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

Finally, derive the following equation for the real internal moment M in segment DE
of the beam:
M
⎛ P ⎞
= ⎜ + 150 kN⎟
x
⎝ 2 ⎠
0 ≤ x ≤ 3 m
4
4
Now differentiate each equation in M with respect to P to obtain ∂M/ ∂P
. Then substitute
P = 0 into each equation in M for the beam. The resulting expressions are summarized
in the following table:
+
Beam
Segment M (kN ⋅ m)
AB
45 ⎛ P ⎞
- x + ⎜ + 300⎟
x
2 ⎝ 2 ⎠
1 2 1
45
⎛ P ⎞
BC - x - (180)( x - 3) + ⎜ + 300⎟
x
2
⎝ 2 ⎠
2 2 2 2
∂M
∂ P
(m)
M
(for P = 0 kN)
(kN ⋅ m)
1
45
x
2 1 - x + 300x
2
1 2 1
1 45
x
2 2 - x -180( x - 3) + 300x
2
2 2 2 2
CD
DE
45
P
x
x
2 ( 3) 2
⎛ ⎞
- 3 - + ⎜ + 150⎟
⎝ 2 ⎠
⎛ P ⎞
⎜ + 150⎟
x4
⎝ 2 ⎠
3
1 45
x
2 3 2
- ( x3
- 3) + 150x
2
1
x
2 4 150x
4
3
Castigliano’s second theorem applied to beam deflections is expressed by Equation
(17.40). Substitute the expressions just derived for ∂M/∂P and M for each beam segment
into Equation (17.40), taking care to note the appropriate limits of integration for each
segment. These expressions, as well as the results of the integration, are summarized in
the following table:
Beam
Segment
x Coordinate
origin
limits
(m)
⎛∂M
⎞
⎜ ⎟
⎝ ∂P
⎠
M
(kN ⋅ m 2 )
⎛ M M
∫ ∂ ⎞
⎜ ⎟ ⎛ ⎝ ∂P
⎠ ⎝ ⎜ ⎞
⎟
EI ⎠
dx
AB A 0–3 - 11.25x + 150x
1 3 1 2
BC A 3–4.5 - 11.25x + 60x + 270x
2 3 2 2 2
CD E 3–4.5 - 11.25x + 142.5x - 101.25x
2
DE E 0–3 75x
4
3 3 3 2 3
1,122.188 kN⋅m 3
EI
1,875.762 kN⋅m 3
EI
1,550.918 kN⋅m 3
EI
675.0 kN⋅m 3
EI
5,223.868 kN⋅m 3
EI
786