Use of the Principle of Least Work and Castigliano's Second Theorem
2a
P
P
2b
Determine the elongation of the frame between the points
where the two loads are applied
For simplicity consider one quarter of the frame
a
u
b
s
F = P/2
M0
there is no shear stress at the cut
because of symmetry and no normal
force, N,
because of symmetry and
↑ ∑ Fy = 0
N
N
We don't know the value of the internal moment M0, so this problem is
statically indeterminant. From the Principle of Least Work
δU =
∂U
δ M0 = 0
∂M 0
for arbitrary δ M 0
U=
Here, we keep only the strain energy due to bending
M (s) =
P
s − M0
2
u
M (u ) =
s
F = P/2
M0
Thus,
1
M 2 ( x ) dx
∫
2 EI
Pb
− M0
2
Pb
− M0
2
∂M ( s )
∂M ( u )
1
1
∂U
M
s
ds
M
u
du = 0
=
+
( )
( )
∫
∫
EI 0
∂M 0 EI 0
∂M 0
∂M 0
b
1 ⎛ Ps
1 ⎛ Pb
⎞
⎞
M
1
ds
M
−
−
+
−
(
)
0⎟
0 ⎟ ( −1) du = 0
⎜
⎜
∫
∫
EI 0 ⎝ 2
EI 0 ⎝ 2
⎠
⎠
b
which gives
a
a
Pba
− Pb 2
+ M 0b −
+ M 0a = 0
4
2
Thus, we obtain
Pb ( 2a + b )
M0 =
4 ( a + b)
Now that we have the moment we can find the displacement ∆ F
u
a
b
s
F = P/2
∆F
M0
Note that
But we have
U = U ( F , M 0 ( F ))
∂U
∆F =
∂F
M0
∂U
+
∂M 0
Thus, even though M 0 = M 0 ( F )
F
∂M 0
∂F
we can treat M0 as if it was a constant
when differentiating the strain energy
1
U=
2 EI
b
∫ ( Fs − M 0 )
0
1
∂U
∆F =
=
∂F EI
2
1
ds +
2 EI
b
1
Fs
−
M
sds
+
(
)
0
∫0
EI
a
∫ ( Fb − M 0 )
2
du
0
a
∫ ( Fb − M ) bdu
0
0
⎞
1 ⎛ Fb3 M 0b 2
2
Fab
M
ab
=
−
+
−
⎜
⎟
0
EI ⎝ 3
2
⎠
Placing in the expressions for F and M0 in terms of P
Pb3 ( 4a + b )
∆F =
24 EI ( a + b )
We need to double this result since this is only the elongation of one
side, giving
Pb3 ( 4a + b )
∆ P = 2∆ F =
12 EI ( a + b )