Respuesta :
Answer:
Answer is part d -736.88 psi
Explanation:
We know that for a bar subjected to pure torsion the shear stresses that are generated can be calculated using the following equation
[tex]\frac{T}{I_{P} } =\frac{t}{r}[/tex]....................(i)
Where
T is applied Torque
[tex]I_{P}[/tex] 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
[tex]I_{P}=\frac{\pi D^{4} }{32}[/tex]
[tex]I_{P}= \frac{\pi\ 1.9^{4} }{32} =1.2794 inch^{4}[/tex]
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
[tex]\frac{992.4 inch*lbs}{1.2794 inch^{4} } \frac{1.9 inches}{2}[/tex] = t
Thus t = 736.88 psi
