A simply supported beam length 3m long is carrying a uniformly distributed load of 10kN/M across entire beam span and concentrated load of 5KN at point C at the center of the beam. Draw the SFD and BMD for the beam and also calculate the maximum bending moment on the section.
Solution
Taking moment about A
-3RB+10*3*3/2+5*1.5=0
-3RB+45+7.5=0
-3RB+52.5=0
RB=17.5KN
Similarly, taking Moment about B
3RA-10*3*3/2-5*1.5=0
3RA-45-7.5=0
3RA-52.5=0
RA=17.5KN
computation for shear force
Shear force at A = reaction at A= 17.5KN
Shear force at C= 17.5-10*1.5-5= -2.5KN
Shear force at B= 17.5-10*1.5-5-10*1.5=-17.5KN
Computation for bending moment
Moment at A and moment at B=0. Because moment at a point is zero. The maximum bending moment will tend to occur at point C. The principle of similarity of triangle will be used.
Manual approach
Using staad- pro vi8
