Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

SolutioN
When we refer to case 4 of Table 7.2, our first instinct might be to represent the linearly
distributed load on the beam with just a single term:
50 kN/m
term 1 = – x – 0 m 1
2.5 m
50 kN/m
wx ( ) =− 〈 x − 0m〉
1
2.5 m
However, this term by itself produces a load that continues to increase as x increases. But
we need to terminate the linear load at B, so we might try adding the algebraic inverse of
the linearly distributed load to the w(x) equation:
50 kN/m
50 kN/m
wx ( ) =− 〈 x − 0m〉 + 〈 x − 2.5 m〉
2.5 m
2.5 m
1 1
The sum of these two expressions represents the loading shown next. While the second
expression has indeed cancelled out the linearly distributed load from B onward, a uniformly
distributed loading remains.
50 kN/m
A
B
x
=
50 kN/m
A
B
x
2.5 m
2.5 m
50 kN/m
term 2 = + x – 2.5 m 1
2.5 m
Adding the inverse of the linearly
increasing load to the beam at B . . .
. . . eliminates the linear portion of the load;
however, a uniformly distributed load remains.
To cancel this uniformly distributed load, a third term that begins at B is required:
50 kN/m
50 kN/m
wx ( ) =− 〈 x − 0m〉 + 〈 x − 2.5 m〉 + 50kN/m〈 x − 2.5 m〉
2.5 m
2.5 m
1 1 0
50 kN/m
A
2.5 m
B
An additional uniform-load term
that begins at B is required in order to
cancel the remaining uniform load.
x
term 3 = +50 kN/m x – 2.5 m 0
=
50 kN/m
2.5 m
therefore, three terms must be
superimposed in order to obtain the desired
linearly distributed load between A and B.
As shown in this example, three terms are required to represent the linearly increasing
load that acts between A and B. Case 6 of Table 7.2 summarizes the general discontinuity
expressions for a linearly increasing load. Similar reasoning is used to develop case
7 of Table 7.2 for a linearly decreasing distributed loading.
A
B
x
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