When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to be 50 MPa. Using the same amount of material, if the dimensions are changed to 50 mm x 250 mm, what will be the shear stress (in MPa)? The breadth and depth of the section are given along the centreline of the wall.

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

= 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

Plugging in then values, gives;

The shear stress will be 80,000,000 Pa or 80 MPa.

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