CONTOH SOAL STATIKA

1 1

Modul 9 – CONTOH SOAL

CONTOH SOAL

CONTOH 1

q = 2 t/m VC

T 2m

P1 = 3 t

V

HB B

HA

VB

1m

A VA 3m

2m

Penyelesaian : 1. Tinjau konstruksi global ABC.

∑M

B

= 0 →

- R. 2

1 − H A .1 + VA .5 2 - 25 – HA + 5 VA 5 VA - HA

=0 =0 = 25 ……… (1)

2. Tinjau Konstruksi Parsial AC:

Pusat Pengembangan Bahan Ajar - UMB

Ir. Edifrizal Darma, MT STATIKA 1

2 2


HC P 1 = 3t

2m

VC

1m

HA A VA

∑M

C

= 0 →

3m

1 − P.2 − H A .3 + VA .3 -R–1 2 =0 -9 – 6 – 3 HA + 3 VA = 0 3VA – 3 HA

= 15

VA – HA

= 5 …….(2)

Persamaan (1) dan (2) dijumlahkan : Dari persamaan (2) didapat ; 5 VA - HA = 25 VA - HA =5 VA – HA = 5 5 - HA =5 HA =0 VA = 20 VA

=5T( )

1. Tinjau konstruksi global ABC :

∑V = 0 →V

A

+ VB − R = 0

5 +VB – 10 = 0 VB

∑

= 5T (

)

H = 0 → H A + P − H B = 0

0 + 3 - HB HB

=0 =3T( )

Periksa :

∑M

A

= 0 →

P.1 + R.2

1 − H B .1 − VB .5 = 0 2 3 + 25 – 3 – 25 = 0 0


= 0 ………… OK!


3 3


CONTOH 2

D P2 = 2 t

q = 3 t/M

1m B

B

C 2m

HA

P1 = 5 t A VA

HE VE 3m

3m

3 m

Ditanya : Reaksi perletakan Penyelesaian : 1. Tinjau konstruksi global ABB’CDD’E.

∑M

A

= 0 → R.4 1 − P2 .3 + H E .1 − VE .9 = 0 2 1 − 2.3 + H E−9 = 0 27.4 2 1 − 6 + H B − 9VE = 0

121 2

1

HE – 9VE = 115 2 ……….. (1)



1m

4 4


2. Tinjau konstruksi parsial CDE : P2 = 2t D 1m m

HC C

D

P1= 5t m HE E

2m m 21m m

5m

∑

0 M C = → R2 .1 1 − P2 .1 + P1.2 + H E .3 − VE .3 = 0 2

1 − 2 + 10 + 3H E − 3VE = 0 9.1 2 3 HE – 3 VE = 0

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

1 HE - 9 VE

= 115

1 HE - VE

=-7

8 VE = - 108

13 VE = 13

He - VE

2 -

6 ……. (2).

Dari persamaan(2) didapat :

1

2

1 -

=-7

=-7

13 HE – 13

3

24 = - 7 3

HE

=6

1

2 2

8T(

)

24 T ( )

3.Tinjau konnstruksi global ABB’CDD’E:



5 5


∑ v = 0 → VA + VE

=0

13 VA+ 13

24 - 27 = 0 13 VA = 13

∑ H = 0 → HA – P2 – P1- HE

24 T ( )

=0

3 HA – 2 – 5 – 6

8

=0

3 HA = 13

8 (

)

Periksa :

∑M

E

= 0 →

1 HA.1 + VA.9 – R.4

2 - P2.4 – P1. = 0

3

1 1 8 8 13 + 121 - 121 8 - 8 – 5 134

1 − 134 1 2 2 0


=0 =0 = 0 …… Ok


6 6


CONTOH 3 q = 2 t/m

P2 = 1

1m

P1 = 3 t HE P3 = 4 t HA

1m

HE E

VA 2Ditanya m 3m : Reaksi perletakan

4m

VE 2m

Penyelesaian : 1. Tinjau konstruksi ABCDE

∑M

A

= 0 → R.2 1 + P1.1 + P2 .9 − H E .1 − VE .7 = 0 2 18.2 1 + 3.1 + 1.9 − H E − 7VE = 0 2

57 – HE – 7 VE

=0

- HE – 7 VE = 0 3. Tinjau konstruksi parsial CDE: P2 = t HE C VA

D

RI = 8 t

1m P3 = 4t

HE

1m 1m

E 4m

VE

2m


1m


7 7


∑M

C

= 0 →

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

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

= - 57 …… x 3

- 3 HE – 4 VE = - 14 …… x 1

4 - HE – 7.9

17 = - 57

- 3 HE – 21 VE = - 171 HE

- 3 HE – 4 VE = - 14 -

11 17 ( =-7

)

-17 VE = - 157

4

VE = 9 17

T

( )

4. Tinjau konstruksi global ABCDE:

∑ H = 0 →

P1 – HA – P3 + H E

=0

11 - 3 - HA – 4 + (7 17 )

=0

HA

=-8

11 17 (

CONTOH 2

)

D P2 = 2 t

q = 3 t/M

1m B

B

C 2m

HA

P1 = 5 t A VA

HE VE 3m

3m


3 m


1m

8 8


Ditanya : Reaksi perletakan Penyelesaian : 5. Tinjau konstruksi global ABB’CDD’E.

∑M

A

= 0 → R.4 1 − P2 .3 + H E .1 − VE .9 = 0 2 1 − 2.3 + H E−9 = 0 27.4 2 1 − 6 + H B − 9VE = 0 121 2 1

HE – 9VE = 115 2 ……….. (1) 6. Tinjau konstruksi parsial CDE : P2 = 2t D 1m m

HC C

D

P1= 5t m HE E

2m m 21m m

5m

∑

0 M C = → R2 .1 1 − P2 .1 + P1.2 + H E .3 − VE .3 = 0 2

1 − 2 + 10 + 3H E − 3VE = 0

9.1 2

3 HE – 3 VE = 0

1 HE=- VE


=-7

6 ……. (2).


9 9


Persamaan (1) dan (2) dijumlahkan :

1 HE - 9 VE

= 115

HE - VE

=-7

1

-

1

2

He - VE

8 VE = - 108

13 VE = 13

=-7

13

2 1

Dari persamaan(2) didapat :

HE – 13

24 = - 7 3

3

HE

=6

1

2 2

8T(

)

24 T ( )

3.Tinjau konnstruksi global ABB’CDD’E:

∑ v = 0 → VA + VE

=0

13 VA+ 13

24 - 27 = 0 13 VA = 13

∑ H = 0 → HA – P2 – P1- HE

24 T ( )

=0

3 HA – 2 – 5 – 6

8

=0

3 HA = 13

8 (

)

Periksa :

∑M

E

= 0 →

1 HA.1 + VA.9 – R.4

3 13

2 - P2.4 – P1. = 0

1 1 8 + 121 8 - 121 8 - 8 – 5


=0


10 10


134

1 − 134 1 2 2 1

=0 = 0 …… Ok

CONTOH 3 q = 2 t/m

P2 = 1

1m

P1 = 3 t HE P3 = 4 t HA

1m

HE E

A Ditanya : V Reaksi perletakan 2Penyelesaian m : 3m

4m

VE 2m

1. Tinjau konstruksi ABCDE

∑M

= 0 → R.2 1 + P1.1 + P2 .9 − H E .1 − VE .7 = 0 2

A

18.2 1 + 3.1 + 1.9 − H E − 7VE = 0 2

57 – HE – 7 VE

=0

- HE – 7 VE = 0 7. Tinjau konstruksi parsial CDE: P2 = t HE C VA

D

RI = 8 t

1m P3 = 4t

HE

1m 1m

E VE Pusat Pengembangan Bahan Ajar - UMB 4m

2m

1m


11 11


∑M

C

= 0 →

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

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

- 3 HE – 4 VE = - 14 …… x 1

4 - HE – 7.9

17 = - 57

- 3 HE – 21 VE = - 171 HE = - 7

- 3 HE – 4 VE = - 14 -

11 17 (

-17 VE = - 157

4

VE = 9 17

T

( )

8. Tinjau konstruksi global ABCDE:

∑ H = 0 →

P1 – HA – P3 + H E

=0

11 17 )

=0

- 3 - HA – 4 + (7

HA

11 = - 8 17 (

)

Contoh 4. Diketahui sebuah portal sederhana, tentukan

gaya-gaya yang bekerja pada

konstruksi tersebut sehingga menjadi stabil.



)

12 12


q= 3 t/m B

C

P2=5t F

E

P1=5t

R=30t

A

D

G

1 2

4

3

3

 Ruas EFG

q= 3t/m E HE

R=9t

F

P2=5t 1

VE 1 G VG

1,5 3,0

ΣME = 0 Rx1,5 – VGx3

=0

9x1,5 – 3VG

=0

13,5 – 3VG

=0

VG

= -13,5/-3

VG

= 4,5 t (↑ )

ΣV = 0 VE – R + VG

=0

VE – 9 + 4,5

=0

VE

= 4,5 t (↑ )

ΣH = 0 HE – P2

=0

HE – 5

=0



13 13


HE = 5 t ( → )

 Ruas ABCDE q=3t/m B P1=4t

VE=4,5t HE=5t E

C R=21t

HA A

D VA

VD

2

4

3

ΣMA = 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 (↑ ) ΣV = 0 VA + VD – R – VE = 0 VA + 25 – 21 – 4,5 = 0 VA = 0,5 t (↑ ) ΣH = 0 HA + P1 – HE

=0

HA + 4 – 5

=0



14 14


HA

=1t(→)

q= 3 t/m B

C

E

P1=5t

R=30t

HA=1t A VA=0,5t 2

P2=5t F

D VD=25t 1 4

3

G VG =4,5t 3

CHECK : ΣV = 0 VA + VD – R + VG = 0 0,5 + 25 – 30 + 4,5 0

=0 = 0 ( OKE )

ΣH = 0 HA + P1 – P2

=0

1+4–5

=0 0

= 0 ( OKE )

ΣMB = 0 VAx2–HAx2–P1x1–VDx4+Rx5–VGx10 = 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.



15 15


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

P1=5t HA

A VA 2

0,5

E

3

2

4

HE VE

 Ruas ABC

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

HC VC

HA A VA 2

4

ΣMC = 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 ΣME = 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 )



16 16


6VA - 2HA – 41

= 0

6x11 - 2HA – 41

= 0

66 – 2HA – 41

= 0

-2HA + 25

= 0

HA

………….( 1 )

= 12,5 t ( → )

ΣV = 0 VA – P2 – R + VE = 0 11 – 5 – 18 + VE = 0 11 – 5 – 18 + VE = 0 -12 + VE = 0 VE = 12 t (↑ ) ΣH = 0 P1 + HA – HE = 0 5 + 12,5 – HE = 0 17,5 – HE

=0

HE

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

P1=5t HA =12,5t

= 17,5 t (←)

A VA=11 2 4

0,5 3

E HE=17,5t VE =12t 2

CHECK : ΣMC = 0 VAx6 – HAx2 – P1x1 - P2x4 + RAx0,5 + HEx2- VEx5 = 0 11x6 – 12,5x2 – 5x1- 5x4 + 18x0,5 + 17,5x2 - 12x5 = 0



17 17


66 – 25 – 5 - 20 + 9 + 35 – 60 = 0 0

= 0 ( OKE )

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

= 0 ( OKE )

ΣH = 0 P1 + HA – HE

=0

5 + 12,5 – 17,5 = 0 0

= 0 ( OKE )

Contoh 6. Hitung: Reaksi Perletakan! P1=ATNIM

1

q=2 t/m

D

C

B P2=ATNIM

1 E

A

4

2

2

3

Penyelesaian: Tinjau Konstruksi Global ABCDE P1=8 t

P2=8 t HA

R = 18 t HE

E

A

Pusat Pengembangan Bahan Ajar - UMB 2 VA

D

C

B

1 1

q=2 t/m

4

Ir. Edifrizal Darma, MT STATIKA 1 2 3 VE

18 18


ΣMA = 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 ΣV = 0 →

=

VE + VA – R – P1

= 0

129/11 + VA – 18 – 8

= 0

-132/11 + VA

= 0

/-11 = 129/11 t

= 132/11 t

VA ΣH = 0 →

-141

P2 + HA – HE

= 0

8 + HA – HE

= 0

HA – HE

= - 8 ……………………(1)

Tinjau Konstruksi Parsial CDE q=2 t/m HC D

C

1

R1 = 10 t VC HE

E 3

1

2 VE

ΣMC = 0 →

R1.21/2 + HE.2 –VE.5


= 0


19 19


(10.21/2) + (2HE) – (129/11)

= 0

25 + 2HE – 641/11

= 0

-391/11 + 2HE

= 0

HE

= 391/11 2 = 196/11 t ¬

HE Dari persamaan (1) didapat: HA – HE

= -8

HA – 196/11

= -8

HA

= 196/11-8

HA

= 116/11 t →

Periksa! P1=8 t

P2=8 t HA=116/11 t

R = 18 t

E

A

2

4

2

3

ΣH = 0 →

HE=196/11 t

2 VE=129/11 t

VA=13 /11t

ΣV = 0 →

D

C

B

1 1

q=2 t/m

VA + VE + R – P1

=

132/11 + 129/11 – 18 – 8

=

26 – 26

= 0… ok!

HA + P2 – HE

=

116/11 + 8 – 196/11

=



20 20


196/11 – 196/11 ΣME = 0 →

= 0… ok!

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!



CONTOH SOAL STATIKA

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