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Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of Newfoundlandspkenny@engr.mun.ca
ENGI 1313 Mechanics I
Lecture 24: 2-D Rigid Body Equilibrium
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Mid-Term Examination
Try to return by end of next weekWill provide hand worked solutionCan review once you receive the mid-term resultsPencil case and water bottle left in 1040
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General Announcement
AllergiesAppreciated if you can refrain from wearing or using scented products
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Lecture 24 Objective
to understand concept of two-force and three-force memberto illustrate application of 2D equations of equilibrium for a rigid body
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Two-Force Member
ConditionsNo couple forces or couple momentsNeglect self-weightCollinear forces
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Two-Force Member (cont.)
What is the Motivation to Recognize that a Rigid Body is a Two Force Member?
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Two-Force Member (cont.)
Motivation?Simplify equilibrium analysis
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Three-Force Member
ConditionsConcurrent force systemParallel force system
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Comprehension Quiz 24-01The three scalar equations ΣFx = ΣFy = ΣMo = 0, are ____ equations of equilibrium in 2-D.
A) incorrectB) the only correctC) the most commonly usedD) not sufficient
Answer: C
∑ =→+ 0F
∑ = 0MB
∑ = 0MA ∑ = 0MB
∑ = 0MA
∑ = 0MC
Alternative Equilibrium Equations
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Statically Determinate Structure
1. Determine Number of Reaction Forces or Unknowns
2. Determine Number of Equilibrium Equations Available
3. Evaluate# Equations ≥ # UnknownsLinear system of equations solved
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Statically Determinate Structure (cont.)
F
Ax Ay By
A B
What are the Support Reactions?What are the Equilibrium Equations?
∑ =→+ 0Fx
∑ =↑+ 0Fy
∑ = 0M
y
x
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Statically Indeterminate Structure
F
Ax Ay By
A B
What are the Support Reactions?What are the Equilibrium Equations?
∑ =→+ 0Fx
∑ =↑+ 0Fy
∑ = 0M
y
x
Cy
C
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Solving 2-D Equilibrium
1. X-Y Coordinate SystemEstablish suitable right, rectangular coordinate system if not provided
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Solving 2-D Equilibrium (cont.)
Establish Suitable Coordinate SystemKey Decision Factors?• Relative orientation of the applied loads and rigid
body (structure)• Simplest or most direct resolution of force
components
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Solving 2-D Equilibrium (cont.)
Example Suitable Coordinate System
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Solving 2-D Equilibrium (cont.)
2. Draw the Rigid Body Free Body Diagram (FBD)
Drill Rig Idealized Model Rigid Body FBD
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Solving 2-D Equilibrium (cont.)
3. Apply the Appropriate Equilibrium Equations
∑ =→+ 0F
∑ = 0MB
∑ = 0MA ∑ = 0MB
∑ = 0MA
∑ = 0MB
Alternative Equilibrium Equations
∑ =→+ 0Fx
∑ =↑+ 0Fy
∑ = 0M
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Important Considerations
Order of Application for Equations of Equilibrium
What are the Support Reactions?
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Important Considerations (cont.)
Order of Application for Equations of Equilibrium
What is the first equilibriumequation to use?
∑ = 0MA
Why?
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Important Considerations (cont.)
Negative Scalar Solutions to Equilibrium Equations?
Force or couple moment opposite to that assumed in the FBD for the designated convention
y
x
∑ =↑+ 0Fy
0N1200Ay =−−
↑∴−= yy AN1200AAy
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Comprehension Quiz 24-02
Which equation of equilibrium allows you to determine the force F immediately?
A) ΣFx = 0B) ΣFy = 0C) ΣMA = 0 D) Any one of
the above.
Answer: C
100 N
Ax Ay
A
y
x
Fθ
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Example 24-01
The lever ABC is pin-supported at A and connected to a short link BD as shown. If the weight of the members is negligible, determine the force of the pin on the lever at A.
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Example 24-01 (cont.)
What Key Features of the Problem Can Be Recognized?
Bracket or link BD is a two-force memberLever ABC is a three-force member
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Example 24-01 (cont.)
o452.02.0tan 1 =⎟
⎠⎞
⎜⎝⎛= −θ
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Example 24-01 (cont.)
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Example 24-01 (cont.)
o3.604.07.0tan 1 =⎟
⎠⎞
⎜⎝⎛= −θ
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Example 24-01 (cont.)
Concurrent Forces
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Example 24-01 (cont.)
∑ =→+ 0Fx
∑ =↑+ 0Fy
045sinF3.60sinFA =− oo
kN07.1FA =
kN32.1F =
AF228.1F =
0N40045cosF3.60cosFA =+− oo
0N40045cosF228.13.60cosF AA =+− oo
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References
Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1