128
Chapter 2
Using Eq. (2.186)
/~s = ~ = 51,250 psi
#s = ~ which is to be solved for
~s = Cvs~ Eq. (2.188)
From Eq. (2.194)
~s = C,,s~
Now substituting in to the coupling Eq. (2.195) and squaring
t 2 _ (~ - ~)2
(Cyst) 2 q- (Cyst)
2
Solving for an unknown ~
Now
Now
(2.196)
t2tc~ss 2-’= + C~2,7~1 = £’ - 2,~ + ~2
(1 -- flC~)~ ~ - 2~ + (~ - tiCks ~) = 0
A~2 + B~ + C = O
solving the quadratic equation
-B + ~/B 2 - 4A C
3=
2A
examine the solutions for S Eq. (2.197)
1.
(2.197)
If A = 0 then s is infinite and the stress due to loading is greater
than ~ which is the stress the material can resist (Example 2.14)
hence a failure.
2 2
A = 1 - t C
w
substitute Eq. (2.194)
1
1
tCvs +0.0266433
t = 37.53
If C = 0 then
C ~2 2 2 -2
=
- t C~sS
then ~z ¢ 0 but
Chapter 2
Using Eq. (2.186)
/~s = ~ = 51,250 psi
#s = ~ which is to be solved for
~s = Cvs~ Eq. (2.188)
From Eq. (2.194)
~s = C,,s~
Now substituting in to the coupling Eq. (2.195) and squaring
t 2 _ (~ - ~)2
(Cyst) 2 q- (Cyst)
2
Solving for an unknown ~
Now
Now
(2.196)
t2tc~ss 2-’= + C~2,7~1 = £’ - 2,~ + ~2
(1 -- flC~)~ ~ - 2~ + (~ - tiCks ~) = 0
A~2 + B~ + C = O
solving the quadratic equation
-B + ~/B 2 - 4A C
3=
2A
examine the solutions for S Eq. (2.197)
1.
(2.197)
If A = 0 then s is infinite and the stress due to loading is greater
than ~ which is the stress the material can resist (Example 2.14)
hence a failure.
2 2
A = 1 - t C
w
substitute Eq. (2.194)
1
1
tCvs +0.0266433
t = 37.53
If C = 0 then
C ~2 2 2 -2
=
- t C~sS
then ~z ¢ 0 but