The channel is subjected to a shear of V = 75 kN . Determine the maximum shear flow in the channel.
Question 7.59: The channel is subjected to a shear of V = 75 kN . Determine...
\bar{y}=\frac{\sum{ \bar{y} A}}{\sum{A}}=\frac{0.015(0.4)(0.03)+2[0.13(0.2)(0.03)]}{0.4(0.03)+2(0.2)(0.03)} \\=0.0725 \mathrm{m}
I=\frac{1}{12}(0.4)\left(0.03^{3}\right)+0.4(0.03)(0.0725-0.015)^{2}
\quad+2\left[\frac{1}{12}(0.03)\left(0.2^{3}\right)+0.03(0.2)(0.13-0.0725)^{2}\right]
=0.12025\left(10^{-3}\right) \mathrm{m}^{4}
Q_{\max }=\bar{y}^{\prime} A^{\prime}=0.07875(0.1575)(0.03)=0.37209\left(10^{-3}\right) \mathrm{m}^{3}
q_{\max }=\frac{75\left(10^{3}\right)(0.37209)\left(10^{-3}\right)}{0.12025\left(10^{-3}\right)}=232 \mathrm{kN/m}
Related Answered Questions
\sigma_{A}=0=\sigma_{a}-\sigma_{b} ...
\bar{y} =\frac{(0.02)(0.2)(0.04)+(0.14)(0.2...
\bar{y}=\frac{(0.375)(4)(0.75)+(3.75)(6)(0....
\bar{y}=\frac{(0.375)(4)(0.75)+(3.75)(6)(0....
Section Properties:
\bar{y} =\frac{\Sigma \...
\bar{y}=\frac{\Sigma \bar{y} A}{\Sigma A}=\...
Shear Center. Referring to Fig.a and summing momen...
Summing moments about A ,
e P=b F_{1} [/lat...