Question

5 Section 2 Problem 5 (25 pts.) A solid shaft 32 inches long must carry a torque of 3600 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. 02:55:27 Allow Shear Stress (ksi) Allow Angle of Density Cost per Modulus of Pound Rigidity ($) (ksi) Twist (deg.) Kib/A3) Aluminum 18 4.5 169 0.75 4000 (in.) (answer should be in decimal format in 1/16th ($) Design Based on Shear Stress: The diameter of the aluminum shaft based on shear stress is: 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 1/16th inch increments) The cost of the aluminum shear based on angle of twist is (in.) (answer should be in demimal format in ($)

Answer 1

Answers are boxed

T= 3600 lb in LE 32 inch Tuon : 18ksi T J old ६ 3600 18x10 1 / d/2 5 TI 32 PX d = 1.00616 in 16.0985 in 16 it in use dia shaft 7 16 Answer = 1.0625 in 2 2 Copt a 0.75 8 169 X X1.0625 X 32 123 Cost C $ 2.08114

Il T & 3600X 32 4.5X1 180 4000x103 x 21 / xd da 1.39 in 22.du3 = 16 Provide 200 23 16 Ansu wer 1.4375 in he Cost ao. 15 x 169 El X 1.4375 x 32 123 Cost $3.80943