Example

Question:

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of \(9.0 \times 10^4\,\mathrm{N}\). The lower edge is riveted to the floor. How much will the upper edge be displaced?

Solution:

The lead slab is fixed and the force is applied parallel to the narrow face.
The area of the face parallel to which this force is applied: \[ A = 50\,\mathrm{cm} \times 10\,\mathrm{cm} = 0.5\,\mathrm{m} \times 0.1\,\mathrm{m} = 0.05\,\mathrm{m}^2 \]
The stress applied is: \[ \text{Stress} = \frac{9.4 \times 10^4\,\mathrm{N}}{0.05\,\mathrm{m}^2} = 1.80 \times 10^6\,\mathrm{N\,m}^{-2} \]
Shearing strain is given by: \[ \frac{\Delta x}{L} = \frac{\text{Stress}}{G} \] Therefore, the displacement: \[ \Delta x = \frac{\text{Stress} \times L}{G} = \frac{1.8 \times 10^6\,\mathrm{N\,m}^{-2} \times 0.5\,\mathrm{m}}{5.6 \times 10^9\,\mathrm{N\,m}^{-2}} = 1.6 \times 10^{-4}\,\mathrm{m} = 0.16\,\mathrm{mm} \]