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07-08-2014
1
These requirements are both necessary and sufficient conditions
for equilibrium
When a body is in equilibrium, the resultant of all forces acting on
it is zero.
Thus the resultant force FR and the resultant couple MR are both
zero.
R
FR = ∑ F = 0
MR = ∑ M = 0
Support Reactions.
General rule.
• If a support prevents the translation of a body in a given direction then a
force is developed on the body in that direction.
• If rotation is prevented, a couple moment is exerted on the body.
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2
Support for rigid bodies subjected to two dimensional force systems
Flexible cable, belt, chain, rope (Weightless)
Smooth Surfaces
Rough Surfaces
Freely sliding guide
Pin Connections Pin free to turn
Pin not free to turn
Roller Support
Built in support of fixed support
Gravitational attraction
Spring actions
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3
To construct free-body diagram for a rigid body or any group of bodies considered
as a single system, the following steps should be performed:
Draw Outlined Shape:
Imagine the body to be isolated or cut "free" from its constraints and connections
and draw (sketch) its outlined shape.
Show All Forces and Couple Moments.
Identify all the known and unknown external forces and couple moments that act
on the body. Those generally are due to
(I) Applied loading. (2) reactions occurring at the supports or at points of contact
with other bodies, and (3) the weight of the body.
Identify Each Loading and Give Dimensions.
The forces and couple moments that are known should be labeled with their proper
magnitudes and directions.
Establish an xy coordinate system so that these unknowns. Ax, Ay can be identified.
Finally, indicate the dimensions of the body necessary for calculating.
Procedure for Drawing a Free - Body Diagram
Examples of Free Body Diagrams
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5
Determine the horizontal and vertical components of reaction on the beam caused
by the pin at B and the rocker at A as shown below. Neglect the weight of the
beam.
Free Body diagram:
Problem:1
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6
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7
Support for rigid bodies subjected to three dimensional force systems
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8
The uniform 7-m steel shaft has a mass of 200 kg and is supported by a ball and
socket joint at A in the horizontal floor. The ball end B rests against the smooth
vertical wall as shown. Compute the forces exerted by the walls and the floor on
the ends of the shaft.
Free Body diagram:
Problem: 2
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9
Vertical position of B
7 = 2� + 6� + ℎ�
h = 3 m
Weight W = mg = 200(9.81)
= 1962 N
A (2,6,0) B(0,0,3) G(1,3,1.5)
The welded tubular frame is secured to
the horizontal x-y plane by a ball and
socket joint at A and receives support
from the loose-fitting ring at B. Under the
action of the 2-kN load, rotation about a
line from A to B is prevented by the cable
CD, and the frame is stable in the position
shown. Neglect the weight of the frame
compared with the applied load and
determine the tension T in the cable, the
reaction at the ring, and the reaction
components at A.
Problem: 3
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10
A(0,0,0) B(0,4.5,6) C(-3,0,6) D (-1,2.5,0) F (2.5,0,6)
Taking moment about A
0)2()()(21
=×+++×++× jrkjirkir zyxzxAB TTTBB
Equating the coefficients of k and j to zero
0125.25.4
055.25.4
=−+
=+−−−
zz
xyx
TB
TTB
On solving
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11
Determine the components of reaction that the ball-and-socket joint at A,
the smooth journal bearing at B , and the roller support at C exert on the
rod assembly.
The reactive forces of the supports will prevent the assembly from rotating
about each coordinate axis, and so the journal bearing at B only exerts
reactive forces on the member.
No couple moments are required.
Free-Body Diagram
Problem: 4
Equations of Equilibrium:
0)900()4.04.0()()2.16.0()(8.0 =×+−−×+−++× kjikjikijCzx
FBB
Taking moment about A
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12
The boom is used to support the 75-lb
flowerpot. Determine the tension
developed in wires AB and AC .
Free-Body Diagram
Problem: 5
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13
Constraints and Statical Determinacy
Redundant Constraints
When a body has redundant supports i.e., more supports than are necessary to hold
it in equilibrium, It becomes statically indeterminate.
Statically indeterminate means that there will be more unknown loadings on the
body than equations of equilibrium available for their solution.
Improper Constraints
Lines of action of the reactive forces are concurrent at A
Lines of action of the reactive forces intersect a common axis
the loading P will rotate the
member about the AB axis
The applied loading P will cause
the beam to rotate slightly
about A
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14
Improper constraining leads to instability when the reactive
forces are parallel