Draw the shear and moment diagrams for the beam. Consider figure. The image depicts a beam scenario in structural engineering. It shows a horizontal beam supported at two points: point A and point B. The beam has a non-uniform distributed load applied to it.### Detailed Description:- **Beam Supports:** - **Point A** is supported by a pin support, which allows rotation but not translation. - **Point B** is supported by a roller support, which allows horizontal movement but restricts vertical movement.- **Distributed Load:** - The load gradually increases from point A to point B. - The maximum load intensity applied is $12 \, \text{kN/m}$, positioned at point B.- **Distance Measurements:** - The beam spans a total length divided into: - $3 \, \text{m}$ from A to an intermediate point C. - $6 \, \text{m}$ from C to B.This type of load distribution creates a shear force and bending moment along the length of the beam, important for analysis in mechanical and structural engineering.

∑MB = 0Ay×6 - 12×3×12×33+3 - 12×3×1.5 = 0Ay = 1266Ay = 21 KN∑FY = 021 - 12×3×12 - 12×3 + By = 0By =…

Answered: Draw the shear and moment diagrams for…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Draw the shear and moment diagrams for the beam. Consider figure.

The image depicts a beam scenario in structural engineering. It shows a horizontal beam supported at two points: point A and point B. The beam has a non-uniform distributed load applied to it.

### Detailed Description:
- **Beam Supports:**
- **Point A** is supported by a pin support, which allows rotation but not translation.
- **Point B** is supported by a roller support, which allows horizontal movement but restricts vertical movement.

- **Distributed Load:**
- The load gradually increases from point A to point B.
- The maximum load intensity applied is $12 \, \text{kN/m}$, positioned at point B.

- **Distance Measurements:**
- The beam spans a total length divided into:
- $3 \, \text{m}$ from A to an intermediate point C.
- $6 \, \text{m}$ from C to B.

This type of load distribution creates a shear force and bending moment along the length of the beam, important for analysis in mechanical and structural engineering.

Expert Solution

Step 1: Reaction

$\begin{array}{rcl} \sum_{} M_{B} & = & 0 \\ A_{y} \times 6 - (\frac{1}{2} \times 3 \times 12) \times (\frac{3}{3} + 3) - (12 \times 3) \times 1.5 & = & 0 \\ A_{y} & = & \frac{126}{6} \\ A_{y} & = & 21 K N \\ \sum_{} F_{Y} & = & 0 \\ 21 - (\frac{1}{2} \times 3 \times 12) - 12 \times 3 + B_{y} & = & 0 \\ B_{y} & = & 33 K N \\ \sum_{} F_{X} & = & 0 \\ B_{x} & = & 0 \end{array}$

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