A point in a loaded part has σₓ = −60 MPa, σᵧ = 40 MPa and τₓᵧ = −10 MPa (tension positive). Find the smallest principal stress σ2.
Mohr’s circle & stress transformation worksheet with answers
A point has σₓ = −50 MPa and σᵧ = 50 MPa, plus some shear stress. On an element rotated by 50° the normal stress in the rotated x′ direction is 38.2 MPa. Find the normal stress σy′ in the rotated y′ direction.
A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
A point has σₓ = 140 MPa, σᵧ = 60 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
A point in a loaded part has σₓ = −60 MPa, σᵧ = −50 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).
A point in a thin plate is in plane stress: σₓ = 130 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa (σz = 0). Find the von Mises equivalent stress σvM.
Answer key
- 1. σ2 = −60.99 MPa
- 2. σy′ = −38.2 MPa
- 3. θp = 13.88 °
- 4. θs = −70.67 °
- 5. τθ = −5 MPa
- 6. σvM = 169.1 MPa
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