Question F12.3: Determine the absolute maximum shear stress in the beam....

V_{\max }=4.5 kip

I=\frac{1}{12}(3)\left(6^{3}\right)=54\ \mathrm{in}^{4}

Take top half of area.

Q_{\max }=y^{\prime} A^{\prime}=1.5(3)(3)=13.5\ \mathrm{in}^{3}

\left(\tau_{\max }\right)_{\mathrm{abs}}=\frac{V_{\max } Q_{\max }}{I t}=\frac{4.5\left(10^{3}\right)(13.5)}{54(3)}=375\ \mathrm{psi}