Chapter 4.9, Problem 148P

A rigid circular plate of 125-mm radius is attached to a solid 150 × 200-mm rectangular post, with the center of the plate directly above the center of the post. If a 4-kN force P is applied at E with θ = 30°, determine (a) the stress at point A, (b) the stress at point B, (c) the point where the neutral axis intersects line ABD.

Fig. P4.148

To determine

Find the value of stress at point A.

Answer to Problem 148P

The stress at point A is $\underset{\_}{{\sigma }_A=633\text{kPa}}$.

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:

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

Refer to Figure 1.

Consider the value of angle $\theta$ is $30°$.

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

$M_x=-PR\sin \theta$

Substitute $4\text{kN}$ for P, $125\text{mm}$ for R, and $30°$ for $\theta$.

$\begin{array}{c}{M}_x=-4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 125\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \sin 30°\\ =-250\text{N}\cdot \text{m}\end{array}$

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

$M_z=-PR\cos \theta$

Substitute $4\text{kN}$ for P, $125\text{mm}$ for R, and $30°$ for $\theta$.

$\begin{array}{c}{M}_z=-4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}\times 125\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\times \cos 30°\\ =-433\text{N}\cdot \text{m}\end{array}$

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.

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

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)

Refer to Figure 2.

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

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

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

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 $z_A$, $-250\text{N}\cdot \text{m}$ for $M_x$, and $-433\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_A=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-250\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 75\text{mm}}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-433\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\left(-100\text{mm}\right)}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0.333+0.433\\ =0.633\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =633\text{kPa}\end{array}$

Hence, the stress at A is $\underset{\_}{{\sigma }_A=633\text{kPa}}$.

To determine

The values of stress at B.

Answer to Problem 148P

The stress at B is $\underset{\_}{{\sigma }_B=-233\text{kPa}}$.

Explanation of Solution

Given information:

Calculation:

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 $z_B$, $-250\text{N}\cdot \text{m}$ for $M_x$, and $-433\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}{\sigma }_B=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-250\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 75\text{mm}}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-433\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 100\text{mm}}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0.333-0.433\\ =-0.233\text{N}/{\text{mm}}^{\text{2}}\times \frac{1,000\text{kPa}}{1\text{N}/{\text{mm}}^{\text{2}}}\\ =-233\text{kPa}\end{array}$

Hence, the stress at B is $\underset{\_}{{\sigma }_B=-233\text{kPa}}$.

To determine

The point where the neutral axis intersect ABD.

Answer to Problem 148P

The point where the neutral axis intersect ABD is $\underset{\_}{\text{146}\text{.2}\text{mm from point }A}$.

Explanation of Solution

Given information:

Calculation:

The stress at G $\left({\sigma }_G\right)$ is zero.

Refer to Figure 2 in part (a).

The location of point G along $x$ direction is $x_g$.

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

Substitute $0$ for ${\sigma }_G$, $4\text{kN}$ 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 $I_z$, $75\text{mm}$ for $z_G$, $-250\text{N}\cdot \text{m}$ for $M_x$, and $-433\text{N}\cdot \text{m}$ for $M_z$ in Equation (1).

$\begin{array}{c}0=\left(\begin{array}{l}-\frac{4\text{kN}\times \frac{1,000\text{N}}{1\text{kN}}}{30\times {10}^3{\text{mm}}^2}-\frac{\left(-250\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times 75\text{mm}}{56.25\times {10}^6{\text{mm}}^4}\\ +\frac{\left(-433\text{N}\cdot \text{m}\times \frac{1,000\text{mm}}{1\text{m}}\right)\times x_g}{100\times {10}^6{\text{mm}}^2}\end{array}\right)\\ =-0.1333+0.333-0.00433x_g\\ x_g=\frac{0.1997}{0.00433}\\ =46.2\text{mm}\end{array}$

Show the distance (d) between the point A and the point G as follows:

$\begin{array}{c}d=x_g+100\text{mm}\\ \text{=46}\text{.2+100}\text{mm}\\ \text{=146}\text{.2}\text{mm}\end{array}$

Thus, The point where the neutral axis intersect ABD is $\underset{\_}{\text{146}\text{.2}\text{mm from point }A}$.