Related Variational and Energy Principles - Mechanics of Structures: Variational and Computational Methods - page 135

FIGURE 3.1
Statically indeterminate bar system.

If the structural system is subjected to a specified temperature distribution, the temper-
atures would have to be kept constant if Ca stigliano's first theorem is to apply. Then Eq.
(3.5) could be written P .

This theorem applies also to structural systems with nonlinear stress-strain laws and also
to structures with large displacements. In the latter case it is necessary to calculate the strain
energy using the expressions for large strains (Chapter 1, Eq. 1.16).
It should be emphasized that by “forces” are mea nt both forces and moments. For ex-
ample, if the rotation corresponding to the moment M k is

(∂ U i /∂ V ) =

T = constant

θ

k , then Castigliano's theorem,

part I would state that

∂ U i

∂θ k =

M k (3.6)

EXAMPLE 3.1 Forc e-D isplacement Relationships for a Structure
Determine the force P that will cause the joint at a of the bar system (truss) of Fig. 3.1 to
displace vertically a certain distance V

.
Castigliano's theorem, part I is represented by Eq. (3.4), where P k is the force in the di-
rection of the displacement V k Here, for this simple structure, we can drop the subscript

k . In order to employ this formula, which leads to a relationship between force and dis-
placement, the strain energy U i must first be expressed in terms of the displacement V

.

The
theorem can be used for statically determinate or indeterminate systems; ours is statically
indeterminate.
From Chapter 2, Eq. (2.5), the strain energy for an extension bar of length L with constant
cross-sectional area A and axial force N can be expressed as
L

.

2

A
2

EA
2

v

2
x

E dx = (1)

L

0

where it is recognized that the axial strain along the bar is constant = ∂ u 0 /∂ x = v/ L,

x

where is the total elongation.

The total strain energy stored in the three rods is the sum of the energy in each rod. This
still must be converted to a function of V, the vertical displacement of point a

v

Since the
displacements are small, it will be assumed that the bars remain at the same angles with
respect to each other throughout the deformation process. Then kinematically admissible
V of any of the bars can be expressed by

.

v = α α

V cos , where is the angle measured from

0 , 60 ◦ , or 60 ◦ ) the vertical ( − , so that

V 2 1

2 V 2

1

2 V 2

3 EAV 2
4 L

EA
2 L

U i = + + = (2)