The T-beam is subjected to a shear of V = 150 kN. Determine the amount of this force that is supported by the web B.

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Accepted Answer

[latex]\bar{y} =\frac{(0.02)(0.2)(0.04)+(0.14)(0.2)(0.04)}{0.2(0.04)+0.2(0.04)}=0.08 \mathrm{m}[/latex]

[latex]I= \frac{1}{12}(0.2)\left(0.04^{3} ight)+0.2(0.04)(0.08-0.02)^{2}[/latex]

[latex]+\frac{1}{12}(0.04)\left(0.2^{3} ight)+0.2(0.04)(0.14-0.08)^{2}=85.3333\left(10^{-6} ight) \mathrm{m}^{4}[/latex]

[latex]A^{\prime} =0.04(0.16-y)[/latex]

[latex]\bar{y}^{\prime} =y+\frac{(0.16-y)}{2}=\frac{(0.16+y)}{2} [/latex]

[latex]Q =\bar{y}^{\prime} A^{\prime}=0.02\left(0.0256-y^{2} ight) [/latex]

[latex] au =\frac{V Q}{I t}=\frac{150\left(10^{3} ight)(0.02)\left(0.0256-y^{2} ight)}{85.3333\left(10^{-6} ight)(0.04)}=22.5\left(10^{6} ight)-878.9\left(10^{6} ight) y^{2} [/latex]

[latex]V =\int au d A, \quad d A=0.04 d y[/latex]

[latex]V =\int_{-0.04}^{0.16}\left(22.5\left(10^{6} ight)-878.9\left(10^{6} ight) y^{2} ight) 0.04 d y[/latex]

[latex]=\int_{-0.04}^{0.16}\left(900\left(10^{3} ight)-35.156\left(10^{6} ight) y^{2} ight) d y[/latex]

[latex]=131250 \mathrm{N}=131 \mathrm{kN}[/latex]

Question 7.72: The T-beam is subjected to a shear of V = 150 kN. Determine...

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