1.Easy
At a point in a loaded part the in-plane principal stresses are σ₁ = 110 MPa and σ₂ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σ1 = 110 MPaσ2 = 30 MPa
τmax = MPa
Mohr’s circle & stress transformation worksheet with answers
2.Easy
A point in a loaded part has σₓ = 40 MPa, σᵧ = −10 MPa and τₓᵧ = 60 MPa (tension positive). Find the largest principal stress σ1.
σx = 40 MPaσy = −10 MPaτxy = 60 MPa
σ1 = MPa
3.Medium
A point has σₓ = 50 MPa and σᵧ = 30 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 110 MPa?
σx = 50 MPaσy = 30 MPaσ1 = 110 MPa
τxy = MPa
4.Medium
A point in a thin plate is in plane stress: σₓ = 50 MPa, σᵧ = 60 MPa and τₓᵧ = 20 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 50 MPaσy = 60 MPaτxy = 20 MPaσz = 0
τabs = MPa
5.Hard
A point in a loaded part has σₓ = 50 MPa, σᵧ = −70 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.
σx = 50 MPaσy = −70 MPaτxy = −50 MPaθ = 25°
σθ = MPa
6.Hard
A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 8 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.
D = 400 mmt = 8 mmp = 1 MPaα = 35°
|τn| = MPa
Answer key
- 1. τmax = 40 MPa
- 2. σ1 = 80 MPa
- 3. τxy = 69.28 MPa
- 4. τabs = 37.81 MPa
- 5. σθ = −9.735 MPa
- 6. |τn| = 5.873 MPa
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