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PROBLEM (8.17) Solve Prob. 8.16 for σ x = 20 ksi, σ y = 5 ksi, τ xy = 4 ksi, and θ = 25o ,
shown in Fig. P8.17.
SOLUTION
1
1
2
2
1
1
= (20 − 5) + (20 − 5) cos 50o + 4sin 50o = 15.38 ksi
2
2
σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ
1
2
1
= − (20 − 5) sin 50o + 4 cos 50o = −3.17 ksi
2
τ x ' y ' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ
Therefore,
σ y ' = σ x + σ y − σ x ' = 20 + 5 − 20.4 = 4.6 ksi
4.6 ksi
y’
3.17 ksi
15.38 ksi
x’
θ = 25o
x
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