Question

The A-36 steel bolt is tightened within a hole so that the reactive torque on the shank AB can be expressed by the equation t = (kx2) N.m/m, where x is in meters. If a torque of T = 50 N.m is applied to the bolt head, determine the constant k and the amount of twist in the 50-mm length of the shank. Assume the shank has a constant radius of 4 mm. The shear modulus of elasticity for A-36 steel is 75 GPa.

Answer 1

Answer:
`k=1.2* 10^6 N/m^2,\theta =3.56^(\circ)`

Step-by-step explanation:

Given

`t=kx^2 Nm/m`

And a torque of 50 N.m is applied thus

`T=\int_(0)^(0.05)kx^2dx`

`50=(1)/(3)\left [ 5* 10^(-2)\right ]^3`

`k=1.2* 10^6 N/m^2`

T at a distance x is given

`T=\int_(0)^(x)1.2* 10^6x^2dx`

`T=0.4* 10^6.x^3`

For Angle of twist

`\theta =(TL)/(GJ)`

`J=(\pi d^4)/(32)`

`J=(\pi * 8^4* 10^(-12))/(32)=0.402176`

G=75 GPa

`\theta =\int_(0)^(0.05)(\left ( 50-0.4* 10^6x^3\right ))/(GJ)`

`\theta =3.56^(\circ)`