Question

A hollow shaft 36 inches long transmits 110 HP at 1300 rpm. The shaft is made out of steel with a modulus of rigidity of 11000 ksi. The shaft has an allowable sear stress of 12 ksi and an allowable angle of twist of 2.5 degrees over the 36 inches length. The inside diameter of the shaft is 0.65 that of the outside diameter of the shaft.

Determine:

a) the required inside and outside diameter of the shaft based on shear stress.

b) the required inside and outside diameter of the shaft based on the angle of twist.

Based on shear stress:

Inside Diameter ______ (in.)

Outside Diameter _____ (in.)

Based on angle of twist

Inside Diameter ______ (in.)

Outside Diameter _____ (in.)

can someone help me with this strengths problem?

Answer 1

L 36" P=110 HP = 110x746 W 32.06 kW N = 1300 22N 20 X1300 136.14 rad Is 60 - 2.5 X 1 0.04363 rad 180 G - 11000 ksi Dall Tall = 12 ksi di = 0.65 d. 3 Now P = Txw => 82.06 X10 = TX136214 T = 602.762 N-m = 5.335é-im a) Now, Tmax Trmax T + 12 = do 50335 Х T 32 x [ do® 40-65}d.)]*] => ido ► di = 0.911 inch 1.402 inch o T - b) Now , Go = 7 TL GJ 0.04363 5.335 X 36 11000 x I [d"-10.65do)'] -) Ho = 10492 inch di = 0.969 inch CS Scanned with CamScanner