Answer:
Answer is part d -736.88 psi
Step-by-step explanation:
We know that for a bar subjected to pure torsion the shear stresses that are generated can be calculated using the following equation
....................(i)
Where
T is applied Torque
is the polar moment of inertia of the shaft
t is the shear stress at a distance 'r' from the center
r is the radial distance
Now in our case it is given in the question T =82.7 ft*lbs
converting T into inch*lbs we have T = 82.7 x 12 inch*lbs =992.4 inch*lbs
We also know that for a circular shaft polar moment of inertia is given by
![I_(P)=(\pi D^(4) )/(32)](https://img.qammunity.org/2020/formulas/engineering/college/ize9jf6555l27p3hmbgckfmbo6wmn4zamf.png)
![I_(P)= (\pi\ 1.9^(4) )/(32) =1.2794 inch^(4)](https://img.qammunity.org/2020/formulas/engineering/college/u7pjsx9f5viulqtthf5xxv1ipt8zvz7bt1.png)
Since we are asked the maximum value of shearing stresses they occur at the surface thus r = D/2
Applying all these values in equation i we get
= t
Thus t = 736.88 psi