Q. Calculate the max bending moment for a simply supported beam with udl of 2 kN/m throughout its span. Length of beam is 4 m.

SOLUTION:
Given loading on the beam is Symmetrical hence
RA = RB= 4 × 2 /2= 4kN

BMD (BENDING MOMENT DIAGRAM)
MA = MB = 0
MC = 4 × 2 – (2 × 2) 2/2 = 4kNm

OR
M max = wl²/ 8 = 2×4²/ 8 =4 kNm.

Q. A simply supported beam of span 6 m carries a udl of 10 kN/m over its entire length. Draw the SF diagram.

SOLUTION:
Total load = 10 × 6 = 60kN
To find the reaction
As the given load on the beam is Symmetrical hence RA = RB = 60/2 = 30kN

SFD (SHEAR FORCE DIAGRAM)
F{A} = 30kN
F{B} = – 30kN
F{C} = 0kN
Max + ve SF occurs at A = 30kN and
max – ve SF occurs at B = – 30 kN