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

[math]f_v = \dfrac{VQ}{Ib}[/math]

Where
[math]Q = \frac{1}{2}xy \, [ \, \frac{2}{3}(h – y) \, ] = \frac{1}{3}xy \, (h – y)[/math]
[math]I = \frac{1}{36}bh^3[/math]
[math]b = x[/math]

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

[math]f_v = \dfrac{12Vy(h - y)}{bh^3}[/math]

[math]f_v = \dfrac{12V}{bh^3}(hy - y^2)[/math]

[math]\dfrac{df_v}{dy} = \dfrac{12V}{bh^3}(h - 2y) = 0[/math]

[math]h = 2y[/math]
[math]y = \frac{1}{2} h[/math]

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

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

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

[math](\,f_v\,)_{max} = \dfrac{3V}{bh} \,\, [/math]            answer