. Be-an ?!esi 2 7m
?s$es" 0.02:=
-s1 - s$es" ?s$es"⋅ A
2⋅:=
-s1 0.84=
kg5m
#. Be-an U-in 2 7m
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?! "n 0.02:=
-!1 - ! "n ?! "n⋅ A
2⋅:=
-!1 0.96=
kg5m
?ota,M1 -$,at1 -$r1+ -s1+ -!1+:=
?ota,M1 561.8=
kg5m
aj$r 2 dan 3
a. Be-an Plat
-$,at2 - eton ?$,at⋅ A
2
A
2+ ⋅:=
-$,at2 960=
kg5m
-. Be-an Pla=on ; /angka
-$r2 - $,afon A
2
A
2+ ⋅:=
-$r2 160=
kg5m
7. Be-an ?!esi 2 7m
-s2 - s$es" ?s$es"⋅ A
2
A
2+ ⋅:=
-s2 1.68=
kg5m
#. Be-an U-in 2 7m
-!2 - ! "n ?! "n⋅ A
2
A
2+ ⋅:=
-!2 1.92=
kg5m
?ota,M2 -$,at2 -$r2+ -s2+ -!2+:=
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?ota,M2 1.124 103×=
kg5m
aj$r '
a. Be-an Plat
-$,at3 - eton ?$,at⋅ A1
4
⋅:=
-$,at3 240=
kg5m
-. Be-an Pla=on ; /angka
-$r3 - $,afon A1
4
⋅:=
-$r3 40=
kg5m
7. Be-an ?!esi 2 7m
-s3 - s$es" ?s$es"⋅ A1
4 ⋅:=
-s3 0.42=
kg5m
#. Be-an U-in 2 7m
-!3 - ! "n ?! "n⋅ A1
4 ⋅:=
-!3 0.48=
kg5m
?ota,M3 -$,at3 -$r3+ -s3+ -!3+:=
?ota,M3 280.9=
kg5m
aj$r B
a. Be-an Plat
-$,at4 - eton ?$,at⋅ A2
2
A1
4+ ⋅:=
-$,at4 +20=
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kg5m
-. Be-an Pla=on ; /angka
-$r4 - $,afon A2
2
A1
4+ ⋅:=
-$r4 120=
kg5m
7. Be-an ?!esi 2 7m
-s4 - s$es" ?s$es"⋅ A2
2
A1
4+ ⋅:=
-s4 1.26=
kg5m
#. Be-an U-in 2 7m
-!4 - ! "n ?! "n⋅ A2
2
A1
4+ ⋅:=
-!4 1.44=
kg5m
?ota,M4 -$r3 -s3+ -!3+ -$,at4+:=
?ota,M4 +60.9=
kg5m
aj$r C
a. Be-an Plat
-$,at5 - eton ?$,at⋅ A2
2
A2
2+ ⋅:=
-$,at5 960=
kg5m
-. Be-an Pla=on ; /angka
-$r5 - $,afon A2
2
A2
2+ ⋅:=
-$r5 160=
kg5m
7. Be-an ?!esi 2 7m
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-s5 - s$es" ?s$es"⋅ A2
2
A2
2+ ⋅:=
-s5 1.68=
kg5m
#. Be-an U-in 2 7m
-!5 - ! "n ?! "n⋅ A2
2
A2
2+ ⋅:=
-!5 1.92=
kg5m
?ota,M5 -$r5 -s5+ -!5+ -$,at5+:=
?ota,M5 1.124 103×=
kg5m
aj$r D
a. Be-an Plat
-$,at6 - eton ?$,at⋅ A2
2 ⋅:=
-$,at6 480=
kg5m
-. Be-an Pla=on ; /angka
-$r6 - $,afon A2
2 ⋅:=
-$r6 80=
kg5m
7. Be-an ?!esi 2 7m
-s6 - s$es" ?s$es"⋅ A2
2 ⋅:=
-s6 0.84=
kg5m
#. Be-an U-in 2 7m
-!6 - ! "n ?! "n⋅ A2
2 ⋅:=
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-!6 0.96=
kg5m
?ota,M6 -$,at6 -$r6+ -s6+ -!6+:=
?ota,M6 561.8=
kg5m
-e)ean MerataA ")at -e)an@"n/"ng
Beban Dinding
a. Lantai 1
Be-an #in#ing #ija#ikan -e-an me ata #ise!anjang -alok lantai
Be-an #in#ing -ata
(Bab 2 tabel 2.1 #al 12 at$ran pembebanan "
- ata 250:=
H7 3:=
-/ - ata H⋅:=
kg5m2
-/ +50=
kg5m
+.1.2 Beban *idup
Beban 4rang Terpusat &P)
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P 250:=
kg5m2
(Bab 3 pasal 3.2 a&at 1 #al 13 at$ran pembebanan "
Be-an #iletakkan !a#a titik go #ing !a#a a=te
Angin *ekan Angin Hisa!
Pada ra ter
-ot P:=
-ot 250=
g
Beban *idup pada Balok
Be-an $i#u! -e u!a -e-an segitiga & Be-an %ati Pa#a Balok
A
B
A A A
A1
1 2 3 4
C
D
A2
A2
- "/!$ - ata:=
- "/!$ 250=
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kg5m2
aj$r 1 dan %
- 1 - "/!$ A
2⋅:=
- 1 500=
kg5m
aj$r 2 dan 3
- 2 - "/!$ A
2
A
2+ ⋅:=
- 2 1 103×=
kg5m
aj$r '
- 3 - "/!$ A1
4 ⋅:=
- 3 250=
kg5m
aj$r B
- 4 - "/!$ A2
2
A1
4+ ⋅:=
- 4 +50=
kg5m
aj$r C
- 5 - "/!$ A2
2
A2
2+ ⋅:=
- 5 1 103×=
kg5m
aj$r D
- 6 - "/!$ A2
2 ⋅:=
- 6 500=
kg5m
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+.1.+ Beban Angin
Pa#a a=te yang #ite!i -e-an angin & C52
Pa#a a=te yang #itenga$ -e-an angin & C
Angin Tekan
Kemi ingan a=te a & 38
Pa 25=
kg5m2
c1 0.2:=
%a1 5=
kg5m2
*ekanan Angin P
%a1 c1 Pa⋅:=
Bi#ang Ke ja D & +a ak anta ku#a'ku#a < ja ak go #ing
go #ing
a=te
-A2
+2 A2
2A A11
!a"ter 1
@1 0.5 A1⋅ ⋅:=
@1 1.155=
m2
!a"ter 2
@2 0.5 A1 A2+( )⋅ ⋅:=
@2 2.309=
m2
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!a"ter 3
@3 0.5 A2 A2+( )⋅ ⋅:=
@3 2.309=
m2
!a"ter #
@4 0.5 A2( )⋅ ⋅:=
@4 1.155=
m2
Beban Angin Tekan
!a"ter 1
#1 0.866 %a1⋅ @1⋅:=
#1 5=
g
!a"ter 2
#2 0.866 %a1⋅ @2⋅:=
#2 10=
g
!a"ter 3
#3 0.866 %a1⋅ @3⋅:=
#3 10=
g
!a"ter #
#4 0.866 %a1⋅ @4⋅:=
#4 5=
g
Angin *isap
*ekanan Angin P
Pa 25=
kg5m2
c2 0.4:=
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%a2 c2 Pa⋅:=
%a2 10=
kg5m2
Ang"n ?e an Ang"n H"sa$
Be-an Angin Hisa!
!a"ter 1
#5 0.866 %a2⋅ @1⋅:=
#5 10=
g
!a"ter 2
#6 0.866 %a2⋅ @2⋅:=
#6 20=
g
!a"ter 3
#+ 0.866 %a2⋅ @3⋅:=
#+ 20=
g
!a"ter #
#8 0.866 %a2⋅ @4⋅:=
#8 10=
g
+.2 Analisa Struktur
Analisa st uktu #engan ?AP 2888 3D
Kom-inasi !em-e-anan yang #i!akai ?K?NI'2882
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1 1.( D
2 1.2 D ; 1.@ L
3 1.2 D ; 1.@ L ; 8.>
( 1.2 D ; 1.@ L ' 8.>
S8M> B>BA'
B>BA' MAT,
Beban mati pada ra ter
-gor/"ng1 15.02=
kg
-ata$1 6.66+=
kg
-gor/"ng2 30.04=
kg
-ata$2 13.334=
kg
-gor/"ng3 30.04=
kg
-ata$3 13.334=
kg
-gor/"ng4 15.02=
kg
-ata$4 6.66+=
kg
beban mati atap pada pin ir ra ter
Rafter1 -ata$1 0.5⋅ 3.3335289003621545+96+→:=
Rafter2 -ata$2 0.5⋅ 6.66+05+800+2430915935→:=
Rafter3 -ata$3 0.5⋅ 6.66+05+800+2430915935→:=
Rafter4 -ata$4 0.5⋅ 3.3335289003621545+96+→:=
6otal beban mati pada ten a# ra ter
-?1 -gor/"ng1 -ata$1+ 21.68+05+800+243091593→:=
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-?2 -gor/"ng2 -ata$2+ 43.3+4115601448618318+→:=
-?3 -gor/"ng3 -ata$3+ 43.3+4115601448618318+→:=
-?4 -gor/"ng4 -ata$4+ 21.68+05+800+243091593→:=
6otal beban mati pada pin ir ra ter
-P1 -gor/"ng1 Rafter1+ 18.3535289003621545+9+→:=
-P2 -gor/"ng2 Rafter2+ 36.+0+05+800+243091593→:=
-P3 -gor/"ng3 Rafter3+ 36.+0+05+800+243091593→:=
-P4 -gor/"ng4 Rafter4+ 18.3535289003621545+9+→:=
Beban mati pada balo*
Laju 1 #an (
?ota,M1 561.8=
kg5m
Laju 2 #an 3
#1 s "n π
3
⋅ 4.33=
#6 s "n π
3 ⋅ 1+.321=
?ota,M2 1.124 103×=
kg5m
Laju A #an D
#1 cos π
3
⋅ 2.5=
?ota,M3 280.9=
kg5m
#6 cos π
3 ⋅ 10=
Laju B #an 0
?ota,M4 +60.9=
kg5m
Beban Dindin
-/ +50=
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kg5m
B>BA' *,D8P
#2 s "n π
3 ⋅ 8.66=
#+ s "n π
3 ⋅ 1+.321=
Beban #id$p pada ra ter
-ot 250=
g
#2 cos π
3
⋅ 5=
#+ cos π
3
⋅ 10=
Beban #id$p pada balo*
Laju 1 #an (
- 1 500=
kg5m
Laju 2 #an 3
Laju A #an D
- 2 1 103×=
kg5m
- 3 250=
kg5m
Laju B #an 0
- 4 +50=
kg5m
B>BA' A'5,'
'n in te*an pada ra ter
/a=te 1
#1 5=
kg
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/a=te 2
#2 10=
kg
/a=te 3
#3 10=
kg
/a=te (
#4 5=
kg
'n in #isap pada ra ter
/a=te 1
#5 10=
kg
/a=te 2
#6 20=
kg
/a=te 3
6atatan " 8ntuk ra=ter tepi beban angin ?%2
8ntuk ra=ter tengah beban angin ?
#+ 20=
kg
/a=te (
#8 10=
kg
AP,T87AS, 5A@A '4
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< 1 &*,SAP T> A')
Tarik Tekan) 2113. ( '
> 2113. ( '
1)23.8 '18 1@>2.3@ '
11 2832.) '
12 2832.) '
13 18@ .)( '
19 32).2> '
1) 8 '
1 33>.@2 '
21 8 '
1( ' '@@3.22
1@ ' '( .>
1> ' '@>@.13
28 ' '91).13
22 ' '233 .8(
23 ' '1 29.82
2( ' '1913.8@
29 ' '1(>>.>@
2@ ' '1 2(.9(
31 ' '[email protected]
ATAS
Posisi Batang5a(a Batang
*4 A' *,SAP)
Tarik Tekan) 2893. ) '
> 2893. ) ' 1@)@.>( '
18 1@[email protected] '
11 282).91 '
12 282).91 '
13 18@ .)9 '
19 33 .1) '
1) 8 '
1 32@.)( '
21 9.@>(E'19 '
1( ' '@>).3(
1@ ' '91). )
1> ' '@@2.8)
28 ' '( >.
22 ' '239@.>
23 ' '1 2(. )
2( ' '1(> .31
29 ' '1912.@(
2@ ' '1 2(.99
31 ' '233>.31
ATAS
Posisi Batang5a(a Batang
*4
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kg
kg
kg
< 2&T> A' *,SAP)
Tarik Tekan) 2(91.@2 '> 2(91.@2 '
1 .3 '18 1 2).>@ '11 2(8(.)) '12 2(8(.)) '13 12)1.>9 '19 (13.29 '1) 8 '
1 3 1.>1 '21 2.9()E'12 '1( ' '>3).(91@ ' '@31.81> ' ') 3. 128 ' '9 >.3922 ' '2>8(.2923 ' '22>>.@12( ' '1)@@.2229 ' '1>8).22@ ' '22>>.1331 ' '2))2.2>
ATAS
Posisi Batang5a(a Batang
*4
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< 2&*,SAP T> A')
Tarik Tekan) [email protected]) '
> [email protected]) '
28>8.39 '18 288>.>( '
11 2(13.>3 '
12 2(13.>3 '
13 12)1.>3 '
19 3 2.(3 '
1) 8 '
1 (12.9 '
21 2.9()E'12 '
1( ' ') 9.29
1@ ' '9 .3
1> ' '>[email protected]
28 ' 8
22 ' '1.13)E'1(
23 ' 8
2( ' 8
29 ' '1.)89E'1(
2@ ' 8
31 ' '3.(11E'1(
ATAS
Posisi Batang5a(a Batang
*4 A' *,SAP)
Tarik Tekan) 231 .8@ '
> 231 .8@ '
1> 1.>@ '
18 1>38.9 '
11 22) .81 '
12 22) .81 '
13 128(.(> '
19 3>>.99 '
1) 8 '
1 3)8.12 '
21 8 '
1( ' ')>).(1
1@ ' '9 3.3>
1> ' ')( . @28 ' '[email protected]
22 ' '[email protected]
23 ' '21@).3
2( ' '1@)3. 1
29 ' '1)8 .8@
2@ ' '21@@. 3
31 ' '2@2).@1
Posisi 5a(a Batang
*4
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< +&*,SAP T> A')
Tarik Tekan) 2(8 .81 '
> 2(8 .81 '
1 @1.2@ '
18 1 88.81 '11 22>@.)> '
12 22>@.)> '
13 128(.(@ '
19 3)8.)1 '
1) 8 '
1 3>). 3 '
21 8 '
1( ' ')91.2(
1@ ' '9@@.13
1> ' ')>@.8@
28 ' '9 2.((
22 ' '2@2>.((
23 ' '21@).(@
2( ' '1)8 .93
29 ' '1@)3.(
2@ ' '21@@. 1
31 ' '2@9(.2>
ATAS
Posisi Batang5a(a Batang
*4).(( [email protected]( 19@9.( *ekan'Hisa! 193.83 1>3 . 2 1 1).((Hisa!'*ekan 193.82 1 1).>3 1>3 .9(
*ekan'Hisa! 131.1) 1)(>.91 1>1(. 3Hisa!'*ekan 131.1@ 1>19.2 1)(>.1@
*ekan'Hisa! 193.83 1>3 . 2 1 1).((Hisa!'*ekan 193.82 1 1).>3 1>3 .9(
*ekan'Hisa! >).(@ 19@9.)2 1@8 . 1Hisa!'*ekan >).(( [email protected]( 19@9.(
/A"*E/ 1
/A"*E/ 2
/A"*E/ 3
/A"*E/ (
/A"*E/ 9
'S9;SI B87'6 S8NDI7I D'7I 9D') 9D'Berat sendiri ra=ter 1
-P1 2⋅( ) 14 -?1⋅( )+ 16 -ot⋅( )+ #1 8⋅( )+ #6 8⋅( )+ B 0.1⋅ 454.033=
kg
Berat sendiri ra=ter 2
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-P2 2⋅( ) 14 -?2⋅( )+ 16 -ot⋅( )+ #2 8⋅( )+ #+ 8⋅( )+ B 0.1⋅ 492.065=
kg
Berat sendiri ra=ter +
-P3 2⋅( ) 14 -?3⋅( )+ 16 -ot⋅( )+ #3 8⋅( )+ #8 8⋅( )+ B 0.1⋅ 484.065=
kg
Berat sendiri ra=ter -
-P4 2⋅( ) 14 -?4⋅( )+ 16 -ot⋅( )+ #4 8⋅( )+ #9 8⋅( )+ B 0.1⋅ =#9
kg
BAB ,C
P>'6A'AA' P>'AMPA'5
-.1 Perencanaan an"
(SNI 2002 #al 11 , tabel 5.3"
P o!e ties Penam!ang
+enis : B+ 3(
14.8:=
7m
ry 2.3+:=
7m
fy 2100:=
kg57m2
f 10:=
7m
138:=
7m3
fr 0.3 fy⋅:=
= & tegangan sisa
tf 0.9:=
7m
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A 26.84:=
7m2
fr 630=
kg57m2
t 0.6:=
7m
f! 3400:=
kg57m2
f 2tf −:=
1020:=
7m(
(SNI 2002 Hal Point 5.1 ? 3"
8.2=
c
y 151:=
7m(
& 2000000:=
kg57m2
r 6.1+:=
7m
( 800000:=
kg57m2
-.1.1 Perencanaan Batang Tekan
, 400:=
7m
7 450:=
7m
&'ilai bentang terpanjang tanpa pengekang lateral)
Bagian ' -agian !enam!ang $a us mem!unyai nilai kelangsingan yang le-i$ ke7il#a i
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(SNI 2002 #al 30 ) 31 , tabel /.5.1"
C#e@* *elan sin an pofl
lens
λf f
tf :=
λf 11.111=
fy 250:=
M$a
λrf 250fy
:=
λrf 15.811=
λf λrf <
11.111 15.811<
...................CD
?eb
λ t
:=
λ 13.66+=
λr 665
fy:=
λr 42.058=
λ λr<
13.66+ 42.058<
.................CD
C#e@* *elan sin an elemen str$*t$r
λ 7 ry
:=
λ 189.8+3=
(SNI 2002 #al 2 , b$tir /.!.%"
λ 200<
..........CD
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0ek kon#isi -atas te $a#a! :
'! 1218.08:=
g
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
Lele$
φ 'n 0.9A fy⋅:=
(SNI 2002 #al /0, b$tir 10.1"
φ 'n 5.0+3 10 4×=
2. " aktu
7 3.465:=
c: 1:=
(SNI 2002 #al /0, b$tir 10.2"
#i en7anakan sam-ungan anta ku#a ' ku#a #engan menggunakan -aut
/"ameter
/ 2.2:=
c
jumla$ -autn 4:=
te-al !enam!ang
t 0.8:=
c
An A n /⋅ t⋅−:=
An 19.8=
(SNI 2002 #al /1, b$tir 10.2.1"
Ae An :⋅:=
Ae 19.8=
φ 'n 0.+5 Ae⋅ f!⋅:=
φ 'n 5.049 104×=
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φ 'n '!≥
(SNI 2002 #al /0, b$tir 10.1.1)1"
9.185 104× 2556.5+≥
.................CD
3. Ke untu$an -lok ujung
a. kon#isi gese mu ni
Bidang geser (v)
Bidang geser (v)
Agv
Anv
s s s
#
s1
t
e 1:=
/).t33s(s1AA
s).tss(s1A
21
nns
g
−+==+++=
s 4:=
c
s1 2:=
c
Ans s1 3s+ 3 /2⋅− t⋅:=
Ans 8.56=
φ 'n 0.+5 0.6 f!⋅( )⋅ Ans⋅:=
φ 'n 1.31 10 4×=
φ 'n '!≥
(SNI 2002 #al /0, b$tir 10.1.1)1"
1.425 104× 2556.5+≥
.................CD
-. Kom-inasi gese ' ta ik
Ant 150 e− 0.5 /⋅−( )t:=
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Ant 118.32=
7m2
An s1 3 s⋅+( )t B 4 /⋅ t⋅( )−:=
Ag s1 3 s⋅+( )t B:=
An 4.16=
7m2
Ag 11.2=
7m2
f! Ant⋅ 4.023 105×=
0.6 f!⋅ An⋅ 8.486 103×=
f! Ant⋅ 0.6 f!⋅ An⋅≥ ma a,φ 'n 0.+5 0.6 fy⋅ Ag⋅ f! Ant⋅+( )⋅:=
φ 'n 3.123 10 5×=
φ 'n '!≥
(SNI 2002 #al /0, b$tir 10.1.1)1"
3.409 105× 2556.5+≥
.................CD A
-.2 Perencanaan Balok
P o l IC" 1)9. 8.@.
(6abel Baja Profl I
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7m3
+enis : B+ 3(
tf 0.8:=
7m
A 23.04:=
7m2
fy 2100:=
kg57m2
t 0.5:=
7m
fr 0.3 fy⋅:=
= & tegangan sisa
(SNI 2002 Hal Point 5.1 ? 3"
1210:=
7m(
fr 630=
kg57m2
& 2000000:=
kg57m2
y 9+.5:=
7m(
800000:=
kg57m2
r +.26:=
7m
-.2.1 6ek Terhadap 7entur
"akto Pengali %omen 0-
1.136:=
&8ntuk balok sederhana)
Nilai 7- #i!e ole$ -e #asa kan tu unan umus :
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2,3
3.M4.M3.MM2,5M12,5
CCBAmax
maxb ≤+++
=
&Sumber " *andout Dosen)
Dimana :
%A & %omen #i 15( -entang & 3532.6.l2
%B & %ma< & %omen #i 152 -entang & 15>.6.l2
%0 & %omen #i 35( -entang & 3532.6.l2
?e$ingga #i!e ole$ nilai 0- & 1.13@
7 400:=
c
&'ilai bentang terpanjang tanpa pengekang lateral)
(SNI 2002 #al 3 ,tabel .3)2"
7$ 1.+6 ry⋅ &
fy⋅:=
7$ 111.888=
c
Dimana + yaitu konstanta !unte to si #engan umus :
3ii .hb3
1J ∑=
&Sumber " *andout Dosen)
?e$ingga umus + menja#i ::
1
3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=
1.558=
7m(
1 π & ⋅ ⋅ A⋅
2⋅:=
1 1.211 105×=
c
tf 3⋅
H2
12⋅
3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=
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kg 7m
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
M! 4383.98:=
kg 7m
6ek 7entur "
M! φMn≤
(SNI 2002 #al 3%, b$tir .1.1"
4383.98 1.4+5 109×≤
.......4 ;;;
-.2.2 Perencanaan Batang Tekan
7 450:=c
&'ilai bentang terpanjang tanpa pengekang lateral)
λc7
fy
&⋅
π ry⋅:=
λc 2.253=
(SNI 2002 #al 5!, b$tir .2"
ω 1.25 λc( ) 2⋅:=
ω 6.346=
(SNI 2002 #al 5 , b$tir .3)!"
;cr fy
ω:=
;cr 330.925=
kg57m2
φ 'n 0.85 A⋅ ;cr ⋅:=
φ 'n 6.481 103×=
g
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
'! 568.44:=
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g
φ 'n '!≥
(SNI 2002 #al ! , b$tir . .2"
8.202 103
× 568.44≥......4 ;;;
ombinasi Momen 7entur Dan 'ormal Tekan
'!
φ 'n0.088=
(SNI 2002 #al 0, b$tir 12.5"
:nt!
'!
φ 'n 0.2≥ma a
'!
φ 'n8
9
M!
φMnM!y
φMny+ ⋅+ 0.+6+=
(SNI 2002 #al 0, b$tir 12.5)2"
0.+6 1≤
......4 ;;;-.2.+ Perencanaan Batang Tarik & 'u Tidak ada batang tarik )
(SNI 2002 #al /0, b$tir 10.1"
'n A fy⋅:=
'n 4.838 104×=
g
"akto /e#uksi
7e,e
φ 'n1 0.9 'n⋅:=
φ 'n1 4.355 104×=
g
(SNI 2002 #al /0, b$tir 10.1.1)2a"
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;ract!re
φ 'n2 0.+5 'n⋅:=
φ 'n2 3.629 10 4×=
g
(SNI 2002 #al /0, b$tir 10.1.1)2b"
Diambil nilai φ 'n (ang terbesar! maka "
φ 'n φ 'n1:=
φ 'n 4.355 104×=
g
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
'! 0:=
g
'! φ 'n≤
(SNI 2002 #al /0, b$tir 10.1.1)1"
0 4.355 104×≤
....4 ;;;
-.2.- Pemeriksaan Terhadap 5eser
2 tf −:=
2 16.+=
c
a 400:=
c
A 2 t−:=
A 16.2=
7m2
2
t33.4=
a
223.952=
Terdiri dari + s(arat! di antaran(a (aitu "
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Dn 5 5
a
2 2
+:=
(SNI 2002 #al %5, b$tir . )2a"
Dn 5.009=
1.1 Dn &⋅
fy⋅ +5.9+3=
ma a
(SNI 2002 #al %5, b$tir . )2a"
2
t1.1
Dn &⋅
;y⋅≤
33.4 +5.9+3≤
....4 ;;;
En 0.6 fy⋅ A⋅:=
En 2.041 104×=
g
(SNI 2002 #al %!, b$tir . )3a"
oefsien . aktor
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E! 6.51 103×=
φMn 1.4+5 10 9×=
φEn 1.83+ 10 4×=
(SNI 2002 #al % , b$tir . )2"
M!
φMn0.625
E!
φEn⋅+ 0.221=
0.221 1≤
....4 ;;;
-.+ Perencanaan olom
P o l IC" 288.198.@.
(6abel Baja Profl I
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7m
fr 0.3 fy⋅:=
= & tegangan sisa
(SNI 2002 Hal Point 5.1 ? 3"
2690:=
7m(
fr 630=
kg57m2
& 2000000:=
kg57m2
y 50+:=
7m(
( 800000:=
kg57m2
r 8.30:=
7m
-.+.1 6ek Terhadap 7entur
"akto Pengali %omen 0-
) 1.136:=
&8ntuk balok sederhana)
Nilai 7- #i!e ole$ -e #asa kan tu unan umus :
2,33.M4.M3.MM2,5
M12,5C
CBAmax
maxb ≤+++
=
&Sumber " *andout Dosen)
Dimana :
%A & %omen #i 15( -entang & 3532.6.l2
%B & %ma< & %omen #i 152 -entang & 15>.6.l2
%0 & %omen #i 35( -entang & 3532.6.l2
?e$ingga #i!e ole$ nilai 0- & 1.13@
7) 400:=
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c
&'ilai bentang terpanjang tanpa pengekang lateral)
(SNI 2002 #al 3 ,tabel .3)2"
7$ 1.+6 ry⋅ &
fy⋅:=
7$ 196.0+6=
c
Dimana + yaitu konstanta !unte to si #engan umus :
3ii .hb3
1J ∑=
&Sumber " *andout Dosen)
?e$ingga umus + menja#i :: 1
3 H 2 ⋅+ 2/+( )⋅ t 3( )⋅:=
3.55+=
7m(
1 π & ⋅ ⋅ A⋅
2⋅:=
1 1.195 105×=
c
tf 3⋅
H2
12⋅
3 ⋅ tf ⋅ 2 H⋅ t⋅+6 ⋅ tf ⋅ H t⋅+ ⋅:=
3.+02 104×=
7m@
2 4
y ⋅
⋅
2⋅:=
2 2.+68 10 6−×=
c
7r ry 1
fy fr − ⋅ 1 1 2 fy fr −( ) 2⋅++⋅:=
7r 560.019=
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c
7 400:=
c
7$ 7< 7r <
(SNI 2002 #al 3 , b$tir .3.5"
196.0+6 400< 560.019<
6ermas$* Bentan ;enen a#
...... 4
(SNI 2002 #al 3/, tabel .3)1"
Mcr & y⋅ ⋅ ⋅ π &⋅
7 2 y⋅ ⋅+⋅:=
Mcr 8.925 109×=
kg 7m
uat 7entur
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-.+.2 Perencanaan Batang Tekan
7 450:=
c
&'ilai bentang terpanjang tanpa pengekang lateral)
λc7
fy
&⋅
π ry⋅:=
λc 1.286=
(SNI 2002 #al 5!, b$tir .2"
ω 1.25 λc( ) 2⋅:=
ω 2.066=
(SNI 2002 #al 5 , b$tir .3)!"
;cr fy
ω:=
;cr 1.016 103×=
kg57m2
φ 'n 0.85 A⋅ ;cr ⋅:=
φ 'n 3.3+ 10 4×=
g
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
'! 12269.33:=
g
φ 'n '!≥
(SNI 2002 #al ! , b$tir . .2"
4.+35 104× 12269.33≥
......4 ;;;
ombinasi Momen 7entur Dan 'ormal Tekan
'!
φ 'n0.364=
(SNI 2002 #al 0, b$tir 12.5"
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:nt!
'!
φ 'n0.2≥
ma a
'!
φ 'n8
9
M!
φMnM!y
φMny+ ⋅+ 1.043=
(SNI 2002 #al 0, b$tir 12.5)2"
0.849 1≤
......4 ;;;
-.+.+ Perencanaan Batang Tarik
(SNI 2002 #al /0, b$tir 10.1" 'n A fy⋅:=
'n 8.192 104×=
g
"akto /e#uksi
7e,e
φ 'n1 0.9 'n⋅:=
φ 'n1 +.3+3 10 4×=
g
(SNI 2002 #al /0, b$tir 10.1.1)2a"
;ract!re
φ 'n2 0.+5 'n⋅:=
φ 'n2 6.144 104×=
g
(SNI 2002 #al /0, b$tir 10.1.1)2b"
Diambil nilai φ 'n (ang terbesar! maka "
φ 'n φ 'n2:=
φ 'n 6.144 10 4×=
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g
(Diperole# Nilai Dari 'nalisa Beban Den an S'P"
'! 286.+2:=
g
'! φ 'n≤
(SNI 2002 #al /0, b$tir 10.1.1)1"
286.+2 6.144 104×≤
....4 ;;;
-.+.- Pemeriksaan Terhadap 5eser
2 tf −:=
2 18.5=c
a 400:=
c
A 2 t−:=
A 1+.9=
7m2
2
t 30.833=
a
221.622=
Terdiri dari + s(arat! di antaran(a (aitu "
Dn 5 5
a
2 2
+:=
(SNI 2002 #al %5, b$tir . )2a"
Dn 5.011=
1.1 Dn &⋅
fy⋅ +5.988=
ma a
(SNI 2002 #al %5, b$tir . )2a"
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2
t1.1
Dn &⋅
;y⋅≤
20 69.613≤
....4 ;;;
En 0.6 fy⋅ A⋅:=
En 2.255 104×=
g
(SNI 2002 #al %!, b$tir . )3a"
oefsien . aktor
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BAB C
P>'6A'AA' SAMB8'5A'
?am-ungan Balok #an Kolom
#;
-a,o
Do,om
%u &dari nilai momen balok (ang di chek)
M! 4383.98:=
kg5mm2
Fu &dari nilai geser balok (ang di check)
E! 62+8.03:=
g
P o l Balok yang #i!akai
IC" 1)9G 8
) 1+5:=
m
90:=
m
Baut yang #igunakan
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-a,o ( #;)
Pe,at
- 34−
f! 340:=
MPa
fy 210:=
MPa
(SNI 2002 #al 11, tabel 5.3"
Diamete Baut =-
φ 58 2.54⋅ 10⋅:=
φ 15.8+5=
m
Luas Baut A-
A 1
4 π⋅ φ 2⋅:=
A 19+.933=
mm2
+umla$ -aut n
n 4:=
6heck 5eser
(SNI 2002 #al 100, b$tir 13.2)%"
+umla$ -i#ang gese m
m 1:=
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Asumsi : Aki-at momen semua -aut mengalami ta ik
/1/2/3/4/
/1/2/3/4/
R$
a
Rn
-a!t
/! & /n
/! & a < - < =y
Rn n ft⋅ 0.+5⋅ A⋅:=
Rn 3.68+ 105×=
a Rn
fy⋅:=
a 10.034=F " a aF /1<
Asumsi semua -aut mengalami ta ik Bena
apasitas Momen
s 4 φ ⋅( )−
5:=
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s 5.3=
/1 s 1
2 φ ⋅+:=
/1 13.238=
/2 2 s⋅ 3
2 φ ⋅+:=
/2 34.413=
/3 3 s⋅ 5
2 φ ⋅+:=
/3 55.58+=
/4 4 s⋅ +
2 φ ⋅+:=
/4 +6.+63=
/ :=
/ 90=
M/ φf 2⋅ ft⋅ 0.+5⋅ A⋅ /1 /2+ /3+ /4+( )⋅ B φy a⋅ ⋅ fy⋅ / a2
− ⋅+:=
M/ 5.309 10+×=
6heck
M/ M!>
5.309 10+× 4563.31>
........4
BAB C,,
>S,MP87A'
DALA% PE/EN0ANAAN BA+A DI,UNAKAN P/ "IL
1. , /DIN,
P o l 0 Lig$t 0$annel : 198 G @9 G 28 G 3.2
H 15:=
c
44.3:=
7m3
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) 6.5:=
c
y 12.2:=
7m3
/ 2:=
c
332:=
7m(
t 0.32:=
c
y 53.8:=
7m(
tf 0.32:=
c
r 5.89:=
c
A 9.56+:=
7m2
ry 2.3+:=
c
%ate ial Baja
& 2000000:=
kg57m2
fy 2100:=
kg57m2
fr 0.3 fy⋅:=
( 800000:=
kg57m2
fr 630=
kg57m2
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2. /A"*E/
P o l IC" 198.188.>.12
P o!e ties Penam!ang
14.8:=
7m
ry 2.3+:=
7m
10:=
7m
138:=
7m3
tf 0.9:=
7m
A 26.84:=
7m2
t 0.6:=
7m
& 2000000:=
kg57m2
1020:=
7m(
800000:=
kg57m2
y 151:=
7m(
fy 2100:=
kg57m2
r 6.1+:=
7m
fr 630:=
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kg57m2
3.-A7CD
P o l IC" 1)9. 8.11.1)
P o!e ties Penam!ang
1+.5:=
7m
ry 2.06:=
7m
) 9:=
7m
139:=
7m3
tf 0.8:=
7m
A 23.04:=
7m2
t 0.5:=
7m
& 2000000:=
kg57m2
1215:=
7m(
( 800000:=
kg57m2
y 9+.5:=
7m(
fy 2100:=
kg57m2
r +.26:=
7m
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fr 630:=
kg57m2
(. K L %
P o l IC" P o l IC" 288.198.11.1)
P o!e ties Penam!ang
19.4:=
7m
ry 3.60:=
7m
15:=
7m
2++:=
7m3
tf 0.9:=
7m
A 39.01:=
7m2
t 0.6:=
7m
& 2000000:=
kg57m2
2690:=
7m(
800000:=
kg57m2
y 509:=
7m(
fy 2100:=
kg57m2
r 8.30:=
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7m
fr 630:=
kg57m2
cos 2 π6
:=
y s"n 2 π6
:=
B>BA' A'5,'
0.5=
y 0.866=
'n in te*an pada ra ter
/a=te 1 #an (/a=te 1 #an (
#1 ⋅ 2.5=
kg
#1 y⋅ 4.33=
kg
/a=te 2 #an 3
/a=te 2 #an 3#2 ⋅ 5=
kg
#2 y⋅ 8.66=
kg
'n in #isap pada ra ter
/a=te 1 #an (
/a=te 1 #an (#6 ⋅ 10=
kg
#6 y⋅ 1+.321=
/a=te 2 #an 3
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/a=te 2 #an 3
#+ ⋅ 10=
kg
#+ y⋅ 1+.321=