For the beam shown below, the distributed load along AB is a= 3 kN/m, the concentrated load at Point C is b = 15 kN.Draw FBD, sove for reaction forces, then draw shear & momment diagram. Input the internal bending moment at the cross-section from Point B: ____ kN m.Input the negative sign "-" if it is negative, and calculate your answer to 1 decimal place. For the beam shown below, the distributed load along AB is a= 3 kN/m, the concentrated load at Point C is b = 15 kN.Draw FBD, sove for reaction forces, then draw shear & momment diagram. Input the internal bending moment at the cross-section from Point B:kN m.Input the negative sign "-" if it is negative, and calculate your answer to 1 decimal place.a kN/m4 mB2 m-b kN

To find : The reaction forces . Given : The distributed load is The concentrated load at point C is Free Body diagram :Calculation : Apply the force equilibrium equation in y direction is, ∑Fy=0RA+RB-3·4-15=0RA+RB=27 kN Apply the moment equation about point A is, ∑MA=03·4·2-RB·4+15·7=0RB=32.25 kN The reaction force at point A is, RA+RB=27RA=27-32.25=-5.25 kN (downward) The shear force from point A in x distance can be obtained as, Vx=RA-3x=-5.25-3x The bending moment from point A in x direction can be obtained as, Mx=-5.25x-3x2 The shear force just after point A is, VA+=-5.25kN The shear force in segment AB is, Vx=-5.25-3x The shear force at just before point B is, VB-=-5.25-3x=-5.25-3·4=-17.25kN The shear force at just after point B is, VB+=-5.25-3x+32.25=-5.25-3·4+32.25=15kN The shear force at point c is, Vc=5.25+12-32.25+15=0 The bending moment at point B is, MB=-5.25x-3x2=-5.25·4-3·16=-69 kN-m The bending moment at point C is , Mc=0To draw : The shear force diagram and bending moment diagram , Diagram is :