Question

The 60-mm-diameter solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the absolute maximum and minimum shear stresses in the shaft's surface and specify their locations, measured from the free end. 10-25. The solid shaft is subjected to the distributed and concentrated torsional loadings shown. Determine the required diameter d of the shaft if the allowable shear stress for the material is Tallow = 60 MPa. 400 N·m 4 kN.m/m А. < 0.2 m 0.2 m X 0.5 m d 0.5 m 800 N·m Probs. 10-24/25

Answer 1

This problem can be solved as follows...

As there is variation of torsional moment is given over the span so to calculate the absolute maximum and minimum shear stress, we have to calculate the maximum and minimum torsional moment.

For this we draw the diagram which show the variation of torsional moment over the span length.

Then we calculate the desired result.

Calculation....

Given. Data in Rukwimm L nego ttttti Soow-m - I no nom o.sm 0.2m 0.2m1 o.sm first we have to draw the variation of b torsional moment over the opon of PORST. fou this we follow the following sign Convention If any Torsional moment give directed / direction toward - the section (by right hand thumb rule), than it will be - treated as positive, because it's effect is similar of tighting ! - of screw.. a upper face act as positive side for showing/ represention. Jon Po-spam - Te = boonm [constant from P to o] Sar span or (a form o) Tx=-800 + 400xx e s pod=0 = To = -800Nm d=0:58 Te = -800+ uoux dos =-6oooom for sparks T = -800 + 200 =-6oonm [constant for Rs7

fov (81) spomen x=-600 -400= 1000Nm constant , diagrame Est Goo Geo fou 800 1ooo (070 - from diagram, absolute maximum absolute minimum Then from Torsion Torsion= Torsion = equation, I= 10000-m To = 6oo Nom ۶ T= Torsionat moment Jp = polar moment of inertia =2x It xD" = Jp = 1272.345 x109 mt ra radiwa Somlo? 30x16?m a= stressa at required readius / radial distance

absolute maximum shear stress - Tmar op or =D J I 1000 x (80x103) 1272.345X169 106 man = 23.578 Mpa required MPa = N/mm2 Answer Now, absolute minimum shear stress at min 1 - - Tmin= . min.yu = 6 cro 1272.345 1169 -X (30x103) Emin. = 146 147 mral required Answas - Foot Solution (10-25): for abosot absolute maximum shear storess 204 e max. - Tallows = I ru 5) 60 = lo Booty

53 16х1 ооо (у ох (ос ГФ = 43.95 тттен — ц3. 5 mm) nywer