1

MINISTRY OF

SCIENCE AND TECHNOLOGY

DEPARTMENT OF

TECHNICAL AND VOCATIONAL EDUCATION

CE – 1013

MECHANICS OF MATERIALS I

SAMPLE QUESTIONS AND ANSWERS

A.G.T.I (First Year)

Civil Engineering

2

1. The screw eye in Fig is subjected to two forces, F1 and F2, Determine the

magnitude and direction of the resultant force.

= 90 – 10 – 15 = 65°

= 360 2(65)

2

= 115°

FR = 2 21 2 1 2F F 2FF cos

= 2 2(100) (150) 2 100 150cos115

= 212.55 N

RF

sin = 2F

sin

212.55

sin115 =

150

sin

sin = 0.6396

= 39.76°

= 39.76° + 15° = 54.76°

2. Resolve the 200lb force shown acting on the pin, fig into components in the (a) x

and y directions and (b) x' and y direction.

F2 = 150 N

F2

15°

F1

F2

10°

F2 = 100 N

15°

10°

30° x'

40°

F = 200 lb

y' y

FR

x

3

(a) x and y directions

y

Fx = F cos 40° = 200 × cos 40°

= 153.21b

Fy = F sin 40° F2

= 128.56 lb

(b) x' and y direction

200

sin 60 =

Fx '

sin 50

Fx' = 176.91 lb

200

sin 60 =

Fy

sin 70

Fy = 200 (sin 70

sin 60)

Fy = 217.01 lb

3. The ring show in Fig is subject to two forces, F1 and F2. If it is required that the

result force have a magnitude of 1 kN and be directed vertically downward,

determine (a) the magnitude of F1 and F2 provided = 30° and (b) the magnitude

of F1 and F2 if F2 is to be a minimum.

Solution

(a)

F2

20°

F1

F1 40°

F = 200

x

y

F2 20° F1

30°

20° 30°

1 kN

Fy 50° 50° F

= 20

0

Fy

6

40° 30° 0° Fx'

x'

4

sine rule 180 30 20 130

1000

sin = 2F

sin 20

F2 = 446.476 N

F2

20° 1000 N

30°

1000

sin130 = 1F

sin 30

F1 = 652.701N

F1 20°

1000 N 30°

(b) F2 is to be minimum

The minimum length or magnitude of F2 will occur when its line of action is

perpendicular to F1.

1000 N

20°

F2

F1 = 180 – 90 – 20 = 70°

Sine Rule

1000

sin 90 = 2F

sin 20

F2 = 342.02 N

1000

sin 90 = 1F

sin

F1 = 939.693 N

4. The end of the boom O in fig is subjected to three concurrent and coplanar

forces. Determine the magnitude and direction of the resultant force.

y

3 4 5 45°

F3 = 200 N

0 F1 = 400 N x

F2 = 250 N

5

Solution

FRx = 250 sin 45° - 400 – 160

= – 383.223 N

= 383.223 N ( )

FRy = 120+ 250 cos 45

= 296.777 N ( )

Resultant forces, FR = 2 2R RF x F y

= 2 2(383.223) (296.777)

= 484.702 N

= Ry1

Rx

Ftan

F

= 1 296.777tan

383.223

= 37.755°

5. The pin in Fig is subjected to two forces, F1 and F2. Determine the magnitude

and direction of the resultant force.

30°

45°

y

F2= 400N F1= 600N

x

O F1 = 400N

F2 Sin 45°

F2 F2 cos 45°

F3

200 4160

5

200 3120

5

6

Soluition

45°

F2 sin 45

F2 = 400N F2 cos 45

F1 sin 30° F1 = 600 N

30° F1 cos 30

FRx = F1 cos 30 – F2 sin 45

= 600 cos 30 – 400 sin 45

= 236.772 N ( )

FRy = F2 cos 45+ F1 sin 30°

= 400 cos 45+ 600 sin 30

= 582.843 N ( )

FR = 2 2Rx RyF F

= 2 2(236.772) (582.843)

= 629.1N

= tan-1 Ry

Rx

F

F

= tan-1582.843

236.772

= 67.891°

Cartesian Vector Notating

F1 = 600 cos 30 i + 600 sin 30 j

F2 = – 400 sin 45 i + 400 cos 45 j

FR = F1 + F2

= (600 cos 30 – 400 sin 45) i + (600 sin 30 + 400 cos 45) j

= {236.772 i + 582.843 j}N

7

6. If the cylinder at A in Fig has a weight of 20 lb, determine the weight of B and

the force in each cord needed to hold the system in the equilibrium.

D

45°

B

C

30°

5

4 3

E

G

A 20lb

Solution

TEC

TEC cos 45°

TEGcos 30°

45°

30°

TEC Sin 45° 20 lb

E

TEG

TEG sin 30°

y

x

= 0 xF

TEG sin 30 - TEC cos 45 = 0

0.5 TEG – 0.707 TEC = 0 ---------------- (1)

Fy = 0

TEG cos 30 – TEC sin 45 – 20 = 0

0.866 TEG – 0.707 TEC – 20 = 0 ---------------- (2)

Eq (1) 0.5 TEG – 0.707 TEC = 0

Eq (2) 0.866 TEG – 0.707 TEC – 20 = 0 + + –

– 0.366 TEG + 20 = 0

TEG = 54.645 lb ( )

Sub; TEG = 54.645 lb in eq (1)

0.5 (54.645) – 0.707 TEC = 0

27.3225 – 0.707 TEC = 0

TEC = 38.646 lb ( )

8

3

Fx = 0 ( )

38.646 cos 45 – TCD × 45 = 0

27.327 – TCD × 45 = 0

TCD = 34.159 lb ( )

Fy = 0 ( )

38.646 sin 45 + TCD × 35 - WB = 0

WB = 47.822 lb

7. The 90 lb cylinder shown in fig is supported by two cables and a spring having a

stiffness k = 500 lb/ft. Determine the force in the cables and the stretch of the

spring for equilibrium. Cable AD lies in the x-y plane and cable AC lies in the

x-z plane.

D

A

B

y K = 500 lb / ft

4

3 5

Z

x

30°

C

90lb

TCD × 38.646 sin 45° 5

TEC = 38.646TCD 5 3

38.646 cos 45° 45°4 TCD ×

4

5

WB

9

Solution

Fz = 0

FC × 35 – 90 = 0

FC = 150 lb

Fx = 0

45 FC – FD sin 30 = 0

FD = 240 lb

Fy = 0

FB – FD cos 30° = 0

FB – 240 cos 30° = 0

FB = 207.846 lb

The stretch of the spring,

FB = k SAB

207.846 = 500 SAB

SAB = 0.416 ft

FB

3

90 lb

FD sin 30° FD

30°

FD cos 30°

Fc × 4

5

Fc

4 5

Fc × 3

5z

x

y

10

8. Determine the required length of cord AC in fig so that the 8-kg lamp is

suspended in the position shown. The unstretched length of the spring AB is

L'AB = 0.4 m and the spring has a stiffness of kAB = 300 N/m.

Solution

C

KAB = 300 N|m B

8 kg

2 m

30° A

TAC cos 30°

30°

TAC TAC sin 30°

TAB

W = 78.48

W = mg

= 8 × 9.81

= 78.48 N

Fy = 0 ( )

– W + TAC sin 30 = 0

TAC sin 30 = 78.48

TAC = 156.96 N

Fx = 0 ( )

TAB –TAC cos 30 = 0

TAB = 135.931 N

TAB = KAB x SAB

SAB = 0.453 m

Stretched length of AB = L'AB + SAB

= 0.4 + 0.453

= 0.853 m

The horizontal distance from C to B,

2 m = LAC cos 30° + 0.853

LAC = 1.324 m

11

9. Determine the horizontal and vertical components of reaction for the beam

loaded as shown in fig. Neglect the weight and thickness of the beam in the

calculations.

600 N

B 2 m

100N

3 m 2 m A

C

By

D

600 600 sin 45°

200 N

100 N Ay

A 600 cos 45° B Bx

45° D C

200N

Solution

Fx = 0 ( )

600 cos 45 – Bx = 0

Bx = 424.264 N ( )

MB = 0 ( ) +

Ay × 7 – 600 sin 45 × 5 – 100 × 2 = 0

Ay = 331.617 N ( )

Fy = 0 ( )

Ay + By – 600 sin 45 – 100 -200 = 0

By = 392.647 N ( )

12

10. The lever ABC is pin supported at A and connected to a short link BD as shown

in fig. If the weight of the members is negligible, determine the force developed

on the lever at A.

D 0.1 m

A

400 N

0.5 m

B

0.2m

C

0.2 m

Solution

= tan-1 0.7

0.4

= 60.255°

Fx = 0

400 + FA cos - FB cos 45 = 0

400 + 0.496 FA – 0.707 FB = 0 --------------- (1)

Fy = 0

FA sin - FB sin 45 = 0

0.868 FA – 0.707 FB = 0 --------------- (2)

0

0.5 m

F 0.2

400 N

FA cos

0.2 2F

0.5

FA sin

45°

FA

FB sin 45°

FB cos 45° 45°B.

0.2 FB

0.1 0.4

13

400 + 0.496 FA – 0.707 FB = 0

0.868 FA – 0.707 FB = 0

400 – 0.372 FA = 0 +–

FA = 1075.269 N

Eq (2) 0.868 FA- 0.707 FB = 0

0.868 (1075.269) – 0.707 FB = 0

930.729 – 0.707 FB = 0

FB = 1320.132 N

11. The beam shown in fig is pin connected at A and rests against a roller at B.

Compute the horizontal and vertical components of reaction at the pin A.

B

A

60 N-m

30°

0.75 m 0.5 m 1 m

60 N

Solution

60 N

Ay 0.5 m 1 m

30°

RB cos30°

Ax

RB sin30°

RB

A

30° R

0.75 m

60 N-m

MA = 0 ( ) +

60 × 1 + 60 – RB × 0.75 = 0

RB = 160 N

Fx = 0 ( )

Ax – RB sin 30 = 0

Ax = 80 N

Fy = 0 ( )

Ay – RB cos 30 – 60 = 0

Ay – 160 cos 30 – 60 = 0

Ay = 198.564 N

14

12. A force of 150 lb acts on the end of the beam shown in fig. Determine the

magnitude and direction of the reaction at the pin A and the tension in the cable.

C

150 lb

B

5

4 3

A

2 ft 8 ft

1.5 ft

3 ft

Solution

B

3 5 4 T × 4/5

T × 3/5

2' 8' A Ax

Ay

T 150

3'

MA = 0 ( ) +

T × 45 × 3 + T × 3

5 × 2 – 150 × 10 = 0

3.6 T – 1500 = 0

T = 416.667 lb

Fx = 0 ( )

T × 45 - Ax = 0

Ax = 333.33 lb

Fy = 0 ( )

- Ay - 150 + T × 35 = 0

- Ay – 150 + 416.667 × 35 = 0

Ay = 100 lb

15

FA = 2 2(Ax) (Ay)

= 2 2(333.33) (100)

= 348.007 lb

= tan-1 Ay

Ax

= tan-1 100

333.33

= 16.699°

13. The 100 kg uniform beam AB shown in Fig is supported at A by a pin and at B

and C by a continuous cable which wraps around a firction less pulley located at

D. If a maximum tension force of 800 N can be developed in the cable before it

breaks, determine the greatest distance d at which the 6 kN force can be placed

on the beam. What are the horizontal and vertical components of reaction at A

just before the cable breaks?

A

D

B

2 m 10 m

60° C

6 kN d

Solution

Ax

Ay 100 × 9.81 = 981 N

2 m 3 m 5 m

60° 800 cos60

800 sin60 800 N 800 N 6000 N

d

MA = 0 ( ) +

6000 × d + 981 ×5 – 800 sin 60 × 8 – 800 × 10 = 0

d = 1.44 m

16

Fx = 0 ( )

800 cos60 – Ax = 0

Ax = 400 N

Fy = 0 ( )

Ay + 800 – 6000 – 981 + 800 sin 60 = 0

Ay = 5488.18 N

= 5.488 kN

14. The uniform beam shown in fig is subjected to a force and couple. If the beam is

supported at A by a smooth wall and at B and C either at the top or bottom by

smooth contacts, determine the reactions at these supports. Neglect the weight

of the beam.

Solution

A

B 2 m

2 m

4000 N-m

4 m C

2 m 30°

300 N

Ax 300

300 cos30°

300 sin30°

30° 4000

A

30°

300

By' sin30°

Cy' sin30°

Cy' cos30° 30°

Cy'

By'cos30° By' x'

y' y

x

Fx = 0 ( )

By' sin 30 + C y' sin 30 – Ax = 0

0.5 (By' + Cy') – Ax = 0 --------------- (1)

Fy = 0 ( )

By' cos 30° + Cy' cos 30° - 300 = 0

0.866 (By' + Cy') = 300

By' + Cy' = 346.420

17

By' = 346.420 – Cy' --------------- (2)

MA = 0 ( ) +

By' × 2 + Cy' × 6 – 300 cos 30 ×8 – 4000 = 0

2 (346.42 – Cy') + 6 Cy' – 6078.461 = 0

692.84 – 2 Cy' + 6 Cy' – 6078.461 = 0

4 Cy' = 5385.621

Cy' = 1346.405 N ( ) Sub; into Eq (2) By' = 346.420 – 1346.405

By' = - 999.985 N

By' = 999. 985 N ( )

Sub; By' & Cy' into Eq (1)

0.5 (By' + Cy') – Ax = 0

0.5 (– 999.985 + 1346.405) – Ax = 0

Ax = 173.209 N ( )

15. Determine the force in each member of the truss shown in fig and indicate

whether the members are in tension or compression.

45°

2 m

C A

B 500 N

2 m

18

Solution

RAx

2 m

A

RAy RCy

2 m 45° C

B 500

MA = 0 ( ) +

500 × 2 – Rcy × 2 = 0

Rcy = 500 N ( )

Fy = 0 ( )

– RAy + RCy = 0

- RAy = - 500

RAy = 500 N ( )

Fx = 0 ( )

500 – RAx = 0

RAx = 500 N ( )

Jonit (A)

Fx = 0 ( )

FAC – 500 = 0

FAC = 500 N (Tension)

yF = 0 ( )

FAB

RAy = 500

RAx = 500 A FAC

FAB – 500 = 0

FAB = 500 N (Tension)

Joint (B)

Fy = 0 ( )

FBC sin 45 – 500 = 0

FBC = 707.107 N (compression)

B

45°

FBC cos 45° FBC

FBC sin45°

500

FBA = 500

19

Fx = 0 ( )

500 – FBC cos 45 = 0

500 – 707.107 cos 45 = 0

500 – 500 = 0 (check)

16. Determine the forces acting in all the members of the truss shown in fig. The

reactions at the supports are shown in the figure.

Solution

Joint C

Fx = 0 ( )

FCB sin 45 – FCD cos 30° = 0

0.707 FCB – 0.866 FCD = 0 -------------- (1)

C

1.5

FCB sin45° FCB cos45° 45°

FCD sin30°

FCD cos30° 30°

FCD

FCB

500

B 500

A C

500 N

707.107 N 50

0 N

500

500

C

2 m 2 m

B

30° A 3 kN

1.5 kN

1.5 kN

45°

30°

3 kN

2 m

20

Fy = 0 ( )

– FCB cos 45 + FCD sin 30° + 1.5 = 0

– 0.707 FCB + 0.5 FCD + 1.5 = 0 -------------- (2)

Eq (1) + Eq (2)

0.707 FCB – 0.866 FCD = 0

– 0.707 FCB + 0.5 FCD + 1.5 = 0

– 0.366 FCD = – 1.5

FCD = 4.098 KN (Tension)

Sub FCD = 4.098 KN in Eq (1)

0.707 FCB – 0.866 FCD = 0

0.707 FCB – 0.866 × 4.098 = 0

FCB = 5.02 kN (compression)

Joint B

B

FBA FBC sin45°

45° FBA sin45°

FBD FBA cos45° 45° FBC cos45°

3 kN

FBC

Fx = 0 ( )

– FBC sin45 + FBA sin45 + 3 = 0

– 5.02 × 0.707 + 0.707 FBA + 3 = 0

– 3.549 + 0.707 FBA + 3 = 0

FBA = 0.776 KN (Compression)

Fy = 0 ( )

FBA cos45 + FBC cos45 – FBD = 0

0.776 × cos45 + 5.02 × cos45 – FBD = 0

FBD = 4.098 KN (T)

21

Joint A

A

45°

FAB 0.776 sin45°

FAD cos30°

0.776 cos45° FAD 45°

30° 3

1.5

Fx = 0 ( )

FAD cos 30 – 0.776 sin 45 – 3 = 0

FAD cos30 = 3.549

FAD = 4.098 KN (T)

Fy = 0 ( )

FAD sin 30 – 1.5 – 0.776 cos 45 = 0

4.098 sin 30 – 1.5 – 0.776 cos 45 = 0

= 0 (check)

17. Determine the force in each member of the truss shown in fugure. Indicate

whether the member are in tension or compression.

3 m 3 m

Ay

600 N

Cx

3 5 4

D A

C

Cy 400 N

4 m

1 4

3tan

53.13

Mc = 0 ( ) +

Ay × 6 – 400 × 3 – 600 × 4 = 0

Ay = 600 N ( )

22

Fy = 0 ( )

Ay – 400 – Cy = 0

600 – 400 – Cy = 0

Cy = 200 N ( )

Fx = 0 ( )

600 – Cx = 0

Cx = 600 N ( )

Joint (A)

Fy = 0 ( )

600 - 45 FAB = 0

FAB = 750 N (compression)

Fx = 0 ( )

A

4/5 FAB

FAD

600

FAB 3/5 FAB

FAD - 35 FAB = 0

FAD = 450 N (tension)

Joint (D)

Fx = 0 ( )

600 – 450 - 35 FDB = 0

FDB = 250 N (tension)

Fy = 0 ( ) 450 D

3/5 FDB

FDB 4/5 FDB

FDC

600

45 FDB- FDC = 0

FDC = 200 N (compression)

Joint (C)

Fx = 0 ( )

FCB – 600 = 0

FCB = 600 N (com;)

Fy = 0 ( )

C FCB

200

200

600

200 – 200 = 0 (Check)

23

200

750 N

450 N

250 N

C 600

200 N

D 600

600 N

400

B

A

600

18. Determine the force in member GE, GC and BC of the bridge truss shown in fig.

Indicate whether the members are in tension or compression.

G

B

Dy Ay 4 m

a

a 4 m

400 N E

3 m

A C

4 m 1200 N

D Ax

Solution

MA = 0 ( ) +

1200 × 8 + 400 × 3 – Dy × 12 = 0

Dy = 900 N ( )

Fx = 0 ( )

400 – Ax = 0

Ax = 400 N ( )

Fy = ( )

Ay + Dy – 1200 = 0

Ay = 300 N ( )

24

Section (a-a)

3

A

300 400

4

5

G

B FGC

FGE

FBC

4/5 FGC

3/5 FGC

MG = 0 ( ) +

300 × 4 + 400 × 3 – FBC × 3 = 0

FBC = 800 N (Tension)

MC = 0 ( ) +

300 × 8 – FGE × 3 = 0

FGE = 800 N (compression)

Fy = 0 ( )

300 - 35 FGC = 0

FGC = 500 N (Tension)

19. Determine the force in member CF of the truss shown in fig. Indicate whether

the member is in tension or compression. The reactions at the supports are

shown in the figure.

25

G

3.25 KN 5 KN 3 KN 4.75 KN

4m

2m

4m

A

B C D

4m 4m 4m

H F

E

2

1

1

a

a

Solution

FFG G

MF = 0 ( )

FCD × 4- 4.75 × 4 = 0

FCD = 4.75 kN (tension)

MG = 0 ( )

FCD × 6 1

2 FCF × 6 + 3 × 4 – 4.75 × 8 = 0

6

2 FCF = – 2.5

FCF = 0.589 kN (compression)

20. Determine the force in member EB of the truss shown in Fig. Indicate whether

the memer is in tension or compression. The reactions of the supports are shown

in figure.

E

4.75

F

FCD

C D

3

FCF 12

FCF 12

+

+

26

A B

C

F D

2000N 4000N

1000N

30°

1000N

3000N 1000N E b

2m 2m 2m 2m

b

a

a

Solution

Section (a – a)

1000

A B

FED sin 30

F

4000

E 3000

30°

1000

a FED cos 30°

MB = 0 ( ) +

4000 × 4 – 1000 × 4 – 300 × 2 – FED sin 30 × 4 = 0

FED = 3000 N (compression)

27

Joint (E)

E

FEF cos 30° 30° 30°

3000 cos 30 FEF

FEF sin 30° 3000 sin 30

1000

FED

FEB

Fx = 0 ( )

FEF cos 30° - 3000 cos 30 = 0

FEF = 3000 N (c)

Fy = 0 ( )

2 (3000 sin 30) – 1000 – FEB = 0

FEB = 2000 N (tension)

21. Determine the horizontal and vertical components of force which the pin at C

exerts on member CB of the frame in Fig.

C

3 m

60°

2 m 2 m

2000 N

B

A

Solution

Cy

Bx

2m 2m

By

2000

Cx

28

MB= 0 ( ) +

2000 × 2 – Cy × 4 = 0

Cy = 1000 N

Fy = 0 ( )

By + Cy- 2000 = 0

By = 1000 N

Fx = 0 ( )

Bx - Cx = 0 --------------- (1)

By = 1000

60°

3 cos 60 Ay

3 sin 60 3m

Bx

Ax

MA = 0 ( ) +

1000 × 3 cos 60 – Bx × 3 sin 60° = 0

Bx = 577.35 N

Fx = 0 ( )

Ax – Bx = 0

Ax = 577.35 N

Fy = 0 ( )

Ay – 1000 = 0

Ay = 1000 N

Sub; Bx = 577.35 N in eq (1)

Bx – Cx = 0

Cx = 577.35 N

The pin at C exert on member CB

Bx = 577.35 N , By = 1000 N , Cx = 577.35 N , Cy = 1000 N

29

22. The beam shown in fig is pin connected at B. Determine the reactions at its

support. Neglect its weight and thickness.

4KN/m 10KN

5 4 3

2m B

2m 2m

A C

Solution

Member BC Member AB

4KN/m

Cy By

8kN

Bx

10× 4/5

10× 3/5 3 5 4

10

By = 4 Ay

Ax A

MA

Bx

Segment BC

Fx = 0

Bx = 0

MB = ( ) +

8 × 1 – Cy × 4 = 0

8 – 4 Cy = 0

Cy = 4 KN

Fy = 0 ( )

By + Cy = 8

By = 4 KN

Segment AB

MA = 0 ( ) +

10 × 45 × 2 + 4 × 4 – MA = 0

MA= 32 KN-m

30

Fx = 0 ( )

Ax – 10 × 35 - 0 = 0

Ax = 6 KN( )

Fy = 0 ( )

Ay – 10 × 45 - 4 = 0

Ay = 4 KN

Ax = 6 KN, Ay = 4 KN, MA = 32 KN-m

Bx = 0, By = 4 KN

Cy = 4 KN

23. Determine the horizontal and vertical components of force which the pin at C

exerts on member ABCD of the frame shown in Fig.

F

100 kg A

B

C

0.4 m 1.6 m D

0.4 m

1.6 m

0.8 m

31

E C F

100 × 9.81 = 981 N

A

Ay

B

Dx 0.4 m 1.6

0.8

1.6

0.4 Solution

Ax

MA = 0 ( ) +

981 × 2 – Dx × 2.8 = 0

Dx = 700. 714 N

Fx = 0 ( )

Ax – Dx = 0

Ax = 700.714 N 1.6

2

45° Fy = 0 ( ) 1.6

Ay – 981 = 0

Ay = 981 N

F

TB

45°

E TB cos 45

TB sin45° Cy

0.4 1.6

C Cx

981 N

MC = 0 ( ) +

981 × 2 – TB sin 45 × 1.6 = 0

TB = 1734 .179 N

Fx = 0 ( )

TB cos 45 – Cx = 0

Cx = 1226.25 N

Fy = 0 ( )

TB sin 45 – 981 – Cy = 0 , Cy = 245.25 N

32

The pin at C exerts on member ABCD

Ax = 700.714 N

Ay = 981 N

TB = 1734.179 N

Cx = 1226.25 N

Cy = 245.25 N

Dx = 700.7 14 N Ay

Cy TB 45°

Cx

Dx

Ax

24. The smooth disk shown in fig is pinned at D and has a weight of 20 lb.

Neglecting the weights of the other members, determine the horizontal and

vertical components of reaction at pins B and D.

3.5 ft

3ft A

D C

B

0.5ft

Solution

A B

C

D

3.5ft

Cx

20lb

3ft Ax

Ay

33

MA = 0

20 × 3 – Cx × 3.5 = 0

Cx = 17.143 lb

Fx = 0 ( )

Ax – Cx = 0

Ax = 17.143 lb

Fy = 0 ( )

Ay – 20 = 0

Ay = 20 lb

Member AB

By

ND

17.143 ND

20 lb

Bx

20

MB = 0 ( ) +

20 × 6 – ND × 3 = 0

ND = 40 lb

Fx = 0 ( )

Bx – 17.143 = 0

Bx = 17.143 lb

Fy = 0 ( )

20 + By – ND = 0

20 + By – 40 = 0 , By = 20 lb

34

Dx

By = 20 lb

Bx = 17.143

17.143 Dy

Fx = 0

Dx = 0

Fy = 0 ( )

Dy – By = 0

Dy = 20 lb

25. A man having a vertical of 150 lb supports himself by means of the cable and

pulley system shown in fig. If the seat has a weight of 15 lb, determines the

equilibrium force that he must exert on the cable at A and the force he exerts on

the seat. Neglect the weight of the cables and pulleys.

C

A

D

E

B

Solution

35

15lb

N

TE

TA

C

TA

TE

N

150lb

TE

Man

Fy = 0 ( )

TA + N – 150 = 0 --------------- (1)

Seat

Fy = 0 ( )

TE – N – 15 = 0 ---------------- (2)

Pulley C

Fy = 0 ( )

2 TE – TA = 0 ----------------- (3)

TA + N – 150 = 0

TE – N – 15 = 0 _______________

TA + TE = 165 ---------------- (4)

– T + 2T = 0 A E __________________

3TE = 165

TE = 55 lb

2TE – TA = 0

TA = 110 lb

36

TE – N – 15 = 0

55 – N – 15 = 0, N = 40 lb

26. Determine the horizontal and vertical force components acting at the pin

connections B and C of the frame shown in figure.

E B

100 kg

0.1 m

0.8 m

0.1 m D

C

0.9 m

1 m

Solution

TC

E

981 N

981 N

981 N

TC D

C

100 × 9.81 = 981 N

45°

0.7

TC sin 45° TC

981

981

Bx

By

D

0.2

= 0 ( ) BM +

981 × 0.7 – TC sin 45 × 0.9 = 0

TC = 1079.045 N

xF 0 ( )

)

Bx – 981 – TC × cos 45 = 0

Bx = 1744.0 N

yF 0 ( By – 981 + TC sin 45 = 0 , By = 217.999 N

37

27. The 100 kg block is held in equilibrium by means of the pulley and continuous

cable system shown in fig. If the cable is attached to the pin at B, compute the

forces which this pin exerts on each of its connecting members.

0.8 m

45°

D

A B

C

0.6 m

100 Kg

Solution

xF 0 ( )

)

Bx – 490.5 cos 45 = 0

Bx = 346.836 N

yF 0 ( By – 490.5 sin 45 – 490.5 = 0

By = 837.336 N

D

490.5N 981490.5N

2 490.5 cos 45°

B

490.5sin 45°

45°

490.5 By 490.5N

Bx

100 × 9.81 = 981 N

38

. .

FCB

4 3

5

FCB

FAB FAB

4/5 FCB

.

490.5 N

3/5 FCB FAB

FCB By = 837. 336

Bx = 346.836

yF 0 ( )

4

FCB – 837.336 – 490.5 = 0 5

)

FCB = 1659.795 N

xF 0 (

FAB – 3

5× FCB -Bx =0

FAB – 3

5× 1659.795 – 346.836 = 0

FAB = 1342.713 N

28. Determine the axial force and shear force and bending moment acting at point E

of the frame loaded as shown in fig.

1 m

A

1 m

1 m

B

0.5 m

0.5 m

100 kg

E

C D

.

.

..

39

Solution

Ay

yF 0 ( )

R sin 45 – 981 = 0

R = 1387.343 N

F 0x ( )

R cos45 – VE = 0

1387.343 cos 45 – VE = 0

VE = 981 N

yF 0 ( )

AE – 1387.343 sin 45 = 0

AE = 981 N

ME = 0 ( )

R cos 45 × 0.5 – ME = 0

ME = 490.4998 N

.

. Ax

Bx

P

P

D C

R A

P

. R C 45°

P 45° R cos45°

R

C

R sin45°

100×9.81 = 981 N

AE

VE E

in45°

ME

C 45°

0.5

R cos45°

R R s

+

40

29. A force of F = {– 3i + 7j – 4k} kN acts at the coner of the beam extended from a

fixed wall as shown in fig. Determine the internal loading at a section passing

through point A.

F = {-3i + 7j -4k}kN

XY

150mm

1.5m

0.5m

100mm

A

0.5m

XY

Z

F

100mm = 0.1 m

100mm

Mx

FA

Mz

My

Solution

FA + F = 0, FA = {3i – 7j + 4k}kN F 0 ( )

MA + r × F = MA + AM 0,i j k

0.5 0.1 0.15

3 7 4

= 0

= MA + i [(0.1 × – 4) – (7 ×- 0.15)] – j[(–0.5 × – 4) –

(– 0.15 × –3)] + k[(– 0.5 × 7) – (– 3 × 0.1)] = 0

= MA + {0.65i – 1.55j – 3.2k} kN = 0

MA = {– 0.65i + 1.55j + 3.2k} kN – m

Fx = {3i} kN (axial force)

Fy = {– 7j} kN (shear force)

Fz = {4k} kN (shear force)

Shear force, V = 2 2y zF F = 2 2(7 j) (4k)

Torsional moment, Mx = {– 0.65i} kN

41

bending moment, My = {1.55j} kN

bending moment, Mz = {3.2k} kN

Mb = 2 2y z(M ) (M )

= 2 2(1.55j) (3.2k)

30. Determine the tension in cords AB and AD for equilibrium of the 10 Kg crate

shown in figure.

D

C

30A

B

30° TD TB cos 30°

TB sin 30°

W = 98.1

TB

W = mg = 10 kg x 9.81 m/sec2

= 98.1 kg-m/sec2 = 98.1 N

42

Fy = 0 ( + ) TB sin 30 – W = 0 TB sin 30 = 98.1 TB = 196.2 N Fx = 0 ( ) TB cos 30 – TD = 0

196.2 cos 30 – TD = 0, TD = 169.914 N

BY TU (Mandalay) aungkyawnyein.mtu@gmail.com aungsanlin@gmail.com 02-36681/38618/36945 09-2016510