Principles of Virtual Work: Integral Form of the Basic Equations - Mechanics of Structures: Variational and Computational Methods - page 128

2.7 Show that the complementary strain energy density for a two-dimensional linearly
elastic isotropic body is

2 E σ

y − E σ

1 1

2 G τ

U 0 =

x xy

+ σ σ +

x y

2.8 Show that the complementary strain energy density for a three-dimensional linearly
elastic isotropic body can be written as

2 E σ

z −

τ

xz

1 2 E (σ x σ y + σ y σ z + σ x σ z ) + 1 + ν

E

U 0 =

2
x

2
y

2 2

xy

2
yz

2

+ σ + σ + τ + τ

1
2 (σ x x + σ y y + σ z z + τ xy γ xy + τ yz γ yz + τ xz γ xz )

=

2.9 Determine an integral expression for the total potential energy of a beam on an elastic
foundati on (modulus k w ) with a compressive axial force ( N ) and a transverse loading

intensity p z .
2.10 In Chapter 13, it is shown that the stresses in a thin elastic plate are

E E E

σ x =

2 ( x + ν y )

, σ y =

2 ( y + ν x )

, τ xy =

+ ν) γ xy

1 − ν 1 − ν 2 ( 1

with

2 2 2

z ∂

w

z ∂

w ∂ w

=−

x 2 ,

=−

y 2 ,

γ =− 2 z

x y xy

∂ ∂ ∂ x ∂ y

where is the transverse displacement and z is the transverse coordinate measured

from the middle plane of the plate. Assume that the contribution to the strain energy
of other stress components is negligible. If t is the plate thickness and A is the area of
the plate, show that the strain energy developed in the plate is

w

∂

y 2 2

∂ ∂ 2 dx dy

Et 3

2 2 2 2 2

x 2 + ∂

w ω

x 2 ∂

w w w

U i = − 2 ( 1 − ν)

y 2 −

24 ( 1 − ν

2

) ∂ ∂ ∂ ∂ ∂ x ∂ y

A

Principle of Virtual Work and Related Theorems
2.11 Find the displacement V of point a and the elongation of each bar of the truss of
Fig. P2.11.

Answer: V = 2 PL /( 3 EA )

2.12 Show that Eq. (2.58a) is equivalent to the displacement form of the governing differ-
ential equations of motion of Chapter 1, Eq. (1.63). To do so it is necessary to utilize
the divergence theorem.
2.13 Verify the identity

σ δ

ij

= σ δ u i, j , which is used in deriving Eq. (2.49).

ij ij

Hint: Since i and j are dummy indices, they can be interchanged in an expression

1

such as σ ij δ u i, j . Hence, σ ij δ u i, j =

2 (σ ij δ

u i, j + σ ji δ u j,i ) and since σ ij = σ ji , σ ij

ij 1
2

δ

ij . Another form of proof would be verification
obtained by simply expanding with the summation convention both sides of the
identity in question.

u i, j = σ (δ u i, j + δ u j,i ) = σ δ

ij