BAB IV Balok Kantilever - umpalangkaraya.ac.id · ini sering dijumpai pada bagian dari sebuah konstruksi gedung. Adapun contoh penerapan atau aplikasi dari balok kantilever dapat

Noviyanthy H. MT.

Statika dan Mekanika Bahan 1 49

BAB IV

Balok Kantilever

Balok kantilever (overstek) adalah balok yang hanya disangga atau dijepit

pada salah satu ujungnya dan ujung lainnya bebas atau tanpa tumpuan. Konstruksi

ini sering dijumpai pada bagian dari sebuah konstruksi gedung. Adapun contoh

penerapan atau aplikasi dari balok kantilever dapat dilihat pada Gambar 4.1

berikut.

Gambar 4.1 Contoh struktur balok kantilever

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

Noviyanthy H. MT.


Contoh Soal 1

Penyelesaian :

1. Perhitungan Reaksi Perletakan

ΣMB = 0

VA . 5,5 – P1 . 3 + P2 . 1,5 = 0

VA . 5,5 – 3 . 3 + 2 . 1,5 = 0

VA = 1,091 ton ( )

ΣMA = 0

-VB . 5,5 + P2 . 7 + P1 . 2,5 = 0

-VB . 5,5 + 2 . 7 + 3 . 2,5 = 0

VB = 3,909 ton ( )

Kontrol :

ΣV = 0

VA + VB = P1 + P2

1,091 + 3,909 = 3,0 + 2,0

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

VA

A 1 B 2

P2 = 2 ton

VB

P1 = 3 ton

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

7,0 m

Noviyanthy H. MT.


2. Perhitungan Gaya-gaya Dalam

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

ΣMx = 0

VA . x – Mx = 0

Mx = 1,091 x

x = 0 ; MA = 0 tm

x = 2,5 ; M1 = 2,728 tm

ΣV = 0

VA – Lx = 0

Lx = 1,091

x = 0 ; LA = 1,091 ton

x = 2,5 ; L1 = 1,091 ton

ΣH = 0

Nx = 0

xVA Lx

Nx

Mx

Noviyanthy H. MT.


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

ΣMx = 0

VA . (2,5 + x) – P1. x – Mx = 0

Mx = 1,091 (2,5 + x) – 3. x

= 2,728 – 1,909x

x = 0 ; M1 = 2,728 tm

x = 3 ; MB = - 2,999 tm

ΣV = 0

VA – P1 – Lx = 0

Lx = 1,091 - 3

= - 1,909

x = 0 ; L1 = - 1,909 ton

x = 3 ; LB = - 1,909 ton

ΣH = 0

Nx = 0

2,5 m x

P1 = 3 ton

Lx

Nx

Mx

VA

Noviyanthy H. MT.


c) Bentang 2 – B ( 0 ≤ x ≤ 1,5 meter )

ΣMx = 0

P . x + Mx = 0

Mx = - 2x

x = 0 ; M2 = 0 tm

x = 1,5 ; MB = - 3,0 tm

ΣV = 0

- P2 + Lx = 0

Lx = 2

x = 0 ; L2 = 2,0 ton

x = 1,5 ; LB = 2,0 ton

ΣH = 0

Nx = 0

P2 = 2 ton

Nx

Mx

x

Lx

Noviyanthy H. MT.


3. Gambar Bidang Momen, Lintang dan Normal

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

7,0 m

VA

A 1 B 2

P2 = 2 ton

1,091 t

1,909 t

2 t

3 tm

2,728 tm

0 t

+

+

+

-

-

VB

0 t

0 t 0 t

0 tmMOMEN

NORMAL

LINTANG

0 tm

P1 = 3 ton

Noviyanthy H. MT.


Contoh Soal 2

Penyelesaian :


ΣMB = 0

VA . 4 – q . 4 (½ . 4) + P sin 60° . 2 = 0

VA . 4 – 4 . 4 (½ . 4) + 2,5 sin 60° . 2 = 0

VA = 6,917 ton ( )

ΣMA = 0

-VB . 4 + P sin 60° . 6 + q . 4 (½ . 4) = 0

-VB . 4 + 2,5 sin 60°. 6 + 4 . 4 (½ . 4) = 0

VB = 11,248 ton ( )

Kontrol :

ΣV = 0

VA + VB = q . L + P sin 60°

6,917 + 11,248 = 4 . 4 + 2,5 sin 60°

18,165 ton = 18,165 ton (oke!!!)

ΣH = 0

HA – P cos 60° = 0

HA = 1,250 t ( )

VA VB

A 1B

q = 4 t/m

4,0 m 2,0 m

6,0 m

P = 2,5 ton

= 60°

P V

P HHA

Noviyanthy H. MT.



a) Bentang A – B ( 0 ≤ x ≤ 4,0 meter )

ΣMx = 0

VA . x – q . x (½. x) - Mx = 0

Mx = 6,917 x – 2x2

x = 0 ; MA = 0 tm

x = 4 ; MB = - 4,332 tm

ΣV = 0

VA – q . x – Lx = 0

Lx = 6,917 – 4x

x = 0 ; LA = 6,917 ton

x = 4 ; LB = - 9,083 ton

ΣH = 0

- HA – Nx = 0

Nx = - 1,250

x = 0 ; NA = - 1,250 ton

x = 4 ; NB = - 1,250 ton

q = 4 t/m

HA

X

VA Lx

Nx

Mx

Noviyanthy H. MT.



ΣMx = 0

P sin 60° . x + Mx = 0

Mx = - 2,165x

x = 0 ; M1 = 0 tm

x = 2 ; MB = - 4,330 tm

ΣV = 0

- P sin 60° + Lx = 0

Lx = 2,165

x = 0 ; L1 = 2,165 ton

x = 2 ; LB = 2,165 ton

ΣH = 0

P cos 60° + Nx = 0

Lx = - 1,250

x = 0 ; L1 = - 1,250 ton

x = 2 ; LB = - 1,250 ton

1

X

Nx

Mx

Lx

P = 2,5 ton

= 60°

V

P H

Noviyanthy H. MT.



q = 4 t/m

4,332 tm

2,165 t

6,917 t

9,083 tm

1,250 t

0 t 0 t

0 t 0 t

0 tm0 tm

++

-

-

-

LINTANG

NORMAL

MOMEN

P = 2,5 ton

= 60°

P V

P H

VA VB

A 1B

4,0 m 2,0 m

6,0 m

HA

Noviyanthy H. MT.


Contoh Soal 3

Penyelesaian :


ΣMB = 0

VA . 5 – ½ q . 3 (13 . 3) + P sin 60° . 2 = 0

VA . 5 – ½ . 3 . 3 (13 . 3) + 2,5 sin 60° . 2 = 0

VA = 0,034 ton ( )

ΣMA = 0

-VB . 5 + P sin 60° . 7 + ½ q . 3 (23 . 3 + 2) = 0

-VB . 5 + 2,5 sin 60°. 7 + ½ 3 . 3 (23 . 3 + 2) = 0

VB = 6,631 ton ( )

Kontrol :

ΣV = 0

VA + VB = ½ q . L + P sin 60°

0,034 + 6,631 = ½ 3 . 3 + 2,5 sin 60°

6,665 ton = 6,665 ton (oke!!!)

ΣH = 0

HA – P cos 60° = 0

HA = 1,250 t ( )

3,0 m 2,0 m

7,0 m

q = 3 t/m

2,0 m

P V

VA VB

A 1 B 2

HA

P = 2,5 ton

= 60°P H

Noviyanthy H. MT.



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

ΣMx = 0

VA . x – Mx = 0

Mx = 0,034 x

x = 0 ; MA = 0 tm

x = 2 ; M1 = 0,068 tm

ΣV = 0

VA – Lx = 0

Lx = 0,034

x = 0 ; LA = 0,034 ton

x = 2 ; L1 = 0,034 ton

ΣH = 0

- HA – Nx = 0

Nx = - 1,250

x = 0 ; NA = - 1,250 ton

x = 2 ; N1 = - 1,250 ton

X

HA

Lx

Nx

Mx

VA

Noviyanthy H. MT.



ΣMx = 0

VA . (2 + x) – ½ qx . x (13 . x) - Mx = 0

Mx = 0,034 (2 + x) – ½ x . x (13 x)

= 0,068 + 0,034x – 1 6 x3

x = 0 ; M1 = 0,068 tm

x = 3 ; MB = - 4,330 tm

ΣV = 0

VA – ½ qx . x – Lx = 0

Lx = 0,034 – ½ x . x

= 0,034 – ½ x2

x = 0 ; L1 = 0,034 ton

x = 3 ; LB = - 4,466 ton

ΣH = 0

HA – Nx = 0

Nx = - 1,250

x = 0 ; N1 = - 1,250 ton

x = 3 ; NB = - 1,250 ton

X

q = 3 t/m

2,0 m

Lx

Nx

MxHA

VA

𝑞𝑥𝑞𝑜

=𝑥

𝐿

𝑞𝑥 =𝑞𝑜𝐿𝑥

𝑞𝑥 = 𝑥

Noviyanthy H. MT.



ΣMx = 0

P sin 60° . x + Mx = 0

Mx = - 2,165x

x = 0 ; M2 = 0 tm

x = 2 ; MB = - 4,330 tm

ΣV = 0

- P sin 60° + Lx = 0

Lx = 2,165

x = 0 ; L2 = 2,165 ton

x = 2 ; LB = 2,165 ton

ΣH = 0

P cos 60° + Nx = 0

Lx = - 1,250

x = 0 ; L2 = - 1,250 ton

x = 2 ; LB = - 1,250 ton

X

Nx

Mx

Lx

P V P = 2,5 ton

= 60°P H

Noviyanthy H. MT.



A 1 B 2

q = 3 t/m

0 t

0 tm

P = 2,5 ton

= 60°

4,332 tm

0,068 tm

4,466 tm

2,165 t

0,034 t

1,250 t

+

+

-

-

+

-

P V

P H

LINTANG

NORMAL

MOMEN0 tm

0 t

0 t0 t

HA

3,0 m 2,0 m

7,0 m

2,0 m

VA VB

Noviyanthy H. MT.


Contoh Soal 4

Penyelesaian :


ΣMB = 0

MB – P2 . 10 – P1 . 8 – q . 5 ( ½ . 5) = 0

MB – 4 . 10 – 6 . 8 – 10 . 5 ( ½ . 5) = 0

MB = 213 ton . meter ( )

ΣV = 0

VB – P1 – P2 – q . L = 0

VB – 4 – 6 – 10 . 5 = 0

VB = 60 ton ( )

ΣH = 0

HB + P3 = 0

HB = - 3 ton ( )

= 3 ton ( )

3,0 m 5,0 m

10,0 m

A 1

2,0 m

2

P1 = 6 tonP2 = 4 ton

P3 = 3 tonB

q = 10 t/m

VB

Noviyanthy H. MT.



a) Bentang A – 1 (0 ≤ x ≤ 2,0 meter )

ΣMx = 0

- P2 . x – Mx = 0

Mx = - 4 x

x = 0 ; MA = 0 tm

x = 2 ; M1 = - 8,0 tm

ΣV = 0

- P2 – Lx = 0

Lx = - 4

x = 0 ; LA = 4,0 ton

x = 2 ; L1 = 4,0 ton

ΣH = 0

P3 – Nx = 0

Nx = 3

x = 0 ; NA = 3,0 ton

x = 2 ; N1 = 3,0 ton

X

P2 = 4 ton

P3 = 3 ton

Lx

Nx

Mx

Noviyanthy H. MT.


b) Bentang 1 – 2 (0 ≤ x ≤ 3,0 meter )

ΣMx = 0

- P2 .(2 + x) – P1 . x – Mx = 0

Mx = - 4 (2 + x) – 6 x

x = 0 ; M1 = - 8,0 tm

x = 3 ; M2 = - 38,0 tm

ΣV = 0

- P2 – P1 – Lx = 0

Lx = - 4 – 6

x = 0 ; L1 = - 10,0 ton

x = 3 ; L2 = - 10,0 ton

ΣH = 0

P3 – Nx = 0

Nx = 3

x = 0 ; N1 = 3,0 ton

x = 3 ; N2 = 3,0 ton

X2,0 m

P1 = 6 ton

Lx

Nx

Mx

P2 = 4 ton

P3 = 3 ton

Noviyanthy H. MT.



ΣMx = 0

- 4 (5 + x) – 6 (3 + x) – q . x ( ½ x) – Mx = 0

Mx = - 20 – 4x – 18 – 6x – 5x2

= - 38 – 10x – 5x2

x = 0 ; M2 = - 38,0 tm

x = 5 ; MB = - 213,0 tm

ΣV = 0

- P2 – P1 – q . x – Lx = 0

Lx = - 4 – 6 – 10x

= - 10 – 10x

x = 0 ; L2 = - 10,0 ton

x = 5 ; LB = - 60,0 ton

ΣH = 0

P3 – Nx = 0

Nx = 3

x = 0 ; N2 = 3,0 ton

x = 5 ; NB = 3,0 ton

3,0 m X2,0 m

q = 10 t/m

Lx

Nx

Mx


P3 = 3 ton

Noviyanthy H. MT.



A 1 2


P3 = 3 tonB

0 tm

0 t

3,0 t

4,0 t

10,0 t

60,0 t

8,0 tm

38,0 tm

213,0 tm

q = 10 t/m

0 t

-

+

-

+LINTANG

NORMAL

MOMEN

0 t

0 tm

0 t

VB

3,0 m 5,0 m

10,0 m

2,0 m

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 2 = 3 ton

= 45°

P1 = 2 ton P3 = 4 ton

2 m 2 m 2 m 2 m

2 m 2 m 2 m 2 m

q = 2 t/mq = 3 t/m

3 m

P1 = 2 ton P2 = 4 ton q = 4 t/m

0,5 m1 m

Documents

BAB IV Balok Kantilever - umpalangkaraya.ac.id · ini sering dijumpai pada bagian dari sebuah konstruksi gedung. Adapun contoh penerapan atau aplikasi dari balok kantilever dapat