Page 273 - Intro to Tensor Calculus

267

EXAMPLE 2.4-4 Assume there exist a state of plane strain with zero body forces. For F 11 ,F 12 ,F 22
constants, consider the function defined by

1 2 2
φ = φ(x, y)= F 22 x − 2F 12 xy + F 11 y .
2
4
This function is an Airy stress function because it satisfies the biharmonic equation ∇ φ = 0. The resulting
stress field is
2
2
2
∂ φ ∂ φ ∂ φ
σ xx = = F 11 σ yy = = F 22 σ xy = − = F 12 .
∂y 2 ∂x 2 ∂x∂y
This example, corresponds to stresses on an infinite flat plate and illustrates a situation where all the stress
components are constants for all values of x and y. In this case, we have σ zz = ν(F 11 +F 22 ). The corresponding
strain field is obtained from the constitutive equations. We find these strains are

1+ ν 1+ ν 1+ ν
e xx = [(1 − ν)F 11 − νF 22 ] e yy = [(1 − ν)F 22 − νF 11 ] e xy = F 12 .
E E E
The displacement field is found to be

1+ ν 1+ ν
u = u(x, y)= [(1 − ν)F 11 − νF 22 ] x + F 12 y + c 1 y + c 2
E E

1+ ν 1+ ν
v = v(x, y)= [(1 − ν)F 22 − νF 11 ] y + F 12 x − c 1 x + c 3 ,
E E
with c 1 ,c 2 ,c 3 constants, and is obtained by integrating the strain displacement equations given in Exercise
2.3, problem 2.

EXAMPLE 2.4-5. A special case from the previous example is obtained by setting F 22 = F 12 =0.
This is the situation of an infinite plate with only tension in the x−direction. In this special case we have
1
2
φ = F 11 y . Changing to polar coordinates we write
2
F 11 2 2 F 11 2
φ = φ(r, θ)= r sin θ = r (1 − cos 2θ).
2 4
The Exercise 2.4, problem 20, suggests we utilize the Airy equations in polar coordinates and calculate the
stresses
2
1 ∂φ 1 ∂ φ 2 F 11
σ rr = + = F 11 cos θ = (1 + cos 2θ)
2
r ∂r r ∂θ 2 2
2
∂ φ 2 F 11
σ θθ = = F 11 sin θ = (1 − cos 2θ)
∂r 2 2
2
1 ∂φ 1 ∂ φ F 11
σ rθ = − = − sin 2θ.
2
r ∂θ r ∂r∂θ 2