Distributed Loads
Decimals have a point.
Distributed Loads
¢  Up
to this point, all the forces we have
considered have been point loads
¢  Single forces which are represented by a vector
¢  Not all loading conditions are of that type
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Distrubuted Loads
Monday, November 5, 2012
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Distributed Loads
¢  Consider
how your ears feel as you go deeper
into a swimming pool.
¢  The deeper you go, the greater the pressure on
your ears.
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Distrubuted Loads
Monday, November 5, 2012
Distributed Loads
¢  If
we consider how this pressure acts on the
walls of the pool, we would have to consider a
force (generated by the pressure) that was
small at the top and increased as we went
down.
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Distrubuted Loads
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Distributed Loads
¢  This
is known as a distributed force or a
distributed load.
¢  It is represented by a series of vectors which
are connected at their tails.
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Distrubuted Loads
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Distributed Loads
¢  One
type of distributed load is a uniformly
distributed load
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Distrubuted Loads
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Distributed Loads
¢  This
load has the same intensity along its
application.
¢  The intensity is given in terms of Force/Length
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Distrubuted Loads
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Distributed Loads
¢  The
total magnitude of this load is the area
under the loading diagram.
¢  So here it would be the load intensity time the
beam length.
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Distrubuted Loads
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Distributed Loads
¢  If,
for analysis purposes, we wanted to replace
this distributed load with a point load, the
location of the point load would be in the center
of the rectangle.
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Distrubuted Loads
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Distributed Loads
¢  We
do this to solve for reactions.
¢  For a uniform load, the magnitude of the
equivalent point load is equal to the area of the
loading diagram and the location of the point
load is at the center of the loading diagram.
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Distrubuted Loads
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Distributed Loads
¢  A
second type of loading we often encounter is
a triangular load
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Distrubuted Loads
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Distributed Loads
¢  A
triangular load has an intensity of 0 at one
end and increases to some maximum at the
other end.
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Distrubuted Loads
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Distributed Loads
¢  You
will often see the intensity represented with
the letter w.
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Distrubuted Loads
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Distributed Loads
¢  The
magnitude of an equivalent point load will
again be the area under the loading diagram.
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Distrubuted Loads
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Distributed Loads
¢  For
a triangle, this would be ½ the base times
the maximum intensity.
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Distrubuted Loads
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Distributed Loads
¢  The
location of the equivalent point load will be
2/3 of the distance from the smallest value in
the loading diagram.
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Distrubuted Loads
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Distributed Loads
¢  There
are other types of loading diagrams but
these will be sufficient for now.
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Distrubuted Loads
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Distributed Loads
¢  You
may see a diagram that appears to be a
trapezoidal loading.
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Distrubuted Loads
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Distributed Loads
¢  In
this case, we can divide the loading diagram
into two parts, one a rectangular load and the
other a triangular load.
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Distrubuted Loads
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Distributed Loads
¢  Now
you have two loads that you already have
the rules for.
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Distrubuted Loads
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Distributed Loads
¢  Take
care to note that the maximum intensity of
the triangular load is now reduced by the
magnitude of the rectangular load.
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Distrubuted Loads
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Example Problem
¢  Given:
The loading and support as shown
¢  Required: Reactions at the supports
200 lb/ft
100 lb/ft
A
B
5 ft
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Distrubuted Loads
4 ft
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Example Problem
¢  Isolate
the selected system from all connections
¢  Start with the pin at A
200 lb/ft
100 lb/ft
A
B
5 ft
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4 ft
Distrubuted Loads
Monday, November 5, 2012
Example Problem
¢  We
have a pin, so we have an x and a y
component of the reaction and we will assume
that both of them are +
200 lb/ft
100 lb/ft
A
Ay
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B
Ax
5 ft
Distrubuted Loads
4 ft
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Example Problem
¢  Now
we can remove the roller support at B
recognizing that the direction of the reaction is
+y
200 lb/ft
100 lb/ft
A
Ay
25
B
Ax
5 ft
4 ft
Distrubuted Loads
By
Monday, November 5, 2012
Example Problem
¢  We
now have all the reactions identified and can
proceed with the analysis
200 lb/ft
100 lb/ft
A
Ay
26
B
Ax
5 ft
Distrubuted Loads
4 ft
By
Monday, November 5, 2012
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Example Problem
¢  The
best idea is to now convert all distributed
loads into point loads
200 lb/ft
100 lb/ft
A
Ay
27
B
Ax
5 ft
4 ft
Distrubuted Loads
By
Monday, November 5, 2012
Example Problem
¢  Break
the load into a rectangular load and a
triangular load
100 lb/ft
100 lb/ft
A
Ay
28
B
Ax
5 ft
Distrubuted Loads
4 ft
By
Monday, November 5, 2012
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Example Problem
¢  For
the rectangular load
900 lb
100 lb/ft
4.5 ft
A
Ay
29
B
Ax
5 ft
4 ft
Distrubuted Loads
By
Monday, November 5, 2012
Example Problem
¢  For
the triangular load
900 lb
200 lb
7.67 ft
4.5 ft
A
Ay
30
B
Ax
5 ft
Distrubuted Loads
4 ft
By
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Example Problem
¢  We
have three unknowns, Ay, Ax, and By
¢  Luckily we have three equilibrium constraints to
solve for them
∑ Fx = 0
∑F =0
∑M = 0
y
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900 lb
200 lb
7.67 ft
4.5 ft
A
B
Ax
Ay
5 ft
4 ft
Distrubuted Loads
By
Monday, November 5, 2012
Example Problem
¢  Summing
the moments about A to solve for By.
∑F =0
∑F =0
∑M = 0
900 lb
x
y
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200 lb
7.67 ft
4.5 ft
A
Ay
B
Ax
Distrubuted Loads
5 ft
4 ft
By
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Example Problem
¢  Writing
the expression for the sum of the
moments around A
∑ M = 0 = − ( 4.5 ft )(900lb) − (7.67 ft )( 200lb) + (9 ft ) ( B )
y
900 lb
200 lb
7.67 ft
4.5 ft
x
A
Ay
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B
Ax
Distrubuted Loads
5 ft
∑F =0
∑F =0
∑M = 0
4 ft
By
y
Monday, November 5, 2012
Example Problem
¢  Isolating
and solving for By
( 4.5 ft )(900lb ) + ( 7.67 ft )( 200lb ) = B
y
(9 ft )
620.37lb = By
900 lb
200 lb
7.67 ft
4.5 ft
x
A
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Ay
B
Ax
Distrubuted Loads
5 ft
∑F =0
∑F =0
∑M = 0
4 ft
By
y
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Example Problem
¢  Sum
∑F
y
of the forces in the y-direction
∑F =0
∑F =0
∑M = 0
x
= 0 = Ay − 900lb − 200lb + By
y
0 = Ay − 900lb − 200lb + 620.37lb
Ay = +900lb + 200lb − 620.37lb
Ay = 479.63lb
200 lb
7.67 ft
4.5 ft
A
Ay
35
900 lb
B
Ax
5 ft
Distrubuted Loads
4 ft
By
Monday, November 5, 2012
Example Problem
¢  Since
the magnitude of our solution came out
positive, we assumed the correct direction
∑F =0
∑F =0
F
=
0
=
A
−
900
lb
−
200
lb
+
B
∑ y
y
y
∑M = 0
900 lb
200 lb
0 = Ay − 900lb − 200lb + 620.37lb
x
y
7.67 ft
4.5 ft
Ay = +900lb + 200lb − 620.37lb
A
Ay = 479.63lb
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Ay
Distrubuted Loads
B
Ax
5 ft
4 ft
By
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Example Problem
¢  Now
our final constraint condition
∑F =0
∑F =0
∑M = 0
∑ Fx = 0 = Ax
x
y
0 = Ax
900 lb
Ax = 0lb
200 lb
7.67 ft
4.5 ft
A
Ay
37
B
Ax
5 ft
4 ft
Distrubuted Loads
By
Monday, November 5, 2012
Example Problem
¢  So
our complete solution to the problem is
∑F =0
∑F =0
∑M = 0
x
Ax = 0lb
y
Ay = 479.63lb
900 lb
By = 620.37lb
200 lb
7.67 ft
4.5 ft
A
38
Ay
Distrubuted Loads
B
Ax
5 ft
4 ft
By
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Distrubuted Loads
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Distrubuted Loads
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Homework
¢  Problem
4-145
¢  Problem 4-148
¢  Problem 4-153
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