Problem 571

For a beam with the same cross section as that in Prob. 570, plot the shearing stress distribution across the section at a section where the shearing force is V = 1800 lb.

Solution 571

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} \,\,            (ok!)

The plot of the shear stress distribution is as shown in the above figure.