CONTOH SOAL STATIKA

(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.

→  =

∑

M_B 0 _{- R. 2}

1 ₂

−

H

A

.

1 +

V

A

.

5

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

(2)

→  =

∑

M_C 0 _{- R – 1}

1 ₂

−

P

.

2 −

H

A

.

3 +

V

A

.

3

= 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 : 0 0→ + − = =

∑

V V_A V_B R 5 +VB – 10 = 0 VB = 5T ( ) 0 0→ + − = =

∑

H HA P HB 0 + 3 - HB = 0 HB = 3 T ( ) Periksa : →  =

∑

M_A 0 _{P.1 + R.2}

1 ₂

−

H

B

.

1 −

V

B

.

5 =

0

3 + 25 – 3 – 25 = 0 0 = 0 ………… OK!

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

STATIKA 1 H C P₁

= 3

_t _V C 2 m 1 m 1 m 2 m H_A A V A _{3 m}

Dari persamaan (2) didapat ; VA - HA = 5

5 - HA = 5

(3)

CONTOH 2

Ditanya : Reaksi perletakan Penyelesaian :

1. Tinjau konstruksi global ABB’CDD’E.

0

9 .

1 .

3 .

2

1

4 .

0 

→

−

₂

+

−

=

∑

M

A

R

P

H

E

V

E 27.4

2

2 .

3

9

0

1 ₋

₊

₋

₌

E

H

121

1

2 −

6 +

H

B

−

9 V

E

=

0

HE – 9VE = 115

2

1

……….. (1) q = 3 t/M D P2

= 2 t

B B C 1m A 1m 2 m 3 m VA HA HE P1 = 5 t 3 m 3 m VE

(4)

2. Tinjau konstruksi parsial CDE :

0

3 .

2 .

1 .

2

1

1 .

₂ ₁ 2 0

₋

₊

₋

₌

→



=

∑

M

C

R

P

H

E

V

E 9.1

1

2 −

2 +

10 +

3 H

E

−

3 V

E

=

0

3 HE – 3 VE = 0 HE=- VE = - 7

6

1

……. (2).

Persamaan (1) dan (2) dijumlahkan :

HE - 9 VE = 115

2

1

HE - VE = - 7

2

1

-- 8 VE = - 108

3

1

VE = 13

24

13

T ( )

3.Tinjau konnstruksi global ABB’CDD’E:

STATIKA 1 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

24

13

= - 7

1

2

HE = 6

8

3

T ( )

(5)

∑

v=0→ _V A + VE = 0 VA+ 13

24

13

- 27 = 0 VA = 13

24

13

T ( ) →  =

∑

H 0 _H A – P2 – P1- HE = 0 HA – 2 – 5 – 6

8

3

= 0 HA = 13

8

3

( ) Periksa : →  =

∑

M_E 0 HA.1 + VA.9 – R.4

2

1

- P2.4 – P1. = 0 13

8

3

+ 121

8

1

- 121

8

1

- 8 – 5 = 0 134

2

1

134

2

1 ₋

= 0 0 = 0 …… Ok

(6)

CONTOH 3

1. Tinjau konstruksi ABCDE

0

7 .

1 .

9 .

1 .

2

1

2 .

0 

→

+

₁

+

₂

−

=

∑

M

A

R

P

H

E

V

E 0 7 9 . 1 1 . 3 2 1 2 . 18 + + −H_E − V_E = 57 – HE – 7 VE = 0 - HE – 7 VE = 0

3. Tinjau konstruksi parsial CDE:

STATIKA 1 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} V A H E P₂ = t C D R_I = 8 t H_E E V E 1 m 1 m 1 m P₃= 4t 4 m 2 m

(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 = - 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

T

17

4

( )

4. Tinjau konstruksi global ABCDE:

∑

H=0→ _P 1 – HA – P3 + HE = 0 - 3 - HA – 4 + (7

17

11

) = 0 HA = - 8

17

11

( ) CONTOH 2

Dari persamaam (1) didapat : - HE – 7 VE = - 57 - HE – 7.9

17

4

= - 57 HE = - 7

17

11

( ) q = 3 t/M D P2

= 2 t

B B C 1m A 1m 2 m 3 m VA HA HE P1 = 5 t 3 m 3 m VE

(8)

5. Tinjau konstruksi global ABB’CDD’E.

0

9 .

1 .

3 .

2

1

4 .

0 

→

−

₂

+

−

=

∑

M

A

R

P

H

E

V

E 27.4

2

2 .

3

9

0

1 ₋

₊

₋

₌

E

H

121

2

6

9

0

1 ₋

₊

₋

₌

E B

V

H

HE – 9VE = 115

2

1

……….. (1)

6. Tinjau konstruksi parsial CDE :

0

3 .

2 .

1 .

2

1

1 .

₂ ₁ 2 0

₋

₊

₋

₌

→



=

∑

M

C

R

P

H

E

V

E 9.1

2

10

3

0

1 ₋

₊

₋

₌

E E

V

H

3 HE – 3 VE = 0 HE=- VE = - 7

6

1

……. (2).

STATIKA 1 HC 5 m P2 = 2t D D C 1m m 2 m m 21m m P1= 5t m HE E

(9)

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

2

1

HE - VE = - 7

2

1

-- 8 VE = - 108

3

1

VE = 13

24

13

T ( )

3.Tinjau konnstruksi global ABB’CDD’E:

∑

v=0→ _V A + VE = 0 VA+ 13

24

13

- 27 = 0 VA = 13

24

13

T ( ) →  =

∑

H 0 _H A – P2 – P1- HE = 0 HA – 2 – 5 – 6

8

3

= 0 HA = 13

8

3

( ) Periksa : →  =

∑

M_E 0 _H A.1 + VA.9 – R.4

2

1

- P2.4 – P1. = 0

3

1

Dari persamaan(2) didapat :

He - VE = - 7

2

1

HE – 13

24

13

= - 7

2

1

HE = 6

8

3

T ( )

(10)

134

1

2 −

134

1

2

= 0

1 = 0 …… Ok

CONTOH 3

1. Tinjau konstruksi ABCDE

0

7 .

1 .

9 .

1 .

2

1

2 .

0 

→

+

₁

+

₂

−

=

∑

M

A

R

P

H

E

V

E 0 7 9 . 1 1 . 3 2 1 2 . 18 + + −HE − VE = 57 – HE – 7 VE = 0 - HE – 7 VE = 0

7. Tinjau konstruksi parsial CDE:

STATIKA 1 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} V A H E P₂ = t C D R_I = 8 t H_E E V E 1 m 1 m 1 m P₃= 4t 4 m 2 m

(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 - 3 HE – 4 VE = - 14 …… x 1 - 3 HE – 21 VE= - 171 - 3 HE – 4 VE = - 14 - -17 VE = - 157 VE = 9

T

17

4

( )

8. Tinjau konstruksi global ABCDE:

∑

H=0→ _P 1 – HA – P3 + HE = 0 - 3 - HA – 4 + (7

17

11

) = 0 HA = - 8

17

11

( ) Contoh 4.

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

Dari persamaam (1) didapat : - HE – 7 VE = - 57 - HE – 7.9

17

4

= - 57 HE = - 7

17

11

( )

(12)

 Ruas EFG 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

STATIKA 1 q= 3 t/m P2=5t B C E F P1=5t R=30t A D G 1 2 4 3 3 q= 3t/m E P2=5t HE R=9t F 1 VE 1 G 1,5 VG 3,0

(13)

HE = 5 t ( → )  Ruas ABCDE 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 q=3t/m VE=4,5t HE=5t B C E P1=4t R=21t HA D A VA VD 2 4 3

(14)

HA = 1 t ( → ) 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.

STATIKA 1 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

(15)

 Ruas ABC 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 P2=5t q=2t/m B C D P1=5t R=18t HA A 0,5 E HE VA VE 2 4 3 2 P2=5t q=2t/m HC B R=8t C P1=5t VC HA A VA 2 4

(16)

6VA - 2HA – 41 = 0 ………….( 1 ) 6x11 - 2HA – 41 = 0 66 – 2HA – 41 = 0 -2HA + 25 = 0 HA = 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 = 17,5 t (←) CHECK : ΣMC = 0

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

STATIKA 1 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

(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!

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 C B P₂=8 t 1 R= 18 t

(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 = -141/-11 = 129/11 t ΣV = 0 → VE + VA – R – P1 = 0 129_/ 11 + VA – 18 – 8 = 0 -132_/ 11 + VA = 0 VA = 132/11 t ΣH = 0 → P2 + HA – HE = 0 8 + HA – HE = 0 HA – HE = - 8 ………(1)

Tinjau Konstruksi Parsial CDE

ΣMC = 0 → R1.21/2 + HE.2 –VE.5 = 0

STATIKA 1 q=2 t/m H_C V_C D C 1 1 E 3 2 H_E V_E R₁= 10 t

(19)

(10.21_/ 2) + (2HE) – (129/11) = 0 25 + 2HE – 641/11 = 0 -391_/ 11 + 2HE = 0 HE = 391/11 2 HE = 196/11 t ¬

Dari persamaan (1) didapat:

HA – HE = -8 HA – 196/11 = -8 HA = 196/11-8 HA = 116/11 t → Periksa! ΣV = 0 → VA + VE + R – P1 = 132_/ 11 + 129/11 – 18 – 8 = 26 – 26 = 0… ok! ΣH = 0 → HA + P2 – HE = q=2 t/m D A C B P₂=8 t 1 1 E 2 4 3 2 H_A=116_/ 11 t V_A=132_/ 11t H_E=196_/ 11 t V_E=129_/ 11 t R= 18 t P₁=8 t

(20)

196_/ 11 – 196/11 = 0… ok! ΣME = 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!