contoh kayu tb ( paku)
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BAB I
PERHITUNGAN GAYA-GAYA BATANG
1.1 Data Struktur
Diperhitungkan Untuk Atap Bangunan Bengkel : Bengkel
Bentang Konstruksi : 20 m
Jarak Antar Kolom : 5 m
Bahan Atap : Seng
Bahan Gording : Bahan Balok Kayu
Kelas Kuat Kayu : I
Konstruksi Kayu Termasuk : A
Tekanan Angin yang bekerja : Terlindung
Sifat beban yang bekerja :
Konstruksi Joint/Sambungan : Paku
Peraturan-peraturan Konstruksi :
1.2 Beban yang terpakai / Beban Standar
Beban yang terpakai dalam perhitungan struktur kuda-kuda kayu yaitu :
1. Beban mati
a. Berat sendiri
b. Berat sendiri kuda-kuda
c. Berat sendiri gording
2. Beban hidup
Beban terpusat (orang + alat), diambil 100 kg
3. Beban Angin (tegak lurus bidang atap)
a. Angin kiri
b. Angin kanan
1
A. Perhitungan Panjang Batang
Kemiringan atap = α = 30 ˚
Panjang Batang AC = CD = DE = EB
Panjang Batang AF = BJ
Panjang batang AC = ¼ x L = ¼ x 10 = 2,5 m
F Aa = 1/6*L Fa = Tan α x Aa
= 1/6*10 = Tan 30˚ x 1,666
= 1,667 m = 0,9618 m
AF = √ Aa 2 + Fa2
= √ 1,6662 + 0.96182
= 1,9232 m
Panjang Batang FG = Batang JI
G X2 = 2*1/6*L Gb = Tan α*X2
= 2*1/6*10 = Tan 30˚*3,333
F b = 3,333 m = 1,924 m
A
X2
Batang FG = Batang JI
Ab = ¼ x L + ( ¼ x L – 1/6 x L )
= 3,3333 m
AG = √ Ab2 + Gb2 FG = AG - AF
= √3,3332 +1,92542 = 3,848-1,923
= 3,848 m = 1,925 m
Batang FG = Batang JI
FG = AG – AF
2
H X3 = 2*1/4*L Y3 = Tan α*X3
G = 2*1/4*10 = Tan 30˚*5
F Y3 = 5 m = 2,887 m
A
X3
AH = √ X32 + Y32 GH = AH - AG
= √ 52 +2,8862 = 5,774-3,848
= 5,773 m = 1,925 m
Bentang AC = CD =DE = EB = 2,5 m
Bentang HD = ½*L*Tan 30˚ = 2,887 m
Bentang CF = EJ
X1 = 1,667* Tan 30˚ = 0,963 m
Y1C = 2,5-1,667 = 0,833 m
CF = EJ = √0,9632 +0,8332
= 1,273 m
Bentang CG = EI
XC = 3,334* Tan 30˚ = 1,925 m
Y2C = 2,5-1,667 = 0,833 m
CG = EI = √1,9252 +0,8332
= 2,098 m
Bentang DG = DI
X2 = 3,334* Tan 30˚ = 1,925 m
X2D = 1,667 m
DG = DI = √1,9252 +1,6672
= 2,546 m
Tabel Panjang Batang
3
No. Batang Panjang Batang (m)
1. S1 = S1’ 1,925
2. S2 = S2’ 1,925
3. S3 = S3’ 1,925
4. S4 = S4’ 2,5
5. S5 = S5’ 2,5
6. S6 = S6’ 1,273
7. S7 = S7’ 2,098
8. S8 = S8’ 2,546
9. S9 2,887
B Perhitungan Sudut
1. Segi Tiga AFC Dan BJE
F
1,273 m b C a
0,963 m
A 30˚ C A B
1,667 m 1,667 m c
Cos A˚ =
< ACF = < BEJ
Cos C˚ =
Cos C˚ = 0,654
C˚ = 49,156˚
< AFC = < BJE
Cos F˚ =
Cos F˚ = -0,189 F˚ = 100,894˚
2. Segi Tiga FCG Dan JEI
4
b2+c2-a2
2*b*c
2,52+1,2732-1,9252
2*2,5*1,273
1,9252+1,2732-2,52
2*1,925*1,273
< FGC = < JIE
G Cos G˚ =
1,925 m
2,098 m Cos G˚ = 0,803
F G˚ = 36,571˚
1,273 m
C
< CFG = < EJI
F˚ = 180˚- < AFC
F˚ = 180˚-100,894˚ = 79,106 ˚
< FCG = < JEI
C˚ = 180˚- (36,571˚+79,106˚)
C˚ = 64,323 ˚
3. Segi Tiga CGD Dan EID
< DCG = < DEI
G Cos C˚ =
2,098 m 2,546 m
Cos C˚ = 0,397
C D C˚ = 66,583˚
2,5m
< CGD = < EID
Cos G˚ =
Cos G˚ = 0,434
G˚ = 64,294˚
< CDG = < EDI
D˚ = 180 – (66,583˚ + 64,294˚)
D˚ = 49,123˚
4. Segi Tiga GDH Dan IDH
< DGH = < DIH
5
1,9252+2,0982-1,2732
2*1,925*2,098
2,0982+2,5462-2,52
2*2,098*2,546
2,0982+2,52-2,5462
2*2,098*2,5
H G˚ = 180˚- (64,249˚+ 36,571˚)
1,925 m G˚ = I˚ =79,135 ˚
< GDH = < IDH
G 2.887 m D˚ = 90˚- 49,123˚
D˚ = 40,877 ˚
2,546 m
< GHD = < IHD
D H˚ = 180 – (79,135˚ + 40,877˚)
H˚ = 59,988˚
C. Perhitungan Dimensi Gording
Data Perencanaan :
Jarak kuda-kuda : 3 m
Jarak gording : 1,925 m : 2 = 0,963 m
Ukuran gording : 10/12
Atap seng : 10 kg/m2
Kelas kuat kayu II (Bengkirai) : 0,91 gr/cm3
1. Perhitungan Beban
Notasi Kelas
Kuat II
Mutu A
f = 1
Terlindungi
f =1
Muatan Angin
+ Angin f =5/4
Tegangan
Perencanaan
σlt (kg/cm²) 100 100 100 125 125
σtk //σtr// (kg/cm²) 85 85 85 106,25 106,25
σtk┴ (kg/cm²) 25 25 25 31,25 31,25
τ // (kg/cm²) 15 15 15 15 15
Dimana :
σlt : Tegangan lentur yang diijinkan
σtk // : Tegangan tekan sejajar serat yang diijinkan
σtr // : Tegangan tarik sejajar serat yang diijinkan
σtk : Tegangan tekan tegak lurus serat yang diijinkan
τ // : Tegangan geser sejajar serat yang diijinkan
6
Gx
G Gy
a. Berat Sendiri
Berat sendiri gording = b*h*B*D
= 0,10 m*0,12 m*0,91 gr/cm3*1000
= 10,92 kg/m
Berat sendiri seng (atap) = Berat sendiri seng *Jarak gording
= 10 kg/m2 * 0,963 m
= 9,630 kg/m
Berat total (G) = Berat sendiri gording + berat sendiri seng (atap)
= 10,92 kg/m + 9,630 kg/m
= 20,55 kg/m
Sehingga Gx = G *sin 30˚ = 20,55*sin 30˚ = 10,275 kg/m
Gy = G *cos 30˚ = 20,55*cos 30˚ = 17,797 kg/m
b. Muatan orang dan lain-lain (P = 100 kg) (Peraturan Muatan Indonesia)
Px = P*sin α = 100* sin 30˚ = 50 kg
Py = P*cos α = 100* cos 30˚ = 86,603 kg
c. Muatan angina (W)
W = 18 kg/m2
Koefisien angin (C) = 0,02 *α-0,4 (PMI)
= 0,02*30˚-0,4 = 0,2
Muatanm angin = C *Muatan angin*Jarak gording
= 0,2*18*0,963
= 3,467 kg/m
Maka qx = Gx = 10,275 kg/m
7
qy = Gy + W = 17,797 + 18 = 35,797 kg/m
2. Penyelesaian terhadap lentur, lendutan dan geser
a. Penyelidikan terhadap lentur
Perletakan sendi-rol
Px = 50 kg Mx = 1/8*qx*L2+ 1/4*Px*L
= 1/8*10,275*32+ 1/4*50*3
= 49,059 kg.m ~ 4905,9 kg.cm
1,5 m 1,5 m
Py = 86,603 kg My = 1/8*qy*L2+ 1/4*Py*L
= 1/8*35,797*32+ 1/4*86,603*3
= 105,224 kg.m ~10522,4 kg.cm
1,5 m 1,5 m
Kontrol Tegangan
y Ukuran gording 10/12
Wx = 1/6*b*h2 = 1/6*10*122 = 240 cm3
Wy = 1/6*b2*h = 1/6*102*12 = 200 cm3
12 cm x Ix = 1/12*b*h3 = 1/12*10*123 = 1440 cm4
Iy = 1/12*b3*h = 1/12*103*12 = 1000 cm4
10 cm
σlt = My/Wx + Mx/Wy = 4905,9/240 + 10522,4/200
= 73,053 kg/cm2 < σlt =125 kg/cm2……….(Aman)
8
qx= 10,275 kg/m
qy= 35,797 kg/m
b. Penyelidikan terhadap lendutan
Untuk kelas kuat II (Mutu A) ~ E = 100.000 kg/cm2
Fy = x + x = x x + x
= 0,262 + 0,341
= 0,603 m
Fx = x + x = x x + x
= 0,108 + 0,284
= 0,392 m
Fy = Fy2 + Fx2 = (0,603)2 + (0,392)2 = 0,517 cm
F max = 1/200*L
= 1/200*10
= 1,5 cm
F = 0,517 cm < 1,5 cm…………………………..(Aman)
c. Penyelidikan terhadap geser
Dy = ½*qy*L +½*Py = ½*35,797*3 + ½*86,603 = 96,998 kg
Dx = ½*qx*L +½*Px = ½*10,275*3 + ½*50 = 40,413 kg
τy = 3/2*Dy/b*h = 3/2*98.998/10*12 = 1,212 kg/cm2
τx = 3/2*Dx/b*h = 3/2*40,413/10*12 = 0,505 kg/cm2
τx total = τy + τx = 1,212 + 0,505 = 1,717 kg/cm2
τ total < τ // ~ 1,717 kg/cm2 < 15 kg/cm2…………………..(Aman)
3. Perhitungan Muatan Tetap (P)
9
5 348
qy*L4 EI*Ix
Py*L3
E* Ix 5 348
0,358*3004 100.000*1440
86,603*3003 100.000*1440
1 48
148
5 348
qx*L4 EI*Iy
Px*L3
E* Iy 5 348
0,103*3004 100.000*1000
50*3003 100.000*1000
1 48
148
1,667 m 1,667 m 1,667 m 1,667 m 1,667 m 1,667 m
P2
P2 H P2
S3 S3’
P2 G I P2
S2 9 S2’
P1 F 8 8’ J P1
S1 6 7 7’ 6’ S1’
A 4 5 5’ 4’ B
r C D E
P3 P4 P4 P4 P3
1 m 10 m 1 m
Jarak antar kuda-kuda = 3 m
Bentang kuda-kuda = 10 m
Berat bentang /konstruksi = Koefisien batang * Jarak kuda-kuda*Bentang
Koefisien Bentang :
C = (L-2) + * ((L+5)-(L-2))
Dimana : L = Bentang kuda-kuda = 10 m
α = Sudut kuda-kuda 30˚
C = (10-2) + * ((10+5)-(10-2))
= 8+0 = 8
Maka :
Berat konstruksi = Koefisien bentang *Bentang*Jarak kuda-kuda
= 8*10*3 = 240 kg
Tiap titik buhul menerima = 240/10 = 24 kg
Cos α = x/r ~ r = x/cos α = 1/cos 30˚ = 1,2 m
Muatan P1
1. Berat konstruksi = 240 kg
10
L-10 α-10
10-10 30˚-10
2. Berat atap(1,2+0,963)*3*10 = 64,89 kg
3. Berat gording (2*0,10*0,12*0,91*1000*3) = 65,52 kg
4. Berat orang dan lain-lain = 100 kg +
Total 752,2 ~ 752 kg
Muatan P2
1. Berat konstruksi = 240 kg
2. Berat atap(0,963+0,963)*3*10 = 57,78 kg
3. Berat gording (2*0,10*0,12*0,91*1000*3) = 65,52 kg
4. Berat orang dan lain-lain = 100 kg +
Total 463,3 ~ 463 kg
Muatan P4
1. Berat hanger = 11 kg/m2
2. Berat Flapond = 7 kg/m2
3. Berat hanger + Berat Flapond = 18 kg/m2
4. Berat hanger + Berat Flapond per titik buhul P4 ada 3
= 18 kg/m2*3*Jarak kuda-kuda
= 18*3*3
=162 kg
Muatan P3
P3 = ½*P4
= ½*162
= 81 kg
4. Perhitungan Muatan Angin
11
½ Wt ½ Wi
H
Wt S3 S3’ Wi
G I Wi
Wt S2 9 S2’
½ Wt F 8 8’ J ½ Wi
S1 6 7 7’ 6’ S1’
A 4 5 5’ 4’ B
C D E
10 m
Koefisien tekanan angin untuk α < 65˚ = 0,02 α-0,4
C1 = 0,02α-0,4 = 0,2*30˚-0,4 = 0,2
Koefisien isapan angin untuk semua α = 0,4 (PMI 83 PASAL 4.3)
C2 = 0,4
Tekanan angin yang bekerja = 18 kg/m2
Angin Tekan (Wt) = Jarak gording*2*Jarak kuda-kuda*C1*Muatan angin
= 0,963*2*3*0,2*18
= 20,801 ~ 21 kg
Angin isap (Wi) = Jarak gording*2*Jarak kuda-kuda*C2*Muatan angin
= 0,963*2*3*0,4*18
= 41,602 ~ 42 kg
5. Perhitungan gaya batang muatan tetap
12
1,667 m 1,667 m 1,667 m 1,667 m 1,667 m 1,667 m
463 kg
463 kg H 463 kg
S3 S3’
463 kg G I 463 kg
S2 9 S2’
752 kg F 8 8’ J 752 kg
S1 6 7 7’ 6’ S1’
A 4 5 5’ 4’ B
C D E
81 kg 81 kg 81 kg 81 kg 81 kg
2,5 m 2,5 m 2,5 m 2,5 m
Reaksi Perletakan :
∑MB = 0
VA*10-463*10-81*10-463*(8,334+6,667+5+3,333+1,666)-162*(7,5+5+2,5) = 0
VA*10-7520-810-11575-2430 = 0
VA = 2233,5 kg (↑) karena konstruksi simetris maka VB = 2233,5 kg (↑)
Kontrol :
∑V = 0
VA+VB-752*2-81*2-162*3-463*5 = 0
2233,5+2233,5-1504-162-486-2315 = 0
0 = 0 OK !!!
Kontrol gaya batang muatan tetap dengan metode ritther
Potongan I-I untuk batang S1 dan S4
1,667 m
13
752 kg I F
S1 0,963 m
α
A S4 I C
81 kg 2,5 m
VA = 2233,5 kg
∑MC = 0
VA*2,5-752*2,5-81*2,5+ S1*sin α*2,5 = 0
5583,75-1880-202,5+ S1*1,25 = 0
S1 = -2801 kg (Tekan)
% kesalahan = * 100% = 0 %
0 % < 3 % OK !!!
∑MF = 0
VB*1,667-752*1,667-81*1,667- S4*0,963 = 0
3723,2445-1253,584-135,027- S4*0,963 = 0
S4 = 2426,854 kg (Tarik)
% kesalahan = * 100% = 0,046 %
0,046 % < 3 % OK !!!
6. Perhitungan gaya batang akibat muatan angin kiri
.
14
(2801-2801) 2801
(2425,7372-2426,854) 2425,7372
∑Wt cos 30 ˚ H ∑Wi cos 30˚
∑Wt S3 S3’ ∑Wi
G I
∑Wt sin 30˚ S2 9 S2’ ∑Wi sin 30˚
F 8 8’ J 2,887 m
1,4435 m S1 6 7 7’ 6’ S1’
A 4 5 5’ 4’ B
C D E
2,5 m 2,5 m 2,5 m 2,5 m
Akibat tekanan angin
∑Wt = 1,4435*Wt = 1,4435*21 = 30,3135 kg
∑Wt*sin 30˚ = 30,3135* sin 30˚ = 15,1568 kg
∑Wt*cos 30˚ = 30,3135*cos 30˚ = 26,2523 kg
Akibat isapan angin
∑Wi = 1,4435*Wi = 1,4435*42 = 60,627 kg
∑Wi*sin 30˚ = 60,627* sin 30˚ = 30,3135 kg
∑Wi*cos 30˚ = 60,627*cos 30˚ = 52,5045 kg
Reaksi Perletakan :
∑MB = 0
VA*10-(∑Wt*sin 30˚)*1,4435+(∑Wi*sin 30˚)*1,4435-(∑Wt*cos 30˚)*7,5+(∑Wi*cos 30˚ )*2,5
= 0
VA*10+21,8788+43,7575-196,8923+131,2613 = 0
VA = -0,0005 kg (↑)
∑MA = 0
-VB*10-(∑Wi*cos 30˚)*7,5+(∑Wi*sin 30˚)*1,4435+(∑Wt*cos 30˚)*2,5+(∑Wt*sin 30˚ )*1,4435
= 0
-VB*10-393,7838+43,7575+65,6308+21,8788 = 0
VA = -26,2517 kg (↑)
Kontrol :
15
∑V = 0
VA+VB-∑Wi cos 30˚-∑Wt*cos 30˚ = 0
-0,0005-26,2517+52,5045-26,2523 = 0
0 = 0 OK !!!
∑H= 0
HA-∑Wt*sin 30˚ -∑Wi sin 30 = 0
HA-15,1568-30,3135 = 0
HA = 45,4703 kg (←)
Kontrol gaya batang muatan tetap dengan metode ritther
Potongan I-I untuk batang S1 dan S4
1,667 m
½ Wt cos 30˚ I F
½ Wt
α S1 0,963 m
HA = 45,4703 kg α
A S4 I C
2,5 m
VA = -0,0005 kg
∑MC = 0
VA*2,5- ½ Wt*cos 30˚*2,5+ S1*sin 30˚*2,5 = 0
-0,0013-32,8154+ S1*1,25 = 0
S1 = 26,2534 kg (Tekan)
% kesalahan = * 100% = 0 %
0 % < 3 % OK !!!
∑MF = 0
VA*2,5+HB*0,963-S4*0,963-½ Wt*cos 30˚*1,667-½ Wt*sin 30˚*0,963 = 0
-0,0008+43,7424-S4*0,963 -21,8813-7,2904= 0
S4 = 15,1454 kg (Tarik)
% kesalahan = * 100% = 0,0686 %
16
(26,2534 -26,2534) 26,2534
(15,1558-15,1454) 15,1558
0,0686 % < 3 % OK !!!
6. Perhitungan gaya batang akibat muatan angin kanan
.
∑Wi cos 30 ˚ H ∑Wt cos 30˚
∑Wt S3 S3’ ∑Wi
G I
∑Wt sin 30˚ S2 9 S2’ ∑Wi sin 30˚
F 8 8’ J 2,887 m
1,4435 m S1 6 7 7’ 6’ S1’
A 4 5 5’ 4’ B
C D E
2,5 m 2,5 m 2,5 m 2,5 m
Akibat tekanan angin
∑Wt = 1,4435*Wt = 1,4435*21 = 30,3135 kg
∑Wt*sin 30˚ = 30,3135* sin 30˚ = 15,1568 kg
∑Wt*cos 30˚ = 30,3135*cos 30˚ = 26,2523 kg
Akibat isapan angin
∑Wi = 1,4435*Wi = 1,4435*42 = 60,627 kg
∑Wi*sin 30˚ = 60,627* sin 30˚ = 30,3135 kg
∑Wi*cos 30˚ = 60,627*cos 30˚ = 52,5045 kg
Reaksi Perletakan :
∑MB = 0
VA*10+(∑Wi*sin 30˚)*7,5-(∑Wi*sin 30˚)*1,4435-(∑Wt*cos 30˚)*2,5-(∑Wt*sin 30˚ )*1,4435 =
0
VA*10+393,7838+43,7575-65,6303-21,8788 = 0
VA = -26,2517 kg (↑)
∑MA = 0
17
-VB*10+(∑Wt*cos 30˚)*7,5-(∑Wt*sin 30˚)*1,4435-(∑Wi*cos 30˚)*2,5-(∑Wi*sin 30˚ )*1,4435
= 0
-VB*10+196,8923-21,8788-131,2613-43,7575 = 0
VA = 0,0005 kg (↑)
Kontrol :
∑V = 0
VA+VB+∑Wi cos 30˚-∑Wt*cos 30˚ = 0
-26,2517+0,0005+52,5045-26,2523 = 0
0 = 0 OK !!!
∑H= 0
HA-∑Wt*sin 30˚ -∑Wi sin 30 = 0
HA-15,1568-30,3135 = 0
HA = 45,470 kg (←)
Kontrol gaya batang muatan tetap dengan metode ritther
Potongan I-I untuk batang S1 dan S4
1,667 m
½ Wi cos 30˚ I F
½ Wi
α S1 0,963 m
HA = 45,470 kg α
A S4 I C
2,5 m
VA = -26,2517 kg
∑MC = 0
VA*2,5+ S1*sin 30˚*2,5+½WI*cos 30˚*2,5 = 0
-65,6293+ S1*1,25+65,6303 = 0
S1 = -0,0013 kg (Tekan)
% kesalahan = * 100% = 0 %
0 % < 3 % OK !!!
18
(0,0013 -0,0013) 0,0013
∑MF = 0
VA*1,667-S4*0,963-HA*0,963+½ Wi*cos 30˚*0,963+½ Wi*cos 30˚*1,667 = 0
-43,7616-S4*0,963 -43,7472+14,5808+43,7625 = 0
S4 = -30,3127 kg (Tekan)
% kesalahan = * 100% = 0,0007 %
0,0007 % < 3 % OK !!!
Perhitungan gaya batang akibat muatan tetap dengan metode titik simpul
a. Titik simpul A
S1 sin α ∑V = 0
VA-P1-P3+S1*sin α = 0
S1 2233,5-752-81-S1*0,5= 0
P1 S1 = -2801 kg (Tekan)
30˚ S4 ∑H = 0
S4-S1*cosα = 0
P3 VA = 2233,5 kg S4-2425,7372 =0
S4 = 2425,7372 kg (Tarik)
b. Titik simpul F
S2 sin α S2
P2
S1 cos α α S2 cos α
30˚ β
S6
S6 sin α 0,963 m
α
S1 0,833
Tan β = 0,963/0,933 = α = 49,1106˚
β = 49,1106˚
19
(30,3125-30,3127) 30,3125
∑V = 0
-S1*sin 30˚-P2+S2*sin α-S6*sin β = 0
+1400,5-752+S2*0,5-S6* 0,7560 = 0
S2*0,5-S6*0,7560 = -648,5 ………………….(1)
∑H = 0
-S1*cos 30˚+S6*cos β+S2*cos 30˚+2425,7372+S1*0,6546+S2*0,866 = 0
S2*0,866+S6*0,6546 = -2425,7372 kg
S2*0,866-S6*0,7560 = -2425,7372 ………….(2)
Diperoleh : S2 = -2299,7705 kg (Tekan)
S6 = -663,2080 kg (Tekan)
c.
S7
S6 S6 sin α S7 sin α
S6 cos α β α S7 cos α
S4 S5
P4
β = 66,6016˚
α = 49,1106˚
∑V = 0
20
S6*sin α +S7*sin α-P4 = 0
-663,2080*sin 49,1106˚+S7*sin 49,1106˚-162 = 0
-501,3684+S7*0,7560-162 = 0
S7 = 877,4714 kg (Tarik)
∑H = 0
S1*cos α+S5-S4˚-S6 cos α = 0
574,3936+S5-2425,7372+434,1366 = 0
S5 = 1417,207 kg (Tarik)
d.
S3 sin α S3
P2
S2 cos α S7 cos α S8 cos α 1,667 m
α S3 cos α α
S8 1,925 m
S8 sin α
S7 sin α
S2 S2 sin α
Tan α = 1,925/1,667 = α = 49,1083˚
∑V = 0
-P2+S3*sin α-S7*sin α+S8 sin α-S2 sin α = 0
21
-463+S3*0,5-805,3132-S8*0,7559+1149,8853 = 0
S3*0,5-S8*0,7559 = 118,4279 ………….….(1)
∑H = 0
S3*cos α+S8*cos α-S7*cos α-S2*cos α = 0
S3*0,866+S8*0,6546-574,3936+1991,6597 = 0
S3*0,866-S8*0,6546 = -1417,2661 .……….(2)
Diperoleh : S2 = -1012,098 kg (Tekan)
S6 = -826,1369 kg (Tekan)_
e.
P4 = 463 kg
S3 cos α S3’ cos α
30˚ 30˚
S3’sin α S3’
S3 S3 sin α
S9
∑V = 0 ∑H = 0
-P2+S3’*sin α-S3*sin α –S9 = 0 -S3*cos α+S3,*cos α = 0
-463-S3’*0,5+506,0490-S9 = 0 876,5026+S3’*0,866 = 0
-S9-S3’-0,5 = -43,0490 …………….(1) S3’ = -1417,2661 (Tekan)
22
Subtitusikan ke pers (1)
-S9+1012,1277*0,5 = -43,049
-S9+506,0639+43,049 = 0
S9 = 549,1129 kg (Tarik)
Daftar gaya batang akibat muatan tetap
No. Batang Tarik Tekan
1. S1=S1’ - 2801
2. S2=S2’ - 2299,7705
3. S3=S3’ - 1012,1277
4. S4=S4’ 2425,7372 -
5. S5=S5’ 1417,207 -
6. S6=S6’ - 663,2080
7. S7=S7’ 877,4714 -
8. S8=S8’ - 826,1369
9. S9 549,1129 -
Perhitungan gaya batang akibat muatan angin kiri dengan metode titik simpul
a. Titik simpul A
S1 sin α ∑V = 0
½ Wt S1 VA-½ Wt cos α+S1*sin α = 0
½ Wt cos α -0,0005-13,1262+S1*0,5= 0
S1 = 26,2534 kg (Tarik)
HA 30˚ S4 ∑H = 0
½ Wt sinα S1 cos α S4+S1*cosα+½ Wt sin α-HA= 0
VA = -0,0005 kg S4+22,7361+7,5784-45,4703 =0
S4 = 15,1558 kg (Tarik)
b. Titik simpul F
S2 sin α S2
23
Wt cos α
Wt
S1 cos α S6 cos α S2 cos α
β
S6
S1 sin α
S1
β = 49,1106˚
∑V = 0
S2*sin 30˚-Wt cos 30 -S6*sin β-S1*sin α = 0
S2*0,5-26,2523-S6* 0,7560-13,1267 = 0
S2*0,5-S6*0,7560 = -39,3790 .…………….(1)
∑H = 0
S2*cos 30˚+S6*cos α+Wt sin α-S1*cos α = 0
S2*0,866+S6*0,6545+15,1568-22,7361 = 0
S2*0,866+S6*0,6545 = 7,5793 ….……….(2)
Diperoleh : S2 = -20,4121 kg (Tekan)
S6 = 38,5886 kg (Tarik)
c.
S6 S6 sin α
S7 sin β S7
S4 α β S5
S6 cos α S7 cos β
24
β = 66,6016˚
α = 49,1106˚
∑V = 0
S6*sin α +S7*sin β = 0
29,1719+S7*0,9178 = 0
-501,3684+S7*0,7560-162 = 0
S7 = -29,1708 kg (Tekan)
∑H = 0
S5+S7*cos β-S6 cos α – S4 = 0
S5-11,5844-25,2601-15,1558 = 0
S5 = 52,0003 kg (Tarik)
d.
S3 sin α S3
Wt Wt cos α
Wt cos α
S2 cos α S7 cos β S8 cos θ
θ S3 cos α
S8
S7 S8 sin θ
S7 sin β
S2 S2 sin α
θ = 49,1083˚
∑V = 0
25
S3*sin α-Wt cos α-S8 sin θ-S7*sin β-S2 sin α = 0
S3*0,5-26,2523-S8*0,7559+26,7720+10,2061 = 0
S3*0,5-S8*0,7559 = -10,7258 .…………….(1)
∑H = 0
Wt sin α-S2*cos α-S7*cos β+S8*cos θ+S3*sin α = 0
15,1568+20,4121+11,5844+S8*0,6546+S3*0,866 = 0
S3*0,866+S8*0,6546 = -47,1533 ………….(2)
Diperoleh : S2 = -43,4503 kg (Tekan)
S6 = -14,5513 kg (Tekan)
e.
½ Wi cos α ½ Wi
½ Wt ½ Wt cos α
S3 cos α S3’ cos α ½ Wi sin α
S3’sin α S3’
S3 S3 sin α
S9
∑V = 0
½ Wi cos α-½Wt cos α –S3’*sin α- S3*sin α = 0
26
26,2523-13,1262-8,5993-21,7252-S9 = 0
S9 = -17,1984 kg (Tekan)
∑H = 0
½ Wi sin α+ S3’*cos α +½Wt cos α –S3*cos α = 0
15,1568- S3’*0,866+7,5784-37,6291 = 0
S3’ = 17,1985 kg (Tarik)
f.
S9
S8
S8 sin α S8’
S8’ sin β
S5 α β S8’ cos β S5’
S8 cos α
β = 49,1083˚
α = 49,1083˚
∑V = 0
S9+S8’*sin β-S8 sin α = 0
-17,1984+S8’*0,7559+11 = 0
S8’ = 8,2 kg (Tarik)
27
∑H = 0
S5’+S8’*cos β–S5 = 0
S5’+5,3680-9,5257-52,0003 = 0
S5’ = 56,1580 kg (Tarik)
g.
S3’ S3’ sin α
Wi cos α Wi
S3’ cos α S8’ cos α Wi sin α S7 cos β S2’ cos α
α
S8’
S8’ sin α
S2’ sin α S2’
S7’ sin β S7’
28
β = 66,6016˚
α = 49,1083˚
∑V = 0
S3’*sin α+Wi cos α-S8’ sin α –S2’*sin α- S7’*sin β = 0
13,0012+52,5045-6,1988- S2’*0,7559-S7’*0,9178 = 0
-S2’*0,7559-S7’*0,9178 = -59,3069……………..(1)
∑H = 0
S2’*cos α+S7’*cos β+ Wi sin α- S8’ cos α –S3’*cos α = 0
S2’*0,6546+S7’*0,3971+30,3135- 5,380-11,2587 = 0
S2’*0,6546+S7’*0,3971 = -13,6869 ……………..(2)
Diperoleh : S2’ = -120,1252 kg (Tekan)
S7’ = 163,5536 kg (Tarik)
h.
S6’ sin α S6’
S7’ sin β
S5’ β α S6’ cos α S4’
S7’ cos β
β = 66,6016˚
α = 49,116˚
∑V = 0
S5’*sin α-S7’*sin β = 0
S6’*0,7560-150,1039 = 0
S6’ = 198,5501 kg (Tarik)
29
∑H = 0
S4’+S6’*cos α+S7’*cos β-S5’ = 0
S4’+129,9569+64,9508-56,1580 = 0
S4’ = -138,7497 kg (Tekan)
i.
S1’ sin α ∑V = 0
S1’ S1’*sin α+½Wi*sin α-26,2517 = 0
½ Wi S1’*0,5+26,2523-26,2517 = 0
½ Wi cos α S1’ = -0,0012 kg (Tekan)
α
S4’ ½ Wi sin α
S1’ cos α
26,2517 kg
Daftar gaya batang akibat muatan angin kiri
No. Batang Tarik Tekan
1. S1 26,2534 -
2. S2 - 20,4121
30
3. S3 - 43,4503
4. S4 15,1558 -
5. S5 52,0003 -
6. S6 38,5885 -
7. S7 - 29,1708
8. S8 - 14,5513
9. S9 - 17,1984
10. S1’ - 0,0012
11. S2’ - 120,1252
12. S3’ 17,1985 -
13. S4’ - 138,7494
14. S5’ 56,1580 -
15. S6’ 198,5501 -
16. S7’ 163,5536 -
17. S8’ 8,2 -
Perhitungan gaya batang akibat muatan angin kanan dengan metode titik simpul
a. Titik simpul A
S1 sin α ∑V = 0
S1 S1*sin α+½ Wi cos α + VA = 0
½ Wi ½ Wi cos α S1*0,5+26,2523-26,2517= 0
S1 = -0,0013 kg (Tekan)
HA 30˚ S4 ∑H = 0
½ Wi sinα S1 cos α S4+S1*cosα-½ Wi sin α+HA= 0
VA S4-0,001-15,1568+45,4703 =0
S4 = -30,3125 kg (Tekan)
b. Titik simpul F
S2 sin α S2
Wi cos α
31
Wi
S1 cos α S6 cos α S2 cos α
Wi sin α β
S6
S6 sin α
S1 sin α
S1
β = 49,1106˚
∑V = 0
S2*sin 30˚+Wi cos 30 -S6*sin β-S1*sin α = 0
S2*0,5+52,5045-S6* 0,7560+0,0007 = 0
S2*0,5-S6*0,7560 = -52,5052 .…………….(1)
∑H = 0
-S1*cos α-Wi sin α +S6*cos β+ S2*0,886 = 0
0,0011-30,3135+S6*0,6545+S2*0,866= 0
S2*0,866+S6*0,6545 = 30,3124 ….……….(2)
Diperoleh : S2 = -11,6590 kg (Tekan)
S6 = 61,7404s kg (Tarik)
c.
S6 S6 sin α
S7 sin β S7
S4 α β S5
S6 cos α S7 cos β
32
β = 66,6016˚
α = 49,1106˚
∑V = 0
S6*sin α +S7*sin β = 0
46,6742+S7*0,9178 = 0
-501,3684+S7*0,7560-162 = 0
S7 = -50,8544 kg (Tekan)
∑H = 0
S5+S7*cos β– S4-S6 cos α = 0
S5-20,1954+30,3125-40,4153 = 0
S5 = 30,2982 kg (Tarik)
d.
S3 sin α S3
Wi Wi cos α
Wi cos α
S2 cos α S7 cos β 30˚ S8 cos θ
30˚ α S3 cos α
S8
S7 S8 sin α
S7 sin β
S2 S2 sin α
β = 66,6016˚
33
α = 49,1083˚
∑V = 0
S3*sin α+Wi cos α-S8 sin α-S7*sin β-S2 sin α = 0
S3*0,5+52,5045-S8*0,7559+46,6724-8,8136 = 0
S3*0,5-S8*0,7559 = -90,3633 .…………….(1)
∑H = 0
S3*cos+ S8*cos α -S7*cos β-Wi sin α-S2*cos α = 0
S3*0,0,6546+ S8*0,7314+20,1954-30,3135-7,6323 = 0
S3*0,6546+S8*0,6546 = -47,1533 ………….(2)
Diperoleh : S2 = -61,2126503 kg (Tekan)
S6 = -79,0541 kg (Tarik)
e.
½ Wi ½ Wi cos α
½ Wt
½ Wt cos α
S3 cos α ½ Wi sin α S3’ cos α ½ Wt sin α
S3’sin α S3’
S3 S3 sin α
S9
34
∑V = 0
½ Wi cos α-½Wt cos α-S3’*sin α–S3*sin α-S9 = 0
26,2523- 13,1262+8,7297+30,6063 = 0
S9 = 52,4621 kg (Tarik)
∑H = 0
S3’*sin α -½ Wt sin α-½Wi sinα –S3*sin α = 0
S3’*0,866-7,5784-30,3135+53,0117 = 0
S3’ = -17,4594 kg (Tekan)
f.
S8 S9
S8 sin α
S8’ sin β S8’
S5 α β S8’ cos β S5’
S8 cos α
α = β = 49,1083˚
∑V = 0
S9+S8*sin α+S8’ sin α = 0
52,4621+S9*0,7608+S8’*0,7559 = 0
S8’ = -148,4626 kg (Tekan)
∑H = 0
S5’+S8’*cos α-S8 cos α –S5 = 0
S5’-97,1883-51,7513-30,2982 = 0
S5’ = 179,2378 kg (Tarik)
g.
S3’ S3’ sin α
Wt cos α Wt
35
S8’ cos α S3’ cos α Wt sin α S7 cos β S2’ cos α
α
S8’ sin α
S2’ sin α S2’
S7’ sin β S7’
S8’ S8’ sin α
β = 66,6016˚
α = 49,1083˚
∑V = 0
S3’*sin α+Wt cos α- S7’*sin β - S2’*sin α -S8’ sin α = 0
-13,1984-26,2523-S7’*0,9178- S2’*0,7559-112,2301 = 0
-S2’*0,7559-S7’*0,9178 = 151,6808……………..(1)
∑H = 0
S2’*cos α+S7’*cos β- Wt sin α- S3’*cos α-S8’ cos α = 0
S2’*0,6546+S7’*0,3971-15,1568+11,4295-97,1883 = 0
S2’*0,6546+S7’*0,3971 = 100,9156 ……………..(2)
Diperoleh : S2’ = 508,4517 kg (Tarik)
S7’ = -584,0264 kg (Tekan)
h.
S7’ S7’ sin β
S6’
S6’ sin β
S5’ β α S6’ cos α S4’
S7’ cos β
36
β = 66,6016˚
α = 49,116˚
∑V = 0
S7’*sin β +S6’*sin α = 0
-441,5091+S6’*0,7560 = 0
S6’ = 584,0068 kg (Tarik)
∑H = 0
S4’+S6’*cos α+S7’*cos β-S5’ = 0
S4’+382,2914-382,3042-179,2378 = 0
S4’ = 179,2506 kg (Tarik)
i.
S1’ sin α ∑V = 0
S1’ S1’*sin α-½Wt*cos α = 0
½ Wt S1’*0,5-13,1262 = 0
½ Wt cos α S1’ = 26,2523 kg (Tarik)
S4’ α ½ Wt sin α
S1’ cos α
Daftar gaya batang akibat muatan angin kanan
No. Batang Tarik Tekan
1. S1 - 0,0013
2. S2 - 11,6590
3. S3 - 61,2126
37
4. S4 - 30,3125
5. S5 30,2982 -
6. S6 61,7404 -
7. S7 - 50,8544
8. S8 79,0541 -
9. S9 52,4621 -
10. S1’ 26,2523 -
11. S2’ 508,4517 -
12. S3’ - 17,454
13. S4’ 179,2506 -
14. S5’ 179,2378 -
15. S6’ 584,0068 -
16. S7’ - 584,0264
17. S8’ - 148,4626
38
No. Batang Muatan tetap Muatan angin kiri Muatan angin kanan Muatan tetap + angin kanan Muatan angin tetap + angin kiri Beban Ekstrim
Tarik (kg) Tekan (kg) Tarik (kg) Tekan (kg) Tarik (kg) Tekan (kg) Tarik (kg) Tekan (kg) Tarik (kg) Tekan (kg) Tarik (kg) Tekan (kg)
S1 - 2801 26,2334 - - 0,0013 - 2801,001 - 2774,747 - 2801,0012
S2 - 2299,7750 - 20,4121 - 11,6590 - 2311,4995 - 2320,1826 - 2320,1826
S3 2425,732 1012,1277 - 43,4503 - 61,2126 - 1073,3403 - 1055,578 - 1073,3403
S4 1417,207 - 15,1558 - - 30,3125 2395,427 - 2440,8930 - 2440,8930 -
S5 - - 52,0003 - 30,2982 - 1447,502 - 1469,2073 - 1469,2073 -
S6 877,4714 663,2080 38,5885 - 61,7404 - - 601,4676 - 624,6195 - 624,6095
S7 - - - 29,1708 - 50,8544 826,6170 - 848,3006 - 848,3006 -
S8 549,1129 826,1369 - 14,5513 79,0541 - - 747,0828 - 840,6882 - 840,6882
S9 - - - 17,1984 52,4621 - 601,5750 - 531,9145 - 601,5750 -
S1’ - 2801 - 0,0012 26,2523 - - 2774,7477 - 2801,0012 - 2801,0012
S2’ - 2299,7750 - 120,1252 508,4517 - - 1791,3188 - 2419,6957 - 2419,8957
S3’ - 1012,1277 17,1985 - - 17,4594 - 1029,8571 - 994,9292 - 1029,5871
S4’ 2425,732 - - 138,7497 179,2506 - 2604,088 - 2286,9875 - 2604,9878 -
S5’ 1417,207 - 56,1580 - 179,2378 - 1596,448 - 1473,3650 - 1596,4448 -
S6’ - 663,2080 198,5501 - 584,0068 - - 79,2012 - 464,6579 - 464,6579
S7’ 877,4714 - 163,5536 - - 584,0264 293,4450 - 1041,0250 - 1041,0250 -
S8’ - 826,1369 8,2 - - 148,4626 - 974,5995 - 817,9369 - 974,5995
Gaya batang ekstrim yang digunakan untuk desain : S1,S2,S3,S1’,S2’,S3’ = 2801,0012 kg,S4,S5,S4’,S5’ = 2604,9878 kg ,S6,S7,S8,S6’,S7’,S8' = 1041,0250 kg,S9 = 601,0250 kg
39
BAB II
DIMENSI BATANG
A. Batang atas (Batang 1,2,3,3’,2’,61’,)
Gaya maksimum ~ P = PK = 2801,0012 kg (Tekan)
L = LK = 1,925 m ~ 192,5 cm
Misalnya digunakan kayu 10/14
Kontrol terhadap tekuk
Ix = 1/12*b.h³ = 1/12*10*143 = 2286,67 cm4
Ix = 1/12*b3.h = 1/12*103*14 = 1167 cm4
Ix = = = = 4,04 cm
Iy = = = = 2,89 cm
Iy < Ix ~ diambil yang terkecil : Iy = I min = 2,89 cm
λ = = = 66,61 ~ 67 cm
λ = 67 ≤ λ max = 150 (PKKI 1961).............OK !!!
Untuk λ = 67 ~ Faktor tekuk (ω) = 1,81 (PKKI 1961)
σtk // = P*w/F br = 2801,0012 *1,81/10*14 = 36,21 kg/cm 2
σtk // = 36,21 cm 2 < σtr // = 106,5 kg/cm 2 ………………(Aman)
Jadi batang atas menggunakan 10/14 memenuhi syarat (Aman)
B. Batang Bawah (Batang 4,5,4’,5’)
Gaya maksimum ~ P = PK = 2604,9878 kg (Tarik)
L = LK = 2,5 m ~ 250 cm
σtr // = 106,5 kg/cm 2
Misal digunakan kayu 10/14
Syarat agar aman :
F = 10*14= 140 >F bruto
σtr // = = F netto =
F bruto = F netto + F perlu
F perlu = 25% F netto ~ F bruto = 1,25*F netto
LK I min
192,5 cm2,89 cm
p F netto
P σtr //
40
F bruto = 1,25*P/ σtr // = 1,25*2604,878/106,25 = 30,647 cm2
F = 140 > F bruto = 30,647 cm2 …………………..Aman !!!
Jadi kayu 10/14 memenuhi syarat (Aman)
C. Batang vertikal (Batang 9)
Gaya maksimum ~ P = PK = 601,5750 kg (Tarik)
σtr // = 106,5 kg/cm 2
Misal digunakan kayu 10/14
F bruto = 1,25*P/ σtr // = 1,25*601,5750/106,25 = 7,007 cm2
F = 140 > F bruto = 7,007 cm2 …………………..Aman !!!
Jadi kayu 10/14 memenuhi syarat (Aman)
D. Batan Diagonal (Batang 6,7,8,8’,7’,6’)
Gaya maksimum ~ P = PK = 1041,0250 kg (Tarik)
L = LK = 2,546 m ~ 254,6 cm
σtr // = 106,5 kg/cm 2
Kontrol terhadap tekuk
Ix = 1/12*b.h³ = 1/12*10*143 = 2286,67 cm4
Ix = 1/12*b3.h = 1/12*103*14 = 1167 cm4
Ix = = = = 4,04 cm
Iy = = = = 2,89 cm
Iy < Ix ~ diambil yang terkecil : Iy = I min = 2,89 cm
λ = LK/ I min = 254,6/2,89 = 88,10 cm~ 88 cm
λ = 88 cm ~ Faktor tekuk (ω) = 2,42 (PKKI 1961)
σtk // = P*w/F br = 1041,0250 *2,42/10*14 = 17,995 kg/cm 2
σtk // = 17,995 cm 2 < σtr // = 106,5 kg/cm 2 ………………(Aman)
Jadi kayu 10/14 untuk batang diagonal memenuhi syarat (Aman)
41
BAB III
PERHITUNGAN SAMBUNGAN
Kayu yang digunakan : Kelas II
Sifat beban yang bekerja : Beban hidup + angin
Dari buku PKKI 1961 terdapat ketentuan untuk kelas II mutu B
σds// = σtr// = 106,25 kg/cm2
σds// ┴ = 31,25 kg/cm2
τ // = 15 kg/cm2
Digunakan sambungan paku sebagai berikut :
1. Plat sambungan ukuran 2 x 3/12
2. Ukuran paku 4” BWE 8 (42/102) untuk sambungan tampang 2
3. Kekuatan paku (P) = 94 kg
4. Kekuatan paku (P) untuk sambungan tampang 2 :
P = 2 x P = 2 x 94 = 188 kg
a. Sambungan titik buhul A
S1 S1 = 2801,0012 kg (Tekan)
S4 = 2440,8930 kg (Tarik)
S4
A
Jumlah paku yang diperlukan (n)
n = S1/P (Koefisien muatan tetap + angin)
= 2801,0012/188*(5/4)
= 2801,0012/235
= 11,9192 ~ 12 paku (Masing-masing sisi 6 paku)
Jarak-jarak paku
5d =5*(0,42 cm) = 2,1 cm ~ 2,1*cos 30 ˚ = 1,8187 cm ~ 2 cm
12d = 12*(0,42 cm)* cos 30 ˚ = 4,3648 ~ 5 cm
10d = 10*(0,42 cm)* cos 30 ˚ = 3,6374 ~ 4 cm
42
b. Sambungan titik buhul F
F S2 S1 = 2801,0012 kg (Tekan)
S1 S2 = 2320,1826 kg (Tekan)
S6 S6 = 624,6195 kg (Tekan)
Jumlah paku yang diperlukan (n)
n = S6/P (Koefisien muatan tetap + angin)
= 624,6195 /188*(5/4)
= 624,6195 /235
= 2,6580 ~ 4 paku (Masing-masing sisi 2 paku)
c. Sambungan titik buhul C
S4 = 2440,8930 kg (Tarik)
S6 S7 S5 = 1469,2073 kg (Tarik)
S6 = 624,6195 kg (Tekan)
S7 = 848,3006 kg (Tekan)
S4 C S5
Jumlah paku yang diperlukan (n)
n = S7/P (Koefisien muatan tetap + angin)
= 848,3006/188*(5/4)
= 848,3006 /235
= 3,6098 ~ 4 paku (Masing-masing sisi 2 paku)
Jumlah paku yang diperlukan (n)
n = S6/P (Koefisien muatan tetap + angin)
= 624,6195 /188*(5/4)
= 624,6195 /235
= 2,6580 ~ 4 paku (Masing-masing sisi 2 paku)
d. Sambungan titik buhul G
G S3 S2 = 2320,1826 kg (Tekan)
S2 S3 = 1073,3403 kg (Tekan)
S8 S7 = 848,3006 kg (Tekan)
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S8 = 840,6862 kg (Tekan)
S7
Jumlah paku yang diperlukan (n)
n = S8/P (Koefisien muatan tetap + angin)
= 840,6862/188*(5/4)
= 840,6862/235
= 3,5774 ~ 4 paku (Masing-masing sisi 2 paku)
Hubungan antara S7 dengan S2 dan S3
n = S7/P (Koefisien muatan tetap + angin)
= 848,3006/188*(5/4)
= 848,3006/235
= 3,6098 ~ 4 paku (Masing-masing sisi 2 paku)
e. Sambungan titik buhul H
H S3 = 1073,3403 kg (Tekan)
S3 S3’ S3’ = 1029,5871 kg (Tekan)
S9 S9 = 601,5750 kg (Tekan)
Jumlah paku yang diperlukan (n)
n = S3’/P (Koefisien muatan tetap + angin)
= 1029,5871/188*(5/4)
= 1029,5871/235
= 4,3812 ~ 6 paku (Masing-masing sisi 3 paku)
Hubungan antara S9 dengan S3 dan S3’
n = S3/P (Koefisien muatan tetap + angin)
= 1073,3403/188*(5/4)
= 1073,3403/235
= 4,5674 ~ 6 paku (Masing-masing sisi 3 paku)
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f. Sambungan titik buhul D
S5 = 1469,2073 kg (Tarik)
S9 S8 = 840,6862 kg (Tekan)
S8 S8’ S9 = 601,5750 kg (Tekan)
S8’ = 840,6882 kg (Tekan)
S5 D S5’ S5’ = 1596,4448 kg (Tarik)
Hubungan antara S5 dengan S5’
n = S8’/P (Koefisien muatan tetap + angin)
= 840,6882/188*(5/4)
= 840,6882/235
= 3,5774 ~ 4 paku (Masing-masing sisi 2 paku)
g. Sambungan titik buhul B
S1’ S4’ = 2604,9878 kg (Tarik)
S1’ = 2801,0012 kg (Tekan)
S4’
B
Jumlah paku yang diperlukan (n)
n = S1’/P (Koefisien muatan tetap + angin)
= 2801,0012/188*(5/4)
= 2801,0012/235
= 11,9192 ~ 12 paku (Masing-masing sisi 6 paku)
h. Sambungan titik buhul J
S2’
J S1’ = 2801,0012 kg (Tekan)
S1’ S2’ = 2419,8957 kg (Tekan)
S6’ S6’ = 464,6579 kg (Tekan)
Jumlah paku yang diperlukan (n)
n = S6’/P (Koefisien muatan tetap + angin)
= 464,6579/188*(5/4)
= 464,6579/235
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= 1,9773 ~ 2 paku (Masing-masing sisi 1 paku)
i. Sambungan titik buhul E
S7’
S6’ S4’ = 2604,9878 kg (Tarik)
S5’ = 1596,4448 kg (Tarik)
S6’ = 464,6579 kg (Tekan)
S5’ E S4’ S7’ = 1041,0250 kg (Tarik)
Jumlah paku yang diperlukan (n)
n = S7’/P (Koefisien muatan tetap + angin)
= 1041,0205/188*(5/4)
= 1041,0205/235
= 4,429 ~ 6 paku (Masing-masing sisi 3 paku)
j. Sambungan titik buhul I
S3’ S2’ = 2419,8957 kg (Tekan)
I S3’ = 1029,5871 kg (Tekan)
S2’ S7’ = 1041,0250 kg (Tarik)
S8’ S7’ S8’ = 974,5995 kg (Tekan)
Jumlah paku yang diperlukan (n)
n = S7’/P (Koefisien muatan tetap + angin)
= 1041,0250/188*(5/4)
= 1041,0250/235
= 4,4299 ~ 6 paku (Masing-masing sisi 3 paku)
Jumlah paku yang diperlukan (n)
n = S8’/P (Koefisien muatan tetap + angin)
= 974,5995/188*(5/4)
= 974,5995/235
= 4,14729 ~ 6 paku (Masing-masing sisi 3 paku
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