The hook is subjected to the force of 80 lb. Determine the state of stress at point A at section a–a. The cross section is circular and has a diameter of 0.5 in. Use the curved-beam formula to compute the bending stress.

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The location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from

[latex]R=\frac{A}{\sum \int_{A} \frac{d A}{r}}[/latex]

where [latex]A=\pi\left(0.25^{2} ight)=0.0625 \pi \mathrm{in}^{2}[/latex]

[latex]\Sigma \int_{A} \frac{d A}{r}=2 \pi\left(\bar{r}-\sqrt{r^{2}-c^{2}} ight)=2 \pi\left(1.75-\sqrt{1.75^{2}-0.25^{2}} ight)=0.11278 \mathrm{in} .[/latex]

Thus,

[latex]R=\frac{0.0625 \pi}{0.11278}=1.74103 \mathrm{in}[/latex]

Then

[latex]e=\bar{r}-R=1.75-1.74103=0.0089746 \mathrm{in}[/latex]

Referring to Fig. b, I and [latex]Q_{A}[/latex] are

[latex]I =\frac{\pi}{4}\left(0.25^{4} ight)=0.9765625\left(10^{-3} ight) \pi \mathrm{in}^{4} [/latex]

[latex]Q_{A} =0[/latex]

Consider the equilibrium of the FBD of the hook's cut segment, Fig. c,

[latex]\overset{+}{\leftarrow } \Sigma F_{x}=0 ; \quad N-80 \cos 45^{\circ}=0 \quad N=56.57 \mathrm{lb}[/latex]

[latex]+\uparrow \Sigma F_{y}=0 ; \quad 80 \sin 45^{\circ}-V=0 \quad V=56.57 \mathrm{lb}[/latex]

[latex]\curvearrowleft +\Sigma M_{o}=0 ; \quad M-80 \cos 45^{\circ}(1.74103)=0 \quad M=98.49 \mathrm{lb} \cdot \mathrm{in}[/latex]

The normal stress developed is the combination of axial and bending stress. Thus,

[latex]\sigma=\frac{N}{A}+\frac{M(R-r)}{A e r}[/latex]

Here, [latex]M=98.49 \mathrm{lb\cdot in}[/latex] since it tends to reduce the curvature of the hook. For point A, r=1.5 in. Then

[latex]\sigma=\frac{56.57}{0.0625 \pi}+\frac{(98.49)(1.74103-1.5)}{0.0625 \pi(0.0089746)(1.5)}[/latex]

[latex]=9.269\left(10^{3} ight) \mathrm{psi}=9.27 \mathrm{ksi}(T)[/latex]

The shear stress in contributed by the transverse shear stress only. Thus

[latex] au=\frac{V Q_{A}}{I t}=0[/latex]

The state of strees of point A can be represented by the element shown in Fig. d.

Question 8.72: The hook is subjected to the force of 80 lb. Determine the s...

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