Chapter 4.9, Problem 129P

4.127 through 4.134 The couple M is applied to a beam of the cross section shown in a plane forming an angle β with the vertical. Determine the stress at (a) point A, (b) point B, (c) point D.

Flg. P4.129

To determine

Find the stress at point A.

Answer to Problem 129P

The stress at point A is $\underset{\_}{{\sigma }_A=-29.3\text{MPa}}$.

Explanation of Solution

Given information:

The couple acts in a vertical plane $M=25\text{kN}\cdot \text{m}$.

The angle is $\theta =15°$.

Calculation:

Sketch the beam cross section as shown in Figure 1.

Refer to Figure 1.

Calculate the moment along $y$ direction $\left(M_y\right)$ as shown below.

$M_y=M\sin \theta$

Substitute $25\text{kN}\cdot \text{m}$ for M and $15°$ for $\theta$.

$\begin{array}{c}{M}_y=25\sin 15°\\ =6.4705\text{kN}\cdot \text{m}\end{array}$

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

$M_z=M\cos \theta$

Substitute $25\text{kN}\cdot \text{m}$ for M and $15°$ for $\theta$.

$\begin{array}{c}{M}_z=25\cos 15°\\ =24.148\text{kN}\cdot \text{m}\end{array}$

Calculate the moment of inertia along y axis $\left(I_y\right)$ as shown below.

$\begin{array}{c}{I}_y=\frac{1}{12}\times 80\times {90}^3+\frac{1}{12}\times 80\times {30}^3\\ =5.04\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =5.04\times {10}^{-6}{\text{m}}^4\end{array}$

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

$\begin{array}{c}{I}_z=\frac{1}{3}\times 90\times {60}^3+\frac{1}{3}\times 60\times {20}^3+\frac{1}{3}\times 30\times {100}^3\\ =16.64\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4\\ =16.64\times {10}^{-6}{\text{m}}^4\end{array}$

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

$\sigma =\frac{M_{y}z}{I_y}-\frac{M_{z}y}{I_z}$ (1)

The location of point A along z axis $z=45\text{mm}$.

The location of point A along y axis $y=60\text{mm}$.

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

Substitute $6.4705\text{kN}\cdot \text{m}$ for $M_y$, $24.148\text{kN}\cdot \text{m}$ for $M_z$, $5.04\times {10}^{-6}{\text{m}}^4$ for $I_y$, $16.64\times {10}^{-6}{\text{m}}^4$ for $I_z$, $45\text{mm}$ for z, and $60\text{mm}$ for y in Equation (1).

$\begin{array}{c}{\sigma }_A=\left(\begin{array}{l}\frac{6.4705\text{kN}\cdot \text{m}\times \left(45\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{5.04\times {10}^{-6}{\text{m}}^4}\\ -\frac{24.148\text{kN}\cdot \text{m}\times \left(60\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{16.64\times {10}^{-6}{\text{m}}^4}\end{array}\right)\\ =57.772\times {10}^3-87.072\times {10}^3\\ =-29.3\times {10}^3\text{kN}/{\text{mm}}^2\times \frac{1\text{MPa}}{1,000\text{kN}/{\text{mm}}^2}\\ =-29.3\text{MPa}\end{array}$

Therefore, the stress at point A is $\underset{\_}{{\sigma }_A=-29.3\text{MPa}}$.

To determine

The stress at point B.

Answer to Problem 129P

The stress at point B is $\underset{\_}{{\sigma }_B=-144.8\text{MPa}}$.

Explanation of Solution

Given information:

The couple acts in a vertical plane $M=25\text{kN}\cdot \text{m}$.

The angle is $\theta =15°$.

Calculation:

Refer to part (a).

The moment along y axis $M_y=6.4705\text{kN}\cdot \text{m}$.

The moment along z axis $M_z=24.148\text{kN}\cdot \text{m}$.

The moment of inertia along y axis $I_y=5.04\times {10}^{-6}{\text{m}}^4$.

The moment of inertia along z axis $I_z=16.64\times {10}^{-6}{\text{m}}^4$.

Refer to Figure 1 in part (a).

The location of point B along z axis $z=-45\text{mm}$.

The location of point B along y axis $y=60\text{mm}$.

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

Substitute $6.4705\text{kN}\cdot \text{m}$ for $M_y$, $24.148\text{kN}\cdot \text{m}$ for $M_z$, $5.04\times {10}^{-6}{\text{m}}^4$ for $I_y$, $16.64\times {10}^{-6}{\text{m}}^4$ for $I_z$, $-45\text{mm}$ for z, and $60\text{mm}$ for y in Equation (1).

$\begin{array}{c}{\sigma }_B=\left(\begin{array}{l}\frac{6.4705\text{kN}\cdot \text{m}\times \left(-45\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{5.04\times {10}^{-6}{\text{m}}^4}\\ -\frac{24.148\text{kN}\cdot \text{m}\times \left(60\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{16.64\times {10}^{-6}{\text{m}}^4}\end{array}\right)\\ =-57.772\times {10}^3-87.072\times {10}^3\\ =-144.844\times {10}^3\text{kN}/{\text{mm}}^2\times \frac{1\text{MPa}}{1,000\text{kN}/{\text{mm}}^2}\\ =-144.8\text{MPa}\end{array}$

Therefore, the stress at point B is $\underset{\_}{{\sigma }_B=-144.8\text{MPa}}$.

To determine

The stress at point D.

Answer to Problem 129P

The stress at point D is $\underset{\_}{{\sigma }_D=125.9\text{MPa}}$.

Explanation of Solution

Given information:

The couple acts in a vertical plane $M=25\text{kN}\cdot \text{m}$.

The angle is $\theta =15°$.

Calculation:

Refer to part (a).

The moment along y axis is $M_y=6.4705\text{kN}\cdot \text{m}$.

The moment along z axis is $M_z=24.148\text{kN}\cdot \text{m}$.

The moment of inertia along y axis is $I_y=5.04\times {10}^{-6}{\text{m}}^4$.

The moment of inertia along z axis is $I_z=16.64\times {10}^{-6}{\text{m}}^4$.

Refer to Figure 1 in part (a).

The location of point D along z axis $z=-15\text{mm}$.

The location of point D along y axis $y=-100\text{mm}$.

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

Substitute $6.4705\text{kN}\cdot \text{m}$ for $M_y$, $24.148\text{kN}\cdot \text{m}$ for $M_z$, $5.04\times {10}^{-6}{\text{m}}^4$ for $I_y$, $16.64\times {10}^{-6}{\text{m}}^4$ for $I_z$, $-15\text{mm}$ for z, and $-100\text{mm}$ for y in Equation (1).

$\begin{array}{c}{\sigma }_D=\left(\begin{array}{l}\frac{6.4705\text{kN}\cdot \text{m}\times \left(-15\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{5.04\times {10}^{-6}{\text{m}}^4}\\ -\frac{24.148\text{kN}\cdot \text{m}\times \left(-100\text{mm}\times \frac{1\text{m}}{1,000\text{mm}}\right)}{16.64\times {10}^{-6}{\text{m}}^4}\end{array}\right)\\ =-19.257\times {10}^3+145.12\times {10}^3\\ =125.863\times {10}^3\text{kN}/{\text{mm}}^2\times \frac{1\text{MPa}}{1,000\text{kN}/{\text{mm}}^2}\\ =125.9\text{MPa}\end{array}$

Therefore, the stress at point D is $\underset{\_}{{\sigma }_D=125.9\text{MPa}}$.