(Solved): (a) While computing the bending moment and the shear force acting on the side walls of the spillwa ...

(a) While computing the bending moment and the shear force acting on the side walls of the spillway chute of Figure $$3\mathbf{a}$$, a novice structural engineer-in-training assumed that the water pressure varies linearly from zero at the surface to $$?gy$$ at the invert of the chute, where $$y$$ is the flow depth measured vertically. What are the computed values for the bending moment and the shear force at the invert level? Are the computed results correct? If not, compute the percent error. Section A-A $$g=32.2\mathrm{ft}/{\mathrm{s}}^2,g=9.81\mathrm{m}/{\mathrm{c}}^2;?{?}^2=\mathrm{eg}=60.4\mathrm{b}/{\mathrm{ft}}^3;1$$ foot=12 inches $$H_L={\left(Y_2?y_1\right)}^3/4y_1{y}_2;y_2/y_1=\frac{1}{2}\left[\sqrt{1+8F_1^2}?1\right];\frac{y_1}{y_2}=\frac{1}{2}\left[\sqrt{1+8F_2^2}?1\right]$$ $$?Q\left(V_2?V_1\right)=F_1?F_2;y_c={\left(q^2/g\right)}^{\frac{1}{3}};H=z+y+\frac{v^2}{2g}$$ $$F_n=V/\sqrt{9D}=V/\sqrt{gy};Q=\frac{1.486}{n}A{R}^{2/3}{s}^{\frac{1}{2}};z=Q/\sqrt{g}\mathrm{and}z=\frac{Q}{\sqrt{gk}}$$ $$V=C\sqrt{RS};C=\frac{1.486}{n}{R}^{1/6};{?}_8={\left[\underset{i=1}{\overset{N}{?}}\left(P_N{n}_N^{1.5}\right)/P\right]}^{2/3}$$ For minimum Wetted Perimeter $$b=2y\left(\sqrt{1+z^2}?z\right)$$ $$?=1+3{?}^2?2{?}^3;?=1+{?}^2;?=\frac{t_{\text{max }}}{V}?1;E=y+\frac{v^2}{2g}$$ $$\frac{V^2}{2g}=\frac{D}{2};\frac{Q^2}{g{A}_1}+{\stackrel{?}{z}}_1{A}_1=\frac{Q^2}{g{A}_2}+{\stackrel{?}{z}}_2{A}_2;y_c^3=\frac{2y^2{y}^2}{y_1+y_2}$$ $$?E={\left(y_2?y_1\right)}^3/4y,y_2;Q=K\sqrt{5};K=\frac{1.486}{n}A{R}^{2/3};q=\frac{Q}{b}$$ $$A{R}^{2/3}=nQ/(1.496\sqrt{5});F=\sqrt{cy};{?}_i=?RS;q_i=0.63q$$ $$K=\sqrt{1?\frac{{\sin }^2?}{{\sin }^2?}};K={?}_0/{?}_L;{?}_L=\frac{W_A}{?}\tan {?}_j\left[L=6y_2\right]k{F}_1<20$$ $${?}_s=\frac{W_{?}}{a}\cos ?\tan ?\sqrt{1?\frac{t_n^2?}{t_n^2?}};S_f=\frac{Q^2}{k^2};H_1=H_2+h_f+h_e$$ $$Q=C_{d}A\sqrt{2gh};HW+S_{2}L+\frac{V_0^2}{2g}=TW+H_L;H_1=HW+S_{1}L+\frac{V_0^2}{2g}$$ $$H_L=HN+S_{0}L?TW;H_L=\left(C_L+1+\frac{g{n}^{2}L}{1.11R^{43}}\right)\frac{8Q^2}{{?}^{2}g{D}^4};H_2=TW$$. $$s_f=\frac{29n^2}{R^{43}}\frac{v^2}{2g}$$ or $$sf=\frac{19.6n^2{v}^2}{R^{4/3}2g};h_E=kE{v}^2/2g$$.