— David Wagner 2009/09/21 07:43
ME 5463-001 Fracture Mechanics, Homework #5
3.10 Prove the series solution in Westergaard form for the traction-free, single-ended crack [Eqs. (3.50)] is equivalent to the Williams solution to the same problem. What is the relationship between the constants C1n in the Williams formulation and Aj and Bm in the Westergaard formulation?
General Westergaard form:
Eqs. (3.50)
,
.
(3.51)
![Z(r,theta) = sum{j=0}{infty}{A_j r^{j-{1/2}}
delim{[}{ cos(j-{1/2})theta + i sin(j-{1/2})theta }{]}
}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_946.5_3f3908e5f810747206ce80fa9509f95a.png "Z(r,theta) = sum{j=0}{infty}{A_j r^{j-{1/2}}
delim{[}{ cos(j-{1/2})theta + i sin(j-{1/2})theta }{]}
}")
Let 
, 
, 
.
;
![Z(r,theta) =
sum{j=0}{infty}{A_j r^{{n_o/2}-{1}} }
delim{[}{ cos({{n_o-2}/2})theta + i sin({{n_o-2}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_953_05021c530645f2d56ed11645dfc4a1dd.png "Z(r,theta) =
sum{j=0}{infty}{A_j r^{{n_o/2}-{1}} }
delim{[}{ cos({{n_o-2}/2})theta + i sin({{n_o-2}/2})theta
}{]}")
=![sum{j=0}{infty}{A_j r^{{n_e -3}/2} }
delim{[}{ cos({{n_e -3}/2})theta + i sin({{n_e -3}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_953_63a77262196ce2ccb6e59909d20f9880.png "sum{j=0}{infty}{A_j r^{{n_e -3}/2} }
delim{[}{ cos({{n_e -3}/2})theta + i sin({{n_e -3}/2})theta
}{]}")
Dropping the terms where the power of r is nonpositive by increasing the index lower value yields,
![sum{j=3}{infty}{} {A_j r^{{n_o-2}/2} }
delim{[}{ cos({{n_o-2}/2})theta + i sin({{n_o-2}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_670bd0d6445ad8fb382ccd9dbe9c4b45.png "sum{j=3}{infty}{} {A_j r^{{n_o-2}/2} }
delim{[}{ cos({{n_o-2}/2})theta + i sin({{n_o-2}/2})theta
}{]}")
-![sum{j=4}{infty}{}{ A_j r^{{n_e -3}/2} }
delim{[}{ cos({{n_e -3}/2})theta + i sin({{n_e -3}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_71729d4edb0f3a6f6128159843ad1b37.png "sum{j=4}{infty}{}{ A_j r^{{n_e -3}/2} }
delim{[}{ cos({{n_e -3}/2})theta + i sin({{n_e -3}/2})theta
}{]}")
=0
Dropping the first index by two requires increasing n_o by 4, and dropping the second by three requires increasing n_e by 6.
![sum{n=1}{infty}{} {A_j r^{{n+2}/2} }
delim{[}{ cos({{n+2}/2})theta + i sin({{n+2}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_90ba8b0256b8144f1486fb7bc08cb913.png "sum{n=1}{infty}{} {A_j r^{{n+2}/2} }
delim{[}{ cos({{n+2}/2})theta + i sin({{n+2}/2})theta
}{]}")
-![sum{j=1}{infty}{}{ A_j r^{{n +3}/2} }
delim{[}{ cos({{n +3}/2})theta + i sin({{n +3}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_c1dd62cd4f9180ab43162f0844ca585f.png "sum{j=1}{infty}{}{ A_j r^{{n +3}/2} }
delim{[}{ cos({{n +3}/2})theta + i sin({{n +3}/2})theta
}{]}")
=0.
Combining the sum and changing the indices,
![sum{n=1}{infty}{} {A_j r^{{n+2}/2} }
delim{[}{
cos({{n+2}/2})theta - cos({{n +3}/2})theta
+ i sin({{n+2}/2})theta - i sin({{n +3}/2})theta
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_956.5_0e0d725420ac91961ea19256af180ca6.png "sum{n=1}{infty}{} {A_j r^{{n+2}/2} }
delim{[}{
cos({{n+2}/2})theta - cos({{n +3}/2})theta
+ i sin({{n+2}/2})theta - i sin({{n +3}/2})theta
}{]}")
=0.
Distributing A_j over each pair of terms, setting A_j to an appropriate linear combination of a C_1n + b C_2n, multiplying this all out, rearranging the terms, and setting both imaginary and real parts separately to zero yields
=
+
![C_{1n}(cos{{n-1}/{2}}theta -cos{{n+3}/{2}}theta)
+C_{2n}(sin{{n-1}/{2}}theta -{{n-1}/{n+3}}sin{{n+3}/{2}}theta)
]](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_971_1f44118381ed42761dab697fc8494012.png "C_{1n}(cos{{n-1}/{2}}theta -cos{{n+3}/{2}}theta)
+C_{2n}(sin{{n-1}/{2}}theta -{{n-1}/{n+3}}sin{{n+3}/{2}}theta)
]")
,
which if separated into odd and even series, is Willams' Airy stress function for a single-ended, traction-free crack:
Eqs. (3.17)
=![F(r,theta) = sum{n=1,3,...}{infty}{r^{1+{n/2}} }
delim{[}{
C_{1n}(cos{{n-2}/{2}}theta -{{n-2}/{n+2}}cos{{n+2}/{2}}theta)
+C_{2n}(sin{{n-2}/{2}}theta -sin{{n+2}/{2}}theta)
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_952_9f60f4be85eed19384dcf2e96ad9de32.png "F(r,theta) = sum{n=1,3,...}{infty}{r^{1+{n/2}} }
delim{[}{
C_{1n}(cos{{n-2}/{2}}theta -{{n-2}/{n+2}}cos{{n+2}/{2}}theta)
+C_{2n}(sin{{n-2}/{2}}theta -sin{{n+2}/{2}}theta)
}{]}")
+![sum{n=2,4,...}{infty}{ r^{1+{n/2}} }
delim{[}{
C_{1n}(cos{{n-2}/{2}}theta -cos{{n+2}/{2}}theta)
+C_{2n}(sin{{n-2}/{2}}theta -{{n-2}/{n+2}}sin{{n+2}/{2}}theta)
}{]}](http://wiki.waggy.org/dokuwiki/lib/exe/fetch.php?w=&h=&cache=cache&media=cache_mathplugin%3amath_952_475fe7fc7413492528d2e94112acfabc.png "sum{n=2,4,...}{infty}{ r^{1+{n/2}} }
delim{[}{
C_{1n}(cos{{n-2}/{2}}theta -cos{{n+2}/{2}}theta)
+C_{2n}(sin{{n-2}/{2}}theta -{{n-2}/{n+2}}sin{{n+2}/{2}}theta)
}{]}")
Something like the following should relate the constants. The exact expression can be found by messing around with the formulas above to find what it takes to make it work.
and
.
Westergaard form for a single-ended, traction-free crack:
Eqs. (3.53)
,
.
