BAB V Balok Gerber A S B C - umpalangkaraya.ac.id · Noviyanthy H. MT. 70 Statika dan Mekanika Bahan 1 BAB V Balok Gerber Pada suatu kondisi kadangkala balok akan dibuat dengan panjang

Noviyanthy H. MT.

Statika dan Mekanika Bahan 1 70

BAB V

Balok Gerber

Pada suatu kondisi kadangkala balok akan dibuat dengan panjang

bentang yang besar sehingga memerlukan cara tersendiri dalam perhitungannya.

Pada kondisi ini maka pilihannya adalah dengan menggunakan kontruksi bersendi

banyak (lebih dari dua sendi) (Wesli, 2012). Balok tersebut umumnya disebut

dengan Balok gerber. Pengertian dari balok gerber adalah suatu konstruksi balok

yang mempunyai jumlah reaksi perletakan lebih dari 3 buah, namun masih bisa

diselesaikan dengan syarat-syarat kesetimbangan (Soelarso).

Penentuan jumlah sendi tambahan dapat menggunakan persamaan 5.1

berikut (Wesli, 2012) :

𝑆 = (𝑛 − 2) ......................................................................................... (5.1)

dimana :

S = Jumlah sendi tambahan

n = jumlah tumpuan

Berikut ini contoh-contoh balok gerber dengan 1 sendi tambahan.

A

B C

S1

A S

S = SENDI GERBER

B C

Noviyanthy H. MT.


Berikut ini contoh-contoh balok gerber dengan 2 sendi tambahan.

Berikut ini beberapa contoh soal dan penyelesaian untuk struktur balok gerber.

Contoh Soal 1

Penyelesaian :

Jumlah sendi tambahan adalah

S = n – 2 = 3 – 2 = 1

A B S S C D

q

q

A B C D

q qS1 S2

A B C D

S1 S2

2,5 m 3,0 m 2,5 m

11,0 m

P2 = 3,5 tonP1 = 3 ton

2,0 m 1,0 m

A 1 S B 2 C

Noviyanthy H. MT.


Tahap pertama : selesaikan dulu bentang A – S seperti perhitungan balok

sederhana, sehingga di dapat nilai VA dan VS.

Tahap kedua : nilai VS yang diperoleh dari bentang A – S, dijadikan beban

terpusat untuk bentang S – B – C, sehingga di dapat nilai VB dan VC.

1. Perhitungan Reaksi Perletakan Bentang A – S

ΣMS = 0

VA . 4,5 – P1 . 2 = 0

VA . 4,5 – 3 . 2 = 0

VA = 1,333 ton ( )

ΣMA = 0

-VS . 4,5 + P1 . 2,5 = 0

-VS . 4,5 + 3 . 2,5 = 0

VS = 1,667 ton ( )

2,5 m 3,0 m 2,5 m

11,0 m

P2 = 3,5 ton

P1 = 3 ton

2,0 m 1,0 m

A 1S

B 2 CS

P1 = 3 ton

A 1S

2,5 m 2,0 m

Noviyanthy H. MT.


Kontrol :

ΣV = 0

VA + VS = P1

1,333 + 1,667 = 3,0

3,0 ton = 3,0 ton (oke!!!)

2. Perhitungan Reaksi Perletakan Bentang S – B – C

ΣMB = 0

-VS . 1 – P2 . 3 – VC . 5,5 = 0

-1,667 . 1 + 3,5 . 3 – VC . 5,5 = 0

VC = 1,606 ton ( )

ΣMC = 0

-VS . 6,5 + VB . 5,5 – P2 . 2,5 = 0

-1,667 . 6,5 + VB . 5,5 – 3,5 . 2,5 = 0

VS = 3,561 ton ( )

Kontrol :

ΣV = 0

VB + VC = P2 + VS

1,606 + 3,561 = 3,5 + 1,667

5, 167 ton = 5, 167 ton (oke!!!)

3,0 m 2,5 m

P2 = 3,5 ton

1,0 m

B 2 CS

Noviyanthy H. MT.


3. Perhitungan Gaya-gaya Dalam

a) Bentang A – 1 ( 0 ≤ x ≤ 2,5 meter )

ΣMx = 0

VA . x – Mx = 0

Mx = 1,333 x

x = 0 ; MA = 0 tm

x = 2,5 ; M1 = 3,333 tm

ΣV = 0

VA – Lx = 0

Lx = 1,333

x = 0 ; LA = 1,333 ton

x = 2,5 ; L1 = 1,333 ton

ΣH = 0

Nx = 0

XVA Lx

Nx

Mx

Noviyanthy H. MT.


b) Bentang 1 – S ( 0 ≤ x ≤ 2 meter )

ΣMx = 0

VA . (2,5 + x) – P1. x – Mx = 0

Mx = 1,333 (2,5 + x) – 3. x

= 3,333 + 1,333x – 3x

x = 0 ; M1 = 3,333 tm

x = 2 ; MS = - 0,001 tm ≈ 0 tm

ΣV = 0

VA – P1 – Lx = 0

Lx = 1,333 – 3

= - 1,667

x = 0 ; L1 = - 1,667 ton

x = 2 ; LS = - 1,667 ton

ΣH = 0

Nx = 0

2,5 m XLx

Nx

Mx

VA

P1 = 3 ton

Noviyanthy H. MT.


c) Bentang S – B ( 0 ≤ x ≤ 1 meter )

ΣMx = 0

- VS . x – Mx = 0

Mx = - 1,667. x

x = 0 ; MS = 0 tm

x = 1 ; MB = - 1,667 tm

ΣV = 0

-VS – Lx = 0

Lx = - 1,667

x = 0 ; LS = - 1,667 ton

x = 1 ; LB = - 1,667 ton

ΣH = 0

Nx = 0

X

VS

Lx

Nx

Mx

Noviyanthy H. MT.


d) Bentang B – 2 ( 0 ≤ x ≤ 3 meter )

ΣMx = 0

-VS . (1,0 + x) + VB. x – Mx = 0

Mx = -1,667 (1,0 + x) + 3,561. x

= -1,667 – 1,667x + 3,561x

x = 0 ; MB = - 1,667 tm

x = 3 ; M2 = 4, 015 tm

ΣV = 0

-VS + VB – Lx = 0

Lx = - 1,667 + 3,561

= 1,894

x = 0 ; LB = 1,894 ton

x = 3 ; L2 = 1,894 ton

ΣH = 0

Nx = 0

X1,0 m

VB

VS

Lx

Nx

Mx

Noviyanthy H. MT.


e) Bentang C – 2 ( 0 ≤ x ≤ 2,5 meter )

ΣMx = 0

- VC . x + Mx = 0

Mx = 1,606. x

x = 0 ; MC = 0 tm

x = 2,5 ; M2 = 4,015 tm

ΣV = 0

VC + Lx = 0

Lx = - 1,606

x = 0 ; LC = - 1,606 ton

x = 2,5 ; L2 = - 1,606 ton

ΣH = 0

Nx = 0

X

VC

Nx

Mx

Lx

Noviyanthy H. MT.


4. Gambar bidang momen, lintang dan normal

2,5 m 3,0 m 2,5 m

11,0 m

VA

P2 = 2 ton

VC

P1 = 3 ton

VB

2,0 m 1,0 m

A 1 S B 2 C

0 tm 0 tmMOMEN

0 tLINTANG 0 t

1,333 t

1,667 t

1,894 t

1,606 t

0 tNORMAL 0 t

++

-

++

- -

3,333 tm

4,015 tm

1,667 tm

P2 = 3,5 ton

P1 = 3 ton

A 1S

B 2 CS

Noviyanthy H. MT.


Contoh Soal 2

Penyelesaian :


S = n – 2 = 3 – 2 = 1

Tahap pertama : selesaikan dulu bentang A – S seperti perhitungan balok

sederhana, sehingga di dapat nilai VA dan VS.

Tahap kedua : nilai VS yang diperoleh dari bentang A – S, dijadikan beban

terpusat untuk bentang S – B – C, sehingga di dapat nilai VB dan VC.

3,5 m 3,0 m 3,0 m

13,5 m

P = 12 ton

2,0 m

A DS B E C

2,0 m

3,5 m 3,0 m 3,0 m

13,5 m

P = 12 ton

2,0 m

A

DB E C

2,0 m

S

S

Noviyanthy H. MT.


1. Perhitungan Reaksi Perletakan Bentang A – S

ΣMS = 0

VA . 3,5 – q . L (½ L) = 0

VA . 3,5 – 2,5 . 3,5 (½ . 3,5) = 0

VA = 4,375 ton ( )

ΣMA = 0

-VS . 3,5 + q . L (½ L) = 0

-VS . 3,5 + 2,5 . 3,5 (½ . 3,5) = 0

VS = 4,375 ton ( )

Kontrol :

ΣV = 0

VA + VS = q . L

4,375 + 4,375 = 2,5 . 3,5

8,750 ton = 8,750 ton (oke!!!)

2. Perhitungan Reaksi Perletakan Bentang S – B – C

ΣMB = 0

-VS . 2 – q . L (½ . 2) + q . L (½ . 2) + P . 5 – VC . 8 = 0

- 4,375 . 2 – 2,5 . 2 (½ . 2) + 2,5 . 2 (½ . 2) + 12 . 5 – VC . 8 = 0

VC = 6,406 ton ( )

3,5 m

AS

3,0 m 3,0 m

P = 12 ton

2,0 m

DB E C

2,0 m

S

Noviyanthy H. MT.


ΣMC = 0

-VS . 10 – q . L (½ . L + 6) – P . 3 + VB . 8 = 0

- 4,375 . 10 – 2,5 . 4 (½ . 4 + 6) – 12 . 3 + VB . 8 = 0

VB = 19,969 ton ( )

Kontrol :

ΣV = 0

VB + VC = q . L + P + VS

19,969 + 6,406 = (2,5 . 4) + 12 + 4,375

26,375 ton = 26,375 ton (oke!!!)


a) Bentang A – S ( 0 ≤ x ≤ 3,5 meter )

ΣMx = 0

VA . x – q . x (½. x) – Mx = 0

Mx = 4,375 x – 1,25 x2

x = 0 ; MA = 0 tm

x = 3,5 ; MS = 0 tm

ΣV = 0

VA – q . x – Lx = 0

Lx = 4,375 – 2,5 x

x = 0 ; LA = 4,375 ton

x = 3,5 ; LS = - 4,375 ton

ΣH = 0

Nx = 0

X

Lx

Nx

Mx

VA

Noviyanthy H. MT.


Terjadi M maks pada jarak :

LX = 0

4,375 – 2,5 x = 0 x = 1,75 meter

M maks = 4,375 x – 1,25 x2

= 4,375 (1,75) – 1,25 (1,75)

2 = 3,828 tm

b) Bentang S – B ( 0 ≤ x ≤ 2,0 meter )

ΣMx = 0

- VS . x – q . x (½. x ) – Mx = 0

Mx = 4,375 . x – 1, 25 x2

x = 0 ; MS = 0 tm

x = 2 ; MB = - 13,75 tm

ΣV = 0

- VS – q . x – Lx = 0

Lx = - 4,375 – 2,5 x

x = 0 ; LS = - 4,375 ton

x = 2 ; LB = - 9,375 ton

ΣH = 0

Nx = 0

X

Lx

Nx

Mx

VS

Noviyanthy H. MT.


c) Bentang B – D ( 0 ≤ x ≤ 2,0 meter )

ΣMx = 0

- VS . (2 + x) – q . 2 (½ . 2 + x) – q . x (1/2. x) + VB . x – Mx = 0

Mx = - 4,375. (2 + x) – 2,5 . 2 (½. 2 + x) – 2,5. x (½ x ) + 19,969 . x

= - 13,750 + 10,594 x – 1,25 x2

x = 0 ; MB = - 13,750 tm

x = 2 ; MD = 2,4375 tm

ΣV = 0

-VS – q . 2 + VB – q . x – Lx = 0

Lx = - 4,375 – 2,5 . 2 + 19,969 – 2,5 . x

= 10,594 – 2,5 x

x = 0 ; LB = 10,594 ton

x = 2 ; LD = 5,594 ton

ΣH = 0

Nx = 0

2,0 m XLx

Nx

Mx

VS

Noviyanthy H. MT.


d) Bentang C – E ( 0 ≤ x ≤ 3,0 meter )

ΣMx = 0

- VC . x + Mx = 0

Mx = 6,406. x

x = 0 ; MC = 0 tm

x = 3 ; ME = 19,219 tm

ΣV = 0

VC + Lx = 0

Lx = - 6,406

x = 0 ; LC = - 6,406 ton

x = 3 ; LE = - 6,406 ton

ΣH = 0

Nx = 0

XVC

Nx

Mx

Lx

Noviyanthy H. MT.


e) Bentang E – D ( 0 ≤ x ≤ 3,0 meter )

X 3,0 m

P = 12 ton

VC

Nx

Mx

Lx

ΣMx = 0

-VC . (3 + x) + P . x + Mx = 0

Mx = - 6, 406 (3 + x) + 12 . x

= 19, 219 – 5,594 x

x = 0 ; ME = 19,219 tm

x = 3 ; MD = 2,4375 tm

ΣV = 0

VC – P + Lx = 0

Lx = - 6,406 + 12

= 5,594

x = 0 ; LE = 5,594 ton

x = 3 ; LD = 5,594 ton

ΣH = 0

Nx = 0

Noviyanthy H. MT.



3,5 m 3,0 m 3,0 m

13,5 m

VA

P = 12 ton

VCVB

2,0 m

A DS B E C

2,0 m

0 tmMOMEN

P = 12 ton

A

DB E C

S

S

0 tm

0 tLINTANG 0 t

0 tNORMAL 0 t

9,375 t

19,219 tm

4,375 t

10,594 t

5,594 t

6,406 tm

4,375 t

++

+

-

--

+ 2,4375 tm

3,828 tm

13,75 tm

Noviyanthy H. MT.


Contoh Soal 3

Penyelesaian :


S = n – 2 = 4 – 2 = 2

Tahap pertama : selesaikan dulu bentang S1 – S2 seperti perhitungan balok

sederhana, sehingga di dapat nilai VS1 dan VS2.

Tahap kedua : nilai VS1 yang diperoleh dari bentang S1 – S2, dijadikan beban

terpusat untuk bentang A – B – S1, sehingga di dapat nilai VA dan VB. Begitu juga

dengan VS2 akan dijadikan beban untuk bentang S2 – C – D, dan diperoleh nilai

VC dan VD.

A B C D

17,0 m

1,0 m5,0 m 4,0 m6,0 m 1,0 m

P = 10 ton

q = 3 t/m q = 3,5 t/m q = 4 t/m

S1 S2

A B C D

17,0 m

1,0 m5,0 m 4,0 m6,0 m 1,0 m

q = 3 t/m q = 4 t/m

S1 S2

P = 10 ton

q = 3,5 t/m

S2S1

Noviyanthy H. MT.


1. Perhitungan Reaksi Perletakan Bentang S1 – S2

ΣMS2 = 0

VS1 . 6 – q . L (½ L) – P . L = 0

VS1 . 6 – 3 . 6 (½ . 6) – 10 . 6 = 0

VS1 = 20,5 ton ( )

ΣMS1 = 0

-VS2 . 6 + q . L (½ L) = 0

-VS2 . 6 + 3 . 6 (½ . 6) = 0

VS2 = 10,5 ton ( )

Kontrol :

ΣV = 0

VS1 + VS2 = q . L + P

20,5 + 10,5 = 3,5 . 6 + 10

31,0 ton = 31,0 ton (oke!!!)

P = 10 ton

q = 3,5 t/m

6,0 m

S1 S2

Noviyanthy H. MT.


2. Perhitungan Reaksi Perletakan Bentang A – B – S1

ΣMB = 0

VA . 5 – q . 5 (½ . 5) + VS1 . 1 = 0

VA . 5 – 3 . 5 (½ . 5) + 20,5 . 1 = 0

VA = 3,40 ton ( )

ΣMA = 0

VS1 . 6 + q . 5 (½ . 5) – VB . 5 = 0

20,5 . 6 + 3 . 5 (½ . 5) – VB . 5 = 0

VB = 32,10 ton ( )

Kontrol :

ΣV = 0

VA + VB = q . L + VS1

3,40 + 32,10 = (3 . 5) + 20,5

35,5 ton = 35,5 ton (oke!!!)

A B

1,0 m5,0 m

q = 3 t/m

S1

Noviyanthy H. MT.


3. Perhitungan Reaksi Perletakan Bentang S2 – C – D

ΣMD = 0

- VS2 . 5 + VC . 4 – q . 4 (½ . 4) = 0

- 10,5 . 5 + VC . 4 – 4 . 4 (½ . 4) = 0

VC = 21,125 ton ( )

ΣMC = 0

- VD . 4 + q . 4 (½ . 4) – VS2 . 1 = 0

-VD . 4 + 4 . 4 (½ . 4) – 10,5 . 1 = 0

VD = 5,375 ton ( )

Kontrol :

ΣV = 0

VC + VD = q . L + VS2

21,125 + 5,375 = 4 . 4 + 10,5

26,5 ton = 26,5 ton (oke!!!)

C D

4,0 m1,0 m

q = 4 t/m

S2

Noviyanthy H. MT.



a) Bentang A – B ( 0 ≤ x ≤ 5,0 meter )

ΣMx = 0

VA . x – q . x (½. x) – Mx = 0

Mx = 3,4 x – 1,5 x2

x = 0 ; MA = 0 tm

x = 5 ; MB = - 20,5 tm

ΣV = 0

VA – q . x – Lx = 0

Lx = 3,4 – 3 x

x = 0 ; LA = 3,4 ton

x = 5 ; LB = - 11,6 ton

ΣH = 0

Nx = 0


LX = 0

3,4 – 3x = 0 x = 1,133 meter

M maks = 3,4 x – 1,5 x2

= 3,4 (1,133) – 1,5 (1,133)

2 = 1,927 tm

X

q = 3 t/m

VA Lx

Nx

Mx

Noviyanthy H. MT.


b) Bentang B – S1 ( 0 ≤ x ≤ 1,0 meter )

ΣMx = 0

VA . (5 + x) – q . 5 (½ .5 + x ) + VB . x – Mx = 0

Mx = 3,4 (5 + x) – 3. 5 (½. 5 + x) + 32,1 x

= - 20,5 + 20,5 x

x = 0 ; MB = - 20,5 tm

x = 1 ; MS1 = 0 tm

ΣV = 0

VA – q . 5 + VB – Lx = 0

Lx = 3,4 – 3 . 5 + 32,1

= 20,5

x = 0 ; LB = 20,5 ton

x = 1 ; LS1 = 20,5 ton

ΣH = 0

Nx = 0

X5,0 m

q = 3 t/m

VA Lx

Nx

Mx

Noviyanthy H. MT.


c) Bentang S1 – S2 ( 0 ≤ x ≤ 6,0 meter )

ΣMx = 0

VS1 . x – q . x (½. x) – P . x – Mx = 0

Mx = 20,5 . x – 3,5 . x (½ x ) – 10 . x

= 10,5 x – 1,75 x2

x = 0 ; MS1 = 0 tm

x = 6 ; MS2 = 0 tm

ΣV = 0

VS1 – P – q . x – Lx = 0

Lx = 20,5 – 10 – 3,5 . x

= 10,5 – 3,5 x

x = 0 ; LS1 = 10,5 ton

x = 6 ; LS2 = -10,5 ton

ΣH = 0

Nx = 0


LX = 0

10,5 – 3,5 x = 0 x = 3,0 meter

M maks = 10,5 x – 1,75 x2

= 10,5 (3) – 1,75 (3)2 = 15,750 tm

P = 10 ton

q = 3,5 t/m

XLx

Nx

Mx

VS1

Noviyanthy H. MT.


d) Bentang VS2 – C ( 0 ≤ x ≤ 1,0 meter )

ΣMx = 0

- VS2 . x – Mx = 0

Mx = - 10,5 x

x = 0 ; MS2 = 0 tm

x = 1 ; MC = - 10,5 tm

ΣV = 0

- VS2 – Lx = 0

Lx = - 10,5

x = 0 ; LS2 = - 10,5 ton

x = 1 ; LC = - 10,5 ton

ΣH = 0

Nx = 0

XLx

Nx

Mx

VS2

Noviyanthy H. MT.


e) Bentang D – C ( 0 ≤ x ≤ 4,0 meter )

ΣMx = 0

-VD . x + q . x (1/2 . x) + Mx = 0

Mx = 5,375 x – 2x2

x = 0 ; MD = 0 tm

x = 4 ; MC = - 10,5 tm

ΣV = 0

VD – q . x + Lx = 0

Lx = - 5,375 + 4x

x = 0 ; LD = 5,375 ton

x = 3 ; LC = 10,625 ton

ΣH = 0

Nx = 0

X

q = 4 t/m

VD

Nx

Mx

Lx

Noviyanthy H. MT.



A B C D

17,0 m

1,0 m5,0 m 4,0 m6,0 m 1,0 m

P = 10 ton

q = 3 t/m q = 3,5 t/m q = 4 t/m

VA VB VC VD

S1 S2

VS1 VS2

A B C D

q = 3 t/m q = 4 t/m

S1 S2

P = 10 ton

q = 3,5 t/m

S2S1

MOMEN

LINTANG

0 tm

0 t 0 t

0 tm

NORMAL 0 t 0 t

20,5 tm

10,5 tm

3,4 t

11,6 t

20,5 t

10,5 t

10,5 t

10,625 t

5,375 t

15,750 tm

+

+

+

+

+

-

-

--

1,927 tm

Noviyanthy H. MT.


Contoh Soal 4

Penyelesaian :


S = n – 2 = 4 – 2 = 2

Tahap perhitungan seperti pada contoh 3.

1. Perhitungan Reaksi Perletakan Bentang S1 – S2

ΣMS2 = 0

VS1 . 5 + M = 0

VS1 . 5 + 5 = 0

VS1 = - 1,0 ton ( )

= 1,0 ton ( )

A B DCS1 S2

3,0 m 3,0 m 1,5 m 2,0 m 3,0 m 1,5 m 3,0 m 2,5 m

P1 = 3 ton P2 = 4 ton

M = 5 tm

19,5 m

1 32

A B DCS1 S2

3,0 m 3,0 m 1,5 m 2,0 m 3,0 m 1,5 m 3,0 m 2,5 m

P1 = 3 ton P2 = 4 ton

19,5 m

M = 5 tm

S1 S2

1 3

2

2,0 m 3,0 m

M = 5 tm

S1 S2

2

Noviyanthy H. MT.


ΣMS1 = 0

-VS2 . 5 + M = 0

-VS2 . 5 + 5 = 0

VS2 = 1,0 ton ( )

Kontrol :

ΣV = 0

VS1 + VS2 = 0

-1,0 + 1,0 = 0 (oke!!!)

2. Perhitungan Reaksi Perletakan Bentang A – B – S1

ΣMB = 0

VA . 6 – P. 3 – VS1 . 1,5 = 0

VA . 6 – 3 . 3 – 1 . 1,5 = 0

VA = 1,75 ton ( )

ΣMA = 0

- VS1 . 7,5 + P. 3 – VB . 6 = 0

- 1 . 7,5 + 3 . 3 – VB . 6 = 0

VB = 0,25 ton ( )

Kontrol :

ΣV = 0

VA + VB + VS1 = P1

1,75 + 0,25 + 1,0 = 3

3,0 ton = 3,0 ton (oke!!!)

A B S1

3,0 m 3,0 m 1,5 m

P1 = 3 ton

1

Noviyanthy H. MT.


3. Perhitungan Reaksi Perletakan Bentang S2 – C – D

ΣMD = 0

- VS2 . 7 + VC . 5,5 – P2. 2,5 = 0

- 1. 7 + VC . 5,5 – 4. 2,5 = 0

VC = 3,0909 ton ( )

ΣMC = 0

- VD . 5,5 + 4 . 3 – VS2 . 1,5 = 0

-VD . 5,5 + 4 . 3 – 1 . 1,5 = 0

VD = 1,9091 ton ( )

Kontrol :

ΣV = 0

VC + VD = P2 + VS2

3,0909 + 1,9091 = 4 + 1,0

5,0 ton = 5,0 ton (oke!!!)

DCS2

1,5 m 3,0 m 2,5 m

P2 = 4 ton

3

Noviyanthy H. MT.



a) Bentang A – 1 ( 0 ≤ x ≤ 3,0 meter )

ΣMx = 0

VA . x – Mx = 0

Mx = 1,75 x

x = 0 ; MA = 0 tm

x = 3 ; M1 = 5,25 tm

ΣV = 0

VA – Lx = 0

Lx = 1,75

x = 0 ; LA = 1,75 ton

x = 3 ; L1 = 1,75 ton

ΣH = 0

Nx = 0

XVA Lx

Nx

Mx

Noviyanthy H. MT.


b) Bentang 1 – B ( 0 ≤ x ≤ 3,0 meter )

ΣMx = 0

VA . (3 + x) – P1. x – Mx = 0

Mx = 1,75 (3 + x) – 3. x

= 5,25 – 1,25 x

x = 0 ; M1 = 5,25 tm

x = 3 ; MB = 1,5 tm

ΣV = 0

VA – P1 – Lx = 0

Lx = 1,75 – 3

= - 1,25

x = 0 ; L1 = - 1,25 ton

x = 3 ; LB = - 1,25 ton

ΣH = 0

Nx = 0

3,0 m XLx

Nx

Mx

VA

P1 = 3 ton

Noviyanthy H. MT.


c) Bentang S1 – B ( 0 ≤ x ≤ 1,5 meter )

ΣMx = 0

- VS1 . x + Mx = 0

Mx = x

x = 0 ; MS1 = 0 tm

x = 1,5 ; MB = 1,5 tm

ΣV = 0

VS1 + Lx = 0

Lx = - 1

x = 0 ; LS1 = -1,0 ton

x = 1,5 ; LB = -1,0 ton

ΣH = 0

Nx = 0

X

VS1

Nx

Mx

Lx

Noviyanthy H. MT.


d) Bentang S1 – 2 ( 0 ≤ x ≤ 2,0 meter )

ΣMx = 0

- VS1 . x – Mx = 0

Mx = - x

x = 0 ; MS1 = 0 tm

x = 2 ; M2 = - 2,0 tm

ΣV = 0

- VS1 – Lx = 0

Lx = - 1

x = 0 ; LS1 = - 1,0 ton

x = 2 ; L2 = - 1,0 ton

ΣH = 0

Nx = 0

XLx

Nx

x

VS1

Noviyanthy H. MT.


e) Bentang S2 – 2 ( 0 ≤ x ≤ 3,0 meter )

ΣMx = 0

-VS2 . x + Mx = 0

Mx = x

x = 0 ; MS2 = 0 tm

x = 3 ; M2 = 3,0 tm

ΣV = 0

VS2 + Lx = 0

Lx = - 1

x = 0 ; LS2 = -1,0 ton

x = 3 ; L2 = - 1,0 ton

ΣH = 0

Nx = 0

XVS2

Nx

Mx

Lx

Noviyanthy H. MT.


f) Bentang S2 – C ( 0 ≤ x ≤ 1,5 meter )

ΣMx = 0

-VS2 . x – Mx = 0

Mx = - x

x = 0 ; MS2 = 0 tm

x = 1,5 ; MC = - 1,5 tm

ΣV = 0

-VS2 – Lx = 0

Lx = - 1

x = 0 ; LS2 = - 1,0 ton

x = 1,5 ; LC = - 1,0 ton

ΣH = 0

Nx = 0

XLx

Nx

Mx

VS2

Noviyanthy H. MT.


g) Bentang C – 3 ( 0 ≤ x ≤ 3,0 meter )

ΣMx = 0

-VS2 . (1,5 + x) + VC . x – Mx = 0

Mx = - 1 (1,5 + x) + 3,0909 x

= - 1,5 + 2,0909 x

x = 0 ; MC = - 1,5 tm

x = 3 ; M3 = 4,7727 tm

ΣV = 0

-VS2 + VC – Lx = 0

Lx = - 1 + 3,0909

= - 2,0909

x = 0 ; LC = - 2,0909 ton

x = 3 ; L3 = - 2,0909 ton

ΣH = 0

Nx = 0

1,5 m X

VC Lx

Nx

Mx

VS2

Noviyanthy H. MT.


h) Bentang D – 3 ( 0 ≤ x ≤ 2,5 meter )

ΣMx = 0

-VD . x + Mx = 0

Mx = 1,9091 x

x = 0 ; MD = 0 tm

x = 2,5 ; M3 = 4,7728 tm

ΣV = 0

VD + Lx = 0

Lx = - 1,9091

x = 0 ; LD = -1,9091 ton

x = 2,5 ; L3 = - 1,9091 ton

ΣH = 0

Nx = 0

XVC

Nx

Mx

Lx

Noviyanthy H. MT.



A

VA

B

VB

D

VD

C

VC

S1

VS1

S2

VS2

3,0 m 3,0 m 1,5 m 2,0 m 3,0 m 1,5 m 3,0 m 2,5 m

P1 = 3 ton P2 = 4 ton

M = 5 tm

19,5 m

1 32

A B DCS1 S2

P1 = 3 ton P2 = 4 ton

M = 5 tm

S1 S2

1 3

2

MOMEN 0 tm 0 tm

LINTANG 0 t 0 t

NORMAL 0 t 0 t

5,25 tm

1,5 tm

2,0 tm

3,0 tm

1,5 tm

4,7727 tm

1,75 t

1,25 t1,0 t

2,0 t 1,9091 t

+

+

+

+

-

--

Noviyanthy H. MT.


SOAL LATIHAN

1. Hitunglah reaksi perletakan, gaya-gaya dalam dari struktur statis tertentu

dibawah ini dan gambarkan bidang momen, lintang dan normal dengan

mencantumkan ordinat-ordinat penting!







P = 5 ton

6 m 1,5 m 1,5 m2,5 m 3,5 m 4 m

q 1 = 3 t/m q 2 = 4 t/m

P1 = 2 ton P2 = 3 ton

5 m 1 m 2 m 2 m 2 m

q = 2,5 t/m

4 m

q1 = 2,0 t/m q 3 = 2,0 t/m

P = 4 ton

q2 = 3,0 t/m

1 m 3 m 4 m1 m3 m

Documents

BAB V Balok Gerber A S B C - umpalangkaraya.ac.id · Noviyanthy H. MT. 70 Statika dan Mekanika Bahan 1 BAB V Balok Gerber Pada suatu kondisi kadangkala balok akan dibuat dengan panjang