modul 9 - ir-darmadi-mt's blog · web viewcontoh 1 q = 2 t/m vc t 2 m p1 = 3 t v hb b ha vb 1...

11

Modul 9 – CONTOH SOAL

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.

- R. 2 = 0

- 25 – HA + 5 VA = 0

5 VA - HA = 25 ……… (1)

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

22


2. Tinjau Konstruksi Parsial AC:

- R – 1 = 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 :

5 +VB – 10 = 0

VB = 5T ( )

0 + 3 - HB = 0

HB = 3 T ( )

Periksa :

P.1 + R.2

3 + 25 – 3 – 25 = 0

0 = 0 ………… OK!


HC

P1 = 3t VC2 m

1 m1 m

2 m

HA

A

VA3 m

Dari persamaan (2) didapat ;

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

33


CONTOH 2

Ditanya : Reaksi perletakan

Penyelesaian :

1. Tinjau konstruksi global ABB’CDD’E.

27.4

121

HE – 9VE = 115 ……….. (1)


q = 3 t/M P2 = 2 tD

BB C1m

A 1m

2 m

3 m

VA

HA

HE

P1 = 5 t

3 m3 mVE

44


2. Tinjau konstruksi parsial CDE :

9.1

3 HE – 3

VE = 0

HE=- VE

= - 7 ……. (2).


HE - 9 VE = 115

HE - VE = - 7 -

- 8 VE = - 108


HC

5 m

P2 = 2tD

DC

1mm

2 mm

21mm

P1= 5tm

HE

E

Dari persamaan(2) didapat :

He - VE = - 7

HE – 13 = - 7

HE = 6 T ( )

55


VE = 13 T ( )

3.Tinjau konnstruksi global ABB’CDD’E:

VA + VE = 0

VA+ 13 - 27 = 0

VA = 13 T ( )

HA – P2 – P1- HE = 0

HA – 2 – 5 – 6 = 0

HA = 13 ( )

Periksa :

HA.1 + VA.9 – R.4 - P2.4 – P1. = 0

13 + 121 - 121 - 8 – 5 = 0

134 = 0

0 = 0 …… Ok


66


CONTOH 3


Penyelesaian :

1. Tinjau konstruksi ABCDE

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

EVE

VA

1 m

1 m

1 m

2 m 3 m 4m 2 m

VA

HE

P2 = t

C DRI = 8 t

HE

E

VE

1 m

1 m

1 m

P3 = 4t

4 m 2 m

77


P2.6 – P3 2 + R1.2 – HE.3 – VE.4 = 0

6 – 8 + 16 – 3HE – 4 VE = 0

- 3 HE – 4 HE = - 14 ……….. (2)


- 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 ( )

4. Tinjau konstruksi global ABCDE:

P1 – HA – P3 + HE = 0

- 3 - HA – 4 + (7 ) = 0

HA = - 8 ( )

CONTOH 2


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

- HE – 7.9 = - 57

HE = - 7

( )

q = 3 t/M P2 = 2 tD

BB C1m

A 1m

2 m

3 mVAHA HE

P1 = 5 t3 m3 m VE

88



Penyelesaian :

5. Tinjau konstruksi global ABB’CDD’E.

27.4

121

HE – 9VE = 115 ……….. (1)

6. Tinjau konstruksi parsial CDE :


HC

5 m

P2 = 2tD

DC

1mm

2 mm

21mm

P1= 5tm

HE

E

99


9.1

3 HE – 3

VE = 0

HE=- VE

= - 7 ……. (2).


HE - 9 VE = 115

HE - VE = - 7 -

- 8 VE = - 108

VE = 13 T ( )

3.Tinjau konnstruksi global ABB’CDD’E:

VA + VE = 0

VA+ 13 - 27 = 0

VA = 13 T ( )


Dari persamaan(2) didapat :

He - VE = - 7

HE – 13 = - 7

HE = 6 T ( )

1010


HA – P2 – P1- HE = 0

HA – 2 – 5 – 6 = 0

HA = 13 ( )

Periksa :

HA.1 + VA.9 – R.4 - P2.4 – P1. = 0

13 + 121 - 121 - 8 – 5 = 0

134 = 0

1 = 0 …… Ok

CONTOH 3


Penyelesaian :


P2 = 1 q = 2 t/m

P1 = 3 t

HE

HA

P3 = 4 t

HE

EVE

VA

1 m

1 m

1 m

2 m 3 m 4m 2 m

1111


1. Tinjau konstruksi ABCDE

57 – HE – 7 VE = 0

- HE – 7 VE = 0

7. Tinjau konstruksi parsial CDE:

P2.6 – P3 2 + R1.2 – HE.3 – VE.4 = 0

6 – 8 + 16 – 3HE – 4 VE = 0

- 3 HE – 4 HE = - 14 ……….. (2)


- 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 ( )


VA

HE

P2 = t

C DRI = 8 t

HE

E

VE

1 m

1 m

1 m

P3 = 4t

4 m 2 m

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

- HE – 7.9 = - 57

HE = - 7

( )

1212


8. Tinjau konstruksi global ABCDE:

P1 – HA – P3 + HE = 0

- 3 - HA – 4 + (7 ) = 0

HA = - 8 ( )

Contoh 4.Diketahui sebuah portal sederhana, tentukan gaya-gaya yang bekerja pada

konstruksi tersebut sehingga menjadi stabil.

Ruas EFG

ΣME = 0

Rx1,5 – VGx3 = 0

9x1,5 – 3VG = 0

13,5 – 3VG = 0


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

1313


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

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


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

HA D A VA VD

2 4 3

1414


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

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


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

1515


0 = 0 ( OKE )

Contoh 5.Diketahui sebuah pelengkung tiga sendi, tentukan gaya-gaya yang bekerja pada

konstruksi tersebut sehingga menjadi stabil.

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


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

1616


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 ( → )

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 (←)


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

1717


CHECK :

ΣMC = 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 )

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/mP1=ATNIM

2

D

A

4

CB

P2=ATNIM1

1

3 2

E

P1=8 tq=2 t/m

D

A

CB

P2=8 t1

1E

2 4 3 2

HA

VA

HE

VE

R = 18 t

1818


SMA = 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

SV = 0 ® VE + VA – R – P1 = 0

129/11 + VA – 18 – 8 = 0

-132/11 + VA = 0

VA = 132/11 t

SH = 0 ® P2 + HA – HE = 0

8 + HA – HE = 0

HA – HE = - 8 ……………………(1)

Tinjau Konstruksi Parsial CDE


q=2 t/m

HC

VC

DC1

1

E

3 2

HE

VE

R1 = 10 t

1919


SMC = 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 ¬

Dari persamaan (1) didapat:

HA – HE = -8

HA – 196/11 = -8

HA = 196/11-8

HA = 116/11 t ®

Periksa!

SV = 0 ® VA + VE + R – P1 =

132/11 + 129/11 – 18 – 8 =

26 – 26 = 0… ok!


q=2 t/m

D

A

CB

P2=8 t1

1E

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

2020


SH = 0 ® HA + P2 – HE =

116/11 + 8 – 196/11 =

196/11 – 196/11 = 0… ok!

SME = 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!


modul 9 - ir-darmadi-mt's blog · web viewcontoh 1 q = 2 t/m vc t 2 m p1 = 3 t v hb b ha vb 1...

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