1.Easy
At a point in a loaded part the in-plane principal stresses are σ₁ = 200 MPa and σ₂ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σ1 = 200 MPaσ2 = 30 MPa
τmax = MPa
Mohr’s circle & stress transformation worksheet with answers
2.Easy
A point on the surface of a shaft in torsion is in pure shear with τₓᵧ = 20 MPa (σₓ = σᵧ = 0). Find the normal stress σθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.
τxy = 20 MPaθ = 25°
σθ = MPa
3.Medium
A point in a loaded part has σₓ = 90 MPa, σᵧ = 80 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 90 MPaσy = 80 MPaτxy = −60 MPa
θp = °
4.Medium
A point has σₓ = 40 MPa, σᵧ = 110 MPa and τₓᵧ = 60 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).
σx = 40 MPaσy = 110 MPaτxy = 60 MPa
θs = °
5.Hard
A point in a loaded part has σₓ = 140 MPa, σᵧ = −50 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.
σx = 140 MPaσy = −50 MPaτxy = 40 MPaθ = 20°
σθ = MPa
6.Hard
A point in a loaded part has σₓ = 50 MPa, σᵧ = −10 MPa and τₓᵧ = −50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 50 MPaσy = −10 MPaτxy = −50 MPaθ = 35°
τθ = MPa
Answer key
- 1. τmax = 85 MPa
- 2. σθ = 15.32 MPa
- 3. θp = −42.62 °
- 4. θs = 15.13 °
- 5. σθ = 143.5 MPa
- 6. τθ = −45.29 MPa
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