Question

Prob 12 4" dia anchored solid steel rod, 10' long. 500 ft-lb T applied at the top. G=12 x 106 psi a Calculate Imax and the twist/distortion b Using the same weight of steel, how could distortion (Y) and shear stress (Tu) be reduced? 1 c. Suppose an axial load is applied. How does what was done for prob 12b also reduce the likelihood of buckling?

Answer 1

using the formula-

$\frac{\tau }{r}=\frac{T}{J}=\frac{G\Theta }{L}$

1 feet = 12inch

now, maximum shear stress will occur at surface of shaft.

$\frac{\tau_{max}}{2}=\frac{500*12}{\frac{\Pi }{32}*4^{4}}$

$\tau _{max}=477.465psi$

& twist is given as

$\Theta =\frac{TL}{GJ}=\frac{500*12*10*12}{12*10^{6}*\frac{\pi }{32}*4^{4}}$

$\Theta =0.002387$ radian

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ANSWER (B)

For reducing the above values we provide a hollow shaft having same weight and material.

ANSWER (C)

When we provide hollow shaft than we increse the moment of inertia as the mass of shaft is centered away from the axis of shaft.

Hence radius of gyration (r) is

$r_{min}=\sqrt{\frac{I}{A}}$

As I increases value of 'r' increases so slenderness ratio decreases-

$\lambda =\frac{l_{eff}}{r_{min}}$

this will result in decrease in buckling of shaft.