Transcript ME13A: CHAPTER FOUR - Faculty of Engineering

ME13A: CHAPTER FOUR
EQUILIBRIUM OF
RIGID BODIES
4.1 INTRODUCTION
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Equilibrium of a rigid body requires both a
balance of forces, to prevent the body from
translating with accelerated motion, and a
balance of moments, to prevent the body
from rotating.
For equilibrium: F = 0 and
Mo = (r x F) = 0
In three dimensions: Fx = 0 Fy = 0
Fz = 0
Mx = 0
My = 0
Mz = 0
4.2 SOLUTION OF REAL
PROBLEMS IN TWO DIMENSIONS
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Step 1: Examine the space diagram and select
the significant body - the free body diagram. Detach
this body from the space diagram and sketch its
contours
Step 2: Identify and indicate all external forces
acting on the body. The external forces consist of
the externally applied forces, the weight and
reactions of the body at the point of contact with
other external bodies.
Step 3: All the important dimensions must be
included in the free body diagram.
Types of Reactions Contd.
Support Reactions
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If a support prevents the
translation of a body in a given
direction, then a force is developed
on the body in that direction.
Likewise, if rotation is prevented, a
couple moment is exerted on the
body.
4.4 STATICALLY DETERMINATE AND
INDETERMINATE REACTIONS.
TYPES OF CONSTRAINTS IN TRUSSES
Example: Find the Reactions in the
Truss Below
Difference Between Improper
and Partial Constraints
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In improper constrained case,
one equation of equilibrium has
been discarded but in the
partial case, there are only two
unknowns so the two equations
were used to determine them.
Conclusion on Supports
Apart from structures designed for movement, partial or improper constraints should be
avoided in design. A rigid body will be improperly constrained when the supports are
such that the reactions are either parallel or concurrent. An improperly constrained body
also has statically independent reactions.
P
Q
R2
R1
Concurrent reactions
R
B
Example
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A truck mounted crane is used to lift a 3
kN compressor. The weights of the
boom AB and of the truck are as
shown, and the angle the boom forms
with the horizontal is
 = 40o.
Determine the reaction at each of the
two (a) rear wheels (b) front wheels D
Solution
4.5 TWO AND THREE-FORCE
MEMBERS
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The solution of some equilibrium problems
are simplified if one is able to recognise
members that are subjected to only two or
three forces.
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4.5.1
Two-Force Members:
member is subject to no couple
and forces are applied at only two
a member, the member is called a
member.
When a
moments
points on
two-force
Two-Force Members
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Consider a two-force member below in (a).
The forces in the two points are first resolved
to get their resultants, FA and FB. For
or force equilibrium:
F translational
F
3
2
FA
F1
FA
FA
FB = -FA
B
FB = - FA
FB = - FA
(a)
(b)
(c)
Two-Force Members Contd.
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( F = 0) to be maintained, FA must be
of equal magnitude and opposite
direction to FB.
For rotational or moment equilibrium:
( Mo = 0) to be satisfied, FA must be
collinear with FB.
See other two force members in (c)
below.
Three Force Members
Example
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One end of rod AB rests in the corner A and
the other is attached to cord BD. If the rod
supports a 200-N load at its midpoint, C, find
the reaction at A and the tension in the cord.
4.6 EQUILIBRIUM IN THREE
DIMENSIONS
4.6.1
Reactions at Supports: See Fig.
4.10 of text book for possible supports and
connections for a three dimensional
structure.
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4.6.2
Equilbrium of a Rigid Body in
Three Dimensions
Six scalar equations are useable
(see
section 4.1)
Equilbrium of a Rigid Body in Three
Dimensions Contd.
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Problems will better be solved if the vector
form of the equilbrium equations is used i.e.
F = 0
Mo =  (r x F) = 0
Express the forces, F and position vectors, r
in terms of scalar components and unit
vectors. Compute all vector products either
by direct calculation or by means of
determinants.
Solution Concluded
Recall that Equation (1) is:
Ax i + (Ay - 2 TB + Dy) j + (Az - 2 TC + DZ ) k = 0 … (1)
From (1), Ax i = 0 i.e. Ax = 0
Ay - 2 TB + Dy = 0 i.e Ay = 2 TB - DY
= (2 x 90) - 60 = 120 N
Az - 2 TC + DZ = 0 i.e. Az = 2 TC - DZ =
(2 x 150) - 166.7 = 133.3 N
A = 120 N j + 133.3 N k
Example
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A 35-kg rectangular plate shown is
supported by three wires. Determine
the tension in each wire.