Answered: 4 in. 30° 50 lb-ft B 6 in.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Determine the maximum bending stress in the beam’s cross section.

Expert Solution

Step 1

The expression for the moment of inertia of the section about x-axis,

$I_{z} = \frac{b d^{3}}{12}$

Here, $b$ is the breadth and $d$ is the depth.

Substitute the values in the above equation,

$\begin{array}{rcl} I_{z} & = & \frac{4 in \times {(6 in)}^{3}}{12} \\ = & 72 i n^{4} \end{array}$

Take moment about y-axis,

$\begin{array}{rcl} M_{y} & = & 50 \sin 30 ° \\ = & 25 lb \cdot ft \times \frac{12 in}{1 ft} \\ = & 300 lb \cdot in \end{array}$

Take moment about z-axis,

$\begin{array}{rcl} M_{z} & = & 50 \cos 30 ° \\ = & 43.3 lb \cdot ft \times \frac{12 in}{1 ft} \\ = & 519.6 lb \cdot in \end{array}$

Step 2

The expression to calculate the stress at point A,

$σ_{A} = - \frac{M_{z} y_{A}}{I_{z}} + \frac{M_{y} z_{A}}{I_{y}}$

Here, $y_{A and} z_{A}$ are the distances of point A from the center along y and z axis, respectively.

Substitute the values in the above equation,

$\begin{array}{rcl} σ_{A} & = & - \frac{519.6 lb \cdot in \times (\frac{6}{2}) in}{72 {in}^{4}} + \frac{(300 lb \cdot in) (\frac{4}{2}) in}{32 {in}^{4}} \\ = & (- 21.65 - 18.75) lb / {in}^{2} \\ σ_{A} & = & - 40.4 psi \end{array}$

The expression to calculate the stress at point B

$σ_{B} = - \frac{M_{z} y_{B}}{I_{z}} + \frac{M_{y} z_{B}}{I_{y}}$

Here, $y_{B and} z_{B}$ are the distances of point A from the center along y and z axis, respectively.

Substitute the values in the above equation,

$\begin{array}{rcl} σ_{B} & = & - \frac{519.6 lb \cdot in \times (\frac{6}{2}) in}{72 {in}^{4}} + \frac{(300 lb \cdot in) (- \frac{4}{2}) in}{32 {in}^{4}} \\ = & (- 21.65 + 18.75) lb / {in}^{2} \\ σ_{B} & = & - 2.9 psi \end{array}$

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