Chapter 4.9, Problem 149P

(a)

To determine

Find the value of $\theta$ for which the stress at D reaches its largest value.

Answer to Problem 149P

The value of $\theta$ for which the stress at D reaches its largest value is $\underset{\_}{53.1°}$.

Explanation of Solution

Given information:

The radius of the circular plate is $R=125\text{mm}$.

The width of the rectangular post is $b=200\text{mm}$ and the height of the rectangular post is $d=150\text{mm}$.

The applied force is $P=4\text{kN}$.

Calculation:

Sketch the cross section of disk as shown in Figure 1.

Refer to Figure 1.

Calculate the moment along $x$ direction $\left(M_x\right)$ as shown below.

$M_x=-PR\sin \theta$

Substitute $4\text{kN}$ for P and $125\text{mm}$ for R.

$\begin{array}{c}{M}_x=-4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 125\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \sin \theta \\ =-500\sin \theta \end{array}$

Calculate the moment along $z$ direction $\left(M_z\right)$ as shown below.

$M_z=-PR\cos \theta$

Substitute $4\text{kN}$ for P and $125\text{mm}$ for R.

$\begin{array}{c}{M}_z=-4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 125\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \cos \theta \\ =-500\cos \theta \end{array}$

Sketch the cross section of the rectangular post as shown in Figure 2.

Refer to Figure 2.

Calculate the area $\left(A\right)$ of the rectangular cross section as shown below.

$A=bd$

Substitute $200\text{mm}$ for b and $150\text{mm}$ for d.

$\begin{array}{c}A=200\times 150\\ =30\times {10}^3{\text{mm}}^2\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^2\\ =30\times {10}^{-3}{\text{m}}^2\end{array}$

Calculate the moment of inertia along $x$ direction $\left(I_x\right)$ as shown below.

$I_x=\frac{b{d}^3}{12}$

Substitute $200\text{mm}$ for b and $150\text{mm}$ for d.

$\begin{array}{c}{I}_x=\frac{1}{12}\times 200\times {150}^3\\ =56.25\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =56.25\times {10}^{-6}{\text{m}}^4\end{array}$

Calculate the moment of inertia along $z$ direction $\left(I_z\right)$ as shown below.

$I_z=\frac{d{b}^3}{12}$

Substitute $200\text{mm}$ for b and $150\text{mm}$ for d.

$\begin{array}{c}{I}_z=\frac{1}{12}\times 150\times {200}^3\\ =100\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =100\times {10}^{-6}{\text{m}}^4\end{array}$

The location of point D along $x$ direction is $x_D=100\text{mm}$.

The location of point D along $z$ direction is $z_D=-75\text{mm}$.

Calculate the stress $\left(\sigma \right)$ as shown below.

$\sigma =-\frac{P}{A}-\frac{M_{x}z}{I_x}+\frac{M_{z}x}{I_z}$ (1)

Substitute $-PR\sin \theta$ for $M_x$ and $-PR\cos \theta$ for $M_z$.

$\begin{array}{c}\sigma =-\frac{P}{A}+\frac{PR\sin \theta \times z}{I_x}-\frac{PR\cos \theta \times x}{I_z}\\ =-P\left(\frac{1}{A}+\frac{Rz\sin \theta }{I_x}-\frac{Rx\cos \theta }{I_z}\right)\end{array}$

The stress $\sigma$ to be maximum, when $\frac{d\sigma }{d\theta }=0$ with $z=z_D$ and $x=x_D$.

Calculate the stress at point D $\left({\sigma }_D\right)$ as shown below.

Substitute $z_D$ for z and $x_D$ for x.

${\sigma }_D=-P\left(\frac{1}{A}+\frac{R{z}_D\sin \theta }{I_x}-\frac{R{x}_D\cos \theta }{I_z}\right)$

Calculate the value of $\theta$ as shown below.

Differentiate both sides of the Equation with respect to $\theta$.

$\begin{array}{c}\frac{d{\sigma }_D}{d\theta }=-P\left(0+\frac{R{z}_D}{I_x}\left(\cos \theta \right)-\frac{R{x}_D}{I_z}\left(-\sin \theta \right)\right)\\ =-P\left(\frac{R{z}_D\cos \theta }{I_x}+\frac{R{x}_D\sin \theta }{I_z}\right)\end{array}$

Substitute $0$ for $\frac{d{\sigma }_D}{d\theta }$.

$\begin{array}{l}-P\left(\frac{R{z}_D\cos \theta }{I_x}+\frac{R{x}_D\sin \theta }{I_z}\right)=0\\ \frac{R{z}_D\cos \theta }{I_x}=-\frac{R{x}_D\sin \theta }{I_z}\\ \frac{\sin \theta }{\cos \theta }=-\frac{I_z{z}_D}{I_x{x}_D}\\ \tan \theta =-\frac{I_z{z}_D}{I_x{x}_D}\end{array}$

Substitute $100\times {10}^6{\text{mm}}^4$ for $I_z$, $56.25\times {10}^6{\text{mm}}^4$ for $I_x$, $-75\text{mm}$ for $z_D$, and $100\text{mm}$ for $x_D$.

$\begin{array}{l}\tan \theta =-\frac{100\times {10}^6\times \left(-75\right)}{56.25\times {10}^6\times 100}\\ =1.3333\\ \theta ={\tan }^{-1}\left(1.3333\right)\\ =53.1°\end{array}$

Therefore, the value of $\theta$ for which the stress at D reaches its largest value is $\underset{\_}{53.1°}$.

(b)

To determine

Find the values of stress at A, B, C, and D.

Answer to Problem 149P

The stress at A is $\underset{\_}{{\sigma }_A=700\text{kPa}}$.

The stress at B is $\underset{\_}{{\sigma }_B=100\text{kPa}}$.

The stress at C is $\underset{\_}{{\sigma }_C=-133.3\text{kPa}}$.

The stress at D is $\underset{\_}{{\sigma }_D=-967\text{kPa}}$.

Explanation of Solution

Given information:

The radius of the circular plate is $R=125\text{mm}$.

The width of the rectangular post is $b=200\text{mm}$ and the height of the rectangular post is $d=150\text{mm}$.

The applied force is $P=4\text{kN}$.

Calculation:

Refer to part (a).

The value of $\theta$ for which the stress at D reaches its largest value is $53.1°$.

The bending moment along x direction is $M_x=-500\sin \theta$.

Substitute $53.1°$ for $\theta$.

$\begin{array}{c}{M}_x=-500\sin 53.1°\\ =-400\text{N}\cdot \text{m}\end{array}$

The bending moment along z direction is $M_z=-500\cos \theta$.

Substitute $53.1°$ for $\theta$.

$\begin{array}{c}{M}_z=-500\cos 53.1°\\ =-300\text{N}\cdot \text{m}\end{array}$

Calculate the stress at A $\left({\sigma }_A\right)$ as shown below.

Refer to Figure 2 in part (a).

The location of point A along $x$ direction is $x_A=-100\text{mm}$.

The location of point A along $z$ direction is $z_A=75\text{mm}$.

Substitute $4\text{kN}$ for $P$, $30\times {10}^3{\text{mm}}^2$ for A, $56.25\times {10}^6{\text{mm}}^4$ for $I_x$, $100\times {10}^6{\text{mm}}^4$ for $I_z$, $-100\text{mm}$ for $x_A$, $75\text{mm}$ for $z_A$, $-400\text{N}\cdot \text{m}$ for $M_x$, and $-300\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_A=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-400\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 75\text{mm}}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-300\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\left(-100\text{mm}\right)}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0.5333+0.3\\ =0.7\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =700\text{kPa}\end{array}$

Hence, the stress at A is $\underset{\_}{{\sigma }_A=700\text{kPa}}$.

Calculate the stress at B $\left({\sigma }_B\right)$ as shown below.

Refer to Figure 2 in part (a).

The location of point B along $x$ direction is $x_B=100\text{mm}$.

The location of point B along $z$ direction is $z_B=75\text{mm}$.

Substitute $4\text{kN}$ for $P$, $30\times {10}^3{\text{mm}}^2$ for A, $56.25\times {10}^6{\text{mm}}^4$ for $I_x$, $100\times {10}^6{\text{mm}}^4$ for $I_z$, $100\text{mm}$ for $x_B$, $75\text{mm}$ for $z_B$, $-400\text{N}\cdot \text{m}$ for $M_x$, and $-300\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_B=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-400\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 75\text{mm}}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-300\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 100\text{mm}}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0.5333-0.3\\ =0.1\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =100\text{kPa}\end{array}$

Hence, the stress at B is $\underset{\_}{{\sigma }_B=100\text{kPa}}$.

Calculate the stress at C $\left({\sigma }_C\right)$ as shown below.

Refer to Figure 2 in part (a).

The location of point C along $x$ direction is $x_C=0$.

The location of point C along $z$ direction is $z_C=0$.

Substitute $4\text{kN}$ for $P$, $30\times {10}^3{\text{mm}}^2$ for A, $56.25\times {10}^6{\text{mm}}^4$ for $I_x$, $100\times {10}^6{\text{mm}}^4$ for $I_z$, $0$ for $x_C$, $0$ for $z_C$, $-400\text{N}\cdot \text{m}$ for $M_x$, and $-300\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_C=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-400\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 0}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-300\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 0}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0+\\ =-0.1333\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =-133.3\text{kPa}\end{array}$

Hence, the stress at C is $\underset{\_}{{\sigma }_C=-133.3\text{kPa}}$.

Calculate the stress at D $\left({\sigma }_D\right)$ as shown below.

Refer to Figure 2 in part (a).

The location of point D along $x$ direction is $x_D=100\text{mm}$.

The location of point D along $z$ direction is $z_D=-75\text{mm}$.

Substitute $4\text{kN}$ for $P$, $30\times {10}^3{\text{mm}}^2$ for A, $56.25\times {10}^6{\text{mm}}^4$ for $I_x$, $100\times {10}^6{\text{mm}}^4$ for $I_z$, $100\text{mm}$ for $x_D$, $-75\text{mm}$ for $z_D$, $-400\text{N}\cdot \text{m}$ for $M_x$, and $-300\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_D=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-400\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times \left(-75\text{mm}\right)}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-300\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 100\text{mm}}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333-0.5333-0.3\\ =-0.9666\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =-967\text{kPa}\end{array}$

Therefore, the stress at D is $\underset{\_}{{\sigma }_D=-967\text{kPa}}$.