contoh Kayu Tb ( PAKU)

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 (←)



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 (←)



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)


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)


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)


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


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)


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)


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


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)


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’)


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)



= 624,6195 /188*(5/4)

= 624,6195 /235


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



= 848,3006/188*(5/4)

= 848,3006 /235




= 624,6195 /188*(5/4)

= 624,6195 /235


d. Sambungan titik buhul G

G S3 S2 = 2320,1826 kg (Tekan)

S2 S3 = 1073,3403 kg (Tekan)

S8 S7 = 848,3006 kg (Tekan)

43

S8 = 840,6862 kg (Tekan)

S7



= 840,6862/188*(5/4)

= 840,6862/235


Hubungan antara S7 dengan S2 dan S3


= 848,3006/188*(5/4)

= 848,3006/235


e. Sambungan titik buhul H

H S3 = 1073,3403 kg (Tekan)

S3 S3’ S3’ = 1029,5871 kg (Tekan)

S9 S9 = 601,5750 kg (Tekan)


n = S3’/P (Koefisien muatan tetap + angin)

= 1029,5871/188*(5/4)

= 1029,5871/235


Hubungan antara S9 dengan S3 dan S3’


= 1073,3403/188*(5/4)

= 1073,3403/235


44

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’


= 840,6882/188*(5/4)

= 840,6882/235


g. Sambungan titik buhul B

S1’ S4’ = 2604,9878 kg (Tarik)

S1’ = 2801,0012 kg (Tekan)

S4’

B



= 2801,0012/188*(5/4)

= 2801,0012/235


h. Sambungan titik buhul J

S2’

J S1’ = 2801,0012 kg (Tekan)

S1’ S2’ = 2419,8957 kg (Tekan)

S6’ S6’ = 464,6579 kg (Tekan)



= 464,6579/188*(5/4)

= 464,6579/235

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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)



= 1041,0205/188*(5/4)

= 1041,0205/235


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)



= 1041,0250/188*(5/4)

= 1041,0250/235




= 974,5995/188*(5/4)

= 974,5995/235

= 4,14729 ~ 6 paku (Masing-masing sisi 3 paku

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contoh Kayu Tb ( PAKU)

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