Principles of Virtual Work: Integral Form of the Basic Equations - Mechanics of Structures: Variational and Computational Methods - page 127

Problems

Work and Energy

T

= σ T

2.1 Verify that U 0 of Eq. (2.10) satisfies (∂ U 0 /∂ ) (Eq. 2.22). Also check Eq. (2.23)

using U 0 of Eq. (2.15).
2.2 For a nonlinear elastic material with an isotropic coefficient of linear thermal expan-
sion, show that the strain energy per unit volume can be written as

ij T

U 0 = σ

ij d

− (σ + σ + σ )α d T

ij x y z

0 0

and the complementary energy density as

T

T d

σ ij σ ii T

U 0 =

ij d

σ − α d σ + (σ + σ + σ ) α d T

ij ii x y z

0 0 0 0

If the material is linearly elastic, show that these reduce to

1

2 σ ij ij − α

T σ ii 1

2 σ ij ij + α

T σ ii

U 0 =

U 0 = and

2 2

2.3 Show that for a linearly elastic solid in plane stress

1
1

xy

E 1

2

2

U 0 =

− ν (

+ ) − 2 +

x y x y

2 ( 1 + ν)

2

− α

E T E (α T )

( + ) +

x y

1 − ν 1 − ν

2 E (σ x + σ y )

+ ν) σ x σ y − σ

xy + α

1

U 0 =

2 2

− 2 ( 1 T (σ x + σ y )

2.4 Write the complementary energy density U 0

in terms of the Airy stress function for a

solid in plane stress.

Answer:

∂

y 2 2

∂

x 2 ∂ y 2

∂ 2

2 2 2 2 2

1
2 E x 2 + ∂

ψ ψ ψ ψ ψ

U 0 = − 2 ( 1 + ν) −

∂ ∂ ∂ ∂ ∂ x ∂ y

T ∂

y 2

2 2

x 2 + ∂

ψ ψ

+ α

∂ ∂

2.5 Show that for a material following a generalized Hooke's law, if a potential function
exists for the internal work then the number of material constants is reduced from 36
to 21.

Hint: Use Eq. (2.22)

2.6 Show that the complementary strain energy for a rod of circular cross section subjected

to torsion is U i = L M t / 2 GJ dx , where M t is the net torque on a cross-section, L is

the length of the rod, G is the shear modulus, and J is the polar moment of inertia.

0