Q. A simply supported beam of span 12 m carries a udl of 5kN / m over a length of 6 m on middle portion of the beam. In addition to that it carries two point loads of 20 kN and 15 kN at 1.5 m and 10 m from left hand support. Draw the SF and BM diagrams and indicate their max values.

SOLUTION:

Total load = 20 + 15 + 5 * 6 = 65kN

To find the reactions

Taking moments about A

R{B} * 12 = 20 * 1.5 + 15 * 10 + (5 * 6)(6/2 + 3)

= 360

R{B} = 360/12 = 30kN

R{A} = 65 – 30 = 35kN

SFD ( SHEAR FORCE DIAGRAM)

F{A} = 35kN

F{B} = – 30kN

F{C} = 35kN

F{D} = 15kN

F{E} = – 15kN

F{F} = – 30kN

F{G} = 0kN

Max +ve SF occurs at support A i.e., 35 kN (same in AC)

Max ve SF occurs at support B i.e. – 30kN (same in FB)

BMD ( BENDING MOMENT DIAGRAM)

M{A} = M{B} = 0

M{C} = 35 * 1.5 = 52.5kNm

M{D} = 35 * 3 – 20 * 1.5 = 75kNm

M{E} = 30 * 3 – 15 * 1 = 75kNm

M{F} = 30 * 2 = 60kNm

M{G} = M max

= 35 * 6 – 20 * 4.5 – 5 * 3 * 3 /2

= 210 – 90 – 22.5 = 97.5KNm

Max B.M occurs at G (where SF changes its sign) i.e., 6 m from A or B=M max =M G =97.5 kNm