The hook is subjected to the force of 80 lb. Determine the state of stress at point B at section . The cross section has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.
Question 8.73: The hook is subjected to the force of 80 lb. Determine the s...
The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from
R=\frac{A}{\sum \int_{A} \frac{d A}{r}}Where A=\pi\left(0.25^{2}\right)=0.0625 \pi \mathrm{in}^{2}
\Sigma \int_{A} \frac{d A}{r}=2 \pi\left(\bar{r}-\sqrt{r^{2}-c^{2}}\right)=2 \pi\left(1.75-\sqrt{1.75^{2}-0.25^{2}}\right)=0.11278 \mathrm{in}
Thus,
R=\frac{0.0625 \pi}{0.11278}=1.74103 \mathrm{in}Then
e=\bar{r}-R=1.75-1.74103=0.0089746 \mathrm{in}Referring to Fig. b, I and Q_{B} are computed as
I=\frac{\pi}{4}\left(0.25^{4}\right)=0.9765625\left(10^{-3}\right) \pi \mathrm{in}^{4}
Q_{B}=\bar{y}^{\prime} A^{\prime}=\frac{4(0.25)}{3 \pi}\left[\frac{\pi}{2}\left(0.25^{2}\right)\right]=0.0104167 \mathrm{in}^{3}
Consider the equilibrium of the FBD of the hook’s cut segment, Fig. c,
\overset{+}{\leftarrow } \Sigma F_{x}=0 ; \quad N-80 \cos 45^{\circ}=0 \quad N=56.57 \mathrm{lb}
+\uparrow \Sigma F_{y}=0 ; \quad 80 \sin 45^{\circ}-V=0 \quad V=56.57 \mathrm{lb}
\curvearrowleft +\Sigma M_{o}=0 ; \quad M-80 \cos 45^{\circ}(1.74103)=0 \quad M=98.49 \mathrm{lb} \cdot \mathrm{in}
The normal stress developed is the combination of axial and bending stress. Thus,
\sigma=\frac{N}{A}+\frac{M(R-r)}{A e r}Here, M=98.49 \mathrm{lb \cdot in} since it tends to reduce the curvature of the hook. For point B, r=1.75 in. Then
\sigma=\frac{56.57}{0.0625 \pi}+\frac{(98.49)(1.74103-1.75)}{0.0625 \pi(0.0089746)(1.75)}
=1.48 \mathrm{psi}(T)
The shear stress is contributed by the transverse shear stress only. Thus,
\tau=\frac{V Q_{B}}{I t}=\frac{56.57(0.0104167)}{0.9765625\left(10^{-3}\right) \pi(0.5)}=384 \mathrm{psi}The state of stress of point B can be represented by the element shown in Fig. d.
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