Example 1

conformal_1.jpg
Along x=0, T=100, which is translated to a + d/y = 100. For this to be independent of y, a = 100, d = 0. Along x2+y2−x=0, T=50, which is translated into 50=100 + c from which c=−50. Thus, we have
T = 100 − 50
x2 + y2
 ,
which of course matches the solution by conformal mapping.

Example 2

Consider a steady-state temperature distribution problem when a cylinder is placed in a constant heat flux as shown below.
uniform_heatflow.jpg

T1
=
2
k1 + k2
 ( hx x + hy y),
(1)
T2
=
⎛
⎝ 		1 − a2 (k1 − k2)
(k1 + k2)
 1
x2 + y2
⎞
⎠ 		hx x + hy y
k2
 .
(2)

2-D elasticity

In state of plane stress (thin plates), the balance of force equations (equilibrium equations) without body force take the following form:

∂σxx
∂x
 + ∂σxy
∂y
 = 0,
(3)
∂σyx
∂x
 + ∂σyy
∂y
 = 0,
(4)
where σxx and σyy are the normal components and σxy is the shear component.
Equations (3) and (4) alone cannot solve for the three unknowns of σxx, σyy and σxy and therefore an additional equation is needed.
For an isotropic solid, the answer is that it is the compatibility condition expressed in the stress components as
∇2 (σxx + σyy ) = 0.
(5)
Using Equation (), one can easily derive the stress components from two independent analytic functions, f(z) and g(z), as

σxx + σyy
=
4 ℜ( f′(z)),
(6)
σyy − σxx + 2 i σxy
=
2 ⎛
⎝
-
z
 
 f"(z) + g"(z)⎞
⎠ 		,
(7)
or

σxx
=
ℜ⎛
⎝ 		2 f′(z) +
-
z
 
 f"(z) + g"(z)⎞
⎠ 		,
(8)
σyy
=
ℜ⎛
⎝ 		2 f′(z) −
-
z
 
 f"(z) − g"(z)⎞
⎠ 		,
(9)
σxy
=
ℑ⎛
⎝
-
z
 
 f"(z) + g"(z)⎞
⎠ 		.
(10)
Therefore, if analytic functions, f(z) and g(z), are chosen, the stress components that satisfy the equilibrium equation are automatically derived. As for what type of f(z) and g(z) one has to choose, it depends on the boundary condition given. In general, if the region considered is singly-connected, f(z) and g(z) can be chosen as Taylor's series on z and if the region is multiply connected (holes), f(z) and g(z) can be chosen as Laurent series (more later).
Examples
Many 2-D elasticity problems (stress analysis) can be solved by assuming f(z) and g(z) to be in series format (Taylor series) and determine the unknown coefficients to satisfy the given boundary conditions.
Assume
f(z) = c1 z + c2 z2 + c3 z3,     g(z) = d1 z + d2 z2 + d3 z3,
(11)
where c1  ∼ d3 are unknown complex coefficients. The unknown coefficients, c1  ∼ d3, can be determined to satisfy the boundary condition which is usually given in terms of traction force at the boundary.

Uniform stress

Consider a rectangular plate (−a ≤ x ≤ a, −b ≤ y ≤ b) subject to a uniform traction force, S, at x = ±a and free at y = ±b.
airy_uniform.jpg
The boundary conditions are stated as:

At x = ±a,     σxx = S,     σxy = 0.

At y = ±b,     σyy = 0,     σxy = 0.
Substituting the conditions above into Equation (11), we can determine the unknown coefficients as

c1 = S
4
 ,     c2 = 0,     c3 = 0,     d1 = − S
2
 ,     d2 = 0,     d3 = 0.
Therefore, the stress functions, f(z) and g(z), are determined as

f(z) = S
4
 z,     g(z) = − S
2
 z.
Finally, the stress components are

σxx = S,     σyy = 0,     σxy = 0.
Of course, the result above is not so much impressive.

Beam

Consider a rectangular bar (−a ≤ x ≤ a, −b ≤ y ≤ b) subject to a bending moment, Mx, at x = ±a and free at y = ±b.
airy_beam.jpg
The proper boundary conditions are stated as

At x = ±a,     ⌠
⌡ 		b

−b 
 y  σxx  d y = Mx,

At y = ±b,     σyy = 0,     σxy = 0.
Substituting the conditions above into Equation (11), we can determine the unknown coefficients as

c1 = 0,     c2 = − 3 Mx
16 b3
 i,     c3 = 0,     d1 = 0,     d2 = 3 Mx
16 b3
 i,     d3 = 0.
Therefore, the stress functions, f(z) and g(z), are determined as

f(z) = − 3 Mx
16 b3
 i z2,     g(z) = 3 Mx
16 b3
 i z2.
Finally, the stress components are

σxx = 3 Mx  y
2 b3
 ,     σyy = 0,     σxy = 0.
This result agrees with those formulae found in strength of materials textbooks.

Shear at end

Consider a rectangular bar (−a ≤ x ≤ a, −b ≤ y ≤ b) subject to a shear traction force , − W, at x = a and free at y = ±b. The other end (x = 0) if fixed.
airy_shear.jpg
The proper boundary conditions are stated as

At x = a,     σxx = 0,     ⌠
⌡ 		b

−b 
  σxy  d y = − W,

At y = ±b,     σyy = 0,     σxy = 0.
Substituting the conditions above into Equation (11), we can determine the unknown coefficients as

c1 = 0,     c2 = − 3 a W
16 b3
 i,     c3 = W
16 b3
 i,     d1 = − 3 W
4 b
 i,     d2 = 3 a W
16 b3
 i,     d3 = − W
8 b3
 i.
Therefore, the stress functions, f(z) and g(z), are determined as

f(z) = − 3 a W
16 b3
 i z2 + W
16 b3
 i z3,     g(z) = − 3 W
4 b
 i z + 3 a W
16 b3
 i z2 − W
8 b3
 i z3.
Finally, the stress components are

σxx = 3 W (a − x) y
2 b3
 ,     σyy = 0,     σxy = 3 W (−b2 + y2)
4 b3
 .
This result agrees with those formulas found in structural mechanics textbooks.

Stress concentration due to a hole

An infinitely extended body (isotropic) with a hole of a radius, a, is subject to the far field stress shown Figure .
hole.jpg
Figure 1: An infinitely extended body with a hole.
The boundary conditions at infinity are
σx → σx∞,     σy → σy∞,    τ→ τ∞     as     x, y → ∞.
Along the peripheral of the hole, there is no traction. Therefore, the following must be satisfied:

tx = 0,     ty = 0     along     x2 + y2 = a2.1
The two functions, f(z) and g(z), can be chosen rather arbitrarily as
f(z)
=
c−3
z3
 + c−2
z2
 + c−1
z
 + c0 + c1 z,
(12)
g(z)
=
d−3
z3
 + d−2
z2
 + d−1
z
 + d0 + d1 z.
(13)
The unknown coefficients, c−3  ∼ d1, are determined to satisfy the above boundary conditions. They can be solved as
c−3 = 0,     c−2 = 0,     c−1 = σx∞ + σy∞
4
 ,    c0 = 0,     c1 = σx∞ − σy∞
2
 a2 + a2 τ∞ i.

d−3 = a4
2
 (σx∞ − σy∞)+ i a4 τ∞,     d−2 = 0,     d−1 = − a2
2
 (σx∞ + σy∞),    d0 = 0,     d1 = σy∞ − σx∞
2
 + i τ∞.
The stress components are
σx
=
3 a4 (σx∞−σy∞) (x4−6 x2 y2+y4)
2 (x2+y2)4
 + 6 a2 x2 y2 (σx∞−σy∞)
(x2+y2)3
 + a2 y4 (σx∞+σy∞)
2 (x2+y2)3
 − a2 x4 (5 σx∞−3 σy∞)
2 (x2+y2)3
− 4 a2 τ∞ x y (−3 a2 x2+3 a2 y2+3 x4+2 x2 y2−y4)
(x2+y2)4
 +σx∞,
σy
=
3 a4 (σy∞−σx∞) (x4−6 x2 y2+y4)
2 (x2+y2)4
 + 6 a2 x2 y2 (σy∞−σx∞)
(x2+y2)3
 − a2 y4 (5 σy∞−3 σx∞)
2 (x2+y2)3
 + a2 x4 (σx∞+σy∞)
2 (x2+y2)3
+ 4 a2 τ∞ x y (−3 a2 x2+3 a2 y2+x4−2 x2 y2−3 y4)
(x2+y2)4
 +σy∞,
τ
=
τ∞ (−3 a4 x4+18 a4 x2 y2−3 a4 y4+2 a2 x6−10 a2 x4 y2−10 a2 x2 y4+2 a2 y6+x8+4 x6 y2+6 x4 y4+4 x2 y6+y8)
(x2+y2)4
− a2 x y (−6 a2 σx∞ x2+6 a2 σy∞ x2+6 a2 σx∞ y2−6 a2 σy∞ y2+5 σx∞ x4−3 σy∞ x4+2 σx∞ x2 y2+2 σy∞ x2 y2−3 σx∞ y4+5 σy∞ y4)
(x2+y2)4
 .
When σy∞ = τ∞ = 0, σx is reduced to
σx = 3 a4 σx∞ (x4−6 x2 y2+y4)−a2 (x2+y2) (5 σx∞ x4−12 σx∞ x2 y2−σx∞ y4)+2 σx∞ (x2+y2)4
2 (x2+y2)4
 .
When x = 0, y = a, the above is reduced to
σx = 3 σx∞,
which shows a stress concentration at the edge.

Footnotes:

1
tx = σx nx + τny = σx x
a
 + τ y
a
 ,     ty = τnx + σy ny = τ x
a
 + σy y
a
 .


File translated from TEX by TTH, version 4.03.
On 18 Jan 2021, 22:13.