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Shear deformation effects, which are frequently small, are easily included in a solution. In
the above solution, the expressions for the moments and axia l f orces derived previously a re
still valid. The internal shear forces can be written as V bar 3
=
P, V bar 2
=
0 , and V bar 1
=−
P
.
Shear deformation contributes terms of the form (see Chapter 2, Example 2.2)
V 2
2 Gk s A dx
to the total strain energy, where k s is the shear shape factor. Then
a
2 a
Gk s A ∂
V
V
P
Gk s A dx
P
Gk s A dx =
3 Pa
Gk s A
V d horizontal
(
) =
dx
=
+
(3)
∂
P
due to shear
0
0
The total horizontal displacement of d is then
2 Pa 3
EI
Pa
EA +
3 Pa
Gk s A
V d horizontal
=
+
(4)
EXAMPLE 3.10 In-Plane Deformation of a Curved Bar
Determine the vertical displacement of point a of the curved bar of Fig. 3.8a. Segment 2 is
a circular arc.
This type of problem is similar to the ones treated in Example 3.9 in that several kinds of
internal forces (stress resultants) occur simultaneously. An axial force acts on element 1; a
bending moment, shear force, and an axial force act on segment 2; and a bending moment
and shear force act on member 3. For this problem, only the bending moment, which is
usually the dominant force, will be considered.
Castigliano's theorem, part II in the form
d
= ∂
U i
∂
EI ∂
M
M
∂
P =
V a vertical
dx
(1)
P
a
provides the vertical movement of point a
Use of formula (1), which has been derived for
straight bars, is justifiable for curved bars if the curved bars cross-sectional dimensions are
very small in comparison with the radius of c ur vature of the middle line. For the various
bar segments, the distribution of M and
.
∂
M
/∂
P are found to be (Fig. 3.8b)
1
∂
M
|
1
M
|
=
0 ,
P =
0
∂
2
∂
M
|
2
M
|
=−
PR
(
1
−
cos
α)
,
P =−
R
(
1
−
cos
α)
(2)
∂
3
∂
M
|
3
Px −
x −
M
|
=−
PR,
P =−
R
∂
|
1
=
|
2
=
|
3
=
where M
M bar 1 ,M
M bar 2 ,M
M bar 3
Then
a 1
π/ 2
a 2
|
1
∂
|
1
|
2
∂
|
2
|
3
∂
|
3
M
M
M
M
M
M
dx
=
+
α +
V a vertical
dx
Rd
EI
∂
P
EI
∂
P
EI
∂
P
0
0
0
PR 3
Pa 2
Pa 2 R
EI
PR 2 a 2
EI
(
3
π −
8
)
=
+
3 EI +
+
(3)
4 EI
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