1. ENGINEERING MECHANICS
ELEMENTS OF CIVIL ENGINEERING
AND
ENGINEERING MECHANICS
SLIDES OF ALL CHAPTERS

3. RESOLUTION F
F Sin θ θ
F cos θ
F cos 0 F Sin 0

4. o θ Resolution of a FORCE A F sin θ F θ B Sin θ =Opp / Hyp = Fy / F
F y = F Sin θ
F cos θ A F
Cos θ = Adj / Hyp
= Fx / F
F x = F Cos θ
Hyp
Opp
F y θ O
Adj
F x B
Reverse Combination = √ ( F sin θ ) ² + ( F cos θ )²
= √ F² ( sin² θ + cos² θ ) = F

5. 75 N
40 N
15 N
35 N
50 N
25 N
40 N
20 N
50 N
30 N
45 N
55 N
∑ F x = – 25 – 50 – 35
= 0
∑ F y = – 45 – 55
= 0

6. +, + - , + - , - + , - Sign Conventions I st Quadrant 2 nd Quadrant
F2 sin θ
I st Quadrant
F1 sin θ
2 nd Quadrant
F2 cos
F1 cos θ θ
F4 cos
F3 cos θ θ
3 rd Quadrant
F3 sin θ
4 th Quadrant
F4 sin θ
+ , -
- , -

7. - 100 cos 60 - 200 cos 65 - 200 sin 65 - 150 sin 25 Y-axis 250 N 100 N45
100 cos 60
250 cos 45
X-axis
150 cos 25
200 cos 65 25 65 25
150 N
200 sin 65
150 sin 25
200 N
- 100 cos 60
- 200 cos 65
∑ F x =
+ 250 cos 45
+ 150 cos 25
= N
- 200 sin 65
- 150 sin 25
∑ F y =
+ 250 sin 45
+ 100 sin 60
= N

8. ∑ F x = N
∑ F y = N
∑ F y R
∑ F y α
Magnitude of the Resultant,
R = √ ∑ F x ² + ∑ F y ²
= √ ² ²
R = N
∑ F x
Direction of Resultant,
tan α = ∑ F y / ∑ F x
= / =
α = 7º 28’ 13’’

9. COMPOSITION α P R Q θ O ----------------------------------------
R = √ p2 + q p q cos θ
Tan α = l q sin θ / p + q cos θ l

10. Q α P tan α = Q / P R = √ P² + Q² Q R 90 MAGNITUDE OF RESULTANT
DIRECTION OF RESULTANT
tan α = Q / P
R = √ P² + Q²

11. STATICS
Statics Deals With the Equilibrium of Bodies, Those That Are Either at Rest or Move With a Constant Velocity.
Dynamics Is Concerned With the Accelerated Motion of Bodies and Will Be Dealt in the Higher Semester.

13. EQUILIBRIUM OF A PARTICLE CONCLUDED
For equilibrium:
 Fx = and
 F y = 0.
Note: Considering Newton’s first law of motion, equilibrium can mean that the particle is either at rest or moving in a straight line at constant speed.

14. PRINCIPLE OF TRANSMISSIVITY OF FORCES
Principle of transmissivity states that the condition of rest or motion of a rigid body is unaffected if a force, F acting on a point A is moved to act at a new point, B provided that the point B lies on the same line of action of that force. F F A B

15. MOMENT OF A FORCE ABOUT A POINT
The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

16. Resolution of a Force into a Force at B and a couple
A Force, F acting at point A on a rigid body can be resolved to the same force acting on another point B and in the same direction as the original force plus a couple M equal to r x F i.e. moment of F about B
i.e Force in (a) equal to that in (b) equal to that in C
M = 25 kN m
F = 5 kN
5 kN
5 kN B A B
5 kN
5 m
(a)
(c)
(b)

17. d d Force replaced by force and couple F A B = B A B = M= F x d F F F
Moment of the pair of forces = F x d F

18. Varignon’s Theorm ( Moment Theorm)
Summation of moments of all forces about one moment centre is equal to the moment due to RESULTANT alone about the same moment centre.
Summation Mc = Rx Perpendicular Distance
from C

19. Sign Conventions Clockwise Moments -- Positive +
Counter Clockwise Negative -

20. For equilibrium of coplanar non- concurrent force system:
 Fx =
 F y =
 M c =
Equations 1 and 2 are called force equations
Equation 3 is called Moment equation

21. LAMI’S THEORM ץ β α P Q P = K sin α Q = K sin β R = K sin ץ Q P P Q R
------
------ = =
------
Sin α
Sin β
Sin ץ R R

22. Lami’s Theorm Statement
When three concurrent forces are acting on a body simultaneously, be in equilibrium then any force is directly proportional to the SINE of the angle subtended by other two forces. P Q R
------
------ =
------
Sin α
Sin β
Sin ץ

23. Find Resultant completely

24. ∑ F x = KN
120 KN
∑ F y = +120 –
= KN
5000 KN
1120 KN
Magnitude of Resultant
R = √ ∑ F x ² + ∑ F y ²
420 KN
R = √ 5000 ² ²
R = kN
5000
1420
tan α =
5000 α
1420 R
α = 15º 51´ 16 ´´
( Fourth Quadrant )

25. α
Applying Varignon’s Theorem
Taking Moments about ‘ o ‘
┴ d
∑ M o = R x ┴ d R kN
x x 2 – 120 x x 5 = x ┴ d
23860
┴ d =
= 4.59 m
5197.7

26. 15 sin 60
40 sin 60
20 kN
15 cos 60
10 sin 30
40 cos 60 60
10 cos 30

27. y
40 sin 60 20
15 sin 60
10 sin 30 x
40 cos 60
10 cos 30
15 cos 60
∑ F x = + 10 cos cos 60 – 40 cos 60
= kN
∑ F y = 10 sin sin sin 60
= kN

28. R α ∑ F x = - 3.84 kN ∑ F y = 72.63 kN ---------------------------
= √ α
R = kN
72.63
tan α =
- 3.84
α = 86 º 58 ´ 24 ´´
Resultant kN acts in second Quadrant at angle of 86 º 58 ´ 24 ´´

29. α R 20 kN 25 kN 50 kN 35 kN ------------------ ∑ F x = 25 – 20 = 5 kN
= √ 7250
∑ F y = – 35 α
R = kN
= - 85 kN 85
tan α = 5
α = 86 º 38 ´ 1 ´´ R
Resultant kN acts in Fourth Quadrant at angle of 86 º 38 ´ 1 ´´

30. X X √ √ α ∑ M o = R x ┴ d R = 85.15 kN Applying Varignon’s Theorem ┴ d
Taking Moments about ‘ o ‘
┴ d α
∑ M o = R x ┴ d
R = kN X X
+ 25 x x 4 = x ┴ d
┴ d = m √ √

31. 1500 sin 60
1805 Sin 33.67
1500 cos 60
1805 cos 33.67
2240 cos 63.40
2240 sin 63.40

32. 1500 sin 60
1805 Sin 33.67
1500 cos 60
1805 cos 33.67
2240 cos 63.40
2240 sin 63.40

33. y
1500 sin 60
1500 cos 60
1805 cos 33.67
2240 cos 63.40 x
1805 Sin 33.67
2240 sin 63.40
∑ F x = cos – 1805 cos – 1500 cos 60
= N
∑ F y = 1500 sin 60 – 1805 sin – 2240 sin 63.40
= N

34. α R 1249.22 1704.58 ----------------------------------
= √
R = N
tan α =
α = 53 º 45 ´ 49´´
Resultant N acts in Third Quadrant at angle of 53 º 45 ´ 49´´

35. α
1500 sin 60
1805 Sin 33.67
1500 cos 60
1805 cos 33.67 √
┴ d
2240 cos 63.40 √ √ √ √ X
2240 sin 63.40
cos 60x 3 – 1500 sin 60 x 2 – 1805 cos x sin x sin 63.4 x 4 = x
┴ d

36. α ∑ M o = R x ┴ d R = 2113.3 N Applying Varignon’s Theorem
Taking Moments about ‘ o ‘
∑ M o = R x ┴ d
┴ d
R = N
0.79m
cos 60x 3 – 1500 sin 60 x 2 – 1805 cos x sin x sin 63.4 x 4 = x
┴ d
Negative sign indicates that the RESULTANT lies below point ‘o’
at same distance 0.79 m
┴ d =
┴ d = m 36 36 36

38. F B D Lami’s thoerem T1 T2 1000 T1 = 500 N = = Sin 90 Sin 150 Sin 120

39. TCB TCA TCB = 7.76 N TCA = 10.98 N T CA T CB F B D 75 150 135 15 N
Lami’s theorem
TCB
TCA 15
TCB = 7.76 N = =
Sin 75
Sin 150
Sin 135
TCA = N

40. Lami’s thoerem T1 T2 5
T1 = 2.89 N = =
Sin 150
Sin 90
Sin 120
T2 = 5.77 N

42. R
5 kN S
10 kN
7 kN

43. R S ∑ F y = 0 ∑ F x = 0 R - 10 – 7 sin 45 - 0.058 sin 30 =0
30º
5 kN
45º S
10 kN
7 kN
∑ F y = 0
∑ F x = 0
R - 10 – 7 sin sin 30 =0
- S cos – 7 cos 45 = 0
R =14.98 kN
S = kN

46. 60 30
105º
120º
135º
At Joint B
∑ Fy = 0
At Joint D
T3 sin 60 – sin 30 – 200 = 0
Lami’s thoerem
T3 = N T1 T2
250 = =
∑ Fx = 0
Sin 120
Sin 135
Sin 105
336.6 cos cos 30 – T4 = 0
T1 = N
T4 =
T2 = N

51. 250
125 mm
125 mm α
110
Cos α =
110
250
α = 63º 53 ‘ 46’’

52. R B R A RA = 49 N R B R A RB = 111.36 N Consider F B D of cylinder o1
116º 6’ 14’’
R A
63º 53 ‘ 46’’ 90
153º 53’ 46’’
100 N
Lami’s theorem
RA = 49 N
R B
R A
100 = =
Sin 153º53’46’’
Sin 90
Sin 116º6’14’’
RB = N

53. R C R D R B R D – 111.36 cos 63º53’46’’ = 0 R c = 200 N R D = 49 N
Consider F B D of cylinder o2
R C
R D
63º 53 ‘ 46’’
R B N
100 N
∑ Fy = 0
∑ Fx = 0
R D – cos 63º53’46’’ = 0
R c – 100 – sin 63º53’46’’ = 0
R c = 200 N
R D = 49 N

54. Q. Find Reactions at all points of contacts, force P required to keep system in equilibrium and force in connecting member A B 60 54

55. 60

56. Lami’s theorem T AB 4000 R 1 R1 = 5464.1 N T AB = - 4898.98 N 240º 45º
Consider F B D of cylinder A
W A
T AB
4000
R 1 = = 30
Sin 45
Sin 75
Sin 240 15 60
T AB R1
R1 = N 60
T AB = N
240º
45º
Negative sign indicates TAB must act in opposite direction. i. e. Acts in Second quadrant 30 15 R1
W A = 4000 N
T AB
75º

57. Lami’s theorem T AB 4000 R 1 R1 = 5464.1 N T AB = 4898.98 N T AB 135
Consider F B D of cylinder A TAB in opposite direction
Lami’s theorem
W A
T AB
4000
R 1 = = 30
Sin 135
Sin 105
Sin 120 15 60
T AB R1
R1 = N 60
W A = 4000 N
T AB = N
135
60º
Here TAB is positive and acts in Second Quadrant with same Magnitude
15º
120 R1
T AB
105 57

58. 60 58 58

59. 2000 N
60º P
15º
T AB 15
30º
15º
45º
R 2 P
45º
T AB
15º
30º
Consider F B D of cylinder B
45º
45º
R 2
2000 N

60. P
T AB
15º
30º
45º
45º
R 2 60
2000 N
Consider F B D of cylinder B

61. R 2 T AB= 4898.98 N P Consider F B D of cylinder B TAB cos 15 15º 30º
R2 sin 45
R 2
P cos 30
R2 cos 45
TAB cos 15
45º
sin 15
15º
30º
P sin 30
T AB= N
2000 N P
Resolving forces vertically
Resolving forces Horizontally
∑ F y = 0
∑ F x = 0
TAB cos 15 – R2 cos 45 – P cos 30 = 0
R2 sin 45- P sin sin 15 – 2000 = 0
cos 15 – R2 cos 45 – P cos 30 = 0
R2 cos 45 + P cos 30 =
R2 sin 45- P sin 30 =

62. R2 cos 45 + P cos 30 = 4732.05 ------Eq 1 x sin 30
R2 sin 45 - P sin 30 = Eq x cos 30
R2 cos 45x P cos 30 x = x
sin 30
sin 30
sin 30
R2 sin P sin =
x cos 30
x cos 30
x cos 30
R1 = N
R2 x =
R2 = N
T AB = N
P = N
R2 = N
P = N

64. Find reactions at all points of contacts RA , RB, RC , and RD
Both cylinders weigh 200 N each
R B
R D
R B
R A
R C

65. Consider F B D of cylinder 1
R B
R A
200 N
30º
60º
R A
R B

66. Consider F B D of cylinder 1
30º
60º
R A
R B 66

67. Lami’s theorem R A R A 200 RA = 173.2 N RB = 100 N R B RB
Consider F B D of cylinder 1
R A
Lami’s theorem
90º
R B
R A RB
200 = =
150º
120º
Sin 90
Sin 120
Sin 150
200 N
RA = N
RB = 100 N 67 67

68. R D R D R C R C R D– R C cos 60 = 150 Rc = 330.94 N
Consider F B D of cylinder 2
200 N
200 N
R B
173.2 N
R B
R D
R D 30 60
R C
R C
Resolving forces Horizontally
Resolving forces vertically
∑ F x = 0
X or Y First
∑ F y = 0
R D – cos 30 – R C cos 60 = 0
Rc sin sin 30 – 200 = 0
R D– R C cos 60 = 150
Rc = N
R D = cos
R D = N

69. R2 45 R1
Lami’s Theorem W1
Lami’s Theorem R1 W1 R2 W2
Resolution R2 R3 R4
Resolution R4 45 R3
Resolution 45 R2 W2

70. Sequence of equilibrium of cylinders to be considered are---
Lami’s Theorem
Lami’s Theorem
Resolution
Lami’s Theorem
Resolution
Resolution
Sequence of equilibrium of cylinders to be considered are---
First Cylinder A * Lami’s Theorm
Second-- Cylinder C * Lami’s Theorem
Third Cylinder B * Resolution
Sequence of equilibrium of cylinders to be considered are---
First Cylinder C * Lami’s Theorm
Second-- Cylinder A or B * Resolution
Third Cylinder B or A * Resolution

71. 80 N B
80 N
100 mm
100 sin 60
100 N
60º
100 mm C
120 cos 30
100 cos 60 A
30º
100 sin 60
120 N
120 sin 30
80 N
120 cos 30
100 cos 60
80 N
120 sin 30

72. ∑ F x = +80 – 100 cos cos 30
= N
∑ F y = 100 sin 60 – 80 – 120 sin 30
= N
100 sin 60
80 N
Magnitude of Resultant
R = √ ∑ F x ² + ∑ F y ²
120 cos 30
100 cos 60
80 N
120 sin 30
R = √ ( ) ² + ( ) ²
R = N
tan α =
- 73.9
α = 35º 50´ 48 ´´
( Third Quadrant )

73. X X X √ √ √ α ∑ M A = R x ┴ d R = 91.17 N 80 N B 80 N
α = 35º 50´ 48 ´´
Applying Varignon’s Theorem
Taking Moments about A
100 mm
100 sin 60
100 N
∑ M A = R x ┴ d
60º
100 mm
R = N C
120 cos 30
100 cos 60 A
30º
┴ d
120 N
120 sin 30 X
R = N
+120 sin 30 x x x 50 = x
┴ d X X
18000
┴ d = √
91.17 √ √
┴ d = mm

74. 2kN
1.2m 2 3 1m θ1 1 θ2
Each Box measures 1 m x 1.2 m 4 5 θ3
2 kN
2 cos 39º 48’ 20’’ = 1.54 kN
2 sin 39º 48’ 20’’ = 1.28 kN 6
15 kN
5 kN
tan θ1 = 1 / 1.2
5 kN
5 cos 78º 13’ 54’’ = 1.02 kN
5 sin 78º 13’ 54’’ = 4.89 kN
θ1 = 39º 48’ 20’’
tan θ2 = 3 / 4.8
θ2 = 78º 13’ 54’’
tan θ2 = 1 / 2.4
15 kN
15 cos 22º 37’ 11’’ = kN
15 sin 22º 37’ 11’’ = 5.77 kN
θ3 = 22º 37’ 11’’

80. . . CENTROID Y 20 mm X2 Y2 X1 50 mm Y1 25 mm X 30 mm 30 1/3x20=6.6715
50 mm Y1 25
25 mm 25 X
30 mm

81. . . CENTROID Y 20 mm X2 Y2 X1 50 mm Y1 25 mm X 30 mm 30 1/3x20=6.6715
50 mm Y1 25
25 mm 25 X
30 mm