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PROBLEM (*8.75) Using Mohr's circle, solve Prob. 8.50.
*SOLUTION
τ (ksi)
R=
σ’=6
2 ksi
1
(11 − 1) = 5 ksi
2
x
τxy
R
σx
4
O 1
C
2θ p '
11
σ (ksi)
y
(2, −τ xy)
1 + 11 = 2 + σ x ;
σ x = 10 ksi
τ xy = 52 − 4 2 = 3 ksi
2θ p ' = tan −1
3
= 36.9o
4
θ p ' = 18.5o
Sketch of results is as shown in solution of Prob. 8.50.
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