baja.docx

Univesitas Negeri Surabaya

Merencana Konstruksi Baja

PERHITUNGAN

A. GORDING

Diketahui :

Bentang Kuda-kuda = 11 m

Panjang gudang = 70 m

Tinggi kolom = 7.25 m

Sudut atap = 20°

Bahan penutup atap = Seng Gelombang

Bahan penutup dinding = Batu bata

Mutu baja = 37 Mpa = 1600 Kg/cm²

Over stek = 3 m

Jarak antar kolom = 5 m

Pondasi = Plat setempat

1. Perhitungan Gording

A. Perhitungan jarak antar gording

a. Panjang lereng atap (S1)

AB=½ x Lcos α

=½ x11mcos 20°

=5.5 m0.94

=5.85 m

b. Banyak lapangan gording (N)

Atap= S 10.95 m

=5.85 m0.95 m

=6.16 ≈ 6lapangan

c. Jarak gording (N)

Atap=S 1N

= 5.85 m6lapangan

=0.975 ≈ 1 m

d. Tinggi Kuda-kuda

sin 20 °= BEAB

BE=sin 20 ° x AB

BE=0.34 x 5.85 m

BE=2m

Indra Christian (105433214) D3 TS B 2010 Page | 1

5.00

q = 35 Kg/m’



B. PEMBEBANAN

a. Beban Mati (Berat Sendiri)

Beban penutup atap Seng 10 Kg/m² (Lihat buku PMI)

Beban atap = jarak gording x jarak kolom

jarak kolomx beban penutupatap

= 1m x5 m

5 mx10 Kg /m ²

= 10 Kg/m

Beban gording ang ditafsir = Bentang jarak antar kolom

= Beban gording x Bentang gording

= 5 Kg/m² x 5 m

= 25 Kg/m

Total beban seluruh = Beban atap + Beban gording

= 10 Kg/m + 25 Kg/m

= 35 Kg/m

Setelah diperoleh besarnya beban mati, maka dapat ditentukan kemudian momen

yang terjadi pada batang ang mana ketahui adalah beban merata. Maka kemudian

momen ang terjadi di tentukan dengan cara berikut.

Momen ang terjadi

Mx = ⅛ x q x cos α x L² My = ⅛ x q x sin α x L/3²

Mx = ⅛ x 35 Kg/m x cos 20° x (5 m)² My = ⅛ x 35 Kg/m x sin 20° x (5/3 m)²

Mx = ⅛ x 35 Kg/m x 0.939 x 25 m² My = ⅛ x 35 Kg/m x 0.342 x 2.77 m²

Mx = 102.703 Kgm My = 4.133 Kgm


P = 100 Kg/m

5.00

Lihat buku PMI

Menghisap C3 = -0.4Menekan C2 = 0.3

C4 = -0.4



b. Beban Hidup

Momen akibat = orang + barang (P) = 100 Kg (PMI 1970 Hal.15)

Momen ang terjadi

Mx = ¼ x P x cos α x L My = ¼ x P x sin α x L/3

Mx = ¼ x 100 Kg/m x cos 20° x (5 m) My = ¼ x 100 Kg/m x sin 20° x (5/3 m)

Mx = ¼ x 100 Kg/m x 0.939 x 5 m My = ¼ x 100 Kg/m x 0.342 x 1.67m

Mx = 117.375 Kgm My = 14.25 Kgm

c. Beban Angin

Berdasarkan buku PMI Pasal 43 ayat 3

C1 = 0.9 C3 = -0.4

C2 = 0.02 x cos α – 0.4 C4 = -0.4

= 0.02 x cos 20° – 0.4 Tekanan angin = 40 Kg/m²

= -0.381

Karena yang digunakan adalah hanya nilai C2 & C3 yaitu gaya yang menekan dan

menghisap, maka yang dicari hanya q2 dan q


5.00

C2 = -0.381

4.00

C3 = -0.4



Angin tekan (q2) = Tekana

n angin xJarak kolom x Jarak gording x Koefisienangintekan

Jarak kolom

= 40 Kg /m ² x5 m x 1m x−0.381

5 m

= -15.24 kg/m

Angin tekan (q2) =

Tekanan angin xJarak kolom x Jarak gording x Koefisien angintekan

Jarak kolom

= 40 Kg /m ² x5m x 1m x (−0.4)

5 m

= -16 kg/m

Momen akibat beban angin

Mx2 = ¼ x q1 x L² My = ¼ x q2 x L²

Mx2 = ¼ x (-15.24 kg/m) x (5 m)² My = ¼ x (-16 kg/m) x (5 m)²

Mx2 = -47.625 Kgm My = -50 Kgm

My = 0 (titik berat pada sumbu gording)




C. MOMEN GABUNGAN

A+B = Beban Mati + Beban Hidup

MxAB = MxA + MxB

= 102.703 Kgm + 117.375 Kgm

= 220.078 Kgm

MyAB = MyA + MyB

= 4.133 Kgm + 14.25 Kgm

= 18.383 Kgm

Banyak Gording = Panjang lereng atap : Jarak gording

= 5.85 m : 1.00 m

= 5.85 m 6 gording + 1 gording = 7 gording


A+C = Beban Mati + Beban Angin

MxAC1 = MxA + MxC1

= 102.703Kgm + 0 Kgm

= 102.703 Kgm

MxAC2 = MxA + MxC2

= 102.703 Kgm + (- 47.625 Kgm)

= 55.078 Kgm

MyAC = MyA + MyC

= 4.133 Kgm + 0 Kgm

= 4.133 KgmJadi Momen yang terbesar adalah :

Mx = 220.078 Kgm 22007.8 Kgcm

My = 18.383 Kgm 1838.3 Kgcm



HITUNGAN PROFIL GORDING

Tegangan Baja 37 Mpa = 1600 kg/cm²

Mx = 22007.8 kgcm

My = 1838.3 kgcm

Persamaan Lendutan

1500

x L=

5384

xqL4

E . Ix18

x q .L2

L500

= 5384

xq . L4

E . Ixx

8q . L4

L500

= 584

xMx . L4

E x Ixx

8qL4

E . Ix=5 .500 . Mx . L²

Ix=5 .500 . Mx . L ²48. E

Ix=5 .500 .22007 .8 Kgcm.(500 cm) ²

48 x 2.1 x106 Kg /cm2

Ix=¿ 272.91 cm4

Maka berdasarkan Tabel Profil Baja dipakai nilai Ix yang mendekatiyaitu Ix = 368 cm4Maka profil yang diperoleh dari Tabel Profil Baja adalah 150 x 50 x 20 x 4.5Dan diperoleh data sebagai berikut.ix = 5.60 cm iy = 1.75 cmIx = 368 cm4 Iy = 35.7 cm4Wx = 49 cm3 F = 11.72 cm2




Wy = 10.5 cm3 g = 9.2 Kg/m

Kontrol LendutanFx’ = 5

48.

Mx . L2

E . Ix Fy’ = 5

48.

My . L2

E . Iy

= 548

.22007.8 Kgcm x (500 cm)2

2,1 x 106 Kg /cm2 x 368 cm4 = 548

.1838.3 Kgcm x (500 cm)2

2,1 x 106 Kg /cm2 x 35.7 cm4

= 548

.27509.75 x106 kgcm3

37094.40 x106 kg cm2 = 548

.2297.875 x106 kgcm3

3598.56 x106 kg cm2 = 0,742 cm = 0,62 cmKontrol Kekuatan

√(F x ' )2+(F y ' )2 < 1500

. L0.966 cm < 1

500. 500 cm

0.966 cm < 1.00 cm Maka Dikatakan Kuat & Layak GunaKontrol Tegangan ( Moment yang Terjadi ) terjadi = Mx

Wx + MyWy < Tegangan Dasar

= 22007.8 Kgcm

49 cm3 + 1838.3 Kgm

10.5 cm3 < 1600 Kg/cm2 = 449.14 Kg/cm2 + 175.10 Kg/cm2 < 1600 Kg/cm2 = 624.24 Kg/cm2 < 1600 Kg/cm2 Maka Dikatakan Kuat & Layak Guna




PERHITUNGAN BEBAN TERPUSAT

A. Beban yang terjadi (P)

a. Beban Mati

Beban atap = (2 x sisi miring) x jarak kolom x berat penutup atap

= (2 x 5.85 m) x 5 m x 10 Kg/cm2

= 585 Kg

Beban Gording kanal C

Berat gording kanal C 150 x 50 x 20 x 4.5 = 9.2 Kg/m (table profil)

Beban gording = berat gording x jarak kolom x (jumlah gording x 2)

= 9.2 Kg/m x 5 m x (7 x 2)

= 644 Kg

Beban Kuda-kuda

Berat tapsir kuda-kuda = bentangan kuda-kuda = 11 Kg/m2

Beban kuda-kuda = bentangan kuda-kuda x jarak kolom x berat kuda-kuda

= 11 m x 5 m x 11 Kg/m2

= 605 Kg

Beban Terpusat (P) =

bebanatap+beban gording+bebankuda} over {jumlah titik simpul¿

= 585 Kg+644 Kg+605 Kg

8

= 141.077 Kg ≈ 141 Kg

Beban ½ P = Beban terpusat

2

= 141 Kg

2 = 70.5 Kg




B. Beban Angin (W)

Koefisien angin tekan = + 0.02 x cos α – 0.4

Tekanan angin = 40 Kg/m2

W tekan = koefisien angin tekan x tekanan angin x jarak antar gording x jarak antar

kolom

= (0.02 x cos 20° - 0.4) x 40 Kg/m2 x 1 m x 5 m

= (0.02 x 0.94 – 0.4) x 40 Kg/m2 x 1 m x 5 m

= - 0.381 x 40 Kg/m2 x 1 m x 5 m

= - 76.2 Kg ≈ - 77 kg




PERHITUNGAN BATANG TARIK DAN TEKAN

Dari metode Chremona didapat :

Batang a2 = -3682.57 Kg Batang b6 = 567.81 Kg

Batang b1 = -222.93 Kg Batang b7 = -557.72 Kg

Batang b2 = 409.30 Kg Batang b8 = 661.22 Kg

Batang b3 = -334.41 Kg Batang b9 = -668.81 Kg

Batang b4 = 482.30 Kg Batang b10 = 915.66 Kg

Batang b5 = -445.87 Kg Batang c1 = 3560.11 Kg

Direncanakan : σ = 1600 Kg/cm2

δ = 10 mm

1. Batang Tekan & Plat simpul

a. Perhitungan Batang a2

Batang a2 : P = 3682.57 Kg Tegangan sementara diambil σs = 1000 Kg/cm2

L = 0.95 m

F = P

σ s = 3682.57 Kg

1000 Kg /cm² = 3.68 cm²

Dicoba profil siku ˩L 35.35.4

ix = iy = 1.05 cm e = 1.00 cm Ix = Iy = 2.96 cm4 d = 11 mm

F = 2.67 cm² F1 = 2.23 cm² ίη = 0.68 cm

F = 2 x F1

= 2 x 2.23 cm² = 4.46 cm²

Lk = ½ x L Jepit - jepit




= ½ x 0.95 m = 0.475 m = 47.5 cm

λ = Lk

imin = 5.85 cm1.05 cm = 45.24 ≈ 45 ( = 1.187 – Lihat tabel profil )

Kontrol : σko = P

2 x F = 3682.57 Kg2 x 2.67 cm ² = 689.62 Kg/cm

2 ( double )

σk = σ❑ =

1600 Kg /cm21.187 = 1347.94 Kg/cm

2

σk > σko Profil baja siku ˩L 35.35.4 bisa di pakai….. Ok!!

Kontrol terhadap sumbu Y

Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 2.96 cm4 + ( 2.23 cm² x ( 1 cm )2 )

= 2 x ( 2.96 cm4 + 2.23 cm4 )

= 2 x 5.19 cm4 = 10.38 cm4

c = 10 mm ; h = 35 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 10.38 cm4 = 9.34 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 9.34 cm4

4.46cm ² = √2.095 cm2 = 1.447 cm

Iy’ > ix Maka Profil baja siku ˩L 35.35.4

1.447 cm > 1.05 cm bisa di pakai……. Ok !!!

Perhitungan Plat Simpul

L’ = i ηiy '

x Lk

= 0.68 cm

1.447 cm x 47.5 cm = 22.32 cm

Jumlah lapangan = LkL'

= 47.5 cm

22.32 cm

= 2.13 ≈ 3

Jarak rata - rata lapangan = Lk

Jumlah lapangan




= 47.5cm

3

= 15.83 cm

Perhitungan Baut

Keterangan : Bila t > 0.785 x Ø Geser

Bila t < 0.785 x Ø Tumpu

t > 0.785 x Ø

10 mm > 0.785 x 11 mm

10 mm > 8.635 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2

Kekuatan satu baut = 0.25 x π x Ø2 x τ

= 0.25 x 3.14 x ( 1.1 cm) 2 x 923.20 Kg/cm2

= 876.90 Kg

ƞg = P

Kekuatansatu baut Maka jumlah baut yang ditentukan

= 3682.57 Kg876.90 Kg = 4.2 ≈ 4 buah adalah 4 Ø 11 mm

b. Perhitungan Batang b1

Batang b1 : P = 222.93 Kg Tegangan sementara diambil σs = 1000 Kg/cm2

L = 0.30 m

F = P

σ s = 222.93 Kg

1000 Kg /cm² = 0.22 cm²





ix = iy = 0.90 cm e = 0.84 cm Ix = Iy = 1.41 cm4 d = 8.5 mm

F = 1.74 cm² F1 = 1.49 cm² ίη = 0.57 cm

F = 2 x F1

= 2 x 1.49 cm² = 2.98 cm²


= ½ x 0.30 m = 0.15 m = 15 cm

λ = Lk

imin = 15 cm

0.90cm = 16.67 ≈ 17 ( = 1.00 – Lihat tabel profil )

Kontrol : σko = P

2 x F = 222.93 Kg

2 x 1.74 cm ² = 64.06 Kg/cm2 ( double )

σk = σ❑ =

1600 Kg /cm21.00 = 1600 Kg/cm

2



Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 1.41 cm4 + ( 1.49 cm² x ( 0.84 cm )2 )

= 2 x ( 1.41 cm4 + 1.25 cm4 )

= 2 x 2.67 cm4 = 5.34 cm4

c = 10 mm ; h = 30 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 5.34 cm4 = 4.8 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 4.8 cm4

2.98 cm² = √1.61 cm2 = 1.27 cm




L’ = i ηiy '

x Lk

= 0.57cm1.27cm

x 15 cm = 6.73 cm





= 15 cm

6.73cm

= 2.23 ≈ 3


Jumlah lapangan

= 15 cm

3 = 5 cm

Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P


= 222.93 Kg523.60 Kg = 0.4 ≈ 2 buah adalah 2 Ø 8.5 mm

c. Perhitungan Batang b3


L = 0.60 m

F = P

σ s = 334.41 Kg

1000 Kg /cm² = 0.33 cm²






F = 1.74 cm² F1 = 1.49 cm² ίη = 0.57 cm

F = 2 x F1

= 2 x 1.49 cm² = 2.98 cm²


= ½ x 0.60 m = 0.30 m = 30 cm

λ = Lk

imin = 30 cm


Kontrol : σko = P

2 x F = 334.41 Kg

2 x 1.74 cm ² = 96.10 Kg/cm2 ( double )

σk = σ❑ =

1600 Kg /cm21.096 = 1459.85 Kg/cm

2



Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 1.41 cm4 + ( 1.49 cm² x ( 0.84 cm )2 )

= 2 x ( 1.41 cm4 + 1.25 cm4 )

= 2 x 2.67 cm4 = 5.34 cm4

c = 10 mm ; h = 30 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 5.34 cm4 = 4.80 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 4.8 cm4

2.98 cm² = √1.61 cm2 = 1.27 cm




L’ = i ηiy '

x Lk

= 0.57cm1.27cm

x 30 cm = 13.46 cm





= 30 cm

13.46 cm

= 2.28 ≈ 3


Jumlah lapangan

= 30 cm

3 = 10 cm

Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



d. Perhitungan Batang b5


L = 0.90 m

F = P

σ s = 445.87 Kg

1000 Kg /cm² = 0.44 cm²






F = 1.74 cm² F1 = 1.49 cm² ίη = 0.57 cm

F = 2 x F1

= 2 x 1.49 cm² = 2.98 cm²


= ½ x 0.90 m = 0.45 m = 45 cm

λ = Lk

imin = 45 cm

0.90cm = 50 ( = 1.234 – Lihat tabel profil )

Kontrol : σko = P

2 x F = 445.87 Kg

2 x 1.74 cm ² = 128.12 Kg/cm2 ( double )

σk = σ❑ =

1600 Kg /cm21.234 = 1296.6 Kg/cm

2



Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 1.41 cm4 + ( 1.49 cm² x ( 0.84 cm )2 )

= 2 x ( 1.41 cm4 + 1.25 cm4 )

= 2 x 2.67 cm4 = 5.34 cm4

c = 10 mm ; h = 30 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 5.34 cm4 = 4.80 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 4.80 cm4

2.98 cm² = √1.61 cm2 = 1.27 cm




L’ = i ηiy '

x Lk

= 0.57cm1.27cm

x 45 cm = 20.20 cm





= 45 cm

20.20cm

= 2.27 ≈ 3


Jumlah lapangan

= 45 cm

3 = 15 cm

Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



e. Perhitungan Batang b7


L = 1.2 m

F = P

σ s = 557.72 Kg

1000 Kg /cm² = 0.56 cm²






F = 1.74 cm² F1 = 1.49 cm² ίη = 0.57 cm

F = 2 x F1

= 2 x 1.49 cm² = 2.98 cm²


= ½ x 1.2 m = 0.6 m = 60 cm

λ = Lk

imin = 60 cm


Kontrol : σko = P

2 x F = 557.72 Kg

2 x 1.74 cm ² = 160.26 Kg/cm2 ( double )

σk = σ❑ =

1600 Kg /cm21.425 = 1122.80 Kg/cm

2



Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 1.41 cm4 + ( 1.49 cm² x ( 0.84 cm )2 )

= 2 x ( 1.41 cm4 + 1.25 cm4 )

= 2 x 2.61 cm4 = 5.34 cm4

c = 10 mm ; h = 30 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 5.34 cm4 = 4.80 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 4.80 cm4

2.98 cm² = √1.61 cm2 = 1.27 cm




L’ = i ηiy '

x Lk

= 0.57cm1.27cm

x 60 cm = 26.93cm





= 60 cm

26.93cm

= 2.23 ≈ 3


Jumlah lapangan

= 60 cm

3 = 20 cm

Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



f. Perhitungan Batang b 9

Batang a2 : P = 668.81 Kg Tegangan sementara diambil σs = 1000 Kg/cm2

L = 1.50 m

F = P

σ s = 668.81 Kg

1000 Kg /cm² = 0.67 cm²






F = 1.74 cm² F1 = 1.49 cm² ίη = 0.57 cm

F = 2 x F1

= 2 x 1.49 cm² = 2.98 cm²


= ½ x 1.5 m = 0.75 m = 75 cm

λ = Lk

imin = 75 cm


Kontrol : σko = P

2 x F = 668.81 Kg

2 x 1.74 cm ² = 192.20 Kg/cm2 ( double )

σk = σ❑ =

1600 Kg /cm21.667 = 959.80 Kg/cm

2



Iy1 = 2 x ( Iy + ( F1 x e2 ) )

= 2 x ( 1.41 cm4 + ( 1.49 cm² x ( 0.84 cm )2 )

= 2 x ( 1.41 cm4 + 1.25 cm4 )

= 2 x 2.61 cm4 = 5.34 cm4

c = 10 mm ; h = 30 mm maka c < h = 0.9 Iy’

Iy” = 0.9 x Iy’

= 0.9 x 5.34 cm4 = 4.80 cm4

Iy’ = √ Iy} over {2 x F1} ¿¿ = √ 4.80 cm4

2.98 cm² = √1.61 cm2 = 1.27 cm




L’ = i ηiy '

x Lk

= 0.57cm1.27cm

x 75 cm = 33.66 cm





= 75 cm

33.66 cm

= 2.23 ≈ 3


Jumlah lapangan

= 75 cm

3 = 25 cm

Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 11 mm

10 mm > 8.635 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P






2. Batang Tarik & Plat Simpul

a. Perhitungan Batang b2 = 3560.11 kg

F netto = Pσ

= 3560.11 Kg

1600 Kg /cm2 = 2.23 cm2

F bruto = 1.15 x F netto

= 1.15 x 2.23 cm2 = 2.56 cm2

F1 = 0.5 x F bruto

= 0.5 x 2.56 cm2 = 1.28 cm2


ix = iy = 0.88 cm Ix = Iy = 2.16 cm4 d = 8.5 mm

F = 2.78 cm² F1 = 2.36 cm²

Fbr tabel dikurangi perlemahan Fn

Fbr tabel = 2 x F1 = 2 x 2.36 cm² = 4.72 cm²

Perlemahan tiap profil

= 2 x Ø x t = 2 x 0.85 cm x 0.5 cm= 0.85 cm² +

Fn = 3.87 cm²

Kontrol

P

Fn < σ

3560.11 Kg

3.87 cm2 < 1600 Kg/cm2

919.92 Kg/cm2 < 1600 Kg/cm

2 Maka Profil baja siku ˩L 30.30.5

bisa di pakai……. Ok !!!

Perhitungan Baut






t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



b. Perhitungan Batang b 2 = 409.30 kg

F netto = Pσ

= 409.30 Kg

1600 Kg /cm2 = 0.26 cm2


= 1.15 x 0.26 cm2 = 0.3 cm2

F1 = 0.5 x F bruto

= 0.5 x 0.3 cm2 = 0.15 cm2



F = 1.74 cm² F1 = 1.49 cm²







= 2 x Ø x t = 2 x 0.85 cm x 0.3 cm= 0.51 cm² +

Fn = 2.47 cm²

Kontrol

P

Fn < σ

409.30 Kg

2.47 cm2 < 1600 Kg/cm2

165.70 Kg/cm2 < 1600 Kg/cm



Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



c. Perhitungan Batang b 4 = 482.3 kg

F netto = Pσ

= 482.3 Kg

1600 Kg /cm2 = 0.3 cm2





= 1.15 x 0.3 cm2 = 0.345 cm2

F1 = 0.5 x F bruto

= 0.5 x 0.345 cm2 = 0.1725 cm2



F = 1.74 cm² F1 = 1.49 cm²




= 2 x Ø x t = 2 x 0.85 cm x 0.3 cm= 0.51 cm² +

Fn = 2.47 cm²

Kontrol

P

Fn < σ

482.3 Kg

2.47 cm2 < 1600 Kg/cm2

195.26 Kg/cm2 < 1600 Kg/cm



Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2




= 523.60 Kg

ƞg = P



d. Perhitungan Batang b 6 = 567.81 kg

F netto = Pσ

= 567.81 Kg

1600 Kg /cm2 = 0.35 cm2


= 1.15 x 0.35 cm2 = 0.40 cm2

F1 = 0.5 x F bruto

= 0.5 x 0.40 cm2 = 0.20 cm2



F = 1.74 cm² F1 = 1.49 cm²




= 2 x Ø x t = 2 x 0.85 cm x 0.3 cm= 0.51 cm² +

Fn = 2.47 cm

Kontrol

P

Fn < σ

567.81 Kg

2.47 cm2 < 1600 Kg/cm2

229.88 Kg/cm2 < 1600 Kg/cm



Perhitungan Baut






t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



e. Perhitungan Batang b 8 = 661.22 kg

F netto = Pσ

= 661.22 Kg

1600 Kg /cm2 = 0.41 cm2


= 1.15 x 0.41 cm2 = 0.475 cm2

F1 = 0.5 x F bruto

= 0.5 x 0.475 cm2 = 0.24 cm2



F = 1.74 cm² F1 = 1.49 cm²

Fbr table dikurangi perlemahan Fn



= 2 x Ø x t = 2 x 0.85 cm x 0.3 cm= 0.51 cm² +

Fn = 2.47 cm²




Kontrol

P

Fn < σ

661.22 Kg

2.47cm2 < 1600 Kg/cm2

267.70 Kg/cm2 < 1600 Kg/cm



Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



f. Perhitungan Batang b 1 0 = 915.66 kg

F netto = Pσ

= 915.66 Kg

1600 Kg /cm2 = 0.57 cm2





= 1.15 x 0.57 cm2 = 0.65 cm2

F1 = 0.5 x F bruto

= 0.5 x 0.65 cm2 = 0.325 cm2



F = 1.74 cm² F1 = 1.49 cm²

Fbr table dikurangi perlemahan Fn



= 2 x Ø x t = 2 x 0.85 cm x 0.3 cm= 0.51 cm² +

Fn = 2.47 cm²

Kontrol

P

Fn < σ

915.66 Kg

2.47cm2 < 1600 Kg/cm2

370.71 Kg/cm2 < 1600 Kg/cm



Perhitungan Baut



t > 0.785 x Ø

10 mm > 0.785 x 8.5 mm

10 mm > 6.67 mm ( Geser )

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2





= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

ƞg = P



3. Perhitungan Baut Tiap Titik

Diketahui :

a. batang A menggunakan profil siku ˩L 35.35.4

Perhitungan baut menggunakan rumus geser dengan diameter baut = 11 mm (tabel profil)

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2

= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 1.1 cm ) 2 x 923.20 Kg/cm2

= 876.90 Kg

b. batang C menggunakan profil siku ˩L 30.30.5

Perhitungan baut menggunakan rumus geser dengan diameter baut = 8.5 mm (tabel

profil)

τ = 0.577 x σ

= 0.577 x 1600 Kg/cm2




= 923.20 Kg/cm2


= 0.25 x 3.14 x ( 0.85 cm ) 2 x 923.20 Kg/cm2

= 523.60 Kg

a. Titik C = Batang a 2−Batanga 1

Kekuatansatu baut

= 3682.57 Kg – 3654.55 Kg

876.90 Kg =

28.02 Kg876.90 Kg

= 0.03 ≈ 2 Ø 11 mm

b. Titik D = Batang c1−Batangc 2

Kekuatan satu baut

= 3560.11 Kg – 3225.01 Kg

523.60 Kg =

335.1 Kg523.60 Kg

= 0.64 ≈ 2 Ø 8.5 mm

c. Titik E = Batang a 2−Batanga 3

Kekuatan satubaut

= 3682.57 Kg – 3353.45 Kg

876.90 Kg =

329.12 Kg876.90 Kg

= 0.37 ≈ 2 Ø 11 mm

d. Titik F = Batang c2−Batang c3

Kekuatan satu baut

= 3225.01 Kg –2889.95 Kg

523.60 Kg =

335.06 Kg523.60 Kg

= 0.64 ≈ 2 Ø 8.5 mm

e. Titik G = Batang a 3−Batanga 4

Kekuatan satu baut

= 3353.45 Kg – 3024.35 Kg

876.90 Kg =

329.10 Kg876.90 Kg

= 0.37 ≈ 2 Ø 11 mm

f. Titik H = Batang c3−Batang c 4

Kekuatan satubaut

= 2889.95 Kg – 2555.46 Kg

523.60 Kg =

334.49 Kg523.60 Kg

= 0.64 ≈ 2 Ø 8.5 mm

g. Titik I = Batang a 4−Batanga 5

Kekuatan satu baut

= 3024.35 Kg−2695.24 Kg

876.90 Kg =

329.11 Kg876.90 Kg

= 0.37 ≈ 2 Ø 11 mm

h. Titik J = Batang c 4−Batang c5

Kekuatan satubaut




= 2555.46 Kg – 2219.77 Kg

523.60 Kg =

335.69 Kg523.60 Kg

= 0.64 ≈ 2 Ø 8.5 mm

i. Titik K = Batang a 5−Batanga 6

Kekuatan satubaut

= 2695.24 Kg−2366.15 Kg

876.90 Kg =

329.09 Kg876.90 Kg

= 0.37 ≈ 2 Ø 11 mm

j. Titik L = Batang c5−Batang c6

Kekuatansatu baut

= 2219.77 Kg – 1816.36 Kg

523.60 Kg =

403.41 Kg523.60 Kg

= 0.77 ≈ 2 Ø 8.5 mm

k. Titik M = Batang a 6−Batanga 7

Kekuatan satu baut

= 2366.15 Kg−1563.75 Kg

876.90 Kg =

802.4 Kg876.90 Kg

= 0.92 ≈ 2 Ø 11 mm

l. Titik N = Batang c6−Batang c7

Kekuatansatu baut

= 1816.36 Kg – 1465.22 Kg

523.60 Kg =

351.14 Kg523.60 Kg

= 0.67 ≈ 2 Ø 8.5 mm

m. Titik O = Batang a 8−Batanga 7

Kekuatan satubaut

= 1787.12 Kg –1563.75 Kg

876.90 Kg =

223.37 Kg876.90 Kg

= 0.25 ≈ 2 Ø 11 mm

n. Titik P = Batang c8−Batang c7

Kekuatansatu baut

= 1675.14 Kg – 1465.22 Kg

523.60 Kg =

209.92 Kg523.60 Kg

= 0.40 ≈ 2 Ø 8.5 mm

o. Titik Q = Batang a 9−Batanga 8

Kekuatan satubaut

= 2010.52 Kg –1787.12 Kg

876.90 Kg =

223.40 Kg876.90 Kg

= 0.25 ≈ 2 Ø 11 mm

p. Titik R = Batang c9−Batang c8

Kekuatansatu baut

= 1883.87 Kg – 1675.14 Kg

523.60 Kg =

208.73 Kg523.60 Kg

= 0.40 ≈ 2 Ø 8.5 mm

q. Titik S = Batang a 10−Batanga 9

Kekuatan satubaut




= 2233.92 Kg –2010.52 Kg

876.90 Kg =

223.40 Kg876.90 Kg

= 0.25 ≈ 2 Ø 11 mm

r. Titik T = Batang c10−Batang c9

Kekuatansatu baut

= 2093.17 Kg – 1883.87 Kg

523.60 Kg =

209.30 Kg523.60 Kg

= 0.40 ≈ 2 Ø 8.5 mm

s. Titik U = Batang a 11−Batanga 10

Kekuatansatu baut

= 2457.32 Kg –2233.92 Kg

876.90 Kg =

223.40 Kg876.90 Kg

= 0.25 ≈ 2 Ø 11 mm

PERHITUNGAN KOLOM

Berat atap H = sisi miring x jarak antar kolom x berat penutup atap x sin α

= 5.85 m x 5 m x 10 Kg/m2 x sin 20°

= 100.04 Kg

V = sisi miring x jarak antar kolom x berat penutup atap x cos α

= 5.85 m x 5 m x 10 Kg/m2 x cos 20°

= 274.86 Kg

Berat gording = berat gording x jumlah gording x jarak antar kolom

= 9.2 Kg/m x 7 buah x 5 m

= 322 Kg

Berat Kuda-kuda

Lereng atap H = berat rangka x 2 x sisi miring x sin α

= 2.10 Kg/m x 2 x 5.85 m x sin 20°




= 8.40 Kg

V = berat rangka x sisi miring x cos α

= 2.10 Kg/m x 2 x 5.85 m x cos 20°

= 23.10 Kg

Emperan H = berat rangka x 2 x sisi emperan x sin α

= 2.10 Kg/m x 2 x 3.19 m x sin 20°

= 4.58 Kg

V = berat rangka x 2 x sisi emperan x cos α

= 2.10 Kg/m x 2 x 3.19 m x cos 20°

= 12.60 Kg

PH = berat atap + berat gording + berat kuda – kuda

= 100.04 Kg + 322 Kg + ( 8.40 Kg + 4.58 Kg )

= 435.02 Kg

PV = berat atap + berat gording + berat kuda – kuda

= 274.86 Kg + 322 Kg + (23.10 Kg +12.60 Kg )

= 632.56 Kg

W = Luasan x tekanan angin x 0.9

W1= ( 2.42 m x 5 m ) x 40 Kg/cm2 x 0.9

= 435.60 Kg




W2= ( 2.42 m x 5 m ) x 40 Kg/cm2 x 0.9

= 435.60 Kg

W3= ( 1.21 m x 5 m ) x 40 Kg/cm2 x 0.9

= 217.80 Kg

M = W x X M total = M1 + M2 + M3

M1= 435.60 Kg x 2.42 m = 1054.15 Kgm + 2108.30 Kgm

+ 1579.05 Kgm

= 1054.15 Kgm = 4741.50 Kgm

M2= 435.60 Kg x 4.84 m

= 2108.30 Kgm

M3= 144 Kg x 7.25 m

= 1579.05 Kgm

Momen akibat tekanan dinding = 4741.50 Kgm = 474150 Kgcm

Wx = Mσ

= 474150 Kgcm

1600 Kg /cm2 = 296.34 cm3

σ = Pσ =

632.56 Kg

1600 Kg /cm2 = 0.395 cm3( dipakai Wx )

Dicoba WF 250 x 250 x 11 x 11

F = 82.06 cm2 tb = 11 mm Wx = 720 cm3

b = 250 mm ts = 11 mm Wy = 233 cm3

h = 250 mm ix = 10.3 cm Ix = 8790 cm4

g = 64.4 Kg/m iy = 5.98 cm Iy = 2940 cm4

P total = P + berat kolom

= 632.56 Kg + ( 64.4 Kg/m x 7.25 m )

= 632.56 Kg + 466.9 Kg




= 1099.46 Kg

σd = MWx =

474150 Kgcm

720 cm3 = 658.54 Kg/cm2

σb = PF =

1099.46 Kg

82.06 cm2 = 13.40 Kg/cm2

σbσd =

13.40 Kg /cm2

658.54 Kg /cm2 = 0.020 < 1.5 VOS

Lk = L Jepit - sendi

λ = Lkiy =

725 cm5.98 cm = 121.24 ≈ 121 ( = 2.873 – Lihat tabel profil )

α = 1ω =

12.873 = 0.35

σdα

+ σb < σ

13.4 Kg /cm2

0.35 + 658.98 Kg/cm

2 < 1600 Kg/cm2

696.83 Kg/cm2 < 1600 Kg/cm

2

WF 250 x 250 x 11 x 11 bisa dipakai………..Ok !!!




PERHITUNGAN PLAT KAKI & BAUT ANGKER

Diketahui :

P total = 1099.46 Kg

M = 474150 Kgcm

Profil = WF 250 x 250 x 11 x 11

σbeton = 70 Kg/cm2

b + 15 cm = 25 cm + 15 cm

= 40 cm

h + 15 cm = 25 cm + 15 cm

= 40 cm

σ+ = PF +

6 x M

b x h2 < σbeton

= 1099.46 Kg

82.06 cm2 + 6 x 474150 Kgcm

40 cm x (40 cm)2 < 70 Kg/cm2




= 57.85 Kg/cm2 < 70 Kg/cm

2

σ- = PF +

6 x M

b x h2 < σbeton

= 1099.46 Kg

82.06 cm2 - 6 x 474150 Kgcm

40 cm x (40 cm)2 < 70 Kg/cm2cm

= - 31.05 Kg/cm2 < 70 Kg/cm

2

1. Mencari Nilai φ

Dengan perbandingan seharga

σ+ x ( b – x ) = σ- x φ

57.85 Kg/cm2 x ( 40 cm – x ) = 31.05 Kg/cm

2 x φ

2314 Kg/cm – 57.85 Kg/cm2 x φ = 31.05 Kg/cm

2 x φ

2314 Kg/cm = 31.05 Kg/cm2 x φ + 57.85 Kg/cm

2 x φ

2314 Kg/cm = 88.90 Kg/cm2 x φ

φ =

2314Kgcm

88.90Kg

cm2

φ = 26 cm




2. Tegangan tarik bautangker

T = φ xσ−¿

2¿ x h

= 26 cm x 31.05

Kgcm

2 x 40 cm

= 16146 Kg

FØ = T10σ ∅

σØ = 980 Kg/cm2

=

16146 Kg10

980Kg

cm2

= 1.6475 Kg/cm2

3. Luas penampang baut

FØ = π x R²

R = √ F Øπ

= √ 1.6475 cm2

3.14 = 0.724 cm

DØ = 2.R = 2 x 0.724 cm

= 1.45 cm = 14.5 mm

4. Panjang baut

F plat tarik = keliling baut x panjang baut x τps

φ x h = 2 x π x L x τps “ τps = 5 – 7 “

L = φ x h

2 x π x R x τ ps

= 26 cm x40 cm

2 x 3.14 x1.025 x5

= 32.31 cm ≈ 33 cm

L = 33 cm – 20 % x L

= 33 cm – 20 % x 33 cm




= 33 cm – 6.6 cm

= 26.4 cm ≈ 264 mm

5. Tebal plat kaki

σ = PF

1666 Kg/cm2 =

1099.46 Kg40 x t

t =

1099.46 Kg

1666Kg

cm2 x 40 cm

t = 0.017 cm

“ Karena terlalu kecil mka dipakai 1 cm “

Sehingga plat kaki t = 1 cm, baut Ø 14.5 cm panjang 26.4 cm


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