The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig. 9.7. The area of the face parallel to which this force is applied is
`" "A=50"cm"xx10"cm"`
`" "` =0.5 m`xx` 0.1 m
`" "` =0.05 `"m"^(2)`
Therefore, the stress applied is
`" "=(9.4xx10^(4)"N"//0.05"m"^(2))`
`" "=1.80xx10^(6)"N.m"^(-2)`
We know that shearing strain = `(Deltax//L)="Stress"//G`
Therefore the displacement `Deltax=("Stress"xxL)//G`
`=(1.8xx10^(6)"N m"^(-2)xx0.5"m")//(5.6xx10^(9)"N m"^(-2))`
`=1.6xx10^(-4)"m"=0.16" mm"`