Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:
Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:
Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:
Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero
Answer: ε₁+ε₂+ε₃ = 0
Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-
l₀l₀l₀=l₁l₂l₃
taking natural log on both sides
Considering the logarithmic Laws of division and multiplication :
ln(AB) = ln(A)+ln(B)
ln(A/B) = ln(A)-ln(B)
Use the image attached to see the definition of true strain defined as
ln(l1/1o)= ε₁
which then proves that ε₁+ε₂+ε₃ = 0
