a) Struktur balok dan tiang b) Portal kaku c) Portal biasa ...€¦ · Statika dan Mekanika Bahan 1 BAB VI Portal Struktur portal yaitu struktur balok yang mempunyai kaki-kaki (Wesli,

Noviyanthy H. MT.

Statika dan Mekanika Bahan 1 111

BAB VI

Portal

Struktur portal yaitu struktur balok yang mempunyai kaki-kaki (Wesli,

2012). Pada portal yang terdiri dari balok dan tiang yang dibebani muatan

diatasnya akan timbul lenturan pada balok saja, dan akan meneruskan gaya-gaya

tersebut ke tiang berupa gaya normal. Balok yang demikian disebut balok

sederhana. Adapun gaya yang bekerja pada tiang pada umumnya berupa gaya

horisontal, tidak berpengaruh pada balok (Mulyati). Struktur portal dengan

kondisi seperti ini dapat diselesaikan dengan prinsip struktur statis tertentu dalam

artian hukum kesetimbangan (ΣM = 0, ΣV = 0, dan ΣH = 0) dapat digunakan.

Bentuk portal yang umumnya dipelajari berupa portal segi empat, segi banyak

atau lengkungan, yang dapat dilihat pada Gambar 6.1 berikut. Pada bab ini hanya

akan dibahas struktur portal segi empat.

Gambar 6.1 Struktur Portal (Mulyati)

a) Struktur balok dan tiang b) Portal kaku

c) Portal biasa d) Portal segi banyak

c) Portal lengkung

Noviyanthy H. MT.


Berikut ini beberapa contoh soal dan penyelesaian untuk struktur portal.

Contoh Soal 1

Penyelesaian :

1. Reaksi Perletakan

ΣMB = 0

VA . 4 + P . L – q . L (½ . L) = 0

VA . 4 + 5 . 2,5 – 2 . 4 (½ . 4) = 0

VA = 0,875 ton ( )

ΣMA = 0

-VB . 4 + P . L + q . L (½. L) = 0

-VB . 4 + 5 . 2,5 + 2 . 4 (½. 4) = 0

VB = 7,125 ton ( )

ΣV = 0

VA + VB = q . L

0,875 + 7,125 = 2 . 4

8,0 ton = 8,0 ton (Oke!!!)

P = 5 ton1 m

2,5 m

4 m

q = 2 t/m

A B

D

C

E

Noviyanthy H. MT.


ΣH = 0

HA + P = 0

HA = - P = - 5,0 ton ( )

= 5,0 ton ( )

2. Perhitungan Gaya-gaya Dalam

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

ΣMx = 0

HA . x – Mx = 0

Mx = 5x

x = 0 ; MA = 0 tm

x = 2,5 ; MC = 12,5 tm

ΣV = 0

- VA – Nx = 0

Nx = - 0,875

x = 0 ; NA = - 0,875 ton

x = 2,5 ; NC = - 0,875 ton

X

HA

VA

LX

NX

MX

Noviyanthy H. MT.


ΣH = 0

HA – Lx = 0

Lx = 5

x = 0 ; LA = 5,0 ton

x = 2,5 ; LC = 5,0 ton

b) Bentang C – D ( 0 ≤ x ≤ 1,0 meter )

ΣMx = 0

HA . (2,5 + x) – P . x – Mx = 0

Mx = 5 (2,5 + x) – 5x

= 12,5 + 5x – 5x

x = 0 ; MC = 12,5 tm

x = 1 ; MD = 12,5 tm

ΣV = 0

- VA – Nx = 0

Nx = - 0,875

x = 0 ; NC = - 0,875 ton

x = 1 ; ND = - 0,875 ton

2,5 m

HA

VA

P = 5 tonX

LX

NX

MX

Noviyanthy H. MT.


ΣH = 0

HA – P – Lx = 0

Lx = 5 – 5

= 0

x = 0 ; LC = 0 ton

x = 1 ; LD = 0 ton

c) Bentang D – E ( 0 ≤ x ≤ 4,0 meter )

ΣMx = 0

VA . x + HA . 3,5 – P . 1 – q . x (½ . x) – Mx = 0

Mx = 0,875 . x + 5 . 3,5 – 5 . 1 – 2 . x (½ . x)

= 12,5 + 0,875x – x2

x = 0 ; MD = 12,5 tm

x = 4 ; ME = 0 tm

ΣV = 0

VA – q . x – Lx = 0

Lx = 0,875 – 2x

x = 0 ; LD = 0,875 ton

x = 4 ; LE = - 7,125 ton

2,5 m

HA

VA

P = 5 ton

q = 2 t/m

Lx

Nx

Mx

X

1 m

Noviyanthy H. MT.


ΣH = 0

- HA + P – Nx = 0

Nx = 5 – 5

= 0

x = 0 ; ND = 0 ton

x = 4 ; NE = 0 ton

terjadi Mmaks pada jarak Lx = 0

0,875 – 2x = 0 x = 0,438 meter

M maks = 12,5 + 0,875x – x2

= 12,5 + 0,875 (0,438) – (0,438)2

= 12,691 tm

d) Bentang E – B ( 0 ≤ x ≤ 3,5 meter )

ΣMx = 0

Mx = 0

x = 0 ; ME = 0 tm

x = 3,5 ; MB = 0 tm

X

VB

LX

NX

MX

Noviyanthy H. MT.


ΣV = 0

- VB – Nx = 0

Nx = - 7,125

x = 0 ; NE = - 7,125 ton

x = 3,5 ; NB = - 7,125 ton

ΣH = 0

Lx = 0

x = 0 ; LE = 0 ton

x = 3,5 ; LB = 0 ton

Noviyanthy H. MT.


3. Gambar bidang Momen, Lintang dan Normal

P = 5 ton1 m

2,5 m

4 m

q = 2 t/m

A B

HA

VA

VB

0 tm 0 tm12,5 tm

12,5 tm

0 tm 0 tm

+

+

+

0 t 0 t

5 t

7,125 t

0,875 t

+

-

+

0 t 0 t

0 t 0 t

7,125 t0,875 t

0 t 0 t

- -

MOMEN

LINTANG

NORMAL

12,5 tm12,691 tm

D

C

E

Noviyanthy H. MT.


Contoh Soal 2

Penyelesaian :


ΣMB = 0

VA . 4 + q . L (½ . L) – P . L = 0

VA . 4 + 2 . 4 (½ . 4) – 2 . 2 = 0

VA = - 3,0 ton ( )

= 3,0 ton ( )

ΣMA = 0

-VB . 4 + P . L + q . L (½. L) = 0

-VB . 4 + 2 . 2 + 2 . 4 (½. 4) = 0

VB = 5,0 ton ( )

ΣV = 0

VA + VB = P

- 3,0 + 5,0 = 2

2,0 ton = 2,0 ton (Oke!!!)

4 m

2 m2 m

P = 2 ton

q = 2 t/m

A B

C ED

Noviyanthy H. MT.


ΣH = 0

q . L + HA = 0

HA = - 2 . 4

= - 8,0 ton ( )

= 8,0 ton ( )



ΣMx = 0

HA . x – q. x (½ x) – Mx = 0

Mx = 8 . x – 2 . x (½ x)

= 8x – x2

x = 0 ; MA = 0 tm

x = 4 ; MC = 16,0 tm

ΣV = 0

VA – Nx = 0

Nx = 3

x = 0 ; NA = 3,0 ton

x = 4 ; NC = 3,0 ton

X q = 2 t/m

LX

NX

MX

VA

Noviyanthy H. MT.


ΣH = 0

HA – q . x – Lx = 0

Lx = 8 – 2x

x = 0 ; LA = 8,0 ton

x = 4 ; LC = 0 ton


ΣMx = 0

-VA . x + HA . 4 – q . 4 (½ . 4) – Mx = 0

Mx = - 3 . x + 8 . 4 – 2 . 4 (½ . 4)

= - 3x + 16

x = 0 ; MC = 16,0 tm

x = 2 ; MD = 10,0 tm

ΣV = 0

- VA – Lx = 0

Lx = - 3

x = 0 ; LC = - 3,0 ton

x = 2 ; LD = - 3,0 ton

4 m q = 2 t/m

C

VA

Lx

Nx

Mx

X

Noviyanthy H. MT.


ΣH = 0

- HA + q . 4 – Nx = 0

Nx = - 8 + 2. 4

= 0

x = 0 ; NC = 0 ton

x = 2 ; ND = 0 ton


ΣMx = 0

- VA . (2 + x) + HA . 4 – P . x – q . 4 (½ . 4) – Mx = 0

Mx = - 3 (2 + x) + 8 . 4 – 2 . x – 2 . 4 (½ . 4)

= - 5x + 10

x = 0 ; MD = 10,0 tm

x = 2 ; ME = 0 tm

ΣV = 0

- VA – P – Lx = 0

Lx = - 3 – 2

= - 5

x = 0 ; LD = - 5,0 ton

x = 2 ; LE = - 5,0 ton

4 m

P = 2 ton

q = 2 t/m

C D

VA

Lx

Nx

Mx

X2 m

Noviyanthy H. MT.


ΣH = 0

- HA + q . x – Nx = 0

Nx = - 8 + 2 . 4

= 0

x = 0 ; ND = 0 ton

x = 2 ; NE = 0 ton

d) Bentang E – B ( 0 ≤ x ≤ 4,0 meter )

ΣMx = 0

Mx = 0

x = 0 ; ME = 0 tm

x = 4 ; MB = 0 tm

ΣV = 0

- VB – Nx = 0

Nx = - 5

x = 0 ; NE = - 5,0 ton

x = 4 ; NB = - 5,0 ton

ΣH = 0

Lx = 0

x = 0 ; LE = 0 ton

x = 4 ; LB = 0 ton

X

VB

LX

NX

MX

Noviyanthy H. MT.



4 m

2 m2 m

P = 2 ton

q = 2 t/m

VA

VB

0 tm 0 tm16 tm

16 tm

10 tm

0 tm 0 tm

0 t 0 t8 t

0 t 0 t

5 t

3 t

3 t

+

+

+

+

-

0 t

0 t

5 t-

0 t

0 t

A B

C ED

MOMEN

LINTANG

NORMAL

Noviyanthy H. MT.


Contoh Soal 3

Penyelesaian :


ΣMB = 0

VA . 4 + P1. L – P2 . L – P3 . L = 0

VA . 4 + 2 . 3 – 2 . 3 – 3 . 1 = 0

VA = 0,75 ton ( )

ΣMA = 0

-VB . 4 + P3 . L + P2 . L + P1 . L = 0

-VB . 4 + 3 . 3 + 2 . 1+ 2 . 3 = 0

VB = 4,25 ton ( )

ΣV = 0

VA + VB = P2 + P3

0,75 + 4,25 = 2 + 3

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

P1 = 2 ton1 m

3 m

P2 = 2 ton P3 = 3 ton

2 m1 m 1 m

A B

D GE F

C

Noviyanthy H. MT.


ΣH = 0

P1 + HA = 0

HA = - 2

= - 2,0 ton ( )

= 2,0 ton ( )



ΣMx = 0

HA . x – Mx = 0

Mx = 2 . x

x = 0 ; MA = 0 tm

x = 3 ; MC = 6,0 tm

ΣV = 0

- VA – Nx = 0

Nx = - 0,75

x = 0 ; NA = - 0,75 ton

x = 3 ; NC = - 0,75 ton

X

HA

VA

LX

NX

MX

Noviyanthy H. MT.


ΣH = 0

HA – Lx = 0

Lx = 2

x = 0 ; LA = 2,0 ton

x = 3 ; LC = 2,0 ton


ΣMx = 0

HA (3 + x) – P1 . x – Mx = 0

Mx = 2 (3 + x) – 2 . x

= 6 + 2x – 2x

x = 0 ; MC = 6,0 tm

x = 1 ; MD = 6,0 tm

ΣV = 0

- VA – Nx = 0

Nx = - 0,75

x = 0 ; NC = - 0,75 ton

x = 1 ; ND = - 0,75 ton

3 m

HA

VA

P = 5 tonX

LX

NX

MX

Noviyanthy H. MT.


ΣH = 0

HA – P1 – Lx = 0

Lx = 2 – 2

= 0

x = 0 ; LC = 0 ton

x = 1 ; LD = 0 ton


ΣMx = 0

VA . x + HA . 4 – P1 . 1 – Mx = 0

Mx = 0,75 . x + 2 . 4 – 2 . 1

= 6 + 0,75 x

x = 0 ; MD = 6 tm

x = 1 ; ME = 6,75 tm

ΣV = 0

VA – Lx = 0

Lx = 0,75

x = 0 ; LD = 0,75 ton

x = 1 ; LE = 0,75 ton

P1 = 2 ton1 m

3 m

X

VA

Lx

Nx

Mx

Noviyanthy H. MT.


ΣH = 0

- HA + P1 – Nx = 0

Nx = - 2 + 2

= 0

x = 0 ; ND = 0 ton

x = 1 ; NE = 0 ton

d) Bentang E – F ( 0 ≤ x ≤ 2,0 meter )

ΣMx = 0

VA . (1 + x) + HA . 4 – P1 . 1 – P2 . x – Mx = 0

Mx = 0,75 (1 + x) + 2 . 4 – 2 . 1 – 2 . x

= 6,75 – 1,25 x

x = 0 ; ME = 6,75 tm

x = 2 ; MF = 4,25 tm

ΣV = 0

VA – P2 – Lx = 0

Lx = 0,75 – 2

= - 1,25

x = 0 ; LE = - 1,25 ton

x = 2 ; LF = - 1,25 ton

P1 = 2 ton1 m

3 m

2

X1 m

VA

Lx

Nx

Mx

Noviyanthy H. MT.


ΣH = 0

- HA + P1 – Nx = 0

Nx = - 2 + 2

= 0

x = 0 ; NE = 0 ton

x = 2 ; NF = 0 ton

e) Bentang F – G ( 0 ≤ x ≤ 1,0 meter )

ΣMx = 0

VA . (3 + x) + HA . 4 – P1 . 1 – P2 (2 + x) – P3 . x – Mx = 0

Mx = 0,75 . (3 + x) + 2 . 4 – 2 . 1 – 2 (2 + x) – 3 . x

= 4,25 – 4,25 x

x = 0 ; MF = 4,25 tm

x = 1 ; MG = 0 tm

ΣV = 0

VA – P2 – P3 – Lx = 0

Lx = 0,75 – 2 – 3

= - 4,25

x = 0 ; LF = - 4,25 ton

x = 1 ; LG = - 4,25 ton

P1 = 2 ton1 m

3 m

P2 = 2 ton P3 = 3 ton

2 m1 m X

VA

Lx

Nx

Mx

Noviyanthy H. MT.


ΣH = 0

- HA + P1 – Nx = 0

Nx = - 2 + 2

= 0

x = 0 ; NF = 0 ton

x = 1 ; NG = 0 ton

f) Bentang G – B ( 0 ≤ x ≤ 4,0 meter )

ΣMx = 0

Mx = 0

x = 0 ; MG = 0 tm

x = 4 ; MB = 0 tm

ΣV = 0

- VB – Nx = 0

Nx = - 4,25

x = 0 ; NG = - 4,25 ton

x = 4 ; NB = - 4,25 ton

ΣH = 0

Lx = 0

x = 0 ; LG = 0 ton

x = 4 ; LB = 0 ton

X

VB

LX

NX

MX

Noviyanthy H. MT.



P1 = 2 ton1 m

3 m

P2 = 2 ton P3 = 3 ton

2 m1 m 1 m

HA

VA

VB

0 tm 0 tm

6 tm

6 tm

6 tm

0 tm 0 tm

6,75 tm

4,25 tm

0 t 0 t2 t

2 t

0 t 0 t

4,25 t0,75 t

0 t0 t

0,75 t

1,25 t

4,25 t

+

+

+

-

+

- -

A B

D GE F

C

MOMEN

LINTANG

NORMAL

Noviyanthy H. MT.


Contoh Soal 4

Penyelesaian :


ΣMB = 0

VA . 4,5 - P. L – q . L (½ . L + L) = 0

VA . 4,5 - 3 . 4 – 2 . 2,5 (½ . 2,5 + 2) = 0

VA = 6,278 ton ( )

ΣMA = 0

- VB . 4,5 - P. L + q . L (½ . L) = 0

- VB . 4,5 - 3 . 4 + 2 . 2,5 (½ . 2,5) = 0

VB = - 1,278 ton ( )

ΣV = 0

VA + VB = q . L

6,278 + (- 1,278) = 2 . 2,5

5 ton = 5 ton (Oke!!!)

P = 3 ton

2,5 m 2 m

q = 2 t/m

4 m

Noviyanthy H. MT.


Konstruksi bentang A – D

ΣMD = 0

HA . 4 - VA . 2,5 – q . L (½ . L) = 0

HA . 4 - 6,278 . 2,5 – 2 . 2,5 (½ . 2,5) = 0

HA = 2,361 ton ( )

Konstruksi bentang B – D

ΣMD = 0

HB . 4 – VB . 2 = 0

HB . 4 + 1,278 . 2 = 0

HB = - 0,639 ton ( )

= 0,639 ton ( )


a) Bentang A – C ( 0 ≤ x ≤ 4 meter )

ΣMx = 0

- HA . x – Mx = 0

Mx = - 2,361 . x

x = 0 ; MA = 0 tm

x = 4 ; MC = - 9,444 tm

X

LX

NX

MX

VA

HA

Noviyanthy H. MT.


ΣV = 0

VA – Nx = 0

Nx = 6,278

x = 0 ; NA = 6,278 ton

x = 4 ; NC = 6,278 ton

ΣH = 0

- HA – Lx = 0

Lx = - 2,361

x = 0 ; LA = - 2,361 ton

x = 4 ; LC = - 2,361 ton


ΣMx = 0

VA . x – HA . L - q . x (½ . x) – Mx = 0

Mx = 6,278 . x – 2,361 . 4 - 2 . x (½ . x)

= 6,278 x – 9,444 - x²

x = 0 ; MC = - 9,444 tm

x = 2,5 ; MD = 0 tm

X

q = 2 t/m

Lx

Nx

Mx

4 m

VA

HA

Noviyanthy H. MT.


ΣV = 0

VA – q . x – Lx = 0

Lx = 6,278 – 2 x

x = 0 ; LC = 6,278 ton

x = 2,5 ; LD = 1,278 ton

ΣH = 0

HA – Nx = 0

Nx = 2,361

x = 0 ; NC = 2,361 ton

x = 2,5 ; ND = 2,361 ton

c) Bentang B – E ( 0 ≤ x ≤ 4 meter )

ΣMx = 0

HB . x + Mx = 0

Mx = - 0,639 x

x = 0 ; MB = 0 tm

x = 4 ; ME = - 2,556 tm

X

LX

NX

MX

VB

HB

Noviyanthy H. MT.


ΣV = 0

-VB + Nx = 0

Nx = 1,278

x = 0 ; NB = 1,278 ton

x = 4 ; NE = 1,278 ton

ΣH = 0

- HB – Lx = 0

Lx = - 0,639

x = 0 ; LB = - 0,639 ton

x = 4 ; LE = - 0,639 ton

d) Bentang E – D ( 0 ≤ x ≤ 2 meter )

ΣMx = 0

- VB . x + HB . 4 + Mx = 0

Mx = 1,278 . x – 0,639 . 4

= 1,278 x – 2,556

x = 0 ; ME = - 2,556 tm

x = 2 ; MD = 0 tm

P = 3 ton

X

Nx

Mx

Lx

VB

HB

4 m

Noviyanthy H. MT.


ΣV = 0

HB – P + Nx = 0

Lx = - 0,639 + 3

x = 0 ; LE = 2,361 ton

x = 2 ; LD = 2,361 ton

ΣH = 0

- VB + Lx = 0

Nx = 1,278

x = 0 ; NE = 1,278 ton

x = 2 ; ND = 1,278 ton

Noviyanthy H. MT.



P = 3 ton

2,5 m 2 m

q = 2 t/m

4 m

0 tm 0 tm

9,444 tm

9,444 tm

2,556 tm0 tm 0 tm

2,556 tm

0 t 0 t

1,278 t

6,278 t

2,361 t

2,361 t

0 t 0 t

6,278 t

6,278 t

1,278 t

1,278 t

+ +

--

+

- -

-

-

MOMEN

LINTANG

NORMAL

+

2,361 t

A B

C D E

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!




2 m

2 m

P = 3 ton

2 m2 m

q = 3 t/m

P = 6 ton

2 m

2 m

4 m

q = 4 t/m

Noviyanthy H. MT.





P = 4 ton

2 m

2 m

4 m

q = 3 t/m

a) Struktur balok dan tiang b) Portal kaku c) Portal biasa ...€¦ · Statika dan Mekanika Bahan 1 BAB VI Portal Struktur portal yaitu struktur balok yang mempunyai kaki-kaki (Wesli,

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