A The beam shown is supported by a roller at A and a pin at B. The support reactions for the beam are Ay - 1000 lb and By = 2500 lb ↑. On your paper: A. Write the equations of the shear (V) and moment (M) as a function of x for the beam. Include complete, clearly labeled free-body diagrams to support your work. B. Sketch the original beam on your paper. C. Graph the shear and moment diagrams for the entire beam under the sketch of the original beam. Be sure to label all key points on your diagrams. Enter the maximum magnitude moment from your diagram in the blank below. 0000 16 in.- 25 lb/in. B 24 in. 500 lb

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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### Beam Analysis Problem for Educational Purposes

#### Problem Statement:
The beam shown is supported by a roller at point A and a pin at point B. The support reactions for the beam are \( A_y = 1000 \text{ lb} \downarrow \) and \( B_y = 2500 \text{ lb} \uparrow \).

#### Tasks:
- **A.** Write the equations for the shear force (\( V \)) and bending moment (\( M \)) as functions of \( x \) for the beam. Include complete, clearly labeled free-body diagrams to support your work.
- **B.** Sketch the original beam on your paper.
- **C.** Graph the shear force and bending moment diagrams for the entire beam under the sketch of the original beam. Be sure to label all key points on your diagrams.

#### Beam Configuration:
- The entire length of the beam from A to C is subjected to a uniformly distributed load of \( 25 \text{ lb/in} \).
- There is also a concentrated load of \( 500 \text{ lb} \) acting downward at point C.
- The distance between points A and B is \( 16 \text{ in} \).
- The distance between points B and C is \( 24 \text{ in} \).

#### Diagram Description:
1. **Support at Point A:**
   - Represented as a roller support.
   - Located at the left end of the beam.
2. **Support at Point B:**
   - Represented as a pin support.
   - Located \( 16 \text{ in} \) to the right of point A.
3. **Distributed Load:**
   - Spans the entire length of the beam from A to C.
   - Distributed load of \( 25 \text{ lb/in} \).
4. **Concentrated Load:**
   - Applied at point C.
   - Magnitude: \( 500 \text{ lb} \) downward.

#### Instructions for Students:
1. **Shear and Moment Equations:**
   - Write the shear force (\( V \)) and bending moment (\( M \)) equations as functions of \( x \).
   - Use free-body diagrams to explain the steps.
   
2. **Beam Sketch:**
   - Draw the original beam with all applied loads and supports labeled.

3. **Shear and Moment Diagrams:**
Transcribed Image Text:### Beam Analysis Problem for Educational Purposes #### Problem Statement: The beam shown is supported by a roller at point A and a pin at point B. The support reactions for the beam are \( A_y = 1000 \text{ lb} \downarrow \) and \( B_y = 2500 \text{ lb} \uparrow \). #### Tasks: - **A.** Write the equations for the shear force (\( V \)) and bending moment (\( M \)) as functions of \( x \) for the beam. Include complete, clearly labeled free-body diagrams to support your work. - **B.** Sketch the original beam on your paper. - **C.** Graph the shear force and bending moment diagrams for the entire beam under the sketch of the original beam. Be sure to label all key points on your diagrams. #### Beam Configuration: - The entire length of the beam from A to C is subjected to a uniformly distributed load of \( 25 \text{ lb/in} \). - There is also a concentrated load of \( 500 \text{ lb} \) acting downward at point C. - The distance between points A and B is \( 16 \text{ in} \). - The distance between points B and C is \( 24 \text{ in} \). #### Diagram Description: 1. **Support at Point A:** - Represented as a roller support. - Located at the left end of the beam. 2. **Support at Point B:** - Represented as a pin support. - Located \( 16 \text{ in} \) to the right of point A. 3. **Distributed Load:** - Spans the entire length of the beam from A to C. - Distributed load of \( 25 \text{ lb/in} \). 4. **Concentrated Load:** - Applied at point C. - Magnitude: \( 500 \text{ lb} \) downward. #### Instructions for Students: 1. **Shear and Moment Equations:** - Write the shear force (\( V \)) and bending moment (\( M \)) equations as functions of \( x \). - Use free-body diagrams to explain the steps. 2. **Beam Sketch:** - Draw the original beam with all applied loads and supports labeled. 3. **Shear and Moment Diagrams:**
### Educational Content: Moments of Inertia for Various Shapes

#### Fundamental Equations

- **Maximum Friction Force:**
  \[
  F_{\text{max}} = \mu_s N
  \]
  where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force.

- **Kinetic Friction Force:**
  \[
  F_k = \mu_k N
  \]
  where \(\mu_k\) is the coefficient of kinetic friction.

- **Angle of Friction:**
  \[
  \tan \Phi = \mu
  \]

#### Parallel Axis Theorem
\[
I = \bar{I} + Ad^2
\]
where \(\bar{I}\) is the moment of inertia about the centroidal axis, \(A\) is the area of the shape, and \(d\) is the distance between the centroidal axis and the axis about which the moment of inertia is being calculated.

#### Centroid Formulas
- **\( \bar{X} \):**
  \[
  \bar{X} = \frac{\sum Ax}{\sum A}
  \]
- **\( \bar{Y} \):**
  \[
  \bar{Y} = \frac{\sum Ay}{\sum A}
  \]

### Moments of Inertia for Common Shapes

#### Rectangular Area
- Diagram: A rectangle with base \(b\) and height \(h\). The centroid \(C\) is at the center.
- **Centroid Coordinates:**
  \[
  \bar{x} = \frac{b}{2}, \quad \bar{y} = \frac{h}{2}
  \]
- **Area Moments of Inertia:**
  \[
  I_x = \frac{bh^3}{3}, \quad \bar{I}_x = \frac{bh^3}{12}, \quad I_z = \frac{bh}{12}(b^2 + h^2)
  \]

#### Triangular Area
- Diagram: A right triangle with base \(b\) and height \(h\). The centroid \(C\) is at \(\frac{1}{3}\) along the base and height.
- **Centroid Coordinates:**
  \[
  \bar{x} = \frac{a + b}{
Transcribed Image Text:### Educational Content: Moments of Inertia for Various Shapes #### Fundamental Equations - **Maximum Friction Force:** \[ F_{\text{max}} = \mu_s N \] where \(\mu_s\) is the coefficient of static friction and \(N\) is the normal force. - **Kinetic Friction Force:** \[ F_k = \mu_k N \] where \(\mu_k\) is the coefficient of kinetic friction. - **Angle of Friction:** \[ \tan \Phi = \mu \] #### Parallel Axis Theorem \[ I = \bar{I} + Ad^2 \] where \(\bar{I}\) is the moment of inertia about the centroidal axis, \(A\) is the area of the shape, and \(d\) is the distance between the centroidal axis and the axis about which the moment of inertia is being calculated. #### Centroid Formulas - **\( \bar{X} \):** \[ \bar{X} = \frac{\sum Ax}{\sum A} \] - **\( \bar{Y} \):** \[ \bar{Y} = \frac{\sum Ay}{\sum A} \] ### Moments of Inertia for Common Shapes #### Rectangular Area - Diagram: A rectangle with base \(b\) and height \(h\). The centroid \(C\) is at the center. - **Centroid Coordinates:** \[ \bar{x} = \frac{b}{2}, \quad \bar{y} = \frac{h}{2} \] - **Area Moments of Inertia:** \[ I_x = \frac{bh^3}{3}, \quad \bar{I}_x = \frac{bh^3}{12}, \quad I_z = \frac{bh}{12}(b^2 + h^2) \] #### Triangular Area - Diagram: A right triangle with base \(b\) and height \(h\). The centroid \(C\) is at \(\frac{1}{3}\) along the base and height. - **Centroid Coordinates:** \[ \bar{x} = \frac{a + b}{
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