6161103 5.7 constraints for a rigid body

5.7 Constraints for a Rigid Body

To ensure the equilibrium of a rigid body, it
is necessary to satisfy the equations
equilibrium and have the body properly held
or constrained by its supports
Redundant Constraints
More support than needed for equilibrium
Statically indeterminate: more unknown
loadings on the body than equations of
equilibrium available for their solution


Example
For the 2D and 3D problems, both are
statically indeterminate because of additional
supports reactions
In 2D, there are 5 unknowns but 3
equilibrium equations can be drawn


Example
In 3D, there are 8 unknowns but 6 equilibrium
equations can be drawn
Additional equations
involving the physical
properties of the body
are needed to solve
indeterminate problems


Improper Constraints
Instability of the body caused by the
improper constraining by the supports
In 3D, improper constraining occur when the
support reactions all intersect a common
axis
In 2D, this axis is perpendicular to the plane
of the forces and appear as a point
When all reactive forces are concurrent at
this point, the body is improperly constrained


Example
From FBD, summation of moments about the x axis
will not be equal to zero, thus rotation occur
In both cases,
impossible to
solve completely
for the unknowns


Instability of the body also can be caused by
the parallel reactive forces
Example
Summation of
forces along the
x axis will not be
equal to zero

Instability of the body also can be caused when
a body have fewer reactive forces than the
equations of equilibrium that must be satisfied
The body become partially constrained
Example
If O is a point not located on line AB, loading
condition and equations of equilibrium are not
satisfied


Proper constraining requires
- lines of action of the reactive forces do not
insect points on a common axis
- the reactive forces must not be all parallel to
one another
When the minimum number of reactive forces is
needed to properly constrain the body, the
problem is statically determinate and equations
of equilibrium can be used for solving


Procedure for Analysis
Free Body Diagram
Draw an outlined shape of the body
Show all the forces and couple moments
acting on the body
Establish the x, y, z axes at a convenient
point and orient the axes so that they are
parallel to as many external forces and
moments as possible
Label all the loadings and specify their
directions relative to the x, y and z axes


Free Body Diagram
In general, show all the unknown
components having a positive sense along
the x, y and z axes if the sense cannot be
determined
Indicate the dimensions of the body
necessary for computing the moments of
forces


Equations of Equilibrium
If the x, y, z force and moment components
seem easy to determine, then apply the six
scalar equations of equilibrium,; otherwise,
use the vector equations
It is not necessary that the set of axes
chosen for force summation coincide with the
set of axes chosen for moment summation
Any set of nonorthogonal axes may be
chosen for this purpose


Choose the direction of an axis for moment
summation such that it insects the lines of
action of as many unknown forces as
possible
In this way, the moments of forces passing
through points on this axis and forces which
are parallel to the axis will then be zero
If the solution yields a negative scalar, the
sense is opposite to that was assumed

Example 5.15
The homogenous plate has a mass of 100kg
and is subjected to a force and couple
moment along its edges. If it is supported in
the horizontal plane by means of a roller at
A, a ball and socket joint
at N, and a cord at C,
determine the components
of reactions at the supports.

Solution
FBD
Five unknown reactions acting on the plate
Each reaction assumed to act in a positive
coordinate direction


Solution
∑ Fx = 0; Bx = 0
∑ Fy = 0; B y = 0
∑ Fz = 0; Az + Bz + TC − 300 N − 981N = 0
Moment of a force about an axis is equal to the
product of the force magnitude and the
perpendicular distance from line of action of the
force to the axis
Sense of moment determined from right-hand
rule


Solution
∑ M x = 0; TC (2m) − 981N (1m) + BZ (2m) = 0
∑ M y = 0;
300 N (1.5m) + 981N (1.5m) − Bz (3m) − Az (3m) − 200 N .m = 0
Components of force at B can be eliminated if x’,
y’ and z’ axes are used
∑ M x ' = 0;981N (1m) + 300 N (2m) − Az (2m) = 0
∑ M y ' = 0;
− 300 N (1.5m) − 981N (1.5m)200 N .m + TC (3m) = 0

Solution
Solving,
Az = 790N Bz = -217N TC = 707N
The negative sign indicates Bz acts downward
The plate is partially constrained since the
supports cannot prevent it from turning about
the z axis if a force is applied in the x-y plane

Example 5.16
The windlass is supported by a thrust
bearing at A and a smooth journal bearing at
B, which are properly aligned on the shaft.
Determine the magnitude of the vertical force
P that must be applied to the
handle to maintain equilibrium
of the 100kg bucket. Also,
calculate the reactions at the bearings.

Solution
FBD
Since the bearings at A and B are aligned
correctly, only force reactions occur at these
supports


Solution
∑ M x = 0;
981N (0.1m) − P(0.3 cos 30o m) = 0
P = 377.6 N
∑ M y = 0;
− 981N (0.5m) + Az (0.8m) + (377.6 N )(0.4m) = 0
Az = 424.3 N
∑ M z = 0;
− Ay (0.8m) = 0
Ay = 0


Solution
∑ Fx = 0;
Ax = 0
∑ Fy = 0;
0 + By = 0
By = 0
∑ Fz = 0;
424.3 − 981 + Bz − 377.6 = 0
Bz = 934 N

Example 5.17
Determine the tension in cables BC and BD
and the reactions at the ball and socket joint
A for the mast.

Solution
FBD
Five unknown force
magnitudes


Solution
r r
F = {−1000 j }N
r r r r
FA = Ax i + Ay j + Az k
r r r
TC = 0.707TC i − 0.707TC k
r
r  rBD  3 r 6 r 6 r
TD = TD 
 r  = − 9 TD i + 9 TD j − 9 TD k

 BD 

r
∑ F = 0;
r r r r
F + FA + TC + TD = 0
3 r 6 r 6 r
( Ax + 0.707TC − TD )i + (−1000 + Ay + TD ) j + ( Az − 0.707TC − TD )k = 0
9 9 9


Solution
3
∑ Fx = 0; Ax + 0.707TC − TD = 0
9
6
∑ Fy = 0;−1000 + Ay + TD = 0
9
6
∑ Fz = 0; Az − 0.707TC − TD = 0
9

r
∑ M A = 0;
r r r r
rB X ( F + TC + TD ) = 0
r r r r 3 r 6 r 6 r
6k X (−1000 j + 0.707TC i − 0.707TC k − TD i + TD j − TD k ) = 0
9 9 9


Solution
r r
(−4TD + 6000)i + ( 4.24TC − 2TD ) j = 0
∑ M x = 0;−4TD + 6000 = 0
∑ M y = 0;4.24TC − 2TD = 0
Solving,
TC = 707 N
TD = 1500 N
Ax = 0 N
Ay = 0 N
Az = 1500 N

Example 5.18
Rod AB is subjected to the 200N force.
Determine the reactions at the ball and
socket joint A and the
tension in cables BD
and BE.

Solution
FBD


Solution
r r r r
FA = Ax i + Ay j + Az k
r r
TE = TE i
r r
TD = TD i
r r
F = {−200k }N

r
∑ F = 0;
r r r r r
FA + FE + TE + TD + F = 0
r r r
( Ax + AE )i + ( Ay + TD ) j + ( Az − 200)k = 0


Solution
∑ Fx = 0; Ax + TE = 0
∑ Fy = 0; Ay + TD = 0
∑ Fz = 0; Az − 200 = 0

r
∑ M A = 0;
r r r r r
rC XF + rB X (TE + TD ) = 0
Since rC = 1/2rB,
r r r
r r r r r r r
(0.5i + 1 j − 1k ) X (200k ) + (1i + 2 j − 2k ) X (TE i + TD j ) = 0


Solution
r r r
(2TD − 200)i + (−2TE + 100) j + (TD − 2TE )k = 0
∑ M x = 0;2TD − 200 = 0
∑ M y = 0;−2TE + 100 = 0
∑ M z = 0;TD − 2TE = 0
Solving,
TD = 100 N
TE = 50 N
Ax = −50 N
Ay = −100 N
Az = 200 N

Example 5.19
The bent rod is supported at A by a
journal bearing, at D by a ball and
socket joint, and at B by means of
cable BC. Using only one equilibrium
equation, obtain a direct solution for
the tension in cable BC. The bearing at
A is capable of exerting force
components only in the z and y
directions since it is properly aligned.

Solution
FBD
Six unknown
Three force components
caused by ball and socket
joint
Two caused by bearing
One caused by cable


Solution
Direction of the axis is defined by the unit
vector
rDA 1 1
u= =− i− j
rDA 2 2
= −0.707 − 0.707j
i

∑MDA = u ⋅ ∑(rXF) = 0


Solution
u ⋅ (rB XTB + rE XW ) = 0
0 .2 0.3 0 .6
( −0.707i − 0.707 j ).[(−1 j ) X ( TB i − TB j + TB k )
0 .7 0 .7 0.7
+ (−0.5 j ) X (−981k )] = 0
( −0.707i − 0.707 j ).[(−0.8577TB + 490.5)i + 0.286TB k ] = 0
490.5
TB = = 572 N
0.857

6161103 5.7 constraints for a rigid body

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6161103 5.7 constraints for a rigid body