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Question

i need help solving this problem ill rate u 5 stars please be clear and show steps. please find fnal answes.

Consider a solid round bar with a diameter of 1 in, under a torque T = 2,000 in#. Consider also a solid square bar with the same area as the round bar. Finally, consider a narrow rectangular bar with the same area as the round bar and having an aspect ratio of 20:1. Calculate the shear stress tau in the round bar due to torsion, expressed in psi Calculate the shear stress tau in the square bar under the same torsion, expressed in psi Calculate the shear stress tau in the narrow rectangular bar under the same torsion, expressed in psi

Explanation / Answer

the given torque = 1000 lbin

diameter of bar = 1in

area of bar =area of square

pi/4*d^2 = l^2

pi/4*1 = l^2

l= 0.88 in

aspect ratio = 20:1

area of rectangular section = cross section area of bar

b * t =pi/4 *d^2

20*b^2 =pi /4

b = 0.1981 in

t = 0.001 in

polar moment of inertia of round bar =J=pi/32 * d^4

= pi /32 = 0.0981 in^4

a) shear stress in round bar = tou = T *d /(2 *J)

= 1000 *1 / (2* 0.0981)

= 5096.83 psi

b) shear stress in square bar = T /(p *q *l) ; here p, q are coefficient depending on size , l= thickness = 0.88 in

p = 0.208

q = 0.141

shear stress = 1000 / (0.208 *0.141 * 0.88)

= 38746.71 psi

c) shear stress in rectangular bar = T /(p *q *l) ; here p, q are coefficient depending on size,t= thickness=0.001 in

here p =0.333 q =0.333

shear stress = 1000/(0.333*0.333*0.001)

=9018027.03 psi

reference if you need :the coefficient table is calculated as per st. venant's principal

and is given in follong link

http://emweb.unl.edu/Mechanics-Pages/J-Elder/torsion.html

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