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ME 5463-001 Fracture Mechanics, Homework #7

Exercise 3.12 The Westergaard stress functions that solve the opening-mode problem of a crack of length, 2a, in an infinite body subjected to four equal point loads, P, on the crack faces, as shown in Figure E3.2, are given by
Z(z) = {{2P}/pi}{ {z sqrt{a^2 - b^2}} / {(z^2 - b^2)sqrt{z^2-a^2}} } and Y(z) = 0.

(a) Derive an exact expression for the Cartesian stress σy valid everywhere in the body. (b) From the results of part (a), derive the geometric stress intensity factors for this combination of geometry and loading.

Y(z) = 0 right Im[Y(z)] = 0 forall z=x+iy (including y=0) so sigma_y = Re[Z] for y=0.

sigma_y = Re[Z] + y(Im[Z prime] +Im[Y prime]) = Re[Z] + y Im[Z prime]

Set up to Find Re[Z]

Define some temporary coordinate systems.

z-a = (x-a) +iy = {r_1}e^{i theta_1} = r_1(cos theta_1 + i sin theta_1), z+a = (x+a) +iy = {r_2}e^{i theta_2} = r_2(cos theta_2 + i sin theta_2), z-b = (x-b) +iy = {r_3}e^{i theta_3} = r_3(cos theta_3 + i sin theta_3), and z+b = (x+b) +iy = {r_4}e^{i theta_4} = r_4(cos theta_4 + i sin theta_4).

⇒ 2z = 2x +2iy = {r_1}e^{i theta_1} + {r_2}e^{i theta_2} ⇒ z = x +iy = { {r_1}e^{i theta_1} + {r_2}e^{i theta_2} }/2 = {1/2} ( r_1 cos theta_1 + i r_1 sin theta_1 + r_2 cos theta_2 + i r_2 sin theta_2 )

z = x +iy = r e^{i theta} = r(cos theta + i sin theta)

Find Re[Z]

Z(z) = {{2P}/pi}{ {z sqrt{a^2 - b^2}} / {(z^2 - b^2)sqrt{z^2-a^2}} } = {{2P}/pi}{ {z sqrt{a^2 - b^2}} / {(z+b)(z-b)sqrt{(z+a)(z-a)}} }

⇒ Z(r_1,theta_1,r_2,theta_2,r_3,theta_3,r_4,theta_4) = {{2P}/pi} sqrt{a^2 - b^2} { { {1/2}(r_1 e^{i theta_1} + r_2 e^{i theta_2}) } / {r_3 e^{i theta_3}r_4 e^{i theta_4} sqrt{r_2 e^{i theta_2} r_1 e^{i theta_1}}} } = {{P}/pi} sqrt{a^2 - b^2} { { r_1 e^{i theta_1} + r_2 e^{i theta_2} } / {r_3 e^{i theta_3}r_4 e^{i theta_4} {r_2 e^{i theta_2/2} r_1 e^{i theta_1/2}}} } = {{P}/pi} sqrt{a^2 - b^2} { { r_1 e^{i theta_1} + r_2 e^{i theta_2} } /{ r_1 e^{i theta_1/2} r_2 e^{i theta_2/2} } } { { 1 } / { r_3 e^{i theta_3}r_4 e^{i theta_4} } }

2009-09-29, 17:36: Well, it looks like in fixing my original error of using powers of r_n instead of e, I didn't follow through and correct all of this manipulation. Unfortunately, I have already spent far too many hours stumbling through this and need to get to work on more tractable assignments.

= {{P}/pi} sqrt{a^2 - b^2} ( { r_1 e^{i theta_1} }{ r_1 e^{-i theta_1/2} r_2 e^{-i theta_2/2} } + { r_2 e^{i theta_2} }{ r_1 e^{-i theta_1/2} r_2 e^{-i theta_2/2} } ) { { 1 } / { r_3 e^{i theta_3}r_4 e^{i theta_4} } } = {{P}/pi} sqrt{a^2 - b^2} { { { r_1 e^{-i theta_1/2} }{ r_2 e^{-i theta_2/2} } + { r_2 e^{-i theta_2/2} }{ r_1 e^{-i theta_1/2} } } / { r_3 e^{i theta_3}r_4 e^{i theta_4} } } = {{2P}/pi} sqrt{a^2 - b^2} { { r_1 e^{-i theta_1/2} r_2 e^{-i theta_2/2} } / { r_3 e^{i theta_3}r_4 e^{i theta_4} } } = {{2P}/pi} sqrt{a^2 - b^2} { r_1 e^{-i theta_1/2} r_2 e^{-i theta_2/2} } { r_3 e^{-i theta_3}r_4 e^{-i theta_4} } = {{2P}/pi} sqrt{a^2 - b^2} { e^{ln r_1 -i theta_1/2} e^{ln r_2 -i theta_2/2} } { e^{ln r_3-i theta_3} e^{ln r_4 -i theta_4} } = {{2P}/pi} sqrt{a^2 - b^2} e^{ln r_1 + ln r_2 + ln r_3 + ln r_4 -i({theta_1/2 +theta_2/2 +theta_3 +theta_4}) } = {{2P}/pi} sqrt{a^2 - b^2} r_1 r_2 r_3 r_4 ( cos({-{theta_1/2} -{theta_2/2} -theta_3 -theta_4}) + i sin({-{theta_1/2} -{theta_2/2} -theta_3 -theta_4}) ) ⇒

Re[r_1,r_2,r_3,r_4,theta_1,theta_2,theta_3,theta_4] ={{2P}/pi} sqrt{a^2 - b^2} r_1 r_2 r_3 r_4 cos({-{theta_1/2} -{theta_2/2} -theta_3 -theta_4}) ⇒

Re[x,y,theta_1, theta_2, theta_3, theta_4]= {{2P}/pi} sqrt{(a^2 - b^2) ((x-a)^2 + y^2) ((x+a)^2 + y^2) ((x-b)^2 + y^2) ((x+b)^2 + y^2) } cos({-{theta_1/2} -{theta_2/2} -theta_3 -theta_4})

Im[r_1, r_2, r_3, r_4, theta_1, theta_2, theta_3, theta_4] ={{2P}/pi} sqrt{a^2 - b^2} r_1 r_2 r_3 r_4 sin({-{theta_1/2} - {theta_2/2} - theta_3 -theta_4})

I still can't get it in terms of z (presumably because it is wrong)so it can be differentiated, to get Im[Z'] or Im[Z]', then x + iy substituted in for z, and the result used to get sigma_y = Re[Z] + y Im[Z'] = Re[Z] + y Im[Z]'.

Finding K

Finding K is a matter of determining the limit as x+ → a.

K = lim{delta + right 0}{Re Z}|^y=0 sqrt{2 pi delta}</m = <m>K = lim{(x-a)+ right 0}{Re Z}|^y=0 sqrt{2 pi (x-a)}

At y=0, all the thetas = 0,

K = lim{(x-a)+ right 0}{Re Z}|^y=0 sqrt{2 pi (x-a)} Re[x,y,theta_1, theta_2, theta_3, theta_4]= {{2P}/pi} sqrt{(a^2 - b^2) (x-a)^2 (x+a)^2 (x-b)^2 (x+b)^2 }=0

This doesn't work since my derivation of Z is incorrect.

However, the given equation works fine. At y=0, z=x

Z(x,y=0) = {{2P}/pi}{ {x sqrt{a^2 - b^2}} / {(x^2 - b^2)sqrt{x^2-a^2}} }

and the limit sqrt{2 pi (x-a)}{{2P}/pi}{ {x sqrt{a^2 - b^2}} / {(x^2 - b^2)sqrt{x^2-a^2}} } = sqrt{2 pi (x-a)}{{2P}/pi}{ {x sqrt{a^2 - b^2}} / {(x^2 - b^2)sqrt{(x-a)(x+a)}} } = sqrt{2 pi}{{2P}/pi}{ {x sqrt{a^2 - b^2}} / {(x^2 - b^2)sqrt{(x+a)}} } = sqrt{2 pi}{{2P}/pi}{ {a sqrt{a^2 - b^2}} / {(a^2 - b^2)sqrt{2a}} } =

K={ 2P sqrt{a}} / {sqrt{pi (a^2 - b^2)} }


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