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 xx 10 ^(4) N.` The lower edge is riveted to the floor. How much will the upper edge be displaced ? Shear modulus of lead `G=5.8 xx 10 ^(9) N//m^(2)`

Explanation

The load 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 is,
`A=l xx b`
`= 50 cm xx 10 cm = 0.5 m xx 0.1 m`
`therefore A = 0.05 m ^(2)`
Stress = `("Force")/("Area")= (9 xx 10 ^(4))/(0.05) =1.8 xx 10 ^(6) Nm ^(-2)`
Shear modulus `(G) =("Stress")/("Shear strain")`
`therefore` Shear strain`= ("Stress")/(G)`
` (Delta x)/(L) = ("Stress")/(G)`
`therefore Deleta x = ("Stress")/(G) xx l`
`therefore Delta x = (1.8 xx 10 ^(6))/(5.6 xx 10 ^(9)) xx 0.5`
`therefore Delta x =0.1607 xx 10 ^(-3)`
`therefore Delta x ~~ 1.6 xx 10 ^(-4) m = 0.16 mm`