Question #120206

For the solid steel shaft shown (G = '77 GPa}, determine the “v angle of twist at A. (b) Solve part (a), assuming that the steel shaft is hollow with a 30-mm enter diameter and a 20-mm inner diameter.

Expert's answer

Given G = 77 GPa

Outer diameter of shaft = 30 mm

Inner diameter of shaft = 20 mm

Length of shaft = 1.8 m

a)

Solution

*d* = 30 mm

Therefore radius *c* = 15 mm

From Torsional equation

*T*/J = *G0*/*I*

For solid steel shaft angle of twist at 0 = *TI/GJ*

* = *(250*1.8)/(77*10^{9}*3.14/2*15^{4}*10^{-12})

= 0.073491*(180/3.14)

= 4.21^{0}

Hence angle of twist = 4.21^{0}

b)

Solution

*d* = 30 mm

*c* = 15 mm

Hence inner radius *c*_{i} = 10 mm

Polar moment of inertia of hollow cylinder *J = *3.14/2(*c*^{4}-*c*^{4}_{i})

Hence; from torsional equation, angle of twist *0*_{H} = (*TI/GJ*)

= - (*TI*)/(*G*3.14/2*(*c*^{2}-*c*^{4}_{i})

= (250*1.8*2)/(77*10^{9}*3.14*(15^{4}-10^{4})*10^{-12}

^{ }= 5.247^{0}

Therefore the angle of twist is 5.247^{0}

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