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