The rod shown below is made of two different materials. Segment AB is made of al

Question

The rod shown below is made of two different materials. Segment AB is made of aluminum (G=3.9*10^6 psi). Segment BC is bonded to segment AB and is made of brass (G=5.6*10^6 psi). Segment BC is hollow and has an inner diameter of di= 1.0in . External torques are applied to the rod at point B and point C. Let LA= 58.0 in , LB= 40.0in , dA= 3.7in , dB= 2.2in , TB= 9.0 kip*in , and Tc = 20.0 kip*in . What is the angle of twist at B and C? What would be the angle of twist at C is BC were solid instead of hollow?

Explanation / Answer

let is the angle of twist, then = (TiLi/(JiGi))

J= polar second moment of area = (d14 -d24)/32 if hollow; otherwise, J = *d4/32

L= length of the rod

T= torque acting on the rod

G = shear modulus of the material.

here,

J for the AB rod =*3.74/32 = 18.4 in4

and J for the BC rod = (2.24 -14)/32 = 2.2 in4

TB = 9.0 kip-in

TC=20.0 kip-in

LA = 58 in

LB = 40 in

GAB = 3.9*106 psi

GBC = 5.6*106 psi

therefore,

angle of twist at B = B = (TALA)/(JABGAB) = 9*58*103/(18.4*3.9*106) = 7.27*10-3 radians

similarily angle of twist at C, C = [(TALA)/(JABGAB) +(TCLC)/(JBCGBC) ] = (9*58*103/(18.4*3.9*106)+ 20*40*103/(2.2*5.6*106))= 0.0722 radians

hence, B =0.00727 radians

C =0.0722 radians

when BC is solid then JBC' = *2.24/32 = 2.3 in4

then angle of twist at C, C' =[(TALA)/(JABGAB) +(TCLC)/(JBC'GBC) ] = (9*58*103/(18.4*3.9*106)+ 20*40*103/(2.3*5.6*106))= 0.0694 radians

hence, when BC is solid instead of hollow, then C' =0.0694 radians

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