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

Note that the second term in this expression is required in order to cancel out the first term
for x > 4 m.
Distributed load between C and E: Again, use case 5 of Table 7.2 to write the following
expression for the 40 kN/m distributed load:
0 0
wx ( ) =−40 kN/m x − 9m + 40 kN/m x − 15 m
The second term in this expression will have no effect, since the beam is only 15 m long;
therefore, that term will be omitted from further consideration.
Reaction forces A y and D y : The upward reaction forces at A and D are expressed by using
case 2 of Table 7.2:
−1 −1
wx ( ) = 200 kN x − 0m + 280 kN x −12 m
Integrate the beam load expression: The load expression w(x) for the beam is thus
−1 0 0
wx ( ) = 200 kN x − 0m − 60 kN/m x − 0m + 60 kN/m x − 4m
0 −1
−40 kN/m x − 9m + 280 kN x −12 m
Integrate w(x) to obtain the shear-force function V(x):
∫
0 1 1
V( x) = wx ( ) dx = 200 kN x − 0m − 60 kN/m x − 0m + 60 kN/m x − 4m
1 0
−40 kN/m x − 9m + 280 kN x −12 m
Then integrate again to obtain the bending-moment function M(x):
60 kN/m
60 kN/m
Mx ( ) = ∫V( x) dx = 200 kN x − 0m − x − 0m + x − 4m
2
2
40 kN/m
2 1
− x − 9m + 280 kN x −12 m
2
1 2 2
The inclusion of the reaction forces in the expression for w(x) has automatically accounted
for the constants of integration up to this point. However, the next two integrations (which
will produce functions for the beam slope and deflection) will require constants of
integration that must be evaluated from the beam boundary conditions.
From Equation (10.1), we can write
EI d 2
v
2
dx
60 kN/m
60 kN/m
= Mx ( ) = 200 kN x − 0m − x − 0m + x − 4m
2
2
40 kN/m
2 1
− x − 9m + 280 kN x −12 m
2
1 2 2
Integrate the moment function to obtain an expression for the beam slope:
EI dv
dx
200 kN
60 kN/m
60 kN/m
= x − 0m − x − 0m + x − 4m
2
6
6
40 kN/m
3 280 kN
2
− x − 9m + x − 12 m + C1
6
2
2 3 3
(a)
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