Question

6 Section 2. Problem 6 (25 pts. A hollow shaft 36 inches long tranmits 105 HP at 1250 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 degreees over the 36 Inche length. The inside diameter of the shaft is 0.8 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. 02:55:20 (in.) Based on shear stress: Inside Diameter Outside Diameter Based on angle of twist Inside Diameter Outside Diameter (in.) (in.)

Answer 1

We know the basic relation of Torque and shear stress 푸 등 je. T J co e where To Torque J> Polar Moment of inertia D. et Shear stress = 0.8D. Given qadiny ca Modulus of Rigidity o Twist angle de length of shaft We know, Power, P = T. w = Tx 2 TIN 60 105xsso lb- lb-ft/2 = TX 2T X 12 SO 608 T =441.2 lb-ft @ Based on shear stress, 441.2x12 lb in la xlo? lb/in ४ TT 609-b) ) 441.2x12x32 12 x103 2 "(-60) = Dex (1-0.84) 441.2x12x39 TT X12 x 2 x103 Do = 1.561 in D = 1.249 in CS Scanned with CamScanner

② Based angle of twist, ליח co l (1000x103)x & sam 441.2X12 36 T (08-0-20.36) 441,2x12x36x32 DY (1-0.89) 3. Sve D; = 1.294 in TTx (11000x103) * D. Do = 1.618 in PANAMA CS Scanned with CamScanner

Please comment if any doubt.