$q(s)=q(0)-\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_z}{I_y}{\displaystyle \underset{0}{\overset{s}{}}}tzds+\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_y}{I_z}{\displaystyle \underset{0}{\overset{s}{}}}tyds$

$q(s)=q(0)-$hat$(q)(s)ds$

hat$(q)(s)=\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_z}{I_y}{\displaystyle \underset{0}{\overset{s}{}}}tzds-\frac{\frac{?}{\text{b}\text{a}\text{r}}(V{)}_y}{I_z}{\displaystyle \underset{0}{\overset{s}{}}}$tydsFor the thin$-$walled single$-$cell closed section shown below, determine $(1)$ the shear flow

distribution for shear force $V_y($positive$)$ acting though the shear center, $(2)$ the maximum

shear stress, and $(3)$ the location of the shear center along the $z-$axis, $e_z,$ from the Center of

Gravity, C$.$G$.$ Draw a diagram of the shear flow $($use symmetry$).($Note: the shear force is

graphically displayed acting through the $C.G.$ since the location of the shear center $(z-$axis

ordinate$)$ is not known$).$ Start $q_{(0)}$ at Point A and flow through B to C to D$.$ All calculations

must be in kilonewtons, $kN,$ and millimeters, $mm.$ Note the coordinate system.