draw a shear and bending moment diagram for the beam shown. Label each diagram with values at locations, minimums and maximuns, changes in slope and where it crosses zero This diagram illustrates a beam supported at two points, labeled B and E, and subjected to various loads. Here's a detailed breakdown:**Description:**- The beam is denoted by points A to E.- It carries a uniform distributed load of 20 lb/ft along its length.- A concentrated load of 120 lb is applied vertically downward at point C.- The beam is also subjected to a moment of 200 lb-ft applied in a clockwise direction at point D.**Measurements:**- The distance between points A and B is 4 ft.- The distance between points B and C is 3 ft.- The distance between points C and D is 1 ft.- The distance between points D and E is 6 ft.**Supports:**- At point B, there is a support in the form of a triangle which indicates a roller support.- At point E, a similar support suggests a pin connection or hinge.The diagram is labeled as "Fig. P8.126". This setup is commonly used in structural analysis to determine reactions at supports and internal forces within the beam.

Given Draw a sheer and bending moment diagram for given figure let RB and RE are the reactions at point B and E repectivally total load due to uniform distribution load W = 20lbft × span length over which UDL actW = 20lbft × (4+3+1+6)W= 280lbfttotal load due to uniform distribution load W will act at the center point of the span (ie. at point C) Anticlock wise moment 200lbft acts at point DAt equilibrium condition moment about the point B will be equal zero take moment clockwise positive and anticlockwise negative280+120 ×3-200-RE×10=0RE×10=280+120 ×3-200RE=100lbAs we know all force must balanced therefore RE+RB=280+120100+RB=280+120RB=300lb Now Draw the shear force diagram Draw bending moment diagram p is the point where bending moment is zero and it is also called contraflexture point.let p is the point at x distance from the left end of the beam and moment about that point is zero 20x22-300(x-4)=010x2-300x+1200=0solve quadratic equation x=4.753ftcontrafluture point from the roller support or point B =x-4=4.753-4=0.753ftpoint P is at 0.753ft from point B or roller support from SFD and BMD diagram we get maximum sheer force 220lb at point Bminimum sheer force is 0 at a point 1ft from the point D maximum bending moment is 440lb-ft at point Dminimum bending moment is 0 at point P 0.753ft from point B