Question

Problem 1. (10 pts) The hollow copper pipe is fixed at the wall. Find the shear stress at points A and B (inside diameter = 2.8” and outside diameter = 3.0”). 125 lbist 500lb.ft 1000ll.ft

Answer 1

(100odbit Āt $ soodb.ft Fasilbuft Torsion in portion 1-2, T12 = 1000 db-ft (6) Torsion in portion 2-3, T23 = 1000-500 = 500 lb. ff G) Torsion in portion 3-4, T34 = 1800-500-725 = -225 lb. ft (2) Thus Torsion at A, TA = T23 = 500 lb.ff Tossion at B, To = T3y = 225 lb ft Shear stress at A is given by: TA - TA x

Here, TA = 500 lb ft = 500x 12 lb, inch => TA = 6000 lb.in b= radius of poße = Do = 3 = 115 in, J = polar moment of inertia of pipe J = F (D.- Di") = (39-2.84) => J= 1091778 iny TA - TAX PO = 6ooo db.in e lisen, 1.91778 ing co FTA = 4692.926 lblin (Psi ) Similarly Shear stress at B s'; To = 78x to = 225x12 lbin, xlisin. - 1.91778 iny 178 = 2111.9167 lb/im? (Ps1') |