**Problem Statement**:Given that \( w = 50 \, \text{N/m} \), \( L = 2 \, \text{m} \), and \( P = 150 \, \text{N} \),**a)** Draw the shear force and bending moment diagrams. **b)** Identify the absolute maximum shear force and bending moment in the beam.**Diagram Explanation**:The diagram shows a beam, ABC, subjected to a uniformly distributed load \( w \) and point loads \( P \) at both ends (points A and C). The beam is supported by a roller at point B, placed centrally between A and C. The overall length of the beam between A and C is \( 2L \). Indicated on the beam are:- The evenly distributed downward force \( w \) across the span \( AC \).- The vertical reactions \( P \) at points A and C.**Diagram Details**:- **Supports**: - Point \( A \) is a fixed support. - Point \( B \) is a roller support. - Point \( C \) is a fixed support.- **Loading Configuration**: - Uniform load \( w = 50 \, \text{N/m} \) extends across the entire beam. - Concentrated reactions \( P = 150 \, \text{N} \) are applied at both A and C pointing upwards.- **Dimensions**: - The length between A and B is \( L = 2 \, \text{m} \). - The length between B and C is also \( L = 2 \, \text{m} \). The task involves drawing shear force and bending moment diagrams to calculate maximum internal forces in the beam.

Objective: To draw the shear force and bending moment diagrams and give the values of absolute maximum shear force and bending moment. Given: w = 50 N/m L = 2 P = 150 N The free body diagram of the beam can be drawn as:∑MA=0CW+ →By×2 m+50 N/m×4 m×42 m-150 N×1 m-150 N×4 m=0 →By=150+150×4-50×42 N ⇒By=175 N ∑Fy=0UP+ →Ay-50 N/m×4 m+150 N-175 NBy+150 N=0 →Ay=50×4+175-150-150 N ⇒Ay=75 NThe shear force equations [V(x)] for the beam can be written as: For 0≤x≤1 Vx=75 N-50 N/m×x m=-50x+75 N Linear Vx=0=-50×0+75 N=75 N Vx=1=-50×1+75 N=25 N For 1≤x≤2 Vx=-50x+75 N+150 N=-50x+225 N Vx=1=-50×1+225 N=175 N Vx=2=-50×2+225 N=125 N For 2≤x≤4 Vx=-50x+225 N-175 N=-50x+50 N Vx=2=-50×2+50 N=-50 N Vx=4=-50×4+50 N=-150 N Using the equations and the calculated values, the shear force diagram can be drawn as: The bending moment equations [M(x)] for the beam can be written as: For 0≤x≤1 Mx=75 N×x m-50 N/m×x m×x2 m=-25x2+75x N.m Parabolic Mx=0=-25×02+75×0 N.m=0 Mx=1=-25×12+75×1 N.m=50 N.m For 1≤x≤2 Mx=-25x2+75x N.m+150 N×x-1 m=-25x2+225x-150 N.m [Parabolic] M(x=1)=-25×12+225×1-150 N.m=50 N.m M(x=2)=-25×22+225×2-150 N.m=200 N.m For 2≤x≤4 Mx=-25x2+225x-150 N.m-175 N×x-2 m=-25x2+50x+200 N.m [Parabolic] Mx=2=-25×22+50×2+200 N.m=200 N.m Mx=4=-25×42+50×4+200 N.m=0 Using the equations and the calculated values, the bending moment diagram can be drawn as: From the diagrams, The absolute maximum shear force is: 175 N The absolute maximum bending moment is: 200 N.m .