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For Circular X-section beam, show that maximum shear stress is 4/3 times of the average shear stress.
1 Answer
written 6.1 years ago by | • modified 6.1 years ago |
a=2R
$\frac{b}{2}=\sqrt {R^2-y^2}$
$b^2=4(R^2-y^2)$ (i)
$A \bar y= \int day$
Now, $I=\frac{\pi R^4}{4}$
$A \bar y= \int_{y}^{R} bdy *y$
$b^2=4(R^2-y^2)$
$2b.db=4*(-2y.dy)$
$b.db=-4y.dy$
$y.dy=\frac{-b.db}{4}$
$A \bar y= \int_{b}^{0} b(-\frac{b.db}{4})$
$A \bar y= \frac{b^3}{12}$
$q_{11}=\frac{F}{Ib}*A \bar y$
$=\frac{F}{Ib}*\frac{b^3}{12}= \frac{Fb^2}{12I}$
$\frac{F}{12I}*4*(R^2-y^2)$ from (i)
q will be minimum (zero) when y=R
q will be maximum when y=0
$q_{max}=\frac{F}{\frac{3\pi R^4}{4}} (R^2-0^2)= \frac{4}{3} \frac{4}{\pi R^2}$
$q_{max}=\frac{4}{3} q_{aug}$