## Analyzing Shear and Moment Diagrams for a Beam### Problem Statement**Exercise 7-56:**Draw the shear and moment diagrams for the beam.### Beam DescriptionThe 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.### TaskThe 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 DiagramThe 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 DiagramThe 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.### ConclusionTo 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

The given figure is shown below:For the span 0≤x≤2m Consider the forces in y-direction and applying equilibrium condition ∑Fy=00.75-V=0V=0.75 kNThe equilibrium of moment is ∑M=0M-0.75x=0M=0.75xFor the span 2m<x<4m Consider the forces in y-direction and applying equilibrium condition ∑Fy=00.75-1.5x-2-V=0V=3.75 -1.5x Consider the condition of equilibrium of moment ∑M=0M+1.52x-22-0.75x=0M=-0.75x2+3.75x-3 The shear force diagram is shown belowThe bending moment diagram is shown below