Draw the shear and bending moment diagrams for the beams below. Label all key Points. The diagram illustrates a beam supported at points B and C, with various loads applied along its length. Here's a detailed explanation of the setup:- **Beam and Supports:** - The beam is horizontal, with supports at points B and C. - Support B is located 5 feet from the left end (point A). - Support C is positioned 20 feet from point B, or 15 feet from the left end (A).- **Loads and Forces:** - A uniform distributed load of 2.5 kips per foot (k/ft) acts between points A and D. - Point D, located 30 feet from point A, is subjected to a clockwise moment of 35 kip-feet (k-ft). - Additionally, there is a concentrated load of 10 kips (k) applied at point D.- **Distances:** - The entire length between points A and D is divided into three segments by supports B and C: - 5 feet from A to B - 15 feet from B to C - 10 feet from C to DThis setup typically represents a common problem in analyzing statically determinate beams in structural engineering, where the goal is to determine reactions at the supports and internal forces within the beam.

Given:- To find:- Shear and bending moment diagrams Concept:- Shear force and bending moment diagram is important analysis aspect to get the idea of reinforcement required for design. For longitudinal reinforcement bending moment give an idea about the top and bottom bar Shear reinforcement provide at the cross section to resist diagonal crack.Support reactions:- Lets assume that reaction at B be RB and reaction at C be RC. Taking sum of moments at C equal to zero ∑MC=0RB×15-2.5×20×10+10×10-35=0RB=29 kN Taking sum of vertical forces equal to zero ∑Fy=0RB+RC-2.5×20-10=029+RC-60=0RC=31 kN Shear force calculations:- Shear force at A isVA=0Shear force at just left of B isVB=-2.5×5=-12.5 kShear force at just right of B isVB=-2.5×5+29=16.5 kShear force at just left of C isVC=-2.5×20+29=-21 kShear force changes sign in span BC, therefore shear force is zero at-2.5x+29=0x=11.6 ftShear force at just right of C isVC=-2.5×20+29+31=10 kShear force at D isVD=10 k Shear force diagram:- Shear force diagram of the beam is given asBending moment calculations:- Bending moment at A isMA=0Bending moment at B isMB=-2.5×5×52=-31.25 k-ftBending moment at C isMC=-2.5×20×2020+29×15=-65 kN-mBending moment at D isMD=-2.5×20×10+10+29×25+31×10=35 kN-mMaximum bending moment at x = 11.6 ft from A isMmax=-2.5×11.6×11.62+29×11.6-5=23.2 kN-mBending moment diagram:- Bending moment diagram of the beam is given as