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)

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 9.0 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)
Answer» Solution : 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`