The eye hook has the dimensions shown. If it supports a cable loading of 80 kN, determine the maximum normal stress at section and sketch the stress distribution acting over the cross section.
Question 8.74: The eye hook has the dimensions shown. If it supports a cabl...
\int \frac{d A}{r}=2 \pi\left(3.125-\sqrt{(3.125)^{2}-(0.625)^{2}}\right)=0.395707
R=\frac{A}{\int \frac{d A}{r}}=\frac{\pi(0.625)^{2}}{0.396707}=3.09343 \mathrm{in} .
M=800(3.09343)=2.475\left(10^{3}\right)
\sigma=\frac{M(R-r)}{A r(\bar{r}-R)}+\frac{P}{A}
\left(\sigma_{t}\right)_{\max }=\frac{2.475\left(10^{3}\right)(3.09343-2.5)}{\pi(0.625)^{2}(2.5)(3.125-3.09343)}+\frac{800}{\pi(0.625)^{2}}=15.8 \mathrm{ksi}
\left(\sigma_{c}\right)_{\max }=\frac{2.475\left(10^{3}\right)(3.09343-3.75)}{\pi(0.625)^{2}(3.75)(3.125-3.09343)}+\frac{800}{\pi(0.625)^{2}}=-10.5 \mathrm{ksi}
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