The document summarizes problems involving calculating reaction forces at supports of structures using Lame's theorem and the principles of equilibrium. It provides 10 example problems showing the application of these principles to determine unknown reaction forces and loads. Diagrams accompany each problem showing the free body diagram and relevant forces and dimensions. Step-by-step solutions are provided for each problem applying the equations of equilibrium.

1. Engineering Mechanics
Faruque Abdullah
Lecturer
Dept. of Civil Engineering
Dhaka International University

2. Free Body

3. Moment:
The bending moment at a section of a structure is the algebraic sum of the
moments produced by all the external forces on one side of the section.
M1
M2
M3M4
P1
P2
P3
P4
P5 P5y
P5x
P6
P6y
P6x
P7
P8
∑ MA = 0
=> P1 x X - RB x L = 0
A B
X
RB
L
X
∑ MA = 0
=> -P2 x X - RB x L = 0
∑ MA = 0
=> P3 x X - RB x L = 0
X
∑ MA = 0
=> P4 x X - RB x L = 0

4. Types of Supports and Reaction Forces (2D)

5. Puzzle Clue
∑ Fx = 0
∑ Fy = 0
∑ M = 0
Lame’s Theory

6. Problem-1: Find the reaction forces at support C and D.
600
300
Q = 12 lb
D
B
C

7. Solution: According to Lame’s theory,
𝑅 𝐶
𝑆𝑖𝑛(900+300)
=
𝑅 𝐷
𝑆𝑖𝑛(900+600)
=
𝑄
𝑆𝑖𝑛(900)
=
12
1
=12
∴ 𝑅 𝐶 = 12 * 𝑆𝑖𝑛(900 + 300) = 6 3 lb
∴ 𝑅 𝐷 = 12 * 𝑆𝑖𝑛(900 + 600) = 6 lb
600300
Q = 12 lb
𝑅 𝐷
𝑅 𝐶
600300
𝑅 𝐶𝑅 𝐷
Q = 12 lb

8. Problem-2: Find the reaction forces at support A if the weight of joist is 20 lb
T
300
450
20 lb
A
RA
RAx
RA𝑦
T Sin300
B

9. Solution: ∑ MA = 0
=> 20 x 6 Cos 450 - T Sin300 x 12 = 0
=> T =
20 x 6 Cos 450
12 x Sin300
=> T = 10 2
6 Cos 450
T
300
450
20 lb
A
RA
RAx
RA𝑦
T Sin300
B

10. ∑ Fx = 0
=> T Cos150 - RAx = 0
=> RAx = 10 2 * Cos150 = 13.66 lb
∑ Fy = 0
=> T Sin150 + 20 - RA𝑦 = 0
=> RA𝑦 = 10 2 * Sin150 +20
=> RA𝑦 = 23.66 lb
𝑅 𝐴 = RAx
2
+ RA𝑦
2
= 13.662
+ 23.662
= 27.32 lb
T 300
450
20 lb
A
RA
RAx
RA𝑦
T Cos150
T Sin150
B
150

11. Problem-3: Find out the value of the forces T1, T2, T3 & the angle θ.
40 N 50 N
A
B
T1
T2
T3
350 θ

12. Solution: According to Lame’s theory,
𝑇1
𝑆𝑖𝑛(900)
=
𝑇2
𝑆𝑖𝑛(900+550)
=
40
𝑆𝑖𝑛(900+350)
= 48.8
∴𝑇1 = 48.83 * 𝑆𝑖𝑛(900
) = 48.83 N
∴𝑇2 = 48.83 * 𝑆𝑖𝑛(900 + 550) =28 N
40 N 50 N
A
B
T1
T2
T3
350
θ350𝑇1
40 N
550
𝑇2

13. Solution: According to Lame’s theory,
𝑇2
𝑆𝑖𝑛(900+900 − 𝜃)
=
𝑇3
𝑆𝑖𝑛(900)
=
50
𝑆𝑖𝑛(900+ 𝜃)
=>
28
𝑆𝑖𝑛𝜃
=
𝑇3
1
=
50
𝐶𝑜𝑠𝜃
Now,
28
𝑆𝑖𝑛𝜃
=
50
𝐶𝑜𝑠𝜃
or,
𝑆𝑖𝑛𝜃
𝐶𝑜𝑠𝜃
=
28
50
or, tanθ =
28
50
or, θ = 𝑡𝑎𝑛−1(
28
50
) = 29.240
∴ 𝑇3 =
50
𝐶𝑜𝑠(29.240)
* 1 = 57.3 N
40 N 50 N
A
B
T1
T2
T3
350
θ
𝑇3
50 N
900
− 𝜃
𝑇2
θ

14. Problem-4: Find out the reaction at A and B

15. ∑ Fy = 0
=> 𝑅 𝐴 + 𝑅 𝐵 - 40 = 0
=> 𝑅 𝐴 + 30 - 40 = 0
=> 𝑅 𝐴 = 10 N
Solution: ∑ MA = 0
=> 40 x
3
4
L - 𝑅 𝐵 x L = 0
=> 𝑅 𝐵 =
40 x 3
4
L
𝐿
=> 𝑅 𝐵 = 40 x
3
4
= 30 N

16. T
2
1
2′ 2′ 2′
200 lb 100 lb
A B
Problem-5: Find out the reaction at A and also find the value of the force T.

17. ∑ Fy = 0
=> 𝑅 𝐴𝑦 + T x
2
5
- 200 -100 = 0
=> 𝑅 𝐴𝑦 + 279.51 x
2
5
- 200 -100 = 0
=> 𝑅 𝐴𝑦 = 50 N
∑ Fx = 0
=> 𝑅 𝐴𝑥 - T x
1
5
= 0
=> 𝑅 𝐴𝑥 + 279.51 x
1
5
= 0
=> 𝑅 𝐴𝑥 = 125 N
Solution: ∑ MA = 0
=> 200 x 2 + 100 x 6 - T x
2
5
x 4 = 0
=> T =
400+600
2
5
x 4
=>T = 279.51 N
T x
2
5
T x
1
5
T
2
1
2′
2′ 2′
200 lb
100 lb
A
B
5
RAx
RA𝑦

18. Problem-6: Find out the reaction at A and also find the value of the force T.
30 N
A
B C
300

19. Solution: According to Lame’s theory,
𝑇
𝑆𝑖𝑛(900)
=
𝑅 𝐵
𝑆𝑖𝑛(900+ 600)
=
30
𝑆𝑖𝑛(900+300)
= 20 3
∴𝑇1 = 20 3 * 𝑆𝑖𝑛(900) = 20 3 N
∴𝑇2 = 20 3 * 𝑆𝑖𝑛(900 + 600) = 10 3 N
30 N
A
B C
300
30 N
T
𝑅 𝐵
C
600
300
T
30 N
600
𝑅 𝐵

20. Problem-7: Find out the reaction at A, B, C & F.
600D
E
100 N
100 N
A
B
CF

21. Solution: According to Lame’s theory,
𝑅 𝐶
𝑆𝑖𝑛(900+ 600)
=
𝑅 𝐹
𝑆𝑖𝑛(900+ 300)
=
100
𝑆𝑖𝑛(900)
= 100
∴𝑅 𝐶 = 100 * 𝑆𝑖𝑛(900
+ 600
) = 50 N
∴𝑅 𝐹 = 100 * 𝑆𝑖𝑛(900 + 300) = 50 3 N
600D
E
100 N
100 N
A
B
CF
𝑅 𝐶
100 N
300
𝑅 𝐹
600

22. 600D
E
100 N
100 N
A
B
CF
∑ Fy = 0
=> 𝑅 𝐵 x Sin300
- 𝑅 𝐹 x Sin600
-100 = 0
=> 𝑅 𝐵 x Sin300 -50 3 x Sin600 -100 = 0
=> 𝑅 𝐵 = 350 N
∑ Fx = 0
=> 𝑅 𝐵 x Cos300 + 𝑅 𝐹 x Cos 600 - 𝑅 𝐴 = 0
=> 𝑅 𝐴 = 350 x Cos300
+50 3 x Cos 600
=> 𝑅 𝐴𝑥 = 346.4 N
𝑅 𝐵
100 N
300
𝑅 𝐴
𝑅 𝐹
600

23. Problem-8: A roller weighting 2000 N rests on a inclined bar weighting 800 N as
shown in Figure. Assuming the weight of bar negligible, determine the reactions
at D and C and reaction in bar AB.
A B
300
C
D
E

24. A B
300
C
D
E
Solution: According to Lame’s theory,
𝑅 𝐴
𝑆𝑖𝑛(900+ 600)
=
𝑅 𝐸
𝑆𝑖𝑛(900)
=
2000
𝑆𝑖𝑛(900+300)
= 2309.4
∴𝑅 𝐴 = 2309.4 * 𝑆𝑖𝑛(900
+ 600
) = 1154.7 N
∴𝑅 𝐸 = 2309.4 * 𝑆𝑖𝑛(900) = 2309.4 N
𝑅 𝐸
2000 N
600
𝑅 𝐴

25. Solution:
∑ Fx = 0
=> 𝑅 𝐵Cos600 - 𝑅 𝐶𝑥 = 0
=>𝑅 𝐶𝑥 = 2309.4 x Cos600
= 1154.7 N
∑ MC = 0
=> 𝑅 𝐷 x 5 Cos300 - 800 x 2.5 Cos300 - 𝑅 𝐸 x 2 = 0
=>𝑅 𝐷 = 1466.67 N
∑ Fy = 0
=> 𝑅 𝐷 - 800 - 𝑅 𝐸 Sin600 + 𝑅 𝐶𝑦 = 0
=>𝑅 𝐶𝑦 = 1333.33 N
300
C
D
E
800 N
𝑅 𝐸
𝑅 𝐷
𝑅 𝐶𝑦
𝑅 𝐶𝑥600
𝑅 𝐸Cos600
𝑅 𝐸
2000 N
𝑅 𝐴
600

26. Problem-9: Determine the horizontal and vertical force components acting at the
pin connections B & C of the loaded frame as shown in the figure.
500 N
8′ 2′
4′
8′
2′
A
B
C
D
E F

27. Solution:
∑ MA = 0
=> 500 x 10 - 𝑅 𝐷 x 14 = 0
=> 𝑅 𝐷 = 357.14 N
∑ Fx = 0
=> 𝑅 𝐴𝑥 - 𝑅 𝐷 = 0
=> 𝑅 𝐴𝑥 = 𝑅 𝐷 = 357.14 N
∑ Fy = 0
=> 𝑅 𝐴𝑦 - 500 = 0
=> 𝑅 𝐴𝑦 = 500 N
500 N
8′ 2′
4′
8′
2′
A
B
C
D
E F
𝑅 𝐴𝑦
𝑅 𝐴𝑥
𝑅 𝐷

28. ∑ MC = 0
=> 500 x 10 - 𝑅 𝐸𝑦 x 8 = 0
=> 𝑅 𝐸𝑦 = 625 N
∑ Fx = 0
=> 𝑅 𝐶𝑥 - 𝑅 𝐸𝑥 = 0
=> 𝑅 𝐶𝑥 = 𝑅 𝐸𝑥
∑ Fy = 0
=> 𝑅 𝐶𝑦 + 𝑅 𝐸𝑦 - 500 = 0
=> 𝑅 𝐶𝑦 + 625 - 500 = 0
=> 𝑅 𝐶𝑦 = -125 N
500 N
C
E
F
𝑅 𝐶𝑥
𝑅 𝐶𝑦
𝑅 𝐸𝑥
𝑅 𝐸𝑦
8′ 2′

29. ∑ MB = 0
=> - 𝑅 𝐴𝑥 x 4 - 𝑅 𝐶𝑥 x 8 - 𝑅 𝐷 x 10 = 0
=> - 357.14 x 4 - 𝑅 𝐶𝑥 x 8 - 357.14x 10 = 0
=> 𝑅 𝐶𝑥 = - 625 N
∑ Fx = 0
=> 𝑅 𝐴𝑥 - 𝑅 𝐵𝑥 - 𝑅 𝐶𝑥 - 𝑅 𝐷 = 0
=> 𝑅 𝐵𝑥 = 357.14 – (-625) -357.14
=> 𝑅 𝐵𝑥 = 625 N
∑ Fy = 0
=> 𝑅 𝐴𝑦 - 𝑅 𝐵𝑦 - 𝑅 𝐶𝑦 = 0
=> 𝑅 𝐵𝑦 = 500 – (-125) = 625 N
4′
8′
2′
A
B
C
D
𝑅 𝐴𝑦
𝑅 𝐴𝑥
𝑅 𝐷
𝑅 𝐵𝑥
𝑅 𝐵𝑦
𝑅 𝐶𝑥
𝑅 𝐶𝑦

30. Problem-10: Determine the horizontal and vertical force components acting at
the pin connections A & C of the loaded frame as shown in the figure.
2′
8′
10 Kip-ft
A
B
C

31. Solution:
∑ MB = 0
=> - 10 - 𝑅 𝐶 x 4 = 0
=> 𝑅 𝐶 = - 1.67 N
∑ Fy = 0
=> 𝑅 𝐴𝑦 + 𝑅 𝐶 = 0
=> 𝑅 𝐴𝑦 = - 𝑅 𝐶 = - (-1.67) = 1.67 N
∑ MB = 0
=> 𝑅 𝐴𝑦 x 2 - 𝑅 𝐴𝑥 x 2 - 𝑅 𝐶 x 4 = 0
=> 1.67 x 2 - 𝑅 𝐴𝑥 x 2 - (-1.67) x 4 = 0
=> 𝑅 𝐴𝑥 = - 1.67 N
2′
4′
10 Kip-ft
A
B
C
𝑅 𝐶
𝑅 𝐴𝑦
𝑅 𝐴𝑥
4′
10 Kip-ft
B
2′
𝑅 𝐵𝑥
𝑅 𝐵𝑦
𝑅 𝐶
C

32. Truss

33. Assumption Made In Truss Frame Analysis:
 Members are connected at the joints through pin connections.
 All the members have only axial force i.e. either tensile or
compressive.
 All the external forces are applied only at the joints.

34. Problem-11: Determine all the member forces by Method of Joint Method.
6′
4 @ 8′
= 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3

35. 6′
4 @ 8′ = 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3
𝑅 𝐿4
𝑅 𝐿𝑜
Solution:
∑ ML0 = 0
=> 10 x 8 + 15 x 16 + 10 x 24 - 𝑅 𝐶 x 32
= 0
=> 𝑅 𝐿4 = 17.5 K = 𝑅 𝐿0
Free Joint of 𝐿0:
∑ Fy = 0
=> 𝐿0 𝑈1 x
6
10
+ 𝑅 𝐿0 = 0
=>𝐿0 𝑈1 = - 29.17 K (C)
∑ FX = 0
=> 𝐿0 𝑈1 x
8
10
+ 𝐿0 𝐿1 = 0
=> 𝐿0 𝐿1 = - (- 29.17 x
8
10
) = 23.33 K (T)
6′
8′
𝐿0
𝐿1
𝑅 𝐿0
𝑈1

36. 6′
4 @ 8′ = 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3
𝑅 𝐿4
𝑅 𝐿𝑜
Free Joint of 𝐿1:
∑ Fy = 0
=> 𝐿1 𝑈1 = 0
∑ FX = 0
=> 𝐿0 𝐿1 - 𝐿1 𝐿2 = 0
=>𝐿1 𝐿2= 𝐿0 𝐿1 = 23.33 K (T)
𝑈1
6′
8′
𝐿0 𝐿2
𝐿1
8′

37. 6′
4 @ 8′ = 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3
𝑅 𝐿4
𝑅 𝐿𝑜
Free Joint of 𝑈1:
∑ Fy = 0
=> - 𝐿0 𝑈1 x
6
10
- 10 - 𝐿2 𝑈1 x
6
10
= 0
=> 𝐿2 𝑈1 x
6
10
= - (-29.17) x
6
10
-10
=> 𝐿2 𝑈1 = 12.5 K (T)
∑ FX = 0
=> 𝑈1 𝑈2 + 𝐿2 𝑈1 x
8
10
- 𝐿0 𝑈1 x
8
10
= 0
=> 𝑈1 𝑈2 = -29.17 x
8
10
- 12.5 x
8
10
=> 𝑈1 𝑈2 = -33.34 K (C)
𝑈2
6′
8′
𝐿1
𝐿0 𝐿2
8′
𝑈1
10 K

38. 6′
4 @ 8′ = 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3
𝑅 𝐿4
𝑅 𝐿𝑜
Free Joint of 𝑈2:
∑ Fy = 0
=> 𝐿2 𝑈2 + 15 = 0
=> 𝐿2 𝑈2 = -15 K (C)
∑ FX = 0
=> 𝑈1 𝑈2 - 𝑈2 𝑈3 = 0
=> 𝑈2 𝑈3 = 𝑈1 𝑈2 = -33.34 K (C)
𝑈1
6′
8′
𝐿2
8′
15 K
𝑈2
𝑈3

39. 6′
4 @ 8′ = 32′
10 K
𝐿0
10 K15 K
𝐿1 𝐿2 𝐿3
𝐿4
𝑈1
𝑈2
𝑈3
𝑅 𝐿4
𝑅 𝐿𝑜
Free Joint of 𝐿2:
∑ Fy = 0
=> 𝐿2 𝑈1 x
6
10
+ 𝐿2 𝑈2 + 𝐿2 𝑈3 x
6
10
= 0
=> 𝐿2 𝑈3 = (- 12.5 x
6
10
- 15) x
10
6
=> 𝐿2 𝑈3 = -37.5 K (C)
∑ FX = 0
=> -𝐿2 𝑈1 x
8
10
- 𝐿1 𝐿2 + 𝐿2 𝑈3 x
8
10
+ 𝐿2 𝐿3= 0
=> 𝐿2 𝐿3 = 12.5 x
8
10
+ 23.33 – (-37.5) x
8
10
=> 𝐿2 𝐿3 = 63.33 K (T)
Proceed the same procedure to find out the
other member forces.
𝑈2
6′
8′
𝐿1 𝐿3
𝐿2
8′
𝑈3𝑈1

40. Problem-12: Determine the force in members DF, DE and CE of the stadium truss
shown.
2 K
1 K
2 K
2 K
A
B
C
D
E
F
G
I
K
M
H
J
L
N
8′8′ 6′
8′ 6′
6′
15′

41. 2 K
1 K
2 K
2 K
A
B
C
D
E
F
G
I
K
M
H
J
L
N
8′
8′ 6′8′ 6′
6′
15′
1
1
2 K
2 K
2 K
A
B
C
D
E
8′
8′
4′
1
F
1

42. Solution:
∑ MD = 0
=> -2 x 16 – 2 x 8 – c x 4 = 0
=> c = - 12 K (C).
∑ MA = 0
=> 2 x 8 + 2 x 16 + b x 16 = 0
=> b = - 3 K (C).
∑ ME = 0
=> a x
8
82+ 22
x 4 - 2 x 16 – 2 x 8 = 0
=> a = - 12.37 K (T). 8′
2′
2 K
2 K
2 K
A
B
C
D
E
8′8′
4′
Fa
b
c
1
1

43. Non Co-Planner Forces

44. X
Y
Z
X′
Y′
Z′

45. It has components along y and z axis
because the member does not require x-
axis to describe it’s position.
Components of reaction at point A is
𝑅 𝐴𝑦 & 𝑅 𝐴𝑧. Components of reaction at
point B is 𝑅 𝐵𝑦 & 𝑅 𝐵𝑧.
X
Y
Z
X′
Y′
Z′
A (4,0)
B (0,5)
𝑅 𝐴𝑧
𝑅 𝐴𝑦
𝑅 𝐵𝑧
𝑅 𝐵𝑦
5′

46. It has components along x, y and z axis
because the member requires x, y & z-axis
to describe it’s position.
Components of reaction at point A is 𝑅 𝐴𝑥,
𝑅 𝐴𝑦 & 𝑅 𝐴𝑧 . Components of reaction at
point B is 𝑅 𝐵𝑥, 𝑅 𝐵𝑦 & 𝑅 𝐵𝑧.
X
Y
Z
X′
Y′
Z′
A (2,4,0)
B (0,0,5)
𝑅 𝐴𝑧
𝑅 𝐴𝑦
𝑅 𝐵𝑦
𝑅 𝐵𝑧
5′
𝑅 𝐴𝑥
𝑅 𝐵𝑥

47. Problem-13: AE & CE steel frame is hanging with a cable DE. From E a load of 1000 lb
has been suspended. The dimension of AB = BC = BD = 6′
& BE = 8′
. Find the tension
in the cable and the force in each timber.
X
Y
Z
X′
Y′
Z′
1000 lb
6′
8′
E
A
B
C
D

48. Solution:
Components of DE,
𝑇𝑥 = T x
8
82+ 62
=
8
10
T
𝑇𝑦 = 0
𝑇𝑧 = T x
6
82+ 62
=
6
10
T
Taking moment about y-axis at A,
𝑇𝑧 x 8 – 1000 x 8 = 0
=>
6
10
T = 1000
=> T = 1666.67 lb.
XY
Z
1000 lb
E (8,0,0)
A (0,6,0)
B (0,0,0)
C (0,6,0)
D (0,0,6)
𝑅 𝐴𝑧
𝑅 𝐴𝑦
𝑅 𝐴𝑥
𝑅 𝐶𝑧
𝑅 𝐶𝑦
𝑅 𝐶𝑥
𝑅 𝐷𝑧
𝑅 𝐷𝑦
𝑅 𝐷𝑥
𝑇𝑧
𝑇𝑥
6′
8′

49. Components of AE,
𝑅 𝐴𝐸𝑥 = 𝑅 𝐴𝐸 x
8
82+ 62
=
8
10
𝑅 𝐴𝐸
𝑅 𝐴𝐸𝑦 = 𝑅 𝐴𝐸 x
6
82+ 62
=
6
10
𝑅 𝐴𝐸
𝑅 𝐴𝐸𝑧 = 0
Components of CE,
𝑅 𝐶𝐸𝑥 = 𝑅 𝐶𝐸 x
8
82+ 62
=
8
10
𝑅 𝐶𝐸
𝑅 𝐶𝐸𝑦 = 𝑅 𝐶𝐸 x
6
82+ 62
=
6
10
𝑅 𝐶𝐸
𝑅 𝐶𝐸𝑧 = 0
XY
Z
1000 lb
E (8,0,0)
A (0,6,0)
B (0,0,0)
C (0,6,0)
D (0,0,6)
𝑅 𝐴𝑧
𝑅 𝐴𝑦
𝑅 𝐴𝑥
𝑅 𝐶𝑧
𝑅 𝐶𝑦
𝑅 𝐶𝑥
𝑅 𝐷𝑧
𝑅 𝐷𝑦
𝑅 𝐷𝑥
𝑇𝑧
𝑇𝑥
6′
8′

50. ∑ Fy = 0
=> 𝑅 𝐶𝐸𝑦 - 𝑅 𝐴𝐸𝑦 = 0
=>
6
10
𝑅 𝐶𝐸 =
6
10
𝑅 𝐴𝐸
=> 𝑅 𝐶𝐸 = 𝑅 𝐴𝐸
∑ Fx = 0
=> 𝑅 𝐴𝐸𝑥 + 𝑅 𝐶𝐸𝑥 - 𝑇𝑥 = 0
=>
8
10
𝑅 𝐴𝐸 +
8
10
𝑅 𝐶𝐸 =
8
10
T
=> 2 x
8
10
𝑅 𝐴𝐸 =
8
10
T
=> 𝑅 𝐴𝐸 = 𝑅 𝐶𝐸 =
𝑇
2
=
1666.67
2
= 833.33 lb.
XY
Z
1000 lb
E (8,0,0)
A (0,6,0)
B (0,0,0)
C (0,6,0)
D (0,0,6)
𝑅 𝐴𝑧
𝑅 𝐴𝑦
𝑅 𝐴𝑥
𝑅 𝐶𝑧
𝑅 𝐶𝑦
𝑅 𝐶𝑥
𝑅 𝐷𝑧
𝑅 𝐷𝑦
𝑅 𝐷𝑥
𝑇𝑧
𝑇𝑥
6′
8′

51. Problem-14: The 1000 lb mass steel pole is supported by a ball and socket joint at A and
is retained by two cables shown. Determine the tensions in the cables and the total force
acting on the joint A.
X Y
Z
𝑇1
𝑇2
10′
10′
20′
16′
2000 lb
A

52. Solution:
Components of 𝑇1,
𝑇1𝑥 = 𝑇1 x
12
122+ 162+ 302
=
12
10 13
𝑇1
𝑇1𝑦 = 𝑇1 x
16
122+ 162+ 302
=
16
10 13
𝑇1
𝑇1𝑧 = 𝑇1 x
30
122+ 162+ 302
=
30
10 13
𝑇1
Components of 𝑇2,
𝑇2𝑥 = 0
𝑇2𝑦 = 𝑇2 x
20
202+ 202
=
20
20 2
𝑇2 =
𝑇2
2
𝑇2𝑧 = 𝑇2 x
20
202+ 202
=
20
20 2
𝑇2 =
𝑇2
2
X Y
Z
𝑇1
𝑇2
10′
10′
20′
16′
2000 lb
A
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
1000 lb
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

53. Taking moment about y-axis at A,
∑ MA−y = 0
=> 2000 x 40 - 𝑇1𝑥 x 30 = 0
=> 2000 x 40 -
12
10 13
𝑇1x 30 = 0
=> 𝑇1 = 8013 lb
Taking moment about x-axis at A,
∑ MA−𝑥 = 0
=> 𝑇1𝑦 x 30 - 𝑇2𝑧 x 20= 0
=>
16
10 13
𝑇1 x 30 -
𝑇2
2
x 20 = 0
=> 𝑇2 = 7543 lb
X Y
Z
𝑇1
𝑇2
10′
10′
20′
16′
2000 lb
A
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
1000 lb
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

54. ∑ F 𝑍 = 0
=> 1000 + 𝑇1𝑧 + 𝑇2𝑧 - 𝑅 𝐴𝑧= 0
=> 1000 +
13
10 13
𝑇1 +
𝑇2
2
- 𝑅 𝐴𝑧= 0
=> 𝑅 𝐴𝑧 = 1000 +
30
10 13
x 8013 +
7543
2
=> 𝑅 𝐴𝑧 = 13001 lb.
∑ F 𝑋 = 0
=> 𝑅 𝐴𝑥 - 𝑇1𝑥 + 2000= 0
=> 𝑅 𝐴𝑥 =
12
10 13
𝑇1 - 2000
=> 𝑅 𝐴𝑥 =
12
10 13
x 8013 - 2000
=> 𝑅 𝐴𝑥 = 666.67 lb
X Y
Z
𝑇1
𝑇2
10′
10′
20′
16′
2000 lb
A
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
1000 lb
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

55. ∑ FY = 0
=> 𝑅 𝐴𝑦 - 𝑇2𝑦 + 𝑇1𝑦= 0
=> 𝑅 𝐴𝑦 =
𝑇2
2
-
16
10 13
𝑇1
=> 𝑅 𝐴𝑦 =
7543
2
-
16
10 13
x 8013
=> 𝑅 𝐴𝑦 = 1778 lb.
∴ 𝑅 𝐴 = 𝑅 𝐴𝑥
2
+ 𝑅 𝐴𝑦
2
+ 𝑅 𝐴𝑧
2
= 666.672 + 17782 + 130012
= 13139 lb.
X Y
Z
𝑇1
𝑇2
10′
10′
20′
16′
2000 lb
A
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
1000 lb
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

56. Problem-15: A 10 m pile is acted upon by a force of 8.4 kN. It is held by a ball and
socket at A and by the two cables BD and BE. Neglecting the weight of the pole,
determine the tension in each cable and the reaction at A.
X Y
Z
𝑇1
𝑇2
7 m
8.4 kN
A
3 m
B
C
D
E

57. Solution:
Components of 𝑇1,
𝑇1𝑥 = 𝑇1 x
6
62+ 62+ 72
=
6
11
𝑇1
𝑇1𝑦 = 𝑇1 x
6
62+ 62+ 72
=
6
11
𝑇1
𝑇1𝑧 = 𝑇1 x
7
62+ 62+ 72
=
7
11
𝑇1
Components of 𝑇2,
𝑇2𝑥 = 𝑇1 x
6
62+ 62+ 72
=
6
11
𝑇1
𝑇2𝑦 = 𝑇2 x
6
62+ 62+ 72
=
6
11
𝑇2
𝑇2𝑧 = 𝑇2 x
7
62+ 62+ 72
=
7
11
𝑇2
X Y
Z
𝑇1
𝑇2
7 m
8.4 kN
A
3 m
B
C
D
E
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
𝑇2𝑥
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

58. Taking moment about y-axis at A,
∑ MA−y = 0
=> 𝑇1𝑥 x 7 - 𝑇2𝑥 x 7 = 0
=>
6
11
𝑇1-
6
11
𝑇2= 0
=> 𝑇1 = 𝑇2
Taking moment about x-axis at A,
∑ MA−𝑥 = 0
=> 𝑇1𝑦 x 7 + 𝑇2𝑦 x 7 - 8.4 x 10= 0
=>
6
11
𝑇1 x 7 +
6
11
𝑇2x 7 -84 = 0
=> 2 x
6
11
x 7 x 𝑇1 = 84
=> 𝑇1 = 11 kN = 𝑇2.
X Y
Z
𝑇1
𝑇2
7 m
8.4 kN
A
3 m
B
C
D
E
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
𝑇2𝑥
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

59. ∑ F 𝑍 = 0
=> 𝑇1𝑧 + 𝑇2𝑧 - 𝑅 𝐴𝑧= 0
=>
7
11
𝑇1+
7
11
𝑇2 - 𝑅 𝐴𝑧= 0
=> 𝑅 𝐴𝑧 = 2 x
7
11
x 11
=> 𝑅 𝐴𝑧 = 14 lb.
∑ F 𝑋 = 0
=> 𝑅 𝐴𝑥 - 𝑇1𝑥 + 𝑇2𝑥= 0
=> 𝑅 𝐴𝑥 = 0
X Y
Z
𝑇1
𝑇2
7 m
8.4 kN
A
3 m
B
C
D
E
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
𝑇2𝑥
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

60. ∑ FY = 0
=> 𝑅 𝐴𝑦 + 𝑇2𝑦 + 𝑇1𝑦 - 8.4 = 0
=> 𝑅 𝐴𝑦 =
6
11
𝑇2 +
6
11
𝑇1 - 8.4
=> 𝑅 𝐴𝑦 = 2 x
6
11
x 11 – 8.4
=> 𝑅 𝐴𝑦 = 3.6 lb.
∴ 𝑅 𝐴 = 𝑅 𝐴𝑥
2
+ 𝑅 𝐴𝑦
2
+ 𝑅 𝐴𝑧
2
= 02 + 3.62 + 142
= 14.46 lb.
X Y
Z
𝑇1
𝑇2
7 m
8.4 kN
A
3 m
B
C
D
E
𝑇1 𝑧
𝑇1𝑥
𝑇1𝑦
𝑇2𝑧
𝑇2𝑦
𝑇2𝑥
𝑅 𝐴𝑧
𝑅 𝐴𝑥
𝑅 𝐴𝑦

61. Centroid

62. Centroid: The center of mass of a geometric object of uniform density.

63. Derivation of the location of the centroids of a plane triangle:
𝑌𝐴 = 𝑌𝑑𝐴 = 𝑦 𝑥2 − 𝑥1 𝑑𝑦
∆ CEF & ∆ CAB similar,
𝑥2 − 𝑥1
𝑏
=
ℎ−𝑦
ℎ
=> 𝑥2 − 𝑥1 =
𝑏
ℎ
(h-y)
𝑌𝐴 = 0
ℎ 𝑏
ℎ
(h−y) ydy =
𝑏
ℎ 0
ℎ
(hy−𝑦2
) dy
=
𝑏
ℎ
[ℎ
𝑦2
2
−
𝑦3
3
]0
ℎ
=
𝑏
ℎ
[ℎ
ℎ2
2
−
ℎ3
3
] =
𝑏
ℎ
x
ℎ3
3
=
𝑏ℎ2
6
𝑌 =
𝑌𝐴
𝐴
=
𝑏ℎ2
6
x
2
𝑏ℎ
=
ℎ
3
A
B
C
dy
𝑥1
𝑥2
h
X
Y
Z
b
y
D E
𝑋 =
𝑋𝐴
𝐴
=
𝑏
2
A =
1
2
bh

64. b
h
𝑋 =
𝑏
2
𝑌 =
ℎ
2
X
Y
X
Y
𝑋 = r 𝑌 =
4𝑟
3𝜋
d = 2r

65. X
Y
𝑋 =
4𝑟
3𝜋
𝑌 =
4𝑟
3𝜋
r
X
Y
α
𝑋 =
2
3
𝑟 𝑆𝑖𝑛∝
∝
r

66. Problem-16: Find out the centroid of the following object.
2.5 m 5 m 2.5 m 2.5 m
5 m
5 m
X
Y

67. Segment Area (𝑚2
) 𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3
) A 𝑦 (𝑚3
)
1 𝜋𝑟2
2
=
𝜋 𝑥 2.52
2
= 9.82 -2.5 - 5 -
4𝑟
3𝜋
= -8.56 0 -84.06 0
2 10 x 5 = 50 -2.5 0 -125 0
3 0.5 x 5 x 5 = 12.5 0 2.5 +
ℎ
3
= 4.17 0 52.125
∑ = 72.32 ∑ = - 209.06 ∑ = 52.125
2.5 m 5 m 2.5 m 2.5 m
5 m
5 m
X
Y
𝑋 =
𝐴 𝑥
𝐴
=
−209.06
72.32
= - 2.89 m 𝑌 =
𝐴 𝑦
𝐴
=
52.125
72.32
= 0.72 m
3
21
Solution:

68. Problem-17: Find out the centroid of the following object.
Y
12 m
24 m
12 m
600600
X
24 x Sin600
= 12 3 m

69. Segment Area (𝑚2
) 𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3
) A 𝑦 (𝑚3
)
Half Circle 𝜋𝑟2
2
=
𝜋 𝑥 242
2
= 904.78 -
4𝑟
3𝜋
= - 10.19 0 - 9219.71 0
Triangle 0.5 x 24 x 12 3 = - 249.42 -
12 3
3
= - 4 3 0 + 1728.03 0
∑ = 655.36 ∑ = - 7491.68 ∑ = 0
𝑋 =
𝐴 𝑥
𝐴
=
− 7491.68
655.36
= - 11.43 m 𝑌 =
𝐴 𝑦
𝐴
= 0 m
Y
12 m
24 m
12 m
600600
X
24 x Sin600
= 12 3 m
Solution:

70. Problem-18: Find out the centroid of the following object.
2 m
8 m
6 m
2 m
9 m
1
2 3
Y
X

71. Segment Area (𝑚2
) 𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3
) A 𝑦 (𝑚3
)
1 -
𝜋𝑟2
2
= -
𝜋 𝑥 42
2
= - 25.13 -
4𝑟
3𝜋
= - 1.7 6 + 2 + 4 = 12 42.72 - 301.56
2 9 x 12 = 108 -
9
2
= - 4.5 6 +
12
2
= 12 -486 1296
3 0.5 x 9 x 6 = 27 -
2
3
x 9 = - 6
2
3
x 6 = 4 -162 108
∑ = 109.87 ∑ = - 600.28 ∑ = 1102.44
𝑋 =
𝐴 𝑥
𝐴
=
− 600.28
109.87
= - 5.46 m 𝑌 =
𝐴 𝑦
𝐴
=
1102.44
109.87
= 10.03 m
2 m
8 m
6 m
2 m
9 m
1
2
3
Y
X
Solution:

72. Problem-19: Find out the centroid of the following object.
600
Y
X
600

73. Segment Area (𝑚2) 𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3) A 𝑦 (𝑚3)
Arc
OAB
𝜋𝑟2
6
=
𝜋 𝑥 62
6
= 18.85
2
3
x
4 𝑥 𝑆𝑖𝑛300
30 𝑥
𝜋
180
= 3.82 0 72 0
Arc
OCD
-
𝜋𝑟2
6
= -
𝜋 𝑥 42
6
= - 8.38
2
3
x
4 𝑥 𝑆𝑖𝑛300
30 𝑥
𝜋
180
= 2.54 0 - 21.29 0
∑ = 10.47 ∑ = 50.71 ∑ =0
𝑋 =
𝐴 𝑥
𝐴
=
50.71
10.47
= 4.84m 𝑌 =
𝐴 𝑦
𝐴
= 0 m
Solution:
300
Y
X
300
O
A
D
C
B

74. Problem-20: Find out the centroid of the following object.
16" 20"
4"
4"
12"
6"
8"10"
Y
X
A
F
H
B
E
D
C
G
J
I

75. Segment Area (𝑖𝑛2) 𝑥 (in) A 𝑥 (𝑖𝑛3)
AKMB 24 x 16 = 384 -12 -
24
2
= - 24 - 9216
AEB - 0.5 x 16 x 10 = - 80 - 26 -
2
3
x 10 = - 32.67 2613.6
CFK & DLM 2 x 0.5 x 6 x 4 = 24 - 12 -
6
3
= - 14 - 336
FGJM 12 x 28 = 336 -
12
2
= - 6 - 2016
HIN -
𝜋𝑟2
2
= -
𝜋 𝑥 102
2
= - 157.08 -
4𝑟
3𝜋
= - 4.24 666.02
∑ = 506.92 ∑ = 8288.38
16" 20"
4"
4"
12"
6"
8"10"
Y
X
A
F
H
B
E
D
C
G
J
I
L
K
M
N
X =
Ax
A
=
8288.38
506.92
= 16.35 ft.
For Symmetry,
Y = 0 ft.
Solution:

76. Problem-21: Find out the centroid of the following object.
X
A B
D C
G
Y
4"
4"
4" 6"
E
F

77. X
A B
D C
G
Y
4"
4"
4" 6"
E
F
Segment Area (𝑖𝑛2
) 𝑥 (in) A 𝑥 (𝑖𝑛3
)
ABCD 4 X 8 = 32 -2 - 64
ABE -
𝜋𝑟2
2
= -
𝜋 𝑥 22
2
= - 6.28 -2 12.56
CDF -
𝜋𝑟2
2
= -
𝜋 𝑥 22
2
= - 6.28 -2 12.56
BCG 0.5 x 6 x 8 = 24 6
3
= 2 48
∑ = 43.44 ∑ = 9.12
X =
Ax
A
=
9.12
43.44
= 0.21 in.
For Symmetry,
Y = 0 ft.
Solution:

78. Problem-22: Find out the centroid of the following object.
A
B
D
C
E
X
Y
4" 3"
2" 3"
2.5"
2.5"
F

79. A
B D
C
E
X
Y
4" 3"
2" 3"
2.5"
2.5"
F
Segment Area (𝑖𝑛2) 𝑥 (in) A 𝑥 (𝑖𝑛3)
ABC 0.5 X 3 X 5 = 7.5 4 +
2
3
X 3 = 6 45
BDEC 5 X 5 = 25 4 + 3 + 2.5 = 9.5 237.5
DEF - 0.5 X 3 X 5 = - 7.5 4 + 5 +
2
3
X 3 = 11 - 82.5
∑ = 25 ∑ = 200
X =
Ax
A
=
200
25
= 8 in.
For Symmetry,
Y = 0 ft.
Solution:

80. Problem-23: Find out the centroid of the following object.
X
Y
40 mm
80 mm
60 mm
120 mm

81. Segment Area (𝑚𝑚2) 𝑥 (mm) 𝑦 (mm) A 𝑥 (𝑚𝑚3) A 𝑦 (𝑚𝑚3)
Semi-Circle 𝜋𝑟2
2
=
𝜋 𝑥 602
2
= 5754.87 60 80 +
4𝑟
3𝜋
=
105.46
345292 606909
Full-Circle -𝜋𝑟2 = -𝜋 𝑥 402 = -5026.55 60 80 - 301593 - 402124
Rectangle 120 x 80 = 9600 60 40 576000 384000
Triangle 0.5 x 120 x 60 = 3600 120
3
= 40 -
60
3
= - 20 144000 - 72000
∑ = 13928 ∑ = 763699 ∑ = 516785
X
Y
40 mm
80 mm
60 mm
120 mm
𝑋 =
𝐴 𝑥
𝐴
=
763699
13928
= 54.83 mm
𝑌 =
𝐴 𝑦
𝐴
=
516785
13928
= 37.10 mm
Solution:

82. Problem-24: Find out the centroid of the following object.
60 mm
60 mm
80 mm
300 mm
400 mm
100 mm
X
Y

83. Segment Area (𝑚𝑚2) 𝑦 (mm) A 𝑦 (𝑚𝑚3)
1 100 x 60 = 6000 30 180000
2 280 x 80 = 22400 60 + 140 = 200 4480000
3 300 x 60 = 18000 400 – 30 = 370 6660000
∑ = 46400 ∑ = 11320000
For Symmetry,
X = 0 mm.
Y =
AY
A
=
11320000
46400
= 244 mm.
y =
60 𝑥 100 𝑥 30+80 𝑥 400−60−60 𝑥 60+
400−60−60
2
+60 𝑥 300 𝑥 [400−30]
60 𝑥 300+80 𝑥 400−60−60 +60 𝑥 100
= 244 mm
60 mm
60 mm
80 mm
300 mm
280mm
100 mm X
2
1
3
400mm
YSolution:

84. Moments of Inertia

85. Moments of Inertia: It is defined as the sum of second moment of area of individual
sections about an axis.
X
Y
𝑋1
𝑋3
𝑎1
𝑋2
𝑌3
𝑌1
𝑌2
𝑎2
𝑎3Total Area, A
Let, 𝑎1, 𝑎2 & 𝑎3 = Small element areas
in total area (A).
Taking moments of all areas about x-
axis once,
𝐼 𝑥 = 𝑎1 𝑦1 + 𝑎2 𝑦2 + 𝑎3 𝑦3
Similarly, Taking moments of area again
about y-axis,
𝐼 𝑥 = 𝑎1 𝑦1
2
+ 𝑎2 𝑦2
2
+ 𝑎3 𝑦3
2
𝐼 𝑥 = ∑ a𝑦2
Similarly,
𝐼 𝑦 = ∑ a𝑥2

86. Derivation of the moment of inertia of a rectangle with respect to centre of gravity of
the rectangle.
dA = bdy
Ix =
−
ℎ
2
+
ℎ
2
𝑦2 𝑑𝐴 =
−
ℎ
2
ℎ
2
𝑦2 𝑏𝑑𝑦 = b
−
ℎ
2
ℎ
2
𝑦2 𝑑𝑦
= b x [
𝑦3
3
]
−
ℎ
2
ℎ
2
=
𝑏
3
x [
ℎ3
8
- (-
ℎ3
8
)]
=
𝑏
3
x
ℎ3
4
=
𝑏ℎ3
12
∴ Ix =
𝑏ℎ3
12
X
A
B C dy
h
Y
b
D
y

87. Y
dA = hdx
Iy =
−
𝑏
2
+
𝑏
2
𝑥2 𝑑𝐴 =
−
𝑏
2
𝑏
2
𝑥2ℎ𝑑𝑥 = h
−
𝑏
2
𝑏
2
𝑥2 𝑑𝑥
= b x [
𝑥3
3
]
−
𝑏
2
𝑏
2
=
𝑏
3
x [
𝑏3
8
- (-
𝑏3
8
)]
=
ℎ
3
x
𝑏3
4
=
ℎ𝑏3
12
∴ Iy =
ℎ𝑏3
12
A
B C
h
X
b
D
dx
x
Derivation of the moment of inertia of a rectangle with respect to centre of gravity of
the rectangle.

88. Y
h
X
b
Y
h
X
b
Origin passes through the c.g. of the rectangle:
Ix =
𝑏ℎ3
12
& Iy =
ℎ𝑏3
12
Origin passes through the corner of the rectangle:
Ix =
𝑏ℎ3
3
& Iy =
ℎ𝑏3
3

89. h
b
X
Y
h
b
X
Y
Origin passes through the c.g. of the triangle:
Ix =
𝑏ℎ3
36
& Iy =
ℎ𝑏3
36
Origin passes through the vertex of triangle:
Ix =
𝑏ℎ3
12
& Iy =
𝑏ℎ3
12

90. Origin passes through the c.g. of the circle:
Ix = Iy =
𝜋𝑟4
4X
Y
r

91. X
Y
𝑋 = r 𝑌 =
4𝑟
3𝜋
r
Origin passes through the centre of the circle:
Ix = Iy =
𝜋𝑟4
8

92. X
Y
𝑋 =
4𝑟
3𝜋
𝑌 =
4𝑟
3𝜋
r
Origin passes through the centre of the circle:
Ix = Iy =
𝜋𝑟4
16

93. 𝑋
𝑌
r
X
Y
Origin passes through the vertex of the quarter
circular spandrel:
Ix = (1 -
5𝜋
16
) 𝑟4 & Iy =(
1
3
-
𝜋
16
) 𝑟4

94. Parallel Axis Theorem:
Origin of axis passes through any ordinate:
Ix = Ix𝑐 + A𝑑2
Iy = Iy𝑐 + A𝑑2
Where,
Ix = Moment of inertia of the object w.r.to the given axis.
Ix𝑐 = Moment of inertia w.r.to centroid of the object.
A = Area of the object.
d = center to center (c/c) distance between the axis and the cantorial axis.

95. Moment of inertial of the rectangle w.r.to the axis X′ & Y′:
IX′ =
𝑏ℎ3
12
+ A𝑑1
2
IY′ =
ℎ𝑏3
12
+ A𝑑2
2
Y
h
X
b
X′
Y′
𝑑1
𝑑2

96. Problem-25: Find out the moment of inertia of the following object w.r.to the x-axis
& y-axis.
X
Y
120 mm
120 mm
120mm
120mm
30 mm
30 mm
30 mm
30 mm

97. For the square section,
Ix =
𝑏ℎ3
12
=
300 𝑥 3003
12
= 675 x 106
𝑚𝑚4
Iy =
ℎ𝑏3
12
=
300 𝑥 3003
12
= 675 x 106 𝑚𝑚4
Solution:
X
120 mm
120 mm
120mm
120mm
30 mm
30 mm
30 mm
30 mm
For the blank square section,
Ix =
𝑏ℎ3
12
+ A𝑑1
2
=
180 𝑥 1203
12
+ (180 x 120) x 902= 200.88 x 106 𝑚𝑚4
Iy =
ℎ𝑏3
12
+ A𝑑2
2
=
120 𝑥 1803
12
+ (180 x 120) x 902= 233.28 x 106 𝑚𝑚4
∴ Ix = 675 x 106 - 2 x 200.88 x 106 = 273.24 x 106 𝑚𝑚4
∴ Iy = 675 x 106 - 2 x 233.28 x 106 = 208.44 x 106 𝑚𝑚4
30 + 60 = 90
30 + 60 = 90
𝑑1
𝑑2

98. 3 m6 m3 m
3 m
Y
X
Problem-26: Find out the moment of inertia of the following object w.r.to the x-axis
& y-axis. Also find out the radius of gyration.

99. Solution:
For the semi-circular section,
Ix = Iy =
𝜋𝑟4
8
=
𝜋 𝑥 64
8
= 508.94 𝑚4
For the triangular section,
Ix = Iy =
𝑏ℎ3
12
=
6 𝑥 33
12
= 13.5 𝑚4
∴ Ix = Iy = 508.94 – 13.5 = 495.44 𝑚4
A =
𝜋𝑟2
2
- 0.5 x b x h =
𝜋 𝑥 62
2
- 0.5 x 6 x 3 = 47.55 𝑚2
Radius of gyration, r =
𝐼 𝑚𝑖𝑛
𝐴
=
495.44
47.55
= 3.22 m.
3 m6 m3 m
3 m
Y
X

100. Problem-27: Find out the moment of inertia of the following object w.r.to the x-axis
& y-axis. Also find out the radius of gyration.
Y
5"
X
10"
6"2"
6"8"

101. Solution: For the semi-circular section,
Ix = Iy =
𝜋𝑟4
8
=
𝜋 𝑥 54
8
= 245.44 𝑖𝑛4
For the rectangular section,
Ix =
𝑏ℎ3
3
=
10 𝑥 63
3
= 720 𝑖𝑛4
Iy =
ℎ𝑏3
12
=
6 𝑥 103
12
= 500 𝑖𝑛4
For the triangular section,
Ix =
𝑏ℎ3
12
+ A𝑑2 =
6 𝑥 63
12
+ 0.5 x 6 x 6 x 22 = 396 𝑖𝑛4
Iy =
ℎ𝑏3
36
=
6 𝑥 63
36
= 36 𝑖𝑛4
Y
5"
X
10"
6"2"
6"
6"

102. Y
5"
X
10"
6"2"
6"
6"
∴ Ix = 245.44 + 720 - 396 = 569.44 𝑖𝑛4
∴ Iy = 245.44 + 500 - 36 = 709.44 𝑖𝑛4
A =
𝜋𝑟2
2
+ b x h - 0.5 x b x h =
𝜋 𝑥 52
2
+ 10 x 6 - 0.5 x 6 x 6 = 81.27 𝑖𝑛2
∴ Radius of gyration, r =
𝐼 𝑚𝑖𝑛
𝐴
=
569.44
81.27
= 2.64 in.

103. Problem-28: Find out the moment of inertia of the following object w.r.to the x-axis.
3"
8"
6"
X

104. Solution:
For the semi-circular section,
Ix =
𝜋𝑟4
8
- A𝑑1
2
=
𝜋 𝑥 34
8
-
𝜋 𝑥 32
2
x (
4 𝑥 3
3 𝑥 𝜋
)2
= 8.87 𝑖𝑛4
Ix = Ix + A𝑑2
2
= 8.87 +
𝜋 𝑥 32
2
x (8 −
4 𝑥 3
3 𝑥 𝜋
)2 = 648.57 𝑖𝑛4
For the rectangular section,
Ix =
𝑏ℎ3
3
=
6 𝑥 83
3
= 1024 𝑖𝑛4
For the triangular section,
Ix =
𝑏ℎ3
12
=
6 𝑥 63
12
= 108 𝑖𝑛4
∴ Ix = 1024 + 108 – 648.57 = 483.43 𝑖𝑛4
3"
8"
6"
X

105. Problem-29: Find out the moment of inertia of the following object w.r.to the
x-axis.
60 mm
60 mm
80 mm
300 mm
400 mm
100 mm
X

106. Segment Area (𝑚𝑚2) 𝑦 (mm) A 𝑦 (𝑚𝑚3)
1 100 x 60 = 6000 30 180000
2 280 x 80 = 22400 60 + 140 = 200 4480000
3 300 x 60 = 18000 400 – 30 = 370 6660000
∑ = 46400 ∑ = 11320000
Y =
AY
A
=
11320000
46400
= 244 mm.
60 mm
60 mm
80 mm
300 mm
280mm
100 mm X
2
1
3
400mm
YSolution:
I =
𝑏1ℎ1
3
12
+ 𝐴1 𝑑1
2
+
𝑏2ℎ2
3
12
+ 𝐴2 𝑑2
2
+
𝑏3ℎ3
3
12
+ 𝐴3 𝑑3
2
=
100 𝑥 603
12
+ (60 x 100) x (244−30) 2
+
80 𝑥 2803
12
+ (80 x 280) x (200−244) 2
+
300 𝑥 603
12
+ (60 x 300) x (400−244−30) 2
= 75.7 x 107 𝑚𝑚4

107. Cables

108. Problem-30: A light cable weighing 40 lb is attached to a support at A and passes over a
small pulley at B and supports a load P. Find out the load P, The slope of the cable at B and
total length of the cable from A to B. Neglect the weight of the portion of cable from B to P.
P
80′
1′
A B

109. P
80′
1′
A B
Solution:
w =
40
80
= 0.5 lb/ft
L = 80 ft
d = 1 ft
𝐹′
Sinθ =
𝑤𝐿
2
=
0.5 𝑥 80
2
= 20 lb
𝐹′Cosθ = Q =
𝑤𝐿2
8𝑑
=
0.5 𝑥 802
8 𝑥 1
= 400 lb
𝐹′
= P = 𝐹′Sinθ2 + 𝐹′Cosθ2
= 202 + 4002 = 400.5 lb
𝜃 = tan−1(
F′Sinθ
F′Cosθ
)
= tan−1(20/400) = 2.860
Total length of the cable from A to B
S = L +
8 𝑑2
3 𝐿
-
32 𝑑4
5 𝐿3
= 80 +
8 𝑥 12
3 𝑥 80
-
32 𝑥 14
5 𝑥 803
= 80.033 ft.

110. Problem-31: Cable AB supports a load distributed uniformly along the horizontal.
Determine the maximum and minimum values of the tension in the cable.
15′
30′
A
B
200′
50 lb/ft.

111. Solution:
∑ MA = 0
=> 50 a x
𝑎
2
- Q x 15 = 0
=> Q =
5
3
𝑎2
∑ M 𝐵 = 0
=> 50 a x
𝑎
2
- Q x 15 = 0
=> Q =
5
3
𝑎2
15′
30′
A
B
200′
50 lb/ft.
Qa
F′
50 a
A
15′

112. ∑ MB = 0
=> 50 (200 – a) x
(200 − 𝑎)
2
- Q x 45 = 0
=>
5
3
𝑎2 x 45 = 25 x (200 − 𝑎)2
=> a = 73.21 ft.
Minimum Tension, Q =
5
3
𝑎2
=
5
3
x 73.212
= 8933 lb
F1
′
Sin𝜃
F1
′
Cos𝜃
𝐹1
′
𝜃
B
Q
30′
(200 – a)
50 (200 – a)
∑ FX = 0
=> F1
′
Cos𝜃 - Q= 0
=> F1
′
Cos𝜃 = Q = 8933 lb.
∑ FY = 0
=> F1
′
Sin𝜃 – 50 (200-a) = 0
=> F1
′
Sin𝜃 = 50 (200 – 73.21)
=> F1
′
Sin𝜃 = 6340 lb
∴ Maximum Tension,
𝐹1
′
= F1
′
Sin𝜃2 + F1
′
Cos𝜃2
= 89332 + 63402
= 10953 lb.

113. Problem-32: Determine the equation of the curve assumed by the cable. T0 is the
tension in the left joint.
A
B
b
h
𝑤0 - kx
x

114. Solution:
Given that, w = w0 - kx
When, x = 0, w = w0
When, x = b, w = 0
Then, w0 = w0 - kx
& 0 = w0 - kb
or, k =
w0
b
From, equation 1 we get,
w = w0 -
w0
b
x = w0 (1 -
x
b
)
T0
b
T
Bh
w
b
3
2
3
b
∑ MB = 0
=> T0 x h -
𝑤𝑏
2
x
2
3
b = 0
=> T0 =
w0 (1 − x
L
) 𝑏2
3ℎ
We know, y =
𝑤𝑥2
2T0
=
w0 (1 − x
b
) 𝑥2 ∗3ℎ
2w0 (1 − x
b
) 𝑏2
=
3ℎ
2
(
𝑥
𝑏
)2

115. Problem-33: If the slope of the cable at point A is zero, determine the deflection
curve and the maximum tension developed in the cable.
A
B
L
h
𝑤0
x

116. Solution:
∑ MB = 0
=> T0 x h -
w0 𝐿
2
x
L
3
= 0
=> T0 =
w0 𝐿2
6 ℎ
∑ FX = 0
=> 𝐹𝐵𝑥 - T0 = 0
=> 𝐹𝐵𝑥 =
w0 𝐿2
6 ℎ
∑ FY = 0
=> 𝐹𝐵𝑦 -
w0 𝐿
2
= 0
=> 𝐹𝐵𝑦 =
w0 𝐿
2
∴ Maximum tension, 𝑇 𝑚𝑎𝑥 = 𝐹𝐵𝑥
2
+ 𝐹𝐵𝑦
2
= (
w0 𝐿2
6 ℎ
)2+ (
w0 𝐿
2
)2 =
w0 𝐿
2
(
𝐿
3ℎ
)2+1
T0
L
𝐹𝐵
Bh
w0 𝐿
2
L
3
2
3
L
𝐹𝐵𝑥
𝐹𝐵𝑦

117. Thank you
For
Taking the Stress