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*7-56. Draw the shear and moment diagrams for the beam. 1.5 kN/m A B 2 m 4 m

*7-56. Draw the shear and moment diagrams for the beam. 1.5 kN/m A B 2 m 4 m

Structural Analysis
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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## Analyzing Shear and Moment Diagrams for a Beam ### Problem Statement **Exercise 7-56:** Draw the shear and moment diagrams for the beam. ### Beam Description The beam is supported at two points, labeled as A and C. It has a point load distributed evenly across a section of the beam: - **Support A**: A pinned support at the left end of the beam. - **Length AB**: 2 meters. - **Length BC**: 4 meters, making the total length of the beam 6 meters. - **Uniformly Distributed Load**: 1.5 kN/m applied from point B to point C. - **Support C**: A roller support at the right end of the beam. ### Task The goal is to determine and sketch the shear force and bending moment diagrams for this particular configuration of the beam. ### Explanation of Diagrams #### Shear Force Diagram The shear force at any section of the beam is influenced by the support reactions and the distributed load. The diagram typically starts from zero at one end and changes at various points due to applied loads and reactions. 1. **From A to B (0 to 2 m)**: - The shear force remains constant since there are no loads acting in this section except for the reaction at A. 2. **From B to C (2 to 6 m)**: - The shear force linearly decreases due to the uniformly distributed load (1.5 kN/m). #### Bending Moment Diagram The bending moment diagram is influenced by the shear force along the beam. 1. **At Point A (0 m)**: - The bending moment is zero as it is a support point. 2. **Between A and B (0 to 2 m)**: - The bending moment increases linearly due to the constant shear force. 3. **Between B and C (2 to 6 m)**: - The moment increases in a parabolic manner due to the linearly varying shear force. 4. **At Point C (6 m)**: - The bending moment returns to zero as it is a support point. ### Conclusion To gain further insights, one should calculate the reactions at the supports by using equilibrium equations (sum of vertical forces and moments). Then, derive precise values for shear forces and bending moments at key locations along the beam. This
Transcribed Image Text:## Analyzing Shear and Moment Diagrams for a Beam ### Problem Statement **Exercise 7-56:** Draw the shear and moment diagrams for the beam. ### Beam Description The beam is supported at two points, labeled as A and C. It has a point load distributed evenly across a section of the beam: - **Support A**: A pinned support at the left end of the beam. - **Length AB**: 2 meters. - **Length BC**: 4 meters, making the total length of the beam 6 meters. - **Uniformly Distributed Load**: 1.5 kN/m applied from point B to point C. - **Support C**: A roller support at the right end of the beam. ### Task The goal is to determine and sketch the shear force and bending moment diagrams for this particular configuration of the beam. ### Explanation of Diagrams #### Shear Force Diagram The shear force at any section of the beam is influenced by the support reactions and the distributed load. The diagram typically starts from zero at one end and changes at various points due to applied loads and reactions. 1. **From A to B (0 to 2 m)**: - The shear force remains constant since there are no loads acting in this section except for the reaction at A. 2. **From B to C (2 to 6 m)**: - The shear force linearly decreases due to the uniformly distributed load (1.5 kN/m). #### Bending Moment Diagram The bending moment diagram is influenced by the shear force along the beam. 1. **At Point A (0 m)**: - The bending moment is zero as it is a support point. 2. **Between A and B (0 to 2 m)**: - The bending moment increases linearly due to the constant shear force. 3. **Between B and C (2 to 6 m)**: - The moment increases in a parabolic manner due to the linearly varying shear force. 4. **At Point C (6 m)**: - The bending moment returns to zero as it is a support point. ### Conclusion To gain further insights, one should calculate the reactions at the supports by using equilibrium equations (sum of vertical forces and moments). Then, derive precise values for shear forces and bending moments at key locations along the beam. This
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