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

[math]f_v = \dfrac{VQ}{Ib} = \dfrac{V}{Ib}Ay'[/math]

From the solution of Problem 570
[math]I = \frac{1232}{3} \, \text{in}^4[/math]
[math]b = 2 \, \text{in}[/math]
[math]Q_{NA} = 52 \, \text{in}^3[/math]

[math]f_{v1} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) \, ][/math]

[math]f_{v1} = 78.90 \, \text{psi}[/math]

[math]f_{v2} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) \, ][/math]

[math]f_{v2} = 94.24 \, \text{psi}[/math]

[math]f_{v3} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) + 1(2)(2.5) \, ][/math]

[math]f_{v3} = 105.19 \, \text{psi}[/math]

[math]f_{v4} = \dfrac{1800}{\frac{1232}{3}(2)} \, [ \, 1(8)(4.5) + 1(2)(3.5) + 1(2)(2.5) + 1(2)(1.5) \, ][/math]

[math]f_{v4} = 111.77 \, \text{psi}[/math]

[math]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) \, ][/math]

[math]f_{v5} = 113.96 \, \text{psi} = ( \, f_v \, )_{max}[/math]

Checking: at the neutral axis
[math]( \, f_v \, )_{max} = \dfrac{1800(52)}{\frac{1232}{3}(2)}[/math]
[math]( \, f_v \, )_{max} = 133.96 \, \text{psi} \,\,[/math]            (ok!)

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