A solid 28indiameter bronze G 7100 ksi shaft is 77ft long T

A solid 2.8-in.-diameter bronze [G = 7100 ksi] shaft is 7.7-ft long. The allowable shear stress in the shaft is 11 ksi and the angle of twist must not exceed 0.049 rad. Determine the maximum horsepower P that this shaft can deliver:
(a) when rotating at 193 rpm.
(b) when rotating at 541 rpm.

(a) P=.... hp

(b) P=.... hp

Solution

Solution:-

Given

G=7100 ksi

? = 0.049 radian

Diameter (D) = 2.8 inch

Length(l) = 7.7 feet

Shear stress = 11 ksi

We know that

T/J = ?/R = G?/l

T = G?/l *J

T =( 7100 *10³*0.049 /(7.7*12))*?/32*(2.8)⁴

= 22.72 *10³ pound –inch

Torque (T) = 1.893 *10³ pound – feet

(a) When N = 193 rpm

Power(P)   = T*N/(5252)

                     = 1.893*10³*193/(5252)

                    = 69.56 hp Answer

(b) when rotation N= 541 rpm

                        P = T*N/(5252)

                       P= 1.893*10³*541/(5252)

                       P = 194.99 hp Answer