Page 274 - Intro to Tensor Calculus

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2
2
EXAMPLE 2.4-6. We now consider an infinite plate with a circular hole x + y = a which is traction
free. Assume the plate has boundary conditions at infinity defined by σ xx = F 11 , σ yy =0, σ xy =0. Find
the stress field.
Solution:
The traction boundary condition at r = a is t i = σ mi n m or

t 1 = σ 11 n 1 + σ 12 n 2 and t 2 = σ 12 n 1 + σ 22 n 2 .

For polar coordinates we have n 1 = n r =1,n 2 = n θ = 0 and so the traction free boundary conditions at
the surface of the hole are written σ rr | r=a =0 and σ rθ | r=a =0. The results from the previous example
are used as the boundary conditions at infinity.
Our problem is now to solve for the Airy stress function φ = φ(r, θ) which is a solution of the biharmonic
equation. The previous example 2.4-5 and the form of the boundary conditions at infinity suggests that we
assume a solution to the biharmonic equation which has the form φ = φ(r, θ)= f 1 (r)+ f 2 (r)cos 2θ, where
f 1 ,f 2 are unknown functions to be determined. Substituting the assumed solution into the biharmonic
equation produces the equation

d 1 d 1 d 1 d 4 1 f 2
2 2
00
0
+ f + f 1 0 + + − f + f − 4 cos 2θ =0.
00
2
1
2
dr 2 r dr r dr 2 r dr r 2 r r 2
We therefore require that f 1 ,f 2 be chosen to satisfy the equations
d 1 d 1 d 1 d 4 1 f 2
2 2
+ f + f 1 0 =0 + − f + f − 4 =0
00
0
00
2
2
1
dr 2 r dr r dr 2 r dr r 2 r r 2
4 (iv) 4 (iv)
2 00
3 000
0
0
3 000
or r f 1 +2r f 1 − r f + rf =0 r f 2 +2r f 2 − 9r f +9rf =0
2 00
1
2
2
1
These equations are Cauchy type equations. Their solutions are obtained by assuming a solution of the form
λ
f 1 = r and f 2 = r m and then solving for the constants λ and m. We find the general solutions of the above
equations are
2 2 2 4 c 7
f 1 = c 1 r ln r + c 2 r + c 3 ln r + c 4 and f 2 = c 5 r + c 6 r + 2 + c 8 .
r
The constants c i ,i =1,... , 8 are now determined from the boundary conditions. The constant c 4 can be
arbitrary since the derivative of a constant is zero. The remaining constants are determined from the stress
conditions. Using the results from Exercise 2.4, problem 20, we calculate the stresses

c 3 c 7 c 8
σ rr = c 1 (1 + 2 ln r)+2c 2 + − 2c 5 +6 +4 cos 2θ
r 2 r 4 r 2

c 3 2 c 7
σ θθ = c 1 (3 + 2 ln r)+2c 2 − 2 + 2c 5 +12c 6r +6 4 cos 2θ
r r
c 7 c 8
2
σ rθ = 2c 5 +6c 6 r − 6 − 2 sin 2θ.
r 4 r 2