Shawn Kenny, Ph.D., P.Eng.Assistant ProfessorFaculty of Engineering and Applied ScienceMemorial University of Newfoundlandspkenny@engr.mun.ca

ENGI 1313 Mechanics I

Lecture 23: Equilibrium of a Rigid Body

2 ENGI 1313 Statics I – Lecture 23© 2007 S. Kenny, Ph.D., P.Eng.

Mid-Term

Thursday October 18Material: Chapter 1 to 4.5 inclusiveTime: 830am-945amLocation: EN 2043, EN 1040, EN 2007,

EN 1001, EN 1003 & EN 1054• Seating arrangements

http://www.engr.mun.ca/undergrad/schedule.php

3 ENGI 1313 Statics I – Lecture 23© 2007 S. Kenny, Ph.D., P.Eng.

Quiz #4

Week of October 22-26Section 4.6 through 4.10

Excluding “Reduction to a Wrench”

4 ENGI 1313 Statics I – Lecture 23© 2007 S. Kenny, Ph.D., P.Eng.

Chapter 5 Objectives

to develop the equations of equilibrium for a rigid bodyto introduce the concept of the free-body diagram for a rigid bodyto show how to solve rigid body equilibrium problems using the equations of equilibrium

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Lecture 23 Objectives

to identify support reactionsto establish the free-body diagram for a rigid body in 2-Dto develop the equations of equilibrium for a 2-D rigid body

6 ENGI 1313 Statics I – Lecture 23© 2007 S. Kenny, Ph.D., P.Eng.

Recall – Particle Equilibrium (L10)

Concurrent Force Systems

F1

F2

F3

V = 0, v+Y

+X

∑ =→+ 0Fx

∑ =↑+ 0Fy

2 Equations ∴ Solve for at most 2 Unknowns

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Rigid Body Equilibrium

Forces are Typically not ConcurrentPotential moment or couple moment

∑ =→+ 0Fx

∑ =↑+ 0Fy

∑ = 0Mo

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Recall – Particle FBD (L10)

+Y

+X θ = 30°

FAB

W = FAC = mg

FAD

W = (255 kg)(9.806m/s2) = 2.5kN

A

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Rigid Body FBD

What is it?Sketch or diagram illustrating all external force and couple vectors acting on a rigid body or group of rigid bodies (system)

Purpose?A visual aid in developing equilibrium equation of motion

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Rigid Body FBD (cont.)What is the procedure?

Draw isolated or “free” outlined shape• Establish idealized model• Establish FBD

Show all forces and couple moments• External applied loads• Rigid body self-weight• Support reactions

Characterize each force and couple• Magnitude• Sense • Direction

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Rigid Body FBD (cont.)

Drilling Rig200 kg suspended platform on derrick tower

Drill Rig Idealized Model Rigid Body FBD

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Rigid Body FBD (cont.)

Cantilever Beam100 kg beam

Idealized Model Rigid Body FBD

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Support Reactions

Newton’s 3rd LawExternal loadsSupport specific characteristicsTranslation prevented

⇒ support reaction forceRotation prevented

⇒ support couple moment

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Common Structural Supports

Cable

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Common Structural Supports (cont.)

Roller

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Common Structural Supports (cont.)

Pin

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Common Structural Supports (cont.)

Fixed

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Example 23-01

Foot Pedal FBDSpring force is 30 lb

Foot Pedal

Idealized Model Rigid Body FBD

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Example 23-02

Dump Truck FBD5000 lb dumpster supported by a pin at A and the hydraulic cylinder BC (short link)

G

W = 5000 lb

20°30°

B

FCB

Ay

Ax

Rigid Body FBD

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Comprehension Quiz 23-01

Internal forces are _________ shown on the free body diagram of a whole body.

A) alwaysB) oftenC) rarelyD) never

Answer: D

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Comprehension Quiz 23-02The beam and the cable (with a frictionless pulley at D) support an 80 kg load at C. In a FBD of only the beam, there are how many unknowns?

A) 2 forces and 1 couple momentB) 3 forces and 1 couple momentC) 3 forcesD) 4 forces

Answer: C Ay

Ax

FBD

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Example 23-03

Draw the free-body diagram of the beam supported at A by a fixed support and at B by a roller. Explain the significance of each force on the diagram.

23 ENGI 1313 Statics I – Lecture 23© 2007 S. Kenny, Ph.D., P.Eng.

Example 23-03 (cont.)w 40

lbft

=

a 3 ft=

b 4 ft=

θ 30 deg=

Ax, Ay, MA effect of wall on beam.

NB force of roller on beam.

wa2

resultant force of distributed load on beam.

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Example 23-04Draw the free-body diagram of the automobile, which is being towed at constant velocity up the incline using the cable at C. The automobile has a mass M and center of mass at G. The tires are free to roll. Explain the significance of each force on the diagram.

Given:

M 5 Mg= d 1.50 m=

a 0.3 m= e 0.6 m=

b 0.75 m= θ1 20 deg=

c 1 m= θ2 30 deg=

g 9.81m

s2=

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Example 23-04 (cont.)

Given:

M 5 Mg= d 1.50 m=

a 0.3 m= e 0.6 m=

b 0.75 m= θ1 20 deg=

c 1 m= θ2 30 deg=

g 9.81m

s2=

NA, NB force of road on car.F force of cable on car.Mg force of gravity on car.

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Textbook Problems

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Textbook Problems

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Textbook Problems

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Textbook Problems

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References

Hibbeler (2007)http://wps.prenhall.com/esm_hibbeler_engmech_1