Problem

# The ring shown in Fig. P5.44 has the following dimensions: r=150 mm, T=50 mm, and h=100...

The ring shown in Fig. P5.44 has the following dimensions: r=150 mm, T=50 mm, and h=100 mm. Taking determine (a) the tangential stress on the inner fiber at and (b) the deflection along the line of action of the load P, considering the effects of the normal and shear forces, as well as bending moment (Sec. 10.4).

#### Step-by-Step Solution

Solution 1

(a)

Calculate the moment at $$\theta=\pi / 4$$.

$$M_{\theta}=0.182 P r-\frac{1}{2} P_{r}(1-\cos \theta)$$

Here, $$M_{\theta}$$ is the moment at angle $$\theta, P$$ is the applied load and $$\bar{r}$$ is centroid and $$\theta \mathrm{s}$$ the angle.

Substitute $$150 \times 10^{-3} \mathrm{~m}$$ for $$\vec{r}$$ and $$\pi / 4$$ for $$\theta$$.

\begin{aligned} M_{\theta} &=(0.182)\left(150 \times 10^{-3}\right) P-\frac{1}{2} P\left(150 \times 10^{-3}\right)\left(1-\cos \left(\frac{\pi}{4}\right)\right) \\ &=0.0273 P-0.021967 P \\ &=0.00533 P \mathrm{~N} \cdot \mathrm{m} \end{aligned}

Calculate the area of section.

$$A=t \times h$$

Here, $$h$$ is width of the riding and $$t$$ is thickness of the ring section.

Substitute $$50 \times 10^{-3} \mathrm{~m}$$ for $$t$$ ad $$100 \times 10^{-3} \mathrm{~m}$$ for $$h$$.

\begin{aligned} A &=\left(50 \times 10^{-3}\right)\left(100 \times 10^{-3}\right) \\ &=0.005 \mathrm{~m}^{2} \end{aligned}

$$R=\frac{h}{\ln \left(\frac{r_{0}}{r_{i}}\right)}$$

Here, $$r_{e}$$ is outer radius and $$r_{i}$$ is inner radius.

Substitute $$100 \mathrm{~mm}$$ for $$h, 200 \mathrm{~mm}$$ for $$r_{e}$$ and $$100 \mathrm{~mm}$$ for $$r_{i}$$.

\begin{aligned} R &=\frac{100}{\ln \left(\frac{200}{100}\right)} \\ &=\frac{100}{0.69315} \\ &=144.2695 \mathrm{~mm} \end{aligned}

Calculate the eccentricity.

$$e=\bar{r}-R$$

Here, $$\bar{r}$$ is centroidal distance and $$R$$ is centroidal Radius.

Substitute $$150 \mathrm{~mm}$$ for $$\bar{r}$$ and $$144.2695 \mathrm{~mm}$$ for $$R$$.

\begin{aligned} e &=150-144.2695 \\ &=5.7355 \mathrm{~mm} \end{aligned}

At inner fiber $$r_{i}=-100 \mathrm{~mm}$$ :

Calculate the tangential stress on the inner fiber at $$\theta=\pi / 4$$.

$$\left(\sigma_{\theta}\right)_{\pi / 4}=-\frac{(P / 2) \cos \theta}{A}+\frac{M_{\theta}}{A}\left(\frac{R-r_{i}}{e r_{i}}\right)$$

Here, $$\sigma_{\theta}$$ is tangential stress on the inner fiber and eccentricity.

Substitute $$\pi / 4$$ for $$\theta, 0.05 \mathrm{~m}^{2}$$ for $$A, 0.00533 P \mathrm{~N} \cdot \mathrm{m}$$ for $$M_{\theta}, 144.2695 \times 10^{-3} \mathrm{~mm}$$ for $$R$$, $$100 \times 10^{-3} \mathrm{~m}$$ for $$r_{j}$$ and $$5.7355 \times 10^{-3} \mathrm{~m}$$ for $$e$$.

\begin{aligned} \left(\sigma_{\theta}\right)_{\frac{r}{4}} &=-\frac{(P / 2) \cos (\pi / 4)}{0.005}+\frac{0.00533 P}{0.005}\left(\frac{144.2695 \times 10^{-3}-100 \times 10^{-3}}{\left(5.7355 \times 10^{-3}\right)\left(-100 \times 10^{-3}\right)}\right) \\ &=-70.71 P+(-1.066 P \times 77.22) \\ &=-(70.71+82.315) P \\ &=-153.03 P \end{aligned}

(b)

Draw the schematic diagram.

At the value of $$\theta=0^{\circ}$$ :

Calculate the moment at $$\theta=0^{\circ}$$.

$$M_{\theta}=0.182 \bar{r} P-\frac{1}{2} P \bar{r}(1-\cos \theta)$$

Here, $$\bar{r}$$ is the centroidal distance.

Substitute $$R$$ for $$\bar{r}$$ and $$0^{\circ}$$ for $$\theta$$.

\begin{aligned} M_{o} &=0.182(R) P-\frac{1}{2} P R\left(1-\cos 0^{\circ}\right) \\ &=0.182 P R-\frac{1}{2} P R(0) \\ &=0.182 P R \end{aligned}

Calculate the normal force at $$\theta=0^{\circ}$$.

$$N_{\theta}=-\frac{P}{2} \cos \theta \ldots \ldots \text { (1) }$$

Here, $$N_{\theta}$$ is the normal force at angle $$\theta$$.

\begin{aligned} &N_{\theta}=\frac{P}{2} \cos \left(0^{\circ}\right) \\ &N_{\theta}=-\frac{P}{2} \end{aligned}

Differentiate the equation (1) with respect to $$\mathrm{P} / 2$$

\begin{aligned} &N_{\theta}=-\frac{P}{2} \cos \theta \\ &\frac{\delta N_{\theta}}{\delta(P / 2)}=\frac{\delta}{\delta(P / 2)}\left(-\frac{P}{2} \cos \theta\right) \\ &\frac{\delta N_{\theta}}{\delta(P / 2)}=-\cos \theta \end{aligned}

Write the shear force relation.

$$V_{\theta}=-\frac{P}{2} \sin \theta$$

Here, $$V_{\theta}$$ is shear force at angle $$\theta$$.

Differentiate with respect to $$P / 2$$.

\begin{aligned} &\frac{\delta}{\delta(P / 2)}\left(V_{\theta}\right)=\frac{\delta}{\delta(P / 2)}\left(-\frac{P}{2} \sin \theta\right) \\ &\frac{\delta V_{\theta}}{\delta(P / 2)}=\sin \theta \end{aligned}

Calculate the moment at any angle $$\theta$$.

\begin{aligned} M_{\theta} &=0.182 r P-\frac{1}{2} P \bar{r}(1-\cos \theta) \\ &=0.182 \bar{r} P-0.5 P r+0.5 P r \cos \theta \\ &=-0.318 P r+0.5 P r \cos \theta \end{aligned}

Here, $$M_{\theta}$$ is the moment at angle $$\theta$$.

Differentiate with respect $$P / 2$$.

\begin{aligned} &\frac{\partial}{\partial(P / 2)}\left(M_{\theta}\right)=\frac{\partial}{\partial(P / 2)}(-0.318 P \bar{r}+0.5 P r \cos \theta) \\ &\frac{\partial M_{\theta}}{\partial(P / 2)}=\frac{\partial}{\partial(P / 2)}\left(-0.636\left(\frac{P}{2}\right)+\frac{P}{2} \cos \theta\right) \\ &\frac{\partial M_{\theta}}{\partial(P / 2)}=-0.636 \bar{r}+\bar{r} \cos \theta \end{aligned}

Calculate the moment of inertia.

$$I=\frac{1}{12} t h^{3}$$

Here, $$t$$ is thickness and $$h$$ is width of section.

Substitute $$50 \times 10^{-3} \mathrm{~m}$$ for $$t$$ and $$100 \times 10^{-3} \mathrm{~m}$$ for $$h$$

\begin{aligned} I &=\frac{1}{12}\left(50 \times 10^{-3}\right)\left(100 \times 10^{-3}\right)^{3} \\ &=4.17 \times 10^{-6} \mathrm{~m}^{4} \end{aligned}

$$r_{i}=\bar{r}-\frac{1}{2} h$$

Here, $$\bar{r}$$ is centroidal distance.

Substitute $$150 \mathrm{~mm}$$ for $$\bar{r}$$ and $$100 \mathrm{~mm}$$ for $$h$$.

\begin{aligned} &r_{i}=150-\frac{1}{2}(100) \\ &r_{i}=100 \mathrm{~mm} \end{aligned}

Write the relation between the young's modulus and rigidity modulus.

\begin{aligned} &E=\frac{5}{2} G \\ &G=0.4 E \end{aligned}

Here, $$G$$ is rigidity modulus and $$E$$ is young's modulus.

Calculate the value of constant $$\alpha$$.

\begin{aligned} \alpha &=\frac{6}{5} \\ &=1.2 \end{aligned}

Here, $$\alpha$$ is the constant value.

$$r_{s}=\bar{r}+\frac{1}{2} h$$

Here, $$\overline{\boldsymbol{r}}$$ is the centroidal distance.

Substitute $$150 \mathrm{~mm}$$ for $$\bar{r}$$ and $$100 \mathrm{~mm}$$ for $$h$$.

\begin{aligned} r_{o} &=150+\frac{1}{2}(100) \\ &=150+50 \\ &=200 \mathrm{~mm} \end{aligned}

Calculate the deflection along the line of section of load $$P$$.

$$\delta_{p}=2 \int_{0}^{\pi / 2}\left[\frac{N_{\theta}}{A E} \frac{\delta N_{\theta}}{\delta(P / 2)}+\frac{M_{\theta}}{E I} \frac{\delta M_{\theta}}{\delta(P / 2)}+\frac{\alpha V_{\theta}}{A G} \frac{\delta V_{\theta}}{\delta(P / 2)}\right] \bar{r} d \theta$$

Here, $$\delta_{p}$$ is the deflection along the line of action of load $$P$$ and $$E$$ is young's modulus, $$\alpha$$ is the constant and $$G$$ is modulus of rigidity.

Substitute $$-0.318 P \bar{r}+0.5 P \bar{r} \cos \theta$$ for $$M_{\theta},-(P / 2) \sin \theta$$ for $$V_{\theta}=-(P / 2) \cos \theta$$ for $$N_{e}$$. $$-\cos \theta$$ for $$\partial N_{\theta} / \partial(P / 2),-0.636 \bar{r}+\bar{r} \cos \theta$$ for $$\partial M_{\theta} / \partial(P / 2)$$ and $$-\sin \theta$$ for $$\partial V_{\theta} / \partial(P / 2)$$

$$\left.\begin{array}{rl} \delta_{P} & =2 \int_{0}^{\pi / 2}\left[\frac{N_{\theta}}{A E} \frac{\delta N_{\theta}}{\delta(P / 2)}+\frac{M_{\theta}}{E I} \frac{\delta M_{\theta}}{\delta(P / 2)}+\frac{\alpha V_{\theta}}{A G} \frac{\delta V_{\theta}}{\delta(P / 2)}\right] \bar{r} d \theta \\ & =2 \int_{0}^{\pi / 2}\left[\frac{(-(P / 2) \cos \theta)}{A E}(-\cos \theta)+\frac{(0.318 P \bar{r}+0.5 P r) \cos \theta}{E I}(-0.636 \bar{r}+\bar{r} \cos \theta)\right] \vec{r} d \theta \\ +\frac{\alpha(-(P / 2) \sin \theta)}{A G}(-\sin \theta) & \end{array}\right] } \end{array}$$

$$=2 \int_{0}^{\pi / 2}\left[\frac{P}{A E} \cos ^{2} \theta+\frac{P r^{2}}{2 E I}(-0.636+\cos \theta)^{2}+\frac{\alpha}{A G} \frac{P}{2} \sin ^{2} \theta\right]-d \theta$$

$$=2\left[\left\{\frac{P \bar{r}}{2 A E}\left|\frac{\theta}{2}+\frac{1}{4} \sin 2 \theta\right|_{0}^{\frac{\pi}{2}}\right\}+\left\{\frac{P r^{3}}{2 E I}\left|0.904 \theta-1.27 \sin \theta+\frac{\sin 2 \theta}{4}\right|_{0}^{\frac{\pi}{2}}\right\}\right]$$

$$+\left\{\frac{\alpha P \bar{r}}{2 A G}\left|\frac{\theta}{2}-\frac{1}{4} \sin 2 \theta\right|_{0}^{\frac{\pi}{2}}\right\}$$

$$\left.=2\left[\begin{array}{l}\frac{P \bar{r}}{2 A E}\left\{\left(\frac{(\pi / 2)}{2}-\frac{0^{\circ}}{2}\right)+\frac{1}{4}\left[\sin 2\left(\frac{\pi}{2}\right)-\sin 2\left(0^{\circ}\right)\right]\right\} \\ +\left\{\frac{P r^{3}}{2 E I} \times\left\{0.904\left(\frac{\pi}{2}-0\right)-1.27\left[\sin \left(\frac{\pi}{2}\right)-\sin 0^{\circ}\right]\right.\right. \\ +\frac{1}{4}\left[\sin 2\left(\frac{\pi}{2}\right)-\sin 2(0)\right] \\ +\frac{\alpha P r}{2 A G}\left\{\frac{1}{2}\left(\frac{\pi}{2}-0\right)-\frac{1}{4}\left(\sin 2\left(\frac{\pi}{2}\right)\right)-\sin 2(0)\right\}\end{array}\right\}\right]$$

$$=2\left[\frac{\operatorname{Pr}}{2 A E}\left(\frac{\pi}{4}\right)+\frac{\operatorname{Pr}^{3}}{2 E I}(0.15)+\frac{\alpha}{A G} \frac{\operatorname{Pr}}{2}\left(\frac{\pi}{2}\right)\right]$$

By solving the preceding equation, the following relation is obtained.

By solving the preceding equation, the following relation is obtained. $$\delta_{p}=2\left[\begin{array}{l}\left.\frac{P \bar{r}}{2 A E}\left\{\frac{\pi}{4}+\frac{1}{4}[0-0]\right\}+\left\{\frac{P r^{3}}{2 E I} \times\left\{0.904\left(\frac{\pi}{2}\right)-1.27[1-0]+\frac{1}{4}[0-0]\right\}\right\}\right] \\ +\frac{\alpha P r}{2 A G}\left\{\frac{1}{2}\left(\frac{\pi}{2}\right)-\frac{1}{4}[0-0]\right\}\end{array}\right.$$ $$\delta_{p}=2\left[\frac{P \bar{r}}{2 A E}\left(\frac{\pi}{4}\right)+\frac{P \bar{r}^{3}}{2 E I}(0.15)+\frac{\alpha}{A G} \frac{P r}{2}\left(\frac{\pi}{4}\right)\right] \ldots \ldots$$ (3)

Substitute $$150 \times 10^{-3} \mathrm{~m}$$ for $$\bar{r}, 1.2$$ for $$\alpha, 0.005 \mathrm{~m}^{2}$$ for $$A, 0.4 E$$ for $$G$$ and $$4.17 \times 10^{-6} \mathrm{~m}^{4}$$ for $$I$$ in the equation (3).

\begin{aligned} \delta_{p} &=2\left[\frac{P \bar{r}}{2 A E}\left(\frac{\pi}{4}\right)+\frac{P \bar{r}^{3}}{2 E I}(0.15)+\frac{\alpha}{A G} \frac{P \bar{r}}{2}\left(\frac{\pi}{4}\right)\right] \\ &=2\left[\frac{P\left(150 \times 10^{-3}\right)}{2 \times(0.005) \times E}\left(\frac{\pi}{4}\right)+\frac{P\left(150 \times 10^{-3}\right)^{3}}{2 E\left(4.17 \times 10^{-6}\right)}(0.15)+\frac{(1.2)}{(0.005)(0.4 E)} \times \frac{P\left(150 \times 10^{-3}\right)}{2}\left(\frac{\pi}{4}\right)\right] \\ &=\frac{2 P}{E}[11.781+60.70+35.343] \\ &=215.706\left(\frac{P}{E}\right) \mathrm{m} \end{aligned}

Therefore, the deflection along the line of section of load $$P$$ is $$215.706\left(\frac{P}{E}\right) \mathrm{m}$$.