1.Easy
A point in a loaded part has σₓ = 100 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa (tension positive). Find the smallest principal stress σ2.
σx = 100 MPaσy = 90 MPaτxy = −10 MPa
σ2 = MPa
Mohr’s circle & stress transformation worksheet with answers
2.Easy
At a point in a loaded part the in-plane principal stresses are σ₁ = 130 MPa and σ₂ = 110 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σ1 = 130 MPaσ2 = 110 MPa
τmax = MPa
3.Medium
A point in a loaded part has σₓ = 140 MPa, σᵧ = −70 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 140 MPaσy = −70 MPaτxy = −10 MPa
τmax = MPa
4.Medium
A point has σₓ = 10 MPa, σᵧ = 30 MPa and τₓᵧ = 30 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 = 10 MPaσy = 30 MPaτxy = 30 MPa
θs = °
5.Hard
A point in a loaded part has σₓ = 80 MPa, σᵧ = 10 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = 80 MPaσy = 10 MPaτxy = −10 MPaθ = 35°
σθ = MPa
6.Hard
A point in a loaded part has σₓ = 160 MPa, σᵧ = 20 MPa and τₓᵧ = 40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.
σx = 160 MPaσy = 20 MPaτxy = 40 MPaθ = -20°
σθ = MPa
Answer key
- 1. σ2 = 83.82 MPa
- 2. τmax = 10 MPa
- 3. τmax = 105.5 MPa
- 4. θs = 9.217 °
- 5. σθ = 47.57 MPa
- 6. σθ = 117.9 MPa
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