bab iii okeeeeeeeee

BAB I

Jurusan Teknik Sipil

Politeknik Negeri Padang Laporan Praproyek Jembatan Rangka Baja

BAB III

ANALISA PERHITUNGAN

C a1 D a2 E a3 F a4 G a5 H a6 I

4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P B 4 m 4 m 4 m 4 m 4 m 4m4m4m

Data Perencanaan :

1. Panjang jembatan (L)

: 32 m

2. Type jembatan

: C3. Tinggi bentang

: 4 m

4. Lebar jembatan (b)

: 5,5 m

5. Beban hidup bergerak untuk jembatan: Kelas II

6. Tegangan izin baja (fu)

: 3200 kg/cm27. Sambungan

: Baut

8. Lebar trotoar

: 0,85 m

9. Beton

: 2,4 t/m310. Trotoar

: 2,2 t/m311. Aspal

: 2,3 t/m312. Tebal beton

: 20 cm

13. Tebal Aspal

: 5 cm

14. Tebal Trotoar

: 20 cm

3.1 Perhitungan Beban Mati


4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 P1 P2 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P B

P1 P2 P3 P4 P5 P6 P7 4 m 4 m 4 m 4 m 4 m 4m4m4m

b1 = b2 = b3 = b4 = b5 = b6 = b7 = b8 = 4 m

tan =

= 45o

Sin 45=

d1 = 5 d1 = 5,7 m

a = b= 4 m

Batang atas ( a )

= 6 x 4 m = 24 m

Batang bawah( b )

= 8 x 4 m = 32 m

Batang diagonal ( d )

= 8 x 5,7m = 45,6 mBatang vertikal (v)

= 7 x 4 m= 28 mPanjang Batang Keseluruhan= 24 m + 32 m + 45,6 m + 28m = 129,6m

Berat Sendiri Batang

Analisa Pembebanan :

1. Berat Sendiri Gelagar ( Q1 )

Dimisalkan pakai profil I WF 350 x 250 x 9 x14Berat = 79,7 kg/m

Q1 = Berat Profil x Panjang gelagar

= 79,7 kg/m x 129,6 m

= 10329,12 kg

= 10,329 ton

2. Beban Terpusat Untuk Setiap Titik Buhul

P1=

=

= 1,291125 ton

3. Berat Sendiri Beton ( Q2 )

Tebal beton ( T )= 20 cm

= 0,2 m

beton= 2,4 t/m3Q2= T x x b x L

= 0,2 m x 2,4 t/m3 x 5,5 m x 32 m

= 84,48 ton

4. Berat Sendiri Aspal (Q3)

Tebal aspal (T)= 5 cm

= 0,05 m

aspal

= 2,4 t/m3Q3

= T x x b x L

= 0,05 m x 2,3 t/ m3 x 3,8 m x 32 m

= 13,984 ton

5. Berat Sendiri Trotoar (Q4)

Tebal trotoar (T)= 20 cm

= 0,2 m

trotoar

= 2,2 t/m3Q4

= T x x b x L

= 0,2 m x 2,2 t/ m3 x 0,85 m x 32 m

= 11,968 ton

Untuk 2 sisi trotoar = 2 x 11,968 ton = 23,936 ton

Berat Sendiri Jembatan di Atas Dua perletakan Q total = Q1 + Q2 + Q3 + Q4

= 10,329 ton + 84,48 ton+ 13,984 ton + 23,936 ton

= 132,729 ton

Q total Pada Gaya Terpusat Untuk Setiap Titik Buhul :

P=

= = 16,5911 ton

Gaya untuk 1 titik buhul tepi

P1= P

= x 16,5911 ton

= 8,2956 ton

P1= P2=P3= P4= P5= P6= P7= P8= P9= P1 = P11= P12= P13= P14= 8,2956 ton

P1= P2 = 8,2956 ton


4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 P1 P2 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P P1 P2 P3 P4 P5 P6 P7 4 m 4 m 4 m 4 m 4 m 4m4m4m

Reaksi Perletakan : MB = 0

VA.32 - P1x 32 - P1 x 28 P2 x 24 P3 x 20 P4 x 16 P5 x 12 P6 x 8 P7 x 4 = 0VA.32 8,2956 x 32 16,5911 x 28 16,5911 x 24 16,5911 x 20 16,5911 x 16 16,5911 x 12 16,5911 x 8 16,5911 x 4 = 032 VA = 2123,6984VA =

VA = 66,3656 ton

VA= VB= 66,3656 tonKontrol :V = 0

VA + VB P1 P2 P3 P4 P4- P5 P6 P7 P1 P2 = 066,3656 66,3656 16,5911 16,5911 16,5911 16,5911 16,5911 16,5911 16,5911 8,2956 8,2956 = 00 = 0 OK !!Perhitungan Gaya Batang Terhadap Titik Kumpul 1. Titik kumpul A = Titik kumpul B

d1

b1

V = 0

VA P1 + d1 sin

= 0

66,3656 8,2956 + d1 sin 450= 0

58,07 + d1 sin 450

= 0

d1 =

d1 = - 83,123 (tekan) H = 0

d1 cos + b1 = 0

- 83,123 cos 450 + b1 = 0

b1 = 58,07 (tarik)

2. Titik kumpul J = titik kumpul P

v1 b1b2

J

P1V = 0

- P1 + v1= 0 v1 = P1 v1

= 16,6911ton (tarik) H = 0

-b1 + b2= 0

b2 = b1 b2 = 58,07 ton (tarik)3. Titik kumpul C = titik kumpul I C

450 a1

d1 v1d2V = 0

- v1 d2 sin 450 d1 sin 45

= 0

-16,5911 16,5911 - d2 sin 450 (-58,070)= 0

d2 = d2 = 58,660 (tarik)

H = 0

d2 cos 450 d1 cos 45+ a1= 0

a1

= d1 sin 450 d2 sin 45

a1

= -82,123 cos 45o 58,660 cos 45o

a1 = -99,548 (tekan)4. Titik kumpul D = titik kumpul H

V = 0

v2 = 0 H = 0

-a1 + a2= 0

a2 = a1 a2 = -99,548 ton (tekan)5. Titik kumpul K = titik kumpul O

45o

V = 0

- P2 + v2 + d2 sin 45 + d3 sin 45

= 0

-16,5911 + 0 + (58,660 sin 45) + d3 sin 45= 0

d3 sin 45= -17,16

d3 = d3 = -35,197 (tekan) H = 0

b3 b2 d2 cos + d3 cos

= 0

b3 58,07 58,660 cos 450 + ( - 35,197 ) cos 450 = 0

b3 58,07 41,48 + (-24,88)

= 0

b3 = 124,437 (tarik)

6. Titik kumpul L = titik kumpul N

v3

b3b4 P10V = 0

V3- p3= 0 v3 = P3 v3 = 16,5911 ton (tarik) H = 0

-b3 + b4= 0

b3 = b4 b3 = 124,437 ton (tarik)7. Titik kumpul E = titik kumpul G a2 E a3

45(v3 d3 d4V = 0

v3 - d3 sin d4 sin

= 0

-16,5911 - (-54,197 sin 45) d4 sin 45= 0

-8,956 d4 sin 45

= 0

d4 =

d4 = 11,734 (tarik)

H = 0

-d3 cos a2 + d4 cos + a3

= 0

- ( - 35,197 ) cos 450 ( - 99,548 ) + 11,734 cos 450 + a2 = 0

a3 = - 82,957 (tekan)8. Titik kumpul F F

a3

a4

v4V = 0

v4 = 0 H = 0

-a3 + a4= 0

a4 = a3 a4 = -82,957 ton (tekan)Kontrol dititik M :

d4 v4d5

b4 45b5M

P11

V = 0

- P4 - v4 + d4 sin 45 + d5 sin 45

= 0

-16,5911 + 0 + 11,734 sin45 + 11,734 sin 45 = 0

0 = 0 OK!

H = 0


= 0

124,437 124,437 11,734 cos 45 + 11,734 cos 45= 0

0 = 0 OK !

Tabel gaya batang akibat beban sendiri :

Nama BatangTarik (+)Tekan (-)

Batang atas

a1 = a6a2 = a5

a3 = a499,54899,54882,957

Batang diagonal

d1 = d8d2 = d7d3 = d6d4 = d5

58,66011,73482,12335,197

Batang bawah

b1 = b8b2 = b7b3 = b6

b4 = b558,0758,07124,437124,437

Batang vertikal

v1 = v7

v2 = v6

v3 = v5

v416,591116,591100

3.2 Perhitungan Garis Pengaruh akibat Beban Bergerak


4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P B

4 m 4 m 4 m 4 m 4 m 4m4m4m

Tinjau Reaksi Perletakan

x P

VA VB

Reaksi Perletakan di titik A

MA= 0

-VB.32 + P.x= 0

VB

= GP VB

=

Reaksi perletakan di titik B

MB = 0

VA.32 P (32 x) = 0

VA =

GP VA =

Potongan I-I

d1

x P

b1

VA

Interval 0 x 4

V = 0

d1 sin P + VA= 0

d1

=

GP d1

=

H = 0

b1 + d1 cos

= 0

b1

= - d1 cos

GP b1

= - GPd1 cos

Interval x > 4V = 0

VA + d1 sin = 0

d1

=

GP d1

=

Potongan II-II

d1

v1 x P R A b2

4 J 4

VA

Interval 0 x 4V = 0

VA P + d1 sin + v1= 0

v1

= -VA + P - d1 sin GP v1

= -GP VA + 1- GP d1 sin Interval 4 x 8

RJ.4 - P (8-x)= 0

RJ

= =

V = 0

VA RJ + d1 sin + v1= 0

v1

= -VA + RJ - d1 sin GP v1

= - GP VA + - GP d1 sin

Interval x > 8

V = 0

VA + d1 sin + v1 = 0

v1

= -VA - d1 sin

GP v1

= - GP VA GP d1 sin H = 0

b2 + d1 cos

= 0

b2

= d1 cos

GP b2

= GP d1 cos Potongan III-III

C a1

x d2 P

b1 VA

Interval 0 x 4

V = 0

VA - d2 sin P= 0

d2

=

GP d2

=

Interval x > 4

V = 0

VA - d2 sin = 0

d2

=

GP d2

= H = 0

b2 + a1 + d2 cos = 0

a1

= - b2 d2 cos

GP a1

= - GP b2 GP d2 cos Potongan IV-IV

C D a2

x d5

P

A b3

4 J 4 K

VA Interval 0 x 8

MK = 0VA.8 + a2.4 P (8-x)= 0

a2

=

GP a2

= V = 0

VA P + d3 sin = 0

d3

=

GP d3

=

Interval x > 8MK = 0VA.8 + a2.4

= 0

a2

= - 2VA

GP a2

= - GP 2VA

V = 0

VA + d3 sin

= 0

d3

=

GP d3

=

Potongan V-V

C a1 d3 x P v2 R A b3

4 J 4 K

VA

Interval 0 x 8V = 0

v2 + VA P + d3 sin = 0

v2


= - GPVA + 1 - GPd3 sin Interval 8 x 12V = 0

v2 + VA RK + d3 sin = 0

v2

= - VA + - d3 sin GP v2

= - VA + - d3 sin Interval x >12

V = 0v2 + VA + d3 sin

= 0

GP v2

= - GPVA - GPd3 sin Potongan VI-VI

C D a2 d3 x v3 R A P b4 4 J 4 K 4 L VA

Interval 0 x 12V = 0

v3 + VA P + d3 sin = 0

v3


= - GPVA + 1 - GPd3 sin Interval 12 x 16V = 0

v3 + VA RL + d3 sin = 0

v3

= - VA + - d3 sin GP v3

= - VA + - d3 sin Interval x >12

V = 0v3 + VA + d3 sin

= 0

GP v3

= - GPVA - GPd3 sin H = 0

b4 + d3 cos

= 0

b4

= d3 cos

GP b4

= GP d3 cos Potongan VII-VII

C D Ea3 x d4 A P b4 4 J 4 K 4 L VA

Interval 0 x 12V = 0

VA P + d4 sin = 0

d4

=

GP d4

=

Interval x > 12V = 0

VA + d4 sin = 0

d4

= GP d4

=

H = 0

b4 + a3 + d4 cos = 0a3

= - b4 d4 cos GP a3

= - GP b4 GP d4 cos

Tabel garis pengaruh akibat beban bergerak :

Potongan Interval Garis pengaruh

GP VA =

GP VB =

Potongan I-I

0 x 4x > 5GP d1 =

GP b1 = -GPd1 sin

GP d1 =

Potongan II - II

0 x 4 4 x 8x > 8GP v1 = -GP VA + 1- GP d1 sin GP v1 = - GP VA + - GP d1 sin GP v1 = - GPV GP d1 sin GP b2 = GP d1 cos

Potongan III - III

0 x 4 x > 8

GP d2 =

GP d2 = GP a1 = - GP b2 GP d2 cos

Potongan IV-IV0 x 8x > 8GP a2 = GP d3 =

GP a2 = - GP 2VA

GP d3 =

Potongan V - V 0 x 8 8 x 12x > 12GP v2 = - GPVA + 1 - GPd3 sin GP v2 = - VA + - d3 sin GP v2 = - GPVA - GPd3 sin

Potongan VI - VI0 x 1212 x 16 x > 16GP v3 = - GPVA + 1 - GPd3 sin GP v3 = - VA + - d3 sin GP v3 = - GPVA - GPd3 sin GP b4 = GP d3 cos

PotonganVII-VII

0 x 8 x > 8GP d4 =

GP d4 =

GP a3 = - GP b4 GPd4 cos

3.3 Pembebanan Akibat Beban HidupBerdasarkan PPJJR 19871. Muatan Merataq = 2,2 t/m 1,1 / 60 (L 30) ....Untuk 30 m < L < 60 m q = 2,2 t/m 1,1 / 60 (32 30) = 2,19 t/m

2. Beban Merataq = (2,19 t/m x 5,5 + 1,5 x t/m ) 70 %

= 9,58 t/mBeban terbagi rata

q=

q= 3,48 t

3. Beban trotoar = 60 % x 0,5 t/m x 0,85 m = 0,255 t q total = 3,48 t + 0,255 t

= 3,74 t4. Beban untuk 1 gelagar =

=

= 1,87 t/m

5. Beban Kejut (K)

K = 1+

= 1+

= 1,246. Muatan Garis P = 12 t / jalur

P =

P = 4,363 t

7. Beban untuk 1 gelagar =

= 2,182 t Menentukan Gaya Batang Akibat Beban Hidup


4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P B 4 m 4 m 4 m 4 m 4 m 4m4m4m

Beban bergerak hanya terdapat pada lantai kendaraan, jadi batang yang menahan beban bergerak adalah batang bawah saja. b1 b8

= (1/2 x 32 x 0,42 x 1870) + (2182 x 0,42) = 13482,84 kg

b2 b7

= (1/2 x 32 x 0,42 x 1870) + (2182 x 0,42) = 13482,84 kg

b3 b6

= (1/2 x 32 x 0,42 x 1870) + (2182 x 0,42) = 13482,84 kg

b4 b5= (1/2 x 32 x 0,42 x 1870) + (2182 x 0,42) = 13482,84 kgTabel gaya batang akibat beban hidup:

Nama BatangTarik (+)Tekan (-)

Batang atas

a1 = a6a2 = a5

a3 = a4

Batang diagonal

d1 = d8d2 = d7d3 = d6d4 = d5

Batang bawah

b1 = b8b2 = b7b3 = b6

b4 = b513,482813,482813,482813,4828

Batang vertikal

v1 = v7

v2 = v6

v3 = v5

v4

3.3 Perhitungan Beban Angin


4m d1 v1 d2 v2 d3 v3 d4 v4 d5 v5 d6 v6 d7 v7 d8 b1 b2 b3 b4 b5 b6 b7 b8 A J K L M N O P B

4 m 4 m 4 m 4 m 4 m 4m4m4m

1. Panjang Jembatan

: 32 m

2. Tinggi Jembatan

: 4 m3. Tinggi Gelagar Melintang: 0,70

4. Tebal Aspal & Beton

: 25 cm = 0,25 m5. Angin Tekan

: 100 kg/m

6. Angin Hisap

: 50 kg/m

7. Angin Total

: 150 kg/mGaya angin akibat beban hidup setinggi 2m (W1)

W1 = 150 kg/m x 32 m x 2 m = 9600 kg

Gaya angin pada sisi rangka jembatan (W2)

W2 = 150 kg/m x 32 m x 4 m = 19200 kg

Gaya angin pada lantai kendaraan (W3)

W3 = 150 kg/m x 32 m x 0,95 = 4560 kg

Reaksi Perletakan

R 7,2 W1 ( 1 + 0,95 ) + W2 (h/2) + W3 ( Y/2 ) = 0

R = R = 8234,167 kg

Gaya batang dengan metode titik kumpulP = P = = 1372,36 kg

P = = = 686,18 kgReaksi Perletakan : MB = 0

VA.32 - P1x 28 - P1 x 24 P2 x 20 P3 x 216 P4 x 12 P5 x 8 P2 x 4 = 0VA.32 (686,18 x 28) (1372,36 x 24) (1372,36 x 20) (1372,36 x 16) (1372,36 x 12) (1372,36 x 8) (686,18 x 4) = 0VA 32= 131746,56VA =

VA = 4117,08 kg

VA= VB= 4117,08 kg

Kontrol :V = 0

VA + VB P1 P2 P3 P4 P4- P5 P1 P2 = 04117,08 + 4117,08 1372,36 1372,36 1372,36 1372,36 1372,36 686,18 686,16 = 00 = 0 OK !!Perhitungan Gaya Batang Terhadap Titik Kumpul 1. Titik kumpul A = Titik kumpul B

d1

b1

V = 0

VA + d1 sin

= 0

4117,08 + d1 sin 450= 0

d1 =

d1 = -5822,43 kg (tekan)

H = 0

d1 cos + b1 = 0

- 5822,43 cos 450 + b1 = 0

b1 = 4117,08 (tarik)

2. Titik kumpul J = titik kumpul P

v1 b1b2

J P1V = 0

v1 = 0

H = 0

-b1 + b2= 0

b2 = b1b2 = 4117,08 kg (tarik)3. Titik kumpul C = titik kumpul I P1

C

450 a1

d1 v1d2V = 0

- v1 - P1 d2 sin 450 d1 sin 45

= 0

0 686,18 - d2 sin 450 (-5822,43 sin 45)= 0

d2 = d2 = 4852,02 (tarik)

H = 0

d2 cos 450 d1 cos 45+ a1= 0

a1

= d1 sin 450 d2 sin 45

a1

= -5822,43 cos 45o 4852,08 cos 45oa1 = -7547,98 (tekan)4. Titik kumpul D = titik kumpul H

V = 0

-v2 P1= 0

v2 = -1372,36 kg (Tekan) H = 0

-a1 + a2= 0

a2 = a1 a2 = -7547,98 kg (tekan)5. Titik kumpul K = titik kumpul O

45o

V = 0

v2 + d2 sin 45 + d3 sin 45

= 0

-1372,36 + (5822,43 sin 45) + d3 sin 45= 0

d3 sin 45= -2058,53

d3 = d3 = -2911,21 (tekan) H = 0


= 0

b3 4117,08 4852,43 cos 450 + ( - 2911,21 ) cos 450= 0

b3 4117,08 4117,08 + (-2744,72)

= 0

b3 = 9606,51 (tarik)

6. Titik kumpul L = titik kumpul N

V = 0

V3= 0 H = 0

-b3 + b4= 0

b4 = b3

b4 = 9606,51 kg (tarik)

7. Titik kumpul E = titik kumpul G

a2 E a3

45(

v3d4 d3V = 0

v3 P2 - d3 sin d4 sin

= 0

-0 1372,36 - (-2911,21 sin 45) d4 sin 45= 0

686,18 d4 sin 45

= 0

d4 =

d4 = 970,40 (tarik)

H = 0

-d3 cos a2 + d4 cos + a3

= 0

- ( - 2911,21 ) cos 450 ( - 7547,98 ) + 970,40 cos 450 + a3 = 0

a3 = - 10292,79 (tekan)

8. Titik kumpul F F

a3

a4

v4V = 0

-v4 P3 = 0

V4 = -1372,36 kg (tarik) H = 0

-a3 + a4= 0

a4 = a3 a4 = -10292,69 kg (tekan)Kontrol dititik M :

d4 v4d5

b4 45b5V = 0

v4 + d4 sin 45 + d5 sin 45

= 0

-1372,36 + 970,40 sin45 + 970,40 sin 45 = 0

0 = 0 OK! H = 0


= 0

9606,51 9606,51 970,40 cos 45 + 970,40 cos 45= 0

0 = 0 OK !Tabel Kombinasi Beban

Nama BatangBerat SendiriBeban AnginBeban HidupKombinasi Beban

Tarik (+)Tekan (-)Tarik (+)Tekan (-)Tarik (+)Tekan (-)Tarik (+)Tekan (-)

Batang Atas

a1 = a699.5487.54798125.5

a2 = a599.5487.54798125.5

a3 = a482.95710.2927107.78

Batang Diagonal

d1 = d882.1235.82243103.21

d2 = d758.664.8520274.274

d3 = d635.1972.9112144.565

d4 = d511.7340.970414.857

Batang Bawah

b1 = b858.074.1170813.482894.55

b2 = b758.074.1170813.482894.55

b3 = b6124.449.6065113.4828178.58

b4 = b5124.449.6065113.4828178.58

Batang Vertikal

v1 = v716.591019.909

v2 = v6 01.372361.0979

v3 = v516.591019.909

v401.372361.0979

Kombinasi beban yang digunakan adalah :

= 1,2 D + 1,6 La + 0,8 W

Dimana :

D = Beban mati yang diakibatkan oleh berat konstruksi permanen

L = Beban hidup yang diakibatkan oleh kendaraan yang lewat diatasnya

W = Beban angin

3.4 Desain Rangka jembatan3.4.1 Batang atas (a1,a2,a3,a4,a5,a6)batang tekanDigunakan profil 350 x 350 x 19 x 19

B= 357 mm

ro= 20 mm

d= 350 mm

h= 350 2 (19+20) = 272 mm

tf= 19 mm

rx= 147 mm

tw= 19 mm

ry= 85,3 mm

Ag= 19840 mmGaya batang maksimum (Nu)= 125,25 tonPanjang batang (L)

= 4000 mmJenis Baja: BJ 37

fy

= 240 Mpa

fu

= 370 Mpa

Periksa Kelangsingan :

Flens

Web

Kondisi tumpuan sendi sendi (k=1)

Arah sumbu kuat (sumbu x)

0,25 < cx < 1,2

x =

x =


0,25 < cx < 1,2

x =

x =

3.4.2 Batang bawah (b1,b2,b3,b4,b5,)batang tarikDigunakan profil 300 x 300 x 15 x 15

Ag= 13480 mm2

rx= 12,6 cm

ry= 7,26 cmGaya batang maksimum (Tu)= 178,58 tonPanjang batang(L)

= 4000 mm

Jenis Baja: BJ 37

fy

= 240 Mpa

fu

= 370 Mpa


Kondisi leleh:

x Tn= x fy x Ag

= 0,90 x 240 x 13480

= 291,17 ton

Kondisi fraktur

Ae= AN . U

Ae = (85% . Ag) . U

Ae = 0,85 . 13480. 0,85

Ae= 9968,5 mm2 x Tn= x fu x Ae

= 0,75 x 370 x 9968,5

= 276,62 ton

x Tn (276,62 ton) > Tu (178,58, ton)ok!!!

3.4.3 Batang diagonal (d2,d7,d4,d5,)batang tarikDigunakan profil 200 x 200 x 12 x 12

Ag= 7153 mm2

rx= 8,35 cm2ry= 4,88 cm2Gaya batang maksimum (Tu)= 74,274 tonPanjang batang(L)

= 4000 mm

Jenis Baja: BJ 37

fy

= 240 Mpa

fu

= 370 Mpa


Kondisi leleh:

x Tn= x fy x Ag

= 0,90 x 240 x 7153

= 154,5 ton

Kondisi fraktur

Ae= AN . U

Ae = (85% . Ag) . U

Ae = 0,85 . 7153. 0,85Ae= 5168,04 mm2 x Tn= x fu x Ae

= 0,75 x 370 x 5168,04

= 133,85 ton

x Tn (133,85 ton) > Tu (74,274 ton)ok!!!

3.4.4 Batang diagonal (d1,d8,d3,d6)batang tekanDigunakan profil 250 x 250 x 9 x 14

B= 250 mm

ro=16 mm

d= 250 mm

h= 250 2 (14+16) = 190 mm

tf= 14 mm

rx= 108 mm

tw= 9 mm

ry= 62,9 mm

Ag= 9218 mm2Gaya batang maksimum (Nu)= 103,21 tonPanjang batang (L)


fy

= 240 Mpa

fu

= 370 MpaPeriksa Kelangsingan :

Flens

Web



0,25 < cx < 1,2

x =

x =


0,25 < cx < 1,2

x =

x =

3.4.5 Batang vertikal (v1,v7,v3,v5,)batang tarikDigunakan profil 100 x 100 x 6 x 8

Ag= 2190 mm2

rx= 4,18 cm2ry= 2,47 cm2Gaya batang maksimum (Tu)= 19,909tonPanjang batang(L)

= 4000 mm

Jenis Baja: BJ 37

fy

= 240 Mpa

fu

= 370 Mpa


Kondisi leleh:

x Tn= x fy x Ag

= 0,90 x 240 x 2190

= 47,34 ton

Kondisi fraktur

Ae= AN . U

Ae = (85% . Ag) . U

Ae = 0,85 . 2190. 0,85

Ae= 1582,28 mm2 x Tn= x fu x Ae

= 0,75 x 370 x 1582,28

= 43,91 ton

x Tn (43,91 ton) > Tu (19,909 ton)ok!!!

3.4.6 Batang vertikal (v2,v6,v4) batang tekanDigunakan profil 100 x100 x 6 x 8

B= 100 mm

ro=10 mm

d= 100 mm

h= 100 2 (8+10) = 64 mm

tf= 8 mm

rx= 41,8 mm

tw= 6 mm

ry= 24,7 mm

Ag= 2190 mm2Gaya batang maksimum (Nu)= 1,0973 tonPanjang batang (L)


fy

= 240 Mpa

fu

= 370 MpaPeriksa Kelangsingan :

Flens

Web



0,25 < cx < 1,2

x =

x =


cx> 1,2

x = 1,25 x cx2

x = 1,25 x 1,782= 3,69

3.5 Perencanaan Baut1. Mutu baut = Bj 37fy

= 240 Mpa

fu

= 370 Mpa

2. Diameter baut () = 22 mm3. Asumsi baut tidak berulir penuh4. Tebal pelat penyambung diambil setebal 1 cm3.5.1 Jumlah Baut

ABaut = . . dn2 = . . 222 = 380,13 mm2 Geser baut

Vd = f . Vn

= f . ri . Fu . ABaut

= (0,75) . (0,5) . (370) . (380,13)

= 5,274 ton Tumpu baut

Rd= f . Rn. 2,4

= f . db . tp . Fu 2,4

= (0,75) . (22) . (10) . (370) .2,4= 14,65 tonDari 2 nilai diatas maka ambil nilai terbesar yaitu 14,65 ton Jumlah baut yang digunakan

a. Batang atas gaya batang maksimum adalah 329,25 ton Maka :

Jumlah Baut = = 8,54 Buah Baut 9 Buah Baut

b. Batang bawah gaya batang maksimum adalah 272,77 ton Maka :


c. Batang diagonal gaya batang maksimum adalah 128,59 ton Maka :

Jumlah Baut = = 7,04 Buah Baut 8 Buah Bautd. Batang Vertikal gaya batang maksimum adalah 1,0973


3.5.2 Mendisain Jarak BautSusunan dan jarak baut direncanakan sebaris dengan syarat :

dn = 22 cm

Jarak antar baut dengan baut ( S )

Minimal S3 dn

S3 . 22 S66 mm

Maksimal S 200 mm

Dari 2 nilai diata maka diambil S = 66 mm 70 mmJarak baut dengan tepi luar (S1)

Minimal S1 = 1,5 dn (untuk tepi dipotong dengan mesin)

= 1,5 (22)

= 33 mm

Maksimal S1150 mm

Dari 2 nilai diatas maka diambil nilai S1 = 33 mm 35 mm U1 = S1 S = U

P1

45

VA = 66,3656

a1

a2

V2

v2

d3

d2

b2

b3

P2

P2

P5

P4

P3

P2

P1

P1

45

VA = 4117,08

P1

a1

a2

V2

v2

d3

d2

b2

b3

V3

b4

b3

P2

P3

70 mm

35 mm

PAGE III-1

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bab iii okeeeeeeeee

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