Question

A horizontal aluminum rod 2.8 cm in diameter projects 6.5 cm from a wall. A 1300 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0·10¹⁰ N/m². Neglecting the rod's mass, find (a) the shear stress on the rod and(b) the vertical deflection of the end of the rod.

I got part A at 2.1*10^7 but can not get part B. I came up with the answer of 4.1*10^-5 m but that is not correct.

Answer 1

(a) The shear stress is given by F/A, where F is the magnitude of the force applied parallel to one face of the aluminum rod and A is the cross-sectional area of the rod. In this case F is the weight of the object hung on the end: F = mg, where m is the mass of the object. If r is the radius of the rod then A = πr2.

Thus, the shear stress is F/ A = mg πr2 = (1300 kg)(9.8 m/s2)/ π(0.014 m)2

= 20 × 10^6 N/m2 .

(b) The shear modulus G is given by G = F/A ∆x/L where L is the protrusion of the rod and ∆x is its vertical deflection at its end. Thus, ∆x = (F/A)L/ G

= (20 × 10^6 N/m2)(0.065 m) /3.0 × 10^10 N/m2

= 4.33 × 10−5 m .