Post on 08-Sep-2014
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