Chapter III a

Equilibrium 2D

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 Lecture Overview
- Body at rest

- equilibrium conditions

- free body diagram

- statical determinancy

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 Equilibrium: Body at Rest
no linear displacement or rotation

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 Action

Reaction

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Force
Components

Action

Reaction

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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

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 M = F·d

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 Equilibrium!

d d

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 Equilibrium!
left side right side

M = +F·d M = -F·d

F F
+ M M -

d 2F d

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 equilibrium conditions

R = ∑F = 0 ∑Fx = 0
∑Fy = 0

M = ∑M = 0 ∑M = 0

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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˚.

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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

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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

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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 y
θ
∑M = 0 x
F2

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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 = F/sinθ y
F1 = 6.0 kN θ
x
F2

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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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 = F/sinθ y
F1 = 6.0 kN θ
x
F2
(2) in (1) : F2 = -F1 cosθ = -5.2 kN

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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
F1 = 6.0 kN F
(2) in (1) : F2 = -F1 cosθ = -5.2 kN
6.0
graphical solution

-5.2
R=0 3.0

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 Free body diagram

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Free body diagram

- define the particular body and its system

(boundary conditions, support)

- isolate the body

- 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

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Free body diagram
- sign convention: assign arbitrarily the sense of force

e.g. tension: away from body

compression: 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
of the coord. system (right hand rule).

positive result: initial sense was correct

negative result: force acts in opposite sense

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 +
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

tension θ compression θ
x x
F2 + F2 +

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 +
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

tension θ compression θ
x x
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 = F/sinθ
F1 = 6.0 kN F1 = 6.0 kN
(1) in (2) : F2 = -F1 cosθ = -5.2 kN (1) in (2) : F2 = +F1 cosθ = +5.2 kN

Sense of F2 opposite to its assignment. Initial sense of both forces correct.

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 +
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
tension F1 + F1 +
tension

tension θ compression θ
F2 + F2 +

Convenient approach: (1) ∑Fx = 0 : +F2 -F1 cosθ = 0

(2) ∑Fy = 0 : F1 sinθ - F = 0 F1 = F/sinθ
- a positive result always indicates F1 = 6.0 kN
+
a tensile force
(1) in (2) : F2 = +F1 cosθ = +5.2 kN
- a negative result always indicates
-
a compressive force Initial sense of both forces correct.

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 Connections and Supports

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Hinge Connection – free rotation

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Free body diagram

Types of Supports

roller 1 support reaction

pin 2 support reactions

fixed 3 support reactions

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 y
Roller Support – 1 Reaction Force Fy
symbol x

1 possible reaction forces

2 degrees of freedom

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 y
Pin Connection – 2 Reaction Forces Fx, Fy
symbol x

2 possible reaction forces

1 degree of freedom

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 y
Fixed Support – 3 Reactions Fx, Fy, M
symbol x

3 possible reaction forces

0 degrees of freedom

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Fixed Foot Support of Steel Column

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Free body diagram – Boundary Conditions

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Free body diagram – Boundary Conditions

symbolic
representation

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Free body diagram – Boundary Conditions

symbolic
representation

symbolic
representation
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Free body diagram – Boundary Conditions

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

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Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

rough pin
connection

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
examples of free body diagrams

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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 General equilibrium conditions

R = ∑F = 0 ∑Fx = 0
∑Fy = 0

M = ∑M = 0 ∑M = 0

three equations for three unknown variables

(forces) for statically determinant systems!

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Alternative
equilibrium conditions y

∑Fx = 0 ∑Fy = 0 ∑MA = 0 F

∑Fx = 0 ∑MA = 0 ∑MB = 0 A B

F
∑Fy = 0 ∑MA = 0 ∑MB = 0 Ay

∑MA = 0 ∑MB = 0 ∑MC = 0 Ax Bx

C

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examples

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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, Engineering Mechanics I

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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.

Weight of Beam: 93.2 N · 5 m = 4.66 kN

Free Body diagram:

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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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.

Weight of Beam: 93.2 N · 5 m = 4.66 kN

Free Body diagram:

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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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

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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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, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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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, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
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, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

Faculty of Technology material by Karsten Schlesier Civil Engineering
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, Engineering Mechanics I

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Faculty of Technology material by Karsten Schlesier Civil Engineering
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, Engineering Mechanics I

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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, Engineering Mechanics I

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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, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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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, Engineering Mechanics I

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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?

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

ADDIS ABABA UNIVERSITY Engineering Mechanics Department of

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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, Engineering Mechanics I

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example 3.4

equilibrium at roller:
F F F F

F = 2Fy
2F
y

x
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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, Engineering Mechanics I

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example 3.4
4 · 76 N
The student pulls on the rope so that the spring
dynamometer B registers 76 N. The scale A reads 268 N. 76 N
What is the weight of the student?

∑Fy = 0 : 268 N + 5 · 76 N = Weight

Weight = 648 N (= 66.1 kg)

Weight

268 N
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I

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 Constraints and Statical Determinacy

Statically
indeterminate
3 equations
4 unknowns

fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I x

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example

Statically Determinate System

- 3 unknown forces
- 3 equations A B

Statically Indeterminate System

- 4 unknown forces
- 3 equations
A B
- possible internal constraint force

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ADDIS ABABA UNIVERSITY Engineering Mechanics Department of
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