Question

To determine the angle of twist on a composite rod given the geometry and externally applied torques, to properly apply a sign convention to determine the angle of twist, to use a torque diagram to aid in determining the angle of twist, and lo determine the maximum applicable lorque given a maximum allowable angle of wist. The rod shown below is made of two different materials. Segment AB is made of aluminum (G= 3.9 x 100 psi). Segment BC is bonded to segment AB and is made of brass (G = 5.6 x 100 psi). Segment BC is hollow and has an inner diameter of d = 1.0 in . Extemal torques are applied to the rod at point B and point C. Let LA = 59.0 in , LB = 50.0 in, da = 4.0 in, dB = 2.4 in. To = 9.0 kip - in , and TC = 19.0 kip - in.

Answer 1

The solution of the given problem is given below

Solution - eriven data; e AB abc = = 3.9 *106 psi 5.6 #10° psi * AB . > Solid da = diameter 2 4.0 in 5910 in * loin & Be Hallow; dia inney diameter - neo diameter = de = outer diameter = 2.4 in to = so in To z a kipin = 9000 .p-in To = 19 kip-in = 19000 ap-in.

OCA - OCB + PSA 04 - Oc = OCB + OBA -O We know that of a fixed support 30 xed C now 0 = 0 - Q = (IL). ( laJ JBA OB Вове 5* 19000 * 50 -oc a 50.6*100* T (2.44_,4) 3.97106 * *44 . 32 -00 = 0.0537 + nap to top Oc = -0.07855 radians Port A)... Os = ? 19000 * 50 Dev = ( 0 comentario para SAW 08- Oc = 0.0537 Op = 0 + 0.0537 = -0.07055 +0.0537 I OB = -0.0168 radians

Part ② - - - • o cross 0.0 10S LOVE todians Park @ BC also solia portion 19000 & 50 Pa-oca I Bocops A- oc = sit * 100 * 4 244 3.88108 I. 32

ca ao$308 0.01605 z = 0. 68954 va otland