The assembly is subjected to a vertical shear of V = 7 kip . Determine the shear flow at points A and B and the maximum shear flow in the cross section.

Question

Accepted Answer

[latex]\bar{y}=\frac{\Sigma \bar{y} A}{\Sigma A}=\frac{(0.25)(11)(0.5)+2(3.25)(5.5)(0.5)+6.25(7)(0.5)}{0.5(11)+2(0.5)(5.5)+7(0.5)}=2.8362 \mathrm{in.} [/latex]

[latex]I=\frac{1}{12}(11)\left(0.5^{3}\right)+11(0.5)(2.8362-0.25)^{2}+2\left(\frac{1}{12}\right)(0.5)\left(5.5^{3}\right)+2(0.5)(5.5)(3.25-2.8362)^{2}[/latex]

[latex]\quad+\frac{1}{12}(7)\left(0.5^{3}\right)+(0.5)(7)(6.25-2.8362)^{2}=92.569 \mathrm{in}^{4}[/latex]

[latex]Q_{A}=\bar{y}_{1}^{\prime} A_{1}^{\prime}=(2.5862)(2)(0.5)=2.5862 \mathrm{in}^{3}[/latex]

[latex]Q_{B}=\bar{y}_{2}^{\prime} A_{2}^{\prime}=(3.4138)(7)(0.5)=11.9483 \mathrm{in}^{3}[/latex]

[latex]Q_{\max }=\Sigma \bar{y}^{\prime} A^{\prime}=(3.4138)(7)(0.5)+2(1.5819)(3.1638)(0.5)=16.9531 \mathrm{in}^{3}[/latex]

[latex]q=\frac{V Q}{I}[/latex]

[latex]q_{A}=\frac{7\left(10^{3}\right)(2.5862)}{92.569}=196 \mathrm{lb/in} .[/latex]

[latex]q_{B}=\frac{1}{2}\left(\frac{7\left(10^{3}\right)(11.9483)}{92.569}\right)=452 \mathrm{lb/in} . [/latex]

[latex]q_{\max }=\frac{1}{2}\left(\frac{7\left(10^{3}\right)(16.9531)}{92.569}\right)=641 \mathrm{lb/in}.[/latex]

Question 7.61: The assembly is subjected to a vertical shear of V = 7 kip ....

Related Answered Questions