Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
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