1.Easy
At a point in a loaded part the in-plane principal stresses are σ₁ = 150 MPa and σ₂ = −40 MPa (tension positive). Find the normal stress σavg that acts on the planes of maximum in-plane shear stress.
σ1 = 150 MPaσ2 = −40 MPa
σavg = 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 45° counter-clockwise from the x-axis.
τxy = 20 MPaθ = 45°
σθ = MPa
3.Medium
A point in a loaded part has σₓ = 150 MPa, σᵧ = −60 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 150 MPaσy = −60 MPaτxy = 40 MPa
τmax = MPa
4.Medium
A point in a loaded part has σₓ = −30 MPa, σᵧ = 120 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −30 MPaσy = 120 MPaτxy = −40 MPa
θp = °
5.Hard
A point in a thin plate is in plane stress: σₓ = 90 MPa, σᵧ = 100 MPa and τₓᵧ = 40 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = 90 MPaσy = 100 MPaτxy = 40 MPa
σvM = MPa
6.Hard
A point in a thin plate is in plane stress: σₓ = 50 MPa, σᵧ = 90 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 300 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = 50 MPaσy = 90 MPaτxy = 10 MPaσY = 300 MPa
n =
Answer key
- 1. σavg = 55 MPa
- 2. σθ = 20 MPa
- 3. τmax = 112.4 MPa
- 4. θp = −75.96 °
- 5. σvM = 117.9 MPa
- 6. n = 3.248
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