Question

A solid shaft 32 inches long must carry a torque of 5600 lb-in. The shaft is made out of aluminum. Aluminum shafts come in increments of 1/16th inch increments between diameter of 0.25 and 2.5 inches. Determine:

a) the required size and cost of an aluminum shaft based on the below design criteria.

	Allow Shear    Stress (ksi)	Allow Angle of Twist (deg.)	Density (lb/ft3)	Cost per Pound ($)	Modulus of Rigidity    (ksi)
Aluminum	18	4.5	169	0.75	4000

Design Based on Shear Stress:

The diameter of the aluminum shaft based on shear stress is: _____ (in.) (answer should be in decimal format in 1/16th inch increments)

The cost of the aluminum shaft based on shear stress is : ______ ($)

Design Based on Twist

The diameter of the aluminum shaft based on the angle of twist is _____ (in.) (answer should be in demimal format in 1/16th inch increments)

The cost of the aluminum shear based on angle of twist is _____ ($)

can someone help me with this strengths problem please?

Answer 1

Given Datar Given Length (L) = 32 inch Torque (T) 15600 lb-in Allowable shear streus (Tallow) = 18 ksi Angle of twist = 4.50 = 4.5XT 120 =0.07854 rad density 169 (blft² Cost por Pound $0.75 Modulus of Rigidity (G) 4000 ksi Y Torsion 17 Equation GO L r J “) Design based Shear stress? & 3 al2 Go r [r= 2 18 4000 X 0.07854 미 32

oleto 9.8175 DE 36 3. 6669 in 9.8175 21 3.6875 in DE 3.8675 in volume of Solid Shaft XL EF (3.8675) X32 376 in 3 weight OH Solid shaft 1 Density & volume 376 169* 123 36.77 Lo. : Cost of the Shaf 2 weight Cost/pound 36.77 x 0.75 Cost = $ 27.58 Based twist om - CH Go GO I 5600 Lorin 400 x 0.07854 32 хүч z

- (0.3639 5.6 kip-in 9.8175 Түч 2% 5.6 TX 9.8175 Yo 0.176 in ry 11 21- 2% 0-776 = 1.552 ~ 1.5625 in. volume IxDx2 = I x 1.5625² 61.36 in3 weight Density & volume 169 x 61:36 6.00 lb 123 Cost weight & Cost por pound = 6x0.75= $4.5 ( a) Design based on shears Diameter CD) - 3.6875 in Cust C) $. 27.58 b Desis based on Torque's on a Diametur (D) - 1.5625 in Cost (C) = $4.5 m