Question

Problem 4: Design the diameter of A992 steel shaft that is subjected to the torques as shown in the figure; provided that the angle of twist of end B with respect to end A should not exceed 1 degree. 1200 N.m 400 Nm 600 N.m 600 mm 500 mm 200 Nm 400 mm

Answer 1

mument 14 # A 992 Steel shaft d=3 shear modulus of rigidity CG 12 75 G Pa [Assumed) Torsional diagram sign convection Torsional tuel » « moment Torsional X Ott X X Die [-ve

free bally diagram. + 2000m +600nm) 400NM -1200Nm *** A 0 C porsional moment dicteram +400 mm tRoo NM с B -200 NI -1240 Nm +400 = -800NM gonm [soonm equation from to usional GO T - Ho L To Torsional moment TL JG t e Length J + poley moment of Inertig G A shear modulus

AC - Sechion TAC = 200 Nms 2x105 Nmm LAC = 400mm G = 75 G Pa 2 758163 Mpa I sud Q 10 XA sad 180 180 max. } TD4 32 OF TAC LAC AC JAC G. 2x105 x 400 CAC C Jae x 75x103 2x105 X400 JAC ADI 32 11 lxn 180 X75X103 2x105x400 04 = 5622517:35 32 x 11 Gl#5X103 DAC = 28.089 mm

Section CD TED 800 Nm = 8x105 Nom LCD = 500 mm 69 75x105 mpa Aman T sad 180 JCD Tot 32 = Tep to Јер G7 Tcoteo JCD = ах Сл 324,8X105x 1500 04 = 32 Ten Lio Ох М 75 Go) x 75x103 302586762 mm D CDF 42-00 section DB TOB 400 um= 4x105 Nam Los = Goomm 75x105 MPa sad Amcx =D Tos

TOBLDB D*= x I Toeloe amaxxG 4X105 x 600 32 x D Du I Geox 756165 DE 11:69 mm DA calculated diameter DAC = 28-08 mm DCD = 42.00 mm la 69 mm DDB we have Herve uniform har (diameter) is used so are be to choose minimum dia meteo for the designing of the shaft. since If we taking thes it will maximum lalue for diameter Violate the given cenclition Amax greater than One. so Design diameter = 11.69mm vlimm

Design dia = 11.69 mm

If we choose D =42 mm then shaft AC & shaft DB will have angular twist greater than one degree.

So for design purpose we choose minimum diameter =11.69 mm