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Modul 9 CONTOH SOAL

Pusat Pengembangan Bahan Ajar - UMB Ir. Edifrizal Darma, MT STATIKA 1

1 1 CONTOH SOAL

CONTOH 1 q = 2 t/m

VC T

2 m

P1 = 3 t V HB

B

HA VB 1 m

A

VA

3 m 2 m

Penyelesaian :

1. Tinjau konstruksi global ABC. = 0BM - R. 2 5.1.2

1AA VH + = 0

- 25 HA + 5 VA = 0 5 VA - HA = 25 (1)

Modul 9 CONTOH SOAL


2 2 2. Tinjau Konstruksi Parsial AC:

= 0CM - R 1 3.3.2.21

AA VHP + = 0

-9 6 3 HA + 3 VA = 0 3VA 3 HA = 15 VA HA = 5 .(2) Persamaan (1) dan (2) dijumlahkan : 5 VA - HA = 25 VA HA = 5 - VA = 20 VA = 5 T ( )

1. Tinjau konstruksi global ABC : 00 =+= RVVV BA

5 +VB 10 = 0 VB = 5T ( )

00 =+= BA HPHH

0 + 3 - HB = 0 HB = 3 T ( ) Periksa :

= 0AM P.1 + R.2 05.1.21

= BB VH

3 + 25 3 25 = 0 0 = 0 OK!

HC

P1 = 3t VC 2 m

1 m 1 m

2 m

HA A

VA 3 m

Dari persamaan (2) didapat ;

VA - HA = 5 5 - HA = 5 HA = 0

Modul 9 CONTOH SOAL


3 3 CONTOH 2

Ditanya : Reaksi perletakan

Penyelesaian : 1. Tinjau konstruksi global ABBCDDE.

09.1.3.214.0 2 =+= EEA VHPRM

27.4 093.221

=+ EH

121 09621

=+ EB VH

HE 9VE = 115 21

.. (1)

q = 3 t/M P2 = 2 t D

B B C 1m

A 1m

2 m

3 m

VA

HA

HE

P1 = 5 t

3 m 3 m

VE

Modul 9 CONTOH SOAL


4 4 2. Tinjau konstruksi parsial CDE :

03.3.2.1.211. 122

0=++= EEC VHPPRM

9.1 03310221

=++ EE VH

3 HE 3 VE = 0

HE=- VE = - 7 61

. (2).

Persamaan (1) dan (2) dijumlahkan : HE - 9 VE = 115 2

1

HE - VE = - 7 21

-

- 8 VE = - 108 31

VE = 13 2413

T ( )

3.Tinjau konnstruksi global ABBCDDE: = 0v VA + VE = 0

VA+ 13 2413

- 27 = 0

HC

5 m

P2 = 2t D

D C

1m m

2 m m

21m m

P1= 5t m

HE E

Dari persamaan(2) didapat : He - VE = - 7 2

1

HE 13 2413

= - 7 21

HE = 6 83

T ( )

Modul 9 CONTOH SOAL


5 5 VA = 13 24

13 T ( )

= 0H HA P2 P1- HE = 0

HA 2 5 6 83

= 0

HA = 13 83

( )

Periksa :

= 0EM HA.1 + VA.9 R.4 21

- P2.4 P1. = 0

13 83

+ 121 81

- 121 81

- 8 5 = 0

134 211342

1 = 0

0 = 0 Ok

Modul 9 CONTOH SOAL


6 6 CONTOH 3


Penyelesaian : 1. Tinjau konstruksi ABCDE

07.1.9.1.212.0 21 =++= EEA VHPPRM

079.11.3212.18 =++ EE VH

57 HE 7 VE = 0 - HE 7 VE = 0

3. Tinjau konstruksi parsial CDE:

P2 = 1 q = 2 t/m

P1 = 3 t

HE

HA

P3 = 4 t

HE

E VE

VA

1 m

1 m

1 m

2 m 3 m 4m 2 m

VA

HE

P2 = t

C D RI = 8 t

HE E

VE

1 m

1 m

1 m

P3 = 4t

4 m 2 m

Modul 9 CONTOH SOAL


7 7 = 0CM P2.6 P3 2 + R1.2 HE.3 VE.4 = 0

6 8 + 16 3HE 4 VE = 0 - 3 HE 4 HE = - 14 .. (2)

Persamaan (1) dan (2) dijumlahkan : - HE 7 VE = - 57 x 3 - 3 HE 4 VE = - 14 x 1

- 3 HE 21 VE = - 171

- 3 HE 4 VE = - 14 - -17 VE = - 157

VE = 9 T174 ( )

4. Tinjau konstruksi global ABCDE: = 0H P1 HA P3 + HE = 0

- 3 - HA 4 + (7 1711 ) = 0

HA = - 8 1711 ( )

CONTOH 2

Dari persamaam (1) didapat : - HE 7 VE = - 57

- HE 7.9 174 = - 57

HE = - 7 1711

( )

q = 3 t/M P2 = 2 t D

B B C 1m

A 1m

2 m

3 m

VA

HA

HE

P1 = 5 t

3 m 3 m

VE

Modul 9 CONTOH SOAL


8 8 Ditanya : Reaksi perletakan Penyelesaian : 5. Tinjau konstruksi global ABBCDDE.

09.1.3.214.0 2 =+= EEA VHPRM

27.4 093.221 =+ EH

121 09621

=+ EB VH

HE 9VE = 115 21

.. (1)

6. Tinjau konstruksi parsial CDE :

03.3.2.1.211. 122

0=++= EEC VHPPRM

9.1 03310221

=++ EE VH

3 HE 3 VE = 0

HE=- VE = - 7 61

. (2).

HC

5 m

P2 = 2t D

D C

1m m

2 m m

21m m

P1= 5t m

HE E

Modul 9 CONTOH SOAL


9 9 Persamaan (1) dan (2) dijumlahkan : HE - 9 VE = 115 2

1

HE - VE = - 7 21

-

- 8 VE = - 108 31

VE = 13 2413

T ( )

3.Tinjau konnstruksi global ABBCDDE: = 0v VA + VE = 0

VA+ 13 2413

- 27 = 0

VA = 13 2413

T ( )

= 0H HA P2 P1- HE = 0

HA 2 5 6 83

= 0

HA = 13 83

( )

Periksa :

= 0EM HA.1 + VA.9 R.4 21

- P2.4 P1. = 0

13 83

+ 121 81

- 121 81

- 8 5 = 0

134 211342

1 = 0

1 = 0 Ok

Dari persamaan(2) didapat : He - VE = - 7 2

1

HE 13 2413

= - 7 21

HE = 6 83

T ( )

Modul 9 CONTOH SOAL


10 10 CONTOH 3


Penyelesaian : 1. Tinjau konstruksi ABCDE

07.1.9.1.212.0 21 =++= EEA VHPPRM

079.11.3212.18 =++ EE VH

57 HE 7 VE = 0 - HE 7 VE = 0

7. Tinjau konstruksi parsial CDE:

P2 = 1 q = 2 t/m

P1 = 3 t

HE

HA

P3 = 4 t

HE

E VE

VA

1 m

1 m

1 m

2 m 3 m 4m 2 m

VA

HE

P2 = t

C D RI = 8 t

HE E

VE

1 m

1 m

1 m

P3 = 4t

4 m 2 m

Modul 9 CONTOH SOAL


11 11

= 0CM P2.6 P3 2 + R1.2 HE.3 VE.4 = 0

6 8 + 16 3HE 4 VE = 0 - 3 HE 4 HE = - 14 .. (2)

Persamaan (1) dan (2) dijumlahkan : - HE 7 VE = - 57 x 3 - 3 HE 4 VE = - 14 x 1

- 3 HE 21 VE = - 171

- 3 HE 4 VE = - 14 -

-17 VE = - 157

VE = 9 T174 ( )

8. Tinjau konstruksi global ABCDE: = 0H P1 HA P3 + HE = 0

- 3 - HA 4 + (7 1711 ) = 0

HA = - 8 1711 ( )

Contoh 4. Diketahui sebuah portal sederhana, tentukan gaya-gaya yang bekerja pada konstruksi tersebut sehingga menjadi stabil.

q= 3 t/m P2=5t B C E F P1=5t R=30t

A D G

1

2 4 3 3

Dari persamaam (1) didapat : - HE 7 VE = - 57

- HE 7.9 174 = - 57

HE = - 7 1711

( )

Modul 9 CONTOH SOAL


12 12 Ruas EFG

ME = 0

Rx1,5 VGx3 = 0 9x1,5 3VG = 0 13,5 3VG = 0 VG = -13,5/-3

VG = 4,5 t ( )

V = 0

VE R + VG = 0 VE 9 + 4,5 = 0 VE = 4,5 t ( ) H = 0

HE P2 = 0

HE 5 = 0 HE = 5 t ( )

q= 3t/m E P2=5t HE R=9t F 1 VE 1 G 1,5 VG 3,0

Modul 9 CONTOH SOAL


13 13 Ruas ABCDE

MA = 0

VEx9 HEx2 VDx6 + Rx5,5 + P1x1 = 0 4,5x9 5x2 VDx6 + 21x5,5 + 4x1 = 0 40,5 10 6VD + 115,5 + 4 = 0 150 - 6VD = 0

VD = -150 -6

VD = 25 t ( ) V = 0

VA + VD R VE = 0

VA + 25 21 4,5 = 0

VA = 0,5 t ( ) H = 0 HA + P1 HE = 0

HA + 4 5 = 0

HA = 1 t ( )

q=3t/m VE=4,5t HE=5t B C E P1=4t R=21t

HA D A VA VD

2 4 3

Modul 9 CONTOH SOAL


14 14

CHECK : V = 0 VA + VD R + VG = 0

0,5 + 25 30 + 4,5 = 0 0 = 0 ( OKE ) H = 0

HA + P1 P2 = 0

1 + 4 5 = 0 0 = 0 ( OKE )

MB = 0 VAx2HAx2P1x1VDx4+Rx5VGx10 = 0 0,5x2 1x2 4x1 25x4 + 30x5 4,5x10 = 0 1 2 4 100 + 150 45 = 0 0 = 0 ( OKE )

Contoh 5. Diketahui sebuah pelengkung tiga sendi, tentukan gaya-gaya yang bekerja pada konstruksi tersebut sehingga menjadi stabil.

P2=5t q=2t/m B C D P1=5t R=18t

HA A 0,5 E HE VA VE 2 4 3 2

q= 3 t/m P2=5t B C E F P1=5t R=30t

HA=1t D G A VD=25t VG VA=0,5t 1 =4,5t

2 4 3 3

Modul 9 CONTOH SOAL


15 15 Ruas ABC

MC = 0

VAx6 - HAx2 P1x1 P2x4 Rx2 = 0 6VA - 2HA 5x1 5x4 8x2 = 0 6VA - 2HA 5 20 16 = 0 6VA - 2HA 41= 0 .( 1 )

Tinjau keseluruhan

ME = 0

VAx11 + P1x1 P2x9 Rx4,5 = 0 11VA + 5x1 5x9 18x4,5 = 0 11VA + 5 45 81 = 0 11VA 121 = 0

VA = 121/ 11

VA = 11 t ( ) ( 2 )

6VA - 2HA 41 = 0 .( 1 ) 6x11 - 2HA 41 = 0 66 2HA 41 = 0 -2HA + 25 = 0 HA = 12,5 t ( )

P2=5t q=2t/m HC B R=8t C P1=5t VC

HA A VA 2 4

Modul 9 CONTOH SOAL


16 16 V = 0 VA P2 R + VE = 0

11 5 18 + VE = 0 11 5 18 + VE = 0 -12 + VE = 0

VE = 12 t ( )

H = 0

P1 + HA HE = 0

5 + 12,5 HE = 0 17,5 HE = 0

HE = 17,5 t ()

CHECK : MC = 0

VAx6 HAx2 P1x1 - P2x4 + RAx0,5 + HEx2- VEx5 = 0 11x6 12,5x2 5x1- 5x4 + 18x0,5 + 17,5x2 - 12x5 = 0 66 25 5 - 20 + 9 + 35 60 = 0

0 = 0 ( OKE ) V = 0

VA P2 R + VE = 0 11 5 18 + 12 = 0

0 = 0 ( OKE )

P2=5t q=2t/m B C D P1=5t R=18t

HA A 0,5 E HE=17,5t =12,5t VA=11 VE =12t 2 4 3 2

Modul 9 CONTOH SOAL


17 17 H = 0

P1 + HA HE = 0

5 + 12,5 17,5 = 0 0 = 0 ( OKE )

Contoh 6. Hitung: Reaksi Perletakan!

Penyelesaian:

Tinjau Konstruksi Global ABCDE

q=2 t/m P1=ATNIM

2

D

A

4

C B P2=ATNIM

1

1

3 2

E

P1=8 t q=2 t/m

D

A

C B P2=8 t

1

1 E

2 4 3 2

HA

VA

HE

VE

R

= 18 t

Modul 9 CONTOH SOAL


18 18 MA = 0 P2.1 + P1.2+ R.61/2 VE = 0 (8.1) + (8.2) + (18.61/2) (11VE) = 0 8 + 16 + 117 11VE = 0 141 11VE = 0

VE = -141/-11 = 129/11 t

V = 0 VE + VA R P1 = 0

129/11 + VA 18 8 = 0

-132/11 + VA = 0

VA = 132/11 t

H = 0 P2 + HA HE = 0

8 + HA HE = 0

HA HE = - 8 (1)

Tinjau Konstruksi Parsial CDE

MC = 0 R1.21/2 + HE.2 VE.5 = 0

(10.21/2) + (2HE) (129/11) = 0 25 + 2HE 641/11 = 0 -391/11 + 2HE = 0 HE = 391/11 2

HE = 196/11 t

q=2 t/m

HC

VC

D C 1

1

E

3 2

HE

VE

R1 = 10 t

Modul 9 CONTOH SOAL


19 19 Dari persamaan (1) didapat:

HA HE = -8 HA 196/11 = -8 HA = 196/11-8

HA = 116/11 t

Periksa!

V = 0 VA + VE + R P1 =

132/11 + 129/11 18 8 = 26 26 = 0 ok!

H = 0 HA + P2 HE =

116/11 + 8 196/11 = 196/11 196/11 = 0 ok!

ME = 0 VA.11 + P2.1 P1.9 R.41/2 =

(132/11.11) + (8.1) (8.9) (18.41/2) = 145 + 8 72 81 = 153 153 = 0 ok!

q=2 t/m

D

A

C B P2=8 t

1

1 E

2 4 3 2

HA=116/11 t

VA=132/11t

HE=196/11 t

VE=129/11 t

R

= 18 t

P1=8 t

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