Chapter 5, Problem 5.7.4P

a.

To determine

The bending stress ${\sigma }_A$ .

Answer to Problem 5.7.4P

The bending stress${\sigma }_A$is 210 MPa.

Explanation of Solution

Given:

  $\begin{array}{l}P=12\text{ kN}\\ L=1.25\text{ m}\text{.}\\ b_A=60\text{ mm}\\ b_B=80\text{ mm}\\ h_B=\text{120 mm}\\ M_0=10\text{ kN}\cdot \text{m}\end{array}$

Calculate the maximum bending stress.

  $h\left(x\right)=h_A\left(1+\frac{x}{3L}\right)$

  $\begin{array}{l}I\left(x\right)=\frac{\pi b\left(x\right)h{\left(x\right)}^3}{64}\\ S\left(x\right)=\frac{I\left(x\right)}{\frac{h\left(x\right)}{2}}\end{array}$

  $\begin{array}{l}S\left(x\right)=\frac{\pi b\left(x\right)h{\left(x\right)}^2}{32}\\ S\left(x\right)=\frac{\left[\pi b_A{h}_A^2{\left(1+\frac{x}{3L}\right)}^3\right]}{32}\end{array}$

  $M\left(x\right)=Px+M_0$

  $\begin{array}{c}\sigma \left(x\right)=\frac{M\left(x\right)}{S\left(x\right)}\\ =\frac{Px+M_0}{\frac{\left[\pi b_A{h}_A^2{\left(1+\frac{x}{3L}\right)}^3\right]}{32}}\\ =864\left(\frac{Px+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+x\right)}^3}\end{array}$

For maximum condition, we take:

  $\begin{array}{l}\frac{d}{dx}\sigma \left(x\right)=0\\ \frac{d}{dx}\left[864\left(\frac{Px+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+x\right)}^3}\right]=0\\ \left[864\frac{P}{\pi b_A{h}_A^2}\frac{L^3}{{\left(3L+x\right)}^3}-2592\left(\frac{Px+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+x\right)}^4}\right]=0\\ \left[-864L^3\frac{-3PL+2Px+3M_0}{\pi b_A{h}_A^2{\left(3L+x\right)}^4}\right]=0\\ x_{\max }=\frac{3\left(PL-M_0\right)}{2P}\\ x_{max}=0.625\text{ in}\end{array}$

Now, plotting the graph for $\sigma \left(x\right)$ and $M\left(x\right)$ , we get:

  

Now, the maximum bending stress is calculated.

  $\begin{array}{l}{\sigma }_{\max }=864\left(\frac{P{x}_{\max }+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+x_{\max }\right)}^3}\\ {\sigma }_{\max }=231\text{ MPa}\end{array}$

The largest bending stress is calculated as:

  $\begin{array}{l}{\sigma }_A=\sigma \left(0\right)\\ {\sigma }_A=864\left(\frac{P\left(0\right)+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+\left(0\right)\right)}^3}\\ {\sigma }_A=210\text{ MPa}\end{array}$

Conclusion:

Therefore, the bending stress ${\sigma }_A$ is 210 MPa.

b.

To determine

The bending stress ${\sigma }_B$ .

Answer to Problem 5.7.4P

The bending stress${\sigma }_B$is 221 MPa.

Explanation of Solution

The largest bending stress is calculated as:

  $\begin{array}{l}{\sigma }_B=\sigma \left(L\right)\\ {\sigma }_B=864\left(\frac{P\left(0\right)+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+\left(0\right)\right)}^3}\\ {\sigma }_B=221\text{ MPa}\end{array}$

Conclusion:

Therefore, the bending stress ${\sigma }_B$ is 221 MPa.

c.

To determine

The distance of maximum bending stress $x_{\max }$ .

Answer to Problem 5.7.4P

The distance of maximum bending stress$x_{\max }$is 0.625 in.

Explanation of Solution

The distance of maximum bending stress $x_{\max }$ .

  $\begin{array}{l}{x}_{\max }=\frac{3\left(PL-M_0\right)}{2P}\\ x_{\max }=0.625\text{ in}\end{array}$

Conclusion:

Therefore, the distance of maximum bending stress $x_{\max }$ is 0.625 in.

d.

To determine

The magnitude of maximum bending stress ${\sigma }_{\max }$ .

Answer to Problem 5.7.4P

The magnitude of maximum bending stress${\sigma }_{\max }$is 231 MPa.

Explanation of Solution

The magnitude of maximum bending stress ${\sigma }_{\max }$ .

  $\begin{array}{l}{\sigma }_{\max }=864\left(\frac{P{x}_{\max }+M_0}{\pi b_A{h}_A^2}\right)\frac{L^3}{{\left(3L+x_{\max }\right)}^3}\\ {\sigma }_{\max }=231\text{ MPa}\end{array}$

Conclusion:

Therefore, the magnitude of maximum bending stress ${\sigma }_{\max }$ is 231 MPa.

e.

To determine

The magnitude of maximum bending stress.

Answer to Problem 5.7.4P

The magnitude of maximum bending stress is 214 MPa.

Explanation of Solution

Given:

  $\begin{array}{l}P=12\text{ kN}\\ L=1.25\text{ m}\text{.}\\ b_A=60\text{ mm}\\ b_B=80\text{ mm}\\ h_A=90\text{ mm}\\ h_B=\text{120 mm}\\ M_0=10\text{ kN}\cdot \text{m}\end{array}$

Calculate the maximum bending stress.

  $h\left(x\right)=h_A\left(1+\frac{x}{3L}\right)$

  $b\left(x\right)=b_A\left(1+\frac{x}{3L}\right)$

  $\begin{array}{l}I\left(x\right)=\frac{\pi b\left(x\right)h{\left(x\right)}^3}{64}\\ S\left(x\right)=\frac{I\left(x\right)}{\frac{h\left(x\right)}{2}}\end{array}$

  $\begin{array}{l}S\left(x\right)=\frac{\pi b\left(x\right)h{\left(x\right)}^2}{32}\\ S\left(x\right)=\frac{\left[\pi b_A{h}_A^2{\left(1+\frac{x}{3L}\right)}^3\right]}{32}\end{array}$

  $M\left(x\right)=Px+M_0-\frac{10P{x}^2}{6L}$

  $\begin{array}{c}\sigma \left(x\right)=\frac{M\left(x\right)}{S\left(x\right)}\\ =\frac{Px+M_0-\frac{10P{x}^2}{6L}}{\frac{\left[\pi b_A{h}_A^2{\left(1+\frac{x}{3L}\right)}^3\right]}{32}}\\ =-288\left(-3PxL-3M_{0}L+5P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}\end{array}$

For maximum condition, we take:

  $\begin{array}{l}\frac{d}{dx}\sigma \left(x\right)=0\\ \frac{d}{dx}\left[-288\left(-3PxL-3M_{0}L+5P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}\right]=0\\ \left[\left(864PL-2880Px\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}-3\left(864PxL+864M_{0}L-1440P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^4}\right]=0\\ \left[288L^2\frac{9P{L}^2-36PxL+5P{x}^2-9M_{0}L}{\pi b_A{h}_A^2{\left(3L+x\right)}^4}\right]=0\\ 9P{L}^2-36PxL+5P{x}^2-9M_{0}L=0\\ x_{max}=0.105\text{ m}\end{array}$

Now, plotting the graph for $\sigma \left(x\right)$ and $M\left(x\right)$ , we get:

  

Now, the maximum bending stress is calculated.

  $\begin{array}{l}{\sigma }_{\max }=-288\left(-3PxL-3M_{0}L+5P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}\\ {\sigma }_{\max }=214\text{ MPa}\end{array}$

The largest bending stress is calculated as:

  $\begin{array}{l}{\sigma }_A=\sigma \left(0\right)\\ {\sigma }_A=-288\left(-3PxL-3M_{0}L+5P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}\\ {\sigma }_A=210\text{ MPa}\end{array}$

The largest bending stress is calculated as:

  $\begin{array}{l}{\sigma }_B=\sigma \left(L\right)\\ {\sigma }_B=-288\left(-3PxL-3M_{0}L+5P{x}^2\right)\frac{L^2}{\pi b_A{h}_A^2{\left(3L+x\right)}^3}\\ {\sigma }_B=0\text{ MPa}\end{array}$

Conclusion:

Therefore, the magnitude of maximum bending stress is 214 MPa.