In the name of ALLAH, the
Beneficent the Merciful 1
Engineering Mechanics (Statics)
ME-112
Chapter 3-1
Equilibrium
By: Engr. Jan
Department of Mechanical Engineering, IIU Islamabad 2
Equilibrium 2D
3
Review
Force.
• Move
• Rotate
• Deform(bend)
Components of force
Torque/Moment
Couple
Concurrent force
Parallel force
Transmissibility
Tension/compression (String/rope and Bar/strip)
4
Equilibrium: Body at Rest/ static
no acceleration
no linear displacement R = ∑F = 0
No rotation M = ∑M = 0
5
Action
Reaction
6
7
8
Force
Component
s
Action
Reaction
9
Vector Addition – graphical method
e.g.: equilibrium at particle: addition of all
vectors = 0
situation
at particle:
A resultant
B
reaction force A C
B R
R
10
11
M=
F·d
12
13
Equilibrium!
d d
14
Equilibrium!
left side right side
M= M=
+F·d -F·d
F F
+ M M -
d 2F d
15
equilibrium conditions
R = ∑F = 0 ∑Fx =
0
∑Fy =
M = ∑M = 0 ∑M = 0
0
16
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two-force-members
that are attached to a wall. Determine the forces acting in the
members if F = 3 kN and θ = 30˚.
17
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚.
Step 1: Isolation of Particle
assumption of
positive sense
of forces
(tensile)!
F1
θ θ
F2
F F
18
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚.
Step 2: equilibrium conditions ∑Fx = 0
∑Fy = 0
∑M = 0
F1
y
θ θ
x
F2
F F
19
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚.
Step 2: equilibrium conditions
∑Fx = 0 (1) : -F2 -F1 cosθ =
F1
0
∑Fy = 0 y
θ
∑M = 0 x
F2
20
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚.
Step 2: equilibrium conditions
∑Fx = 0 (1) : -F2 -F1 cosθ = 0
F1
∑Fy = 0 (2) : F1 sinθ - F = 0 → F1 = y
F/sinθ θ
x
F1 = 6.0 kN F2
21
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚.
Step 2: equilibrium conditions
∑Fx = 0 (1) : -F2 -F1 cosθ = 0
F1
∑Fy = 0 (2) : F1 sinθ - F = 0 → F1 = y
F/sinθ θ
x
(2) in (1)
F1 = 6.0 kN F2
: F2 = -F1 cosθ = -5.2 kN
F
22
example 3.1 (equilibrium at
particle)
The support for a shop sign consists of two
two-force-members that are attached to a wall. Determine the
forces acting in the members if F = 3 kN and θ = 30˚. y
F1
Step 2: equilibrium conditions
x
θ
∑Fx = 0 (1) : -F2 -F1 cosθ = 0
F2
∑Fy = 0 (2) : F1 sinθ - F = 0 → F1 =
F/sinθ F
(2) in (1)
F1 = 6.0 kN
6.0
: F2 = -F1 cosθ graphical
= -5.2 kN solution
-5.2
R=0 3.0
23
Free body diagram
fig by J.L. Meriam, L.G. Kraige, I
24
Free body diagram (Diagrammatic
representation)
define the particular body and its system
(boundary conditions, support)
- isolate the body (system – surrounding)
- draw all possible external forces acting on it
(action forces: known actions, weight, load concentrated
/ distributed
reaction forces: contact with other bodies /
support)
- choose coordinate system
25
Free body diagram
- sign convention: assign arbitrarily the sense of force
e.g. tension: away from
compression body towards
: body
- solution: always follow the initial sense of
the forces.
apply the algebraic sign due to the
sense of the force and the
directions
positive result: of the sense
initial coord.was
system (right hand rule).
correct
negative force acts in opposite
result: sense
26
+
example on sign convention
y
F = 3 kN and θ = 30˚ 1 F
- +
x
θ
- 2
Assumption: tensile forces in both members Sense of forces assigned due to estimation
F F
F1
+ F1 +
y tension y tension
compression
tension θ
x x θ
F2 + +
F2
27
+
example on sign convention
y
F = 3 kN and θ = 30˚ 1 F
- +
x
θ
- 2
Assumption: tensile forces in both members Sense of forces assigned due to estimation
F F
F1
+ F1 +
y tension y tension
compression
tension θ
x x θ
F2 + +
F2
(1) ∑Fx = 0 : -F2 -F1 cosθ = 0 (1) ∑Fx = 0 : +F2 -F1 cosθ = 0
(2) ∑Fy = 0 : F1 sinθ - F = 0 F1 = F/sinθ (2) ∑Fy = 0 : F1 sinθ - F = 0 F1 =
F1 = 6.0 kN F/sinθ
F1 = 6.0 kN
(1) in (2) : F2 = -F1 cosθ = -5.2 (1) in (2)
kN
: F2 = +F1 cosθ = +5.2 kN
Sense of F2 opposite to its assignment.
Initial sense of both forces correct.
28
+
example on sign convention
y
F = 3 kN and θ = 30˚ 1 F
- +
x
θ
- 2
Assumption: tensile forces in both members Sense of forces assigned due to estimation
tension F F
F1
+ F1 +
tension
tensio
compression
θ
n
θ
F2 + +
F2
Convenient approach: (1) ∑Fx = 0 : +F2 -F1 cosθ = 0
(2) ∑Fy = 0 : F1 sinθ - F = 0 F1 =
-a positive result always indicates + F/sinθ
a tensile force F1 = 6.0 kN
(1) in (2)
-a negative result always indicates : F2 = +F1 cosθ = +5.2 kN
-
a compressive force
Initial sense of both forces correct.
29
Connections and Supports
30
Free body diagram
Types of Supports
roller
1 support reaction
pin
2 support reactions
fixed 3 support reactions
31
32
33
y
Roller Support – 1 Reaction Force Fy
symbol x
1 possible reaction forces
Capable of holding
• Vertical forces only
2 degrees of freedom
34
35
y
Pin/hinge Connection – 2 Reaction Forces Fx, Fy
x
symbol
2 possible reaction forces
Capable of holding
• Vertical forces
• Horizontal forces
1 degree of freedom
36
37
38
y
Fixed Support (Built in/Welded)– 3 Reactions Fx, Fy, M
x
symbol
Capable of holding
3 possible reaction forces
• Vertical forces
• Horizontal forces
• Moment
0 degrees of freedom
39
40
Free body diagram – Action of forces on body
fig by J.L. Meriam, L.G. Kraige, I
41
Free body diagram – Action of forces on body
symbolic
representation
fig by J.L. Meriam, L.G. Kraige, I
42
Free body diagram – Action of forces on body
symbolic
representation
symbolic
representation
fig by J.L. Meriam, L.G. Kraige, I
43
Free body diagram – Action of forces on body
fig by J.L. Meriam, L.G. Kraige, I
44
examples of free body diagrams
45
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
46
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
47
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
48
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
49
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
50
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
51
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
52
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
53
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
54
examples of free body diagrams
rough pin
connectio
n
fig by J.L. Meriam, L.G. Kraige, I
55
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
56
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
57
examples of free body diagrams
fig by J.L. Meriam, L.G. Kraige, I
58
General equilibrium conditions
R = ∑F = 0 ∑Fx = 0
∑Fy = 0
M = ∑M = 0 ∑M = 0
59
60
Alternative
y
equilibrium
conditions x
∑Fx = 0 ∑Fy = 0 ∑MA = F
0
∑Fx = 0 ∑MA = 0 ∑MB = 0 A B
∑Fy = 0 ∑MA = 0 ∑MB = 0 F
Ay
∑MA = 0 ∑MB = 0 ∑MC = 0 Ax Bx
C
61
62
63
example (3/4 )
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
fig by J.L. Meriam, L.G. Kraige, I
64
example 3.4
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
Weight of Beam: 93.2 N · 5 m = 4.66
kN Free Body diagram:
fig by J.L. Meriam, L.G. Kraige, I
65
example 3.4
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
Weight of Beam: 93.2 N · 5 m = 4.66
kN Free Body diagram:
fig by J.L. Meriam, L.G. Kraige, I
66
example 3.4
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
∑MA = 0 : - 4.66(2.5-0.12) - 10(3.5-0.12)
+ T cos25˚ 0.25 + T sin25˚ (5-0.12) = 0
→ T = 19.61 kN
fig by J.L. Meriam, L.G. Kraige, I
67
example 3.2
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
∑MA = 0 : - 4.66(2.5-0.12) - 10(3.5-0.12)
+ T cos25˚ 0.25 + T sin25˚ (5-0.12) = 0
→ T = 19.61 kN
∑Fx = 0 : Ax – T cos25˚ = 0
→ Ax = 17.77 kN
fig by J.L. Meriam, L.G. Kraige, I
68
example 3.2
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
∑MA = 0 : - 4.66(2.5-0.12) - 10(3.5-0.12)
+ T cos25˚ 0.25 + T sin25˚ (5-0.12) = 0
→ T = 19.61 kN
∑Fx = 0 : Ax – T cos25˚ = 0
→ Ax = 17.77 kN
∑Fy = 0 : Ay + T sin25˚ - 4.66 - 10 = 0
→ Ay = 6.37 kN
fig by J.L. Meriam, L.G. Kraige, I
69
example 3.2
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
∑MA = 0 : - 4.66(2.5-0.12) - 10(3.5-0.12)
+ T cos25˚ 0.25 + T sin25˚ (5-0.12) = 0
→ T = 19.61 kN
∑Fx = 0 : Ax – T cos25˚ = 0
→ Ax = 17.77 kN
∑Fy = 0 : Ay + T sin25˚ - 4.66 - 10 = 0
→ Ay = 6.37 kN
A = √Ax2 + Ay2 = 18.88 kN
fig by J.L. Meriam, L.G. Kraige, I
70
example 3.2
Determine the tension T in the supporting cable and the reaction force on pin A for the
jib crane. Beam AB is a standard I-beam with a weight of 93.2 N per meter of length.
fig by J.L. Meriam, L.G. Kraige, I
71
example 3.3
Determine tension P in the cable required to elevate end B of the beam (weight 100 N)
and reaction Ay.
fig by J.L. Meriam, L.G. Kraige, I
72
example 3.3
Determine tension P in the cable required to elevate end B of the beam (weight 100 N)
and reaction Ay.
Free Body diagram:
100 N
Ay
fig by J.L. Meriam, L.G. Kraige, I
73
example 3.3
Determine tension P in the cable required to elevate end B of the beam (weight 100 N)
and reaction Ay.
∑MA = 0 : P cosθ (4+2) - 100 cosθ ( 4) = 0
→ P 6 - 100 4 = 0
→ P = 66.67 N
Free Body diagram:
100 N
Ay
fig by J.L. Meriam, L.G. Kraige, I
74
example 3.3
Determine tension P in the cable required to elevate end B of the beam (weight 100 N)
and reaction Ay.
∑MA = 0 : P cosθ (4+2) - 100 cosθ 4 = 0
→ P 6 - 100 4 = 0
→ P = 66.67 N
∑Fy = 0 : Ay + P - 100 = 0
→ Ay = 33.33 N Free Body diagram:
100 N
Ay
fig by J.L. Meriam, L.G. Kraige, I
75
76
example
fig by J.L. Meriam, L.G. Kraige, I
77
78
example 3.4
The student pulls on the rope so that the spring
dynamometer B registers 76 N. The scale A reads 268 N.
What is the weight of the student?
isolating the student’s body:
76 N
∑Fy = 0 : 268 N + 76 N + F = Weight F
two unknown forces left …
Weight
268 N
fig by J.L. Meriam, L.G. Kraige, I
79
example 3.4
equilibrium at roller:
FF F F
F = 2Fy
2F
y
80
example 3.4
The student pulls on the rope so that the spring
dynamometer B registers 76 N. The scale A reads 268 N.
What is the weight of the student?
isolating the lower part of the diagram by a cut through
the cables and applying the rules of the rollers … 76 N
Weight
268 N
fig by J.L. Meriam, L.G. Kraige, I
81
example 3.4
4 · 76
The student pulls on the rope so that the spring N
dynamometer B registers 76 N. The scale A reads 268 N. 76 N
What is the weight of the student?
268 N + 5 · 76 N = Weight
∑Fy = 0 :
Weight = 648 N (= 66.1 kg)
Weight
268 N
fig by J.L. Meriam, L.G. Kraige, I
82
Constraints and Statical Determinacy
Statically
indeterminate
3 equations
4 unknowns
fig by J.L. Meriam, L.G. Kraige, I x
83
Assignment
• Exercise Problems of Chapter 3 (5th edition)
• 2, 3, 4, 6, 7, 8, 10, 12, 14, 16, 24, 25, 36, 44
• Due date
• Quiz on
• No Submission after due date
Thanks
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