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

segment AB of the beam. The second free-body diagram cuts through the beam at a
distance x′ from roller support C. From this diagram, derive the bending-moment
equation M(x′) for segment BC of the beam. Substitute two moment expressions into
Equation (17.20) to determine the elastic strain energy for the complete beam.
P
SolutioN
From the free-body diagram for the entire beam, determine
the vertical reaction force at A:
A
Pb
L
a
L
B
b
C
Pa
L
x
A
y =
Pb
L
Also, determine the reaction force at C:
C
y =
Pa
L
A
Pb
L
x
a
a
V
M
Note that the horizontal reaction force at A has been omitted, since A x = 0.
Now draw a free-body diagram that cuts through the beam between A and B at a
distance x from pin support A. Then, sum moments about section a–a to derive the
bending-moment equation for segment AB of the beam:
∑ M M Pb
a−a
= −
L x = 0
∴ M = Pb
L x (0 ≤ x ≤ a )
M
V
b
b
x'
C
Pa
L
Similarly, draw a free-body diagram that cuts through the beam between B and C at
a distance x′ from roller support C. Then, sum moments about section b–b to derive the
bending-moment equation for segment BC of the beam:
∑ M M Pa
b−b
=− +
L x ′ = 0
∴ M = Pa
L x ′ (0 ≤ x ′ ≤ b )
The total elastic strain energy in the beam is the sum of the elastic strain energies in
segments AB and BC. By Equation (17.20),
U = U + U
AB
BC
2
1 a⎛
Pb
EI L x ⎞
dx 1 b Pa
= ∫ ⎜ ∫
2
EI L x dx
0 ⎝ ⎠
⎟ + ⎛
2
′ ⎞
⎜ ⎟ ′
0 ⎝ ⎠
2 2
Pb
LEI a
2 2
3
Pa
6 6LEI b 3
= +
2
2
2 2 2
Pab
= ( a + b)
2
6LEI
2
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