sigma/4 r^2 -{sigma a^2}/2 ln r +C_4 +(sigma/4 r^2 + {{sigma a^4}/12}{1/{r^2}} -sigma/2)cos 2theta

sigma*r^2/4 - sigma* a^2 /2*log( r ) +C_4 +(-sigma*r^2/4- sigma*a^4 /(4*r^2) +sigma*a^2/2)*cos(2*theta)

Oops

FIXME OOPS! I went and started solving the wrong problem!

FIXME Is this asking about be a finite or semi-infinite (fixed width, infinite length) plate? (An infinite plate with a hole under uniform remote stress was worked in class.)

FIXME What is a “simple tensile load” in this context? (A point load results in infinite stress where it's applied.)

Assume this is a semi-infinite plate of thickness t and width b = 2c along x2, with a central hole of radius ar, and a load P applied uniformly across the finite width resulting in a stress σ = P/(b*t) far from the hole as x1 → ±∞.

  • At x2 = ±c, σ12 = σ22 = 0

Polar to Cartesian coordinates:

  • x1 = r cos θ;
  • x2 = r sin θ.

Like the infinite case, this also has two-fold symmetry, so assume an Airy function.

  • Φ(r,θ) = f1® + f2® cos 2θ.
  • ∇4Φ = ({partial^2}/{partial r^2} + 1/r {partial}/{partial r} + {1}/{r^2} {partial^2}/{partial theta^2})^2 Phi = 0.
  • ({d^2}/{dr^2} + 1/r {d}/{dr}) ({d^2}/{dr^2} + 1/r {d}/{dr}) f_1(r) = 0.
  • ({d^2}/{dr^2} + 1/r {d}/{dr} -{4}/{r^2}) ({d^2}/{dr^2} + 1/r {d}/{dr} -{4}/{r^2}) f_2(r) = 0.

Let ζ = lnr, r≠0:

  • zeta = ln r; d zeta = 1/r dr; d zeta^2 = -{1/{r^2}} dr^2.
  • e^zeta = r; r d zeta = dr; -r^2 d zeta^2 = dr^2.
  • ({d^2}/{-r^2 d zeta^2} + 1/r {d}/{r d zeta}) ({d^2}/{-r^2 d zeta^2} + 1/r {d}/{r d zeta}) f_1(r) = 0.
  • = -{1/{r^2}} ({d^2}/{d zeta^2} - {d}/{d zeta}) ({d^2}/{d zeta^2} - {d}/{d zeta}) f_1(r) = 0.
  • = e^{-2 zeta} ({d^2}/{d zeta^2} - {d}/{d zeta}) ({d^2}/{d zeta^2} - {d}/{d zeta}) f_1(e^zeta) = 0.

Assuming f_1(e^zeta)≠0, and since e^{-2 zeta}≠0, solutions are the same as an infinite plane.

  • f_1(r) = C_1 r^2 ln r + C_2 r^2 +C_3 ln r +C_4
  • f_2(r) = C_5 r^2 +C_6 r^4 + {1/{r^2}}C_7 +C_8
  • Phi = f_1(r) +f_2(r)cos 2theta = C_1 r^2 ln r + C_2 r^2 +C_3 ln r +C_4 +(C_5 r^2 +C_6 r^4 + {1/{r^2}}C_7 +C_8)cos 2theta

Assuming f_1(e^zeta)≠0, and trying f_1=ln x,

  • ({d^2 zeta}/{d zeta^2} - {d zeta}/{d zeta}) ({d^2 zeta}/{d zeta^2} - {d zeta}/{d zeta}) = 0
  • = {d^4 zeta}/{d zeta^4} - 2 {d^2 zeta}/{d zeta^2} +1=0

Assuming f_1(e^zeta)≠0, and since e^{-2 zeta}≠0, then

  • {d^2}/{d zeta^2} - {d}/{d zeta}=0

FIXME

  • zeta prime prime = -{1/{r^2}};
  • r = e^{zeta};
  • r prime = e^{zeta};
  • r prime prime = e^{zeta}.

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