Question

solve with explanasion please

1. The gear shaft shown below is a solid cylinder made of a material that has an allowable shear stress of 75 MPa and a shear modulus of 25 GPa. Calculate: 600 Nm a. The minimum shaft diameter (to nearest mm) required in sections AB and CD (10 pts) b. The resulting angle of twist of gear A relative to gear B (10 pts) 200 mm 900 Nm 500 Nm 200 mm 300 N·m SOON.m 200 mm 200 mm

Answer 1

Given Given allowable ahear stren T = 75 MPa Shear Modulus G = 25 6 Pa = 25x103 MPa writing Right the direction of applied torque as per the hand thumb Rule :- 300 Nm GooN.ml 300 Nm 509 km SOON D E 200 mm 200 mm 200 วก 200 mm I 1 T we know the Tossion formula: 1. Getty where = sheor stress Generated in shaft 8= radius of shaft T= applied taque J= polos moment of prestia 1/2 t for circular shaft G= Angle of twist b length of shaft (a) Minimum shaft diameter required in section AB and CD for section AB : we cut section somewhere between AB them from left side we Can See net torque is GOD Nam on we can see from Right also from ean o = if AB T T H T= 600 00 T A 7 32 04 1S putting the values olu 600x133 7 D= 34.4 mm D = 35mm

-> Similarly for section CD if we cut the section & somewhere between co then either from left Right we get torque applied el 900 Nom so 3 75 D4 22 200x10 을 매 33 32 hence, D = 23.85 mm ~ 24 mm (b) Resulting angle of twist of gear A Relative to gean B:- length of shaft Between AaB Given G=250103 MPa from formular L= 200mm 7 = co TL T= Goox 103 H-mm J104 32 (35)" J=1178 588.119 mm Х 32 so hoox163x 200 25X103 x 1178588-119 4.072x103 Radian OP 0.23 degree