sm8 17

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PROBLEM (8.17) Solve Prob. 8.16 for
x
σ
= 20 ksi, y
σ
= 5 ksi,
x
y
τ
= 4 ksi, and
θ
= 25o ,
shown in Fig. P8.17.
SOLUTION
'11
()()cos2sin
22
xxyxy xy
2
σ
σσ σσ θτ θ =++− +
11
(20 5) (20 5)cos50 4sin50 15.38
22 oo
ksi=−+− + =
'' 1()sin2cos
2
xy x y xy 2
τ
σσ θτ θ =− − +
1(20 5)sin50 4cos50 3.17
2oo
ksi=− − + =−
Therefore,
''
20 5 20.4 4.6
yxyx ksi
σ
σσσ =+− =+− =
x
’
3.17 ksi
y
’
θ
=
25o
15.38 ksi
x
4.6 ksi
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