Solution to Problem 573 | Horizontal Shearing Stress | Strength of Materials Review at MATHalino

Problem 573
The cross-section of a beam is an isosceles triangle with vertex uppermost, of altitude h and base b. If V is the vertical shear, show that the maximum shearing stress is 3V / bh located at the midpoint of the altitude.

Solution 573

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$f_v = \dfrac{VQ}{Ib}$

Location of maximum horizontal shear of triangular beam

Where
$Q = \frac{1}{2}xy \, [ \, \frac{2}{3}(h - y) \, ] = \frac{1}{3}xy \, (h - y)$

$I = \frac{1}{36}bh^3$

$b = x$

Thus,
$f_v = \dfrac{V \, [ \,\frac{1}{3}xy \, (h - y) \, ]}{(\frac{1}{36}bh^3)(x)}$

$f_v = \dfrac{12Vy(h - y)}{bh^3}$

$f_v = \dfrac{12V}{bh^3}(hy - y^2)$

$\dfrac{df_v}{dy} = \dfrac{12V}{bh^3}(h - 2y) = 0$

$h = 2y$

$y = \frac{1}{2} h$

Maximum shearing stress occurs at the mid-height of the altitude, h
$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} [ \, h(\frac{1}{2} h) - (\frac{1}{2} h)^2 \, ]$

$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} [ \, \frac{1}{2}h^2 - \frac{1}{4}h^2 \, ]$

$(\,f_v\,)_{max} = \dfrac{12V}{bh^3} (\frac{1}{4}h^2)$

$(\,f_v\,)_{max} = \dfrac{3V}{bh}$ answer

Solution to Problem 573 | Horizontal Shearing Stress

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