Question 3.40:  A pin in a knuckle joint carrying a tensile load F deflects...

A pin in a knuckle joint carrying a tensile load  F deflects somewhat on account of this loading, making the distribution of reaction and load as shown in part (  b ) of the figure. A common simplification is to assume uniform load distributions, as shown in part (  c ). To further simplify, designers may consider replacing the distributed loads with point loads, such as in the two models shown in parts  d and  e . If  a=0.5 in,  b=0.75 in,  d=0.5 in, and  F=1000 \mathrm{lbf} estimate the maximum bending stress and the maximum shear stress due to  V for the three simplified models. Compare the three models from a designer’s perspective in terms of accuracy, safety, and modeling time.

 

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I=\frac{\pi}{64}\left(0.5^{4}\right)=3.068\left(10^{-3}\right) \mathrm{in}^{4}, A=\frac{\pi}{4}\left(0.5^{2}\right)=0.1963 \mathrm{in}^{2}

 

Model  (c)

M=\frac{500(0.5)}{2}+\frac{500(0.75 / 2)}{2}=218.75 \mathrm{lbf} \cdot \mathrm{in}

 

\sigma=\frac{M c}{I}=\frac{218.75(0.25)}{3.068\left(10^{-3}\right)} \\\sigma=17825 \mathrm{psi}=17.8 \mathrm{kpsi} \quad .

\tau_{\max }=\frac{4}{3} \frac{V}{A}=\frac{4}{3} \frac{500}{0.1963}=3400 \mathrm{psi}=3.4 \mathrm{kpsi} \quad

Model  (d)

\begin{aligned}&M=500(0.625)=312.5 \mathrm{lbf} \cdot \mathrm{in} \\&\sigma=\frac{M c}{I}=\frac{312.5(0.25)}{3.068\left(10^{-3}\right)} \\&\sigma=25464 \mathrm{psi}=25.5 \mathrm{kpsi} \quad \text {   } \\&\tau_{\max }=\frac{4}{3} \frac{V}{A}=\frac{4}{3} \frac{500}{0.1963}=3400 \mathrm{psi}=3.4 \mathrm{kpsi} \quad  \end{aligned}\\

Model  (e)

\begin{aligned}&M=500(0.4375)=218.75 \mathrm{lbf} \cdot \mathrm{in} \\&\sigma=\frac{M c}{I}=\frac{218.75(0.25)}{3.068\left(10^{-3}\right)}\end{aligned} \\\begin{aligned}&\sigma=17825 \mathrm{psi}=17.8 \mathrm{kpsi} \quad \text {  . } \\&\tau_{\max }=\frac{4}{3} \frac{V}{A}=\frac{4}{3} \frac{500}{0.1963}=3400 \mathrm{psi}=3.4 \mathrm{kpsi} \quad \text {   }\end{aligned}

 

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