$f_v = \dfrac{VQ}{Ib} = \dfrac{V}{Ib}Ay'$
From the solution of Problem 570
$I = \frac{1232}{3} \, \text{in}^4$
$b = 2 \, \text{in}$
$Q_{NA} = 52 \, \text{in}^3$
$f_{v1} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) \, ]$
$f_{v1} = 78.90 \, \text{psi}$
$f_{v2} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) \, ]$
$f_{v2} = 94.24 \, \text{psi}$
$f_{v3} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) + 1(2)(2.5) \, ]$
$f_{v3} = 105.19 \, \text{psi}$
$f_{v4} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) + 1(2)(2.5) + 1(2)(1.5) \, ]$
$f_{v4} = 111.77 \, \text{psi}$
$f_{v5} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) + 1(2)(2.5) + 1(2)(1.5) + 1(2)(0.5) \, ]$
$f_{v5} = 113.96 \, \text{psi} = ( \, f_v \, )_{max}$
Checking: at the neutral axis
$( \, f_v \, )_{max} = \dfrac{1800(52)}{\frac{1232}{3}(2)}$
$( \, f_v \, )_{max} = 133.96 \, \text{ psi}$ (okay!)
The plot of the shear stress distribution is as shown in the above figure.
