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 ?

A square lead slab of side 50 cm and thickness 10 cm is subject to a s

1.	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 ?
Answer» Solution :The leadslab is fixed and the force is appliedparallelto the narrowface as shownin figure. The areaof the face parallelto which this force is appliedis `A=50cm xx 10 CM =0.05 m^(2)` THEREFORE the stressapplied is `=(9.4xx10^(4) //0.05)=1.80 xx 10^(6) N.m^(-2)` We know thatshearing strain `=(DELTAX//L)="Stress"//G.` Therefore the dispalcement `Deltax=("Stress"xxL)//G` `=(1.8xx10^(6)xx0.5)//(5.6xx10^(9))=0.16mm`