Draw the shear and moment diagrams forthe beam. ### Diagram Description:The diagram illustrates a horizontal beam that is fixed at point A, with a length extending to point C. The beam interacts with two forces:- **Point A:** The left end of the beam is embedded into a wall or support structure, providing a fixed support.- **Point B:** Located at the midpoint of the beam, 5 feet from both ends. Here, the beam experiences a moment of 800 lb-ft, depicted by a curved arrow.- **Point C:** The right end of the beam, 5 feet from point B, has a downward force of 100 lb applied.### Measurements:- The total length of the beam, from point A to point C, is 10 feet.- The distance from point A to point B is 5 feet.- The distance from point B to point C is 5 feet.This setup is representative of a typical problem in statics, where equilibrium of moments and forces is analyzed to determine reactions at supports.

Draw the free body diagram of the given member and calculate the reactions: Consider the equilibrium of vertical forces: RA-100 lb=0RA=100 lb The moment reaction at A will be equal to the moment about point A in magnitude and opposite in direction. Calculate moment about point A: M=-800 lb-ft-100 lb10 ftM=-1800 lb-ft=1800 lb-ftclockwise Therefore the reaction moment about point A is given by: MA=1800 lb-ft anticlockwise Calculate the shear force at the points specified on the beam: SFA=RA=100 lbSFB=RA=100 lbShear force to the left of point C:SFLC=RA=100 lbShear force to the right of point C:SFRC=RA-100 lbSFRC=100 lb-100 lb=0 Calculate bending moments t different points of the beam: BMA=MA=-1800 lb-ftBending moment to the left of point B:BMLB=MA+RA5 ft=-1800lb-ft+100 lb5 ft=-1300 lb-ftBending moment to the right of point B:BMRB=MA+RA5 ft+800 lb-ft=-1800 lb-ft+100 lb5 ft+800 lb-ft=-500 lb-ftBMC=MA+RA10 ft+800 lb-ft=-1800 lb-ft+100 lb10 ft+800 lb-ft=0Draw the shear force and bending moment diagram: