Elasticity_ Barber

446 28 The prismatic bar
σ zz (0, ζ, ¯ζ) = S(2 + ν)ζ ¯ζ(ζ + ¯ζ)
4a 2 (1 + ν)
−
S(3 − 2ν)(ζ + ¯ζ)
6
Ψ(0, ζ, ¯ζ) = Ā1[(3 + 2ν)a 2 − 2ζ ¯ζ] − A 1 (1 + 2ν)ζ 2
4(1 + ν)
+ A 2 ζ + Ā2 ¯ζ + B 2
. (28.70)
Clearly the shear tractions Ψ = σ zx + ıσ zy on z = 0 can be set to zero in the
strong sense by choosing A 1 = 0. For the normal tractions, we use (28.3) to
calculate the force and moment resultants
F z =
∫ a ∫ 2π
0
0
σ zz (0, r, θ)rdθdr = πB 2 a 2
M x + ıM y =
∫ a ∫ 2π
0
0
σ zz (0, r, θ)r 2 e ıθ drdθ = − πS(1 − 2ν2 )a 4
12(1 + ν)
Thus, the weak traction-free conditions are satisfied by setting
+ πĀ2a 4
2
.
A 2 = Ā2 = S(1 − 2ν2 )
6(1 + ν)
; B 2 = 0 .
Substituting these constants into the potentials and simplifying, we obtain
S(1 − 2ν) (
φ = − χ
1
a 2 (1 − ν 2 ) 4 + χ 1 S(7 + ν − 8ν
4) 2 − 8ν 3 ) (
− χ
1
24(1 − ν 2 )(1 + ν) 2 + χ 1 2)
ψ = − 2Sχ0 4
a 2 (1 − ν 2 ) + S(1 − 2ν)χ2 2
4a 2 (1 − ν 2 ) − S(1 + 3ν + 4ν2 )χ 0 2
12(1 − ν 2 )(1 + ν) − S(1 + 7ν)χ2 0
96(1 − ν 2 )(1 + ν) ,
which defines the complete solution of the problem. The complex stress components
can be recovered by substitution into equations (28.16–28.19) as
Θ =
S(ζ + ¯ζ)
3
− Sζ ¯ζ(ζ + ¯ζ)
4(1 + ν)a 2
Φ = Sζ[2a2 − 3ζ ¯ζ − (1 + 4ν)ζ 2 + 2(1 − 2ν)a 2 ]
12(1 + ν)a 2
σ zz = S(2 + ν)(3ζ ¯ζ − 2a 2 )(ζ + ¯ζ)
12(1 + ν)a 2 − Sz2 (ζ + ¯ζ)
a 2
Ψ = Sz [ (1 + 2ν)ζ 2 + 2ζ ¯ζ − a 2]
2(1 + ν)a 2 .
We have deliberately given a very detailed and hence lengthy solution of
this example problem in order to clarify the steps involved, but the astute
reader will realize that many steps might reasonably have been omitted as
trivial. For example, the original problem is even in z, which immediately
allows us to omit the terms involving the constants A 1 , Ā1, B 1 , causing the inplane
correction to be trivial in problems P j where j is odd and the antiplane
correction to be trivial when j is even.