A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of `9.0xx10^(4)N`. The lower edge is riveted to the floor How much will the upper edge be displaced ?

Explanation

The lead slab is fixed and the force is applied parallel to the narrow face as shown in figure. The area of the face parallel to which this force is applied is
`A = 50 cm xx10 cm = 0.05 m^2`
Therefore, the stress applied is
` = (9.4 xx 10^4 // 0.05) = 1.80 xx 10^6 Nm^(-2)`

We know that shearing strain =`(Delta x//L)` = Stress/G.
Therefore the displacement `Delta x` = (Stress x L)/G
` = (1.8 xx 10^6 xx 0.5)//(5.6 xx 10^9) = 0.16mm`