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ENGI 1313 Mechanics I
Lecture 25:
Equilibrium of a Rigid Body
Shawn Kenny, Ph.D., P.Eng.
Assistant Professor
Faculty of Engineering and Applied Science
Memorial University of Newfoundland
[email protected]
Lecture Objective
to illustrate application of 2D equations of
equilibrium for a rigid body
 to examine concepts for analyzing
equilibrium of a rigid body in 3D

2
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Example 25-01

3
Determine the force
P needed to pull the
50-kg roller over the
smooth step. Take θ
= 60°.
© 2007 S. Kenny, Ph.D., P.Eng.
=
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

What XY-coordinate
System be Established?
=
4
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

Establish FBD

=
w = mg = (50 kg)(9.807 m/s2) = 490 N
5
© 2007 S. Kenny, Ph.D., P.Eng.
NB
NA
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

Determine Force Angles

Roller self-weight


70
 = 20
 = 20
=
w = 490 N
6
© 2007 S. Kenny, Ph.D., P.Eng.
NB
NA
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

Determine Force Angles

Normal reaction force at A

90
=
NA
7
© 2007 S. Kenny, Ph.D., P.Eng.
w = 490 N
NB
NA
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

Determine Force Angles

Normal reaction force at B

yB = (0.6 m – 0.1 m)
= 0.5 m
  a cos
1
r = 0.6 m
NB
 0 .5 m 

  33 . 56
 0 .6 m 
=
x B  r sin   0 . 6 m sin 33 . 56
8
© 2007 S. Kenny, Ph.D., P.Eng.
w = 490 N


NB
NA
 0 . 3317 m
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

Draw FBD
w = 490 N
 = 20
P
 = 60

NB
NA= 0 N
9
© 2007 S. Kenny, Ph.D., P.Eng.
=
w = 490 N
NB
NA
ENGI 1313 Statics I – Lecture 25
Example 25-01 (cont.)

What Equilibrium Equation
should be Used to Find P?

w = 490 N
 = 20
MB = 0
P
 = 60
w sin   y B   w cos   x B   
P cos   y B   P sin   x B   0
yB = 0.5 m xB = 0.3317 m
490 N sin 20  0 . 5 m   490 N cos 20  0 . 3317
P cos 60  0 . 5 m   P sin 60  0 . 3317   0



m

P  6 . 35 kN
10
© 2007 S. Kenny, Ph.D., P.Eng.

ENGI 1313 Statics I – Lecture 25
NA
NB
Comprehension Quiz 25-01

If a support prevents rotation
of a body about an axis,
then the support exerts a
________ on the body
about that axis.





11
A) Couple moment
B) Force
C) Both A and B
D) None of the above.
Answer: A
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
3-D Equilibrium

Basic Equations






F
x
0
M
x
0
F
y
0
M
y
0
M
z
0
F
z
0
Moment equations can also be
determined about any point on
the rigid body. Typically the
point selected is where the
most unknown forces are
applied. This procedure helps
to simplify the solution.
12
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Application to 3D Structures (cont.)

Engineering Design
Basic analysis
 Check more rigorous methods

13
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Application to 3D Structures (cont.)
Axial Forces

Design of Experimental
Test Frame
Lateral
Loads
14
© 2007 S. Kenny, Ph.D., P.Eng.
Couple Forces
For Bending
ENGI 1313 Statics I – Lecture 25
3-D Structural Connections

Ball and Socket

15
Three orthogonal forces
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
3-D Structural Connections (cont.)

Single Journal Bearing

Two forces and two couple moments
• Frictionless
• Circular shaft

16
Orthogonal to longitudinal bearing axis
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
3-D Structural Connections (cont.)

Journal Bearing (cont.)

17
Two or more (properly aligned) journal
bearings will generate only support reaction
forces
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
3-D Structural Connections (cont.)

Single Hinge
Three orthogonal forces
 Two couple moments
orthogonal to hinge axis

18
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
3-D Structural Connections (cont.)

Hinge Design

19
Two or more
(properly aligned)
hinges will
generate only
support
reaction
forces
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Rigid Body Constraints

What is the
Common
Characteristic?

20
Statically
determinate
system
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Redundant Constraints

Statically Indeterminate System

21
Support reactions > equilibrium equations
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Improper Constraints

Rigid Body Instability

2-D problem
• Concurrent reaction forces

22
Intersects an out-of-plane axis
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Improper Constraints (cont.)

Rigid Body Instability

3-D problem
• Support reactions intersect a common axis
23
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
Improper Constraints (cont.)

Rigid
Body
Instability

24
Parallel
reaction
forces
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
References
Hibbeler (2007)
 http://wps.prenhall.com/esm_hibbeler_eng
mech_1

25
© 2007 S. Kenny, Ph.D., P.Eng.
ENGI 1313 Statics I – Lecture 25
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