Data:
L = 3m
W1 = 10kN
W2 = 20kN
W3 = 20kN
Solution:
Maximum S.F = 10 + 20 + 20 = +50kN
Shear at support AB = -10kN
Shear at support BC = -(10 + 20) = -30kN
Shear at support CD = -(10 + 20 + 20) = -50kN
Shear Force Diagram
Maximum B.M = -WL
Moment at support A, MA = 0kN.m
Moment at support B, MB = -(10 x 1) = -10kN.m
Moment at support C, MC = -[(10 x 2) + (20 x 1)] = -40kN.m
Moment at support D, MD = -[(10 x 3) + (20 x 2) + (20 x 1)] = -90kN.m
Bending Moment Diagram
Result:
Max S.F = +50kN
S.F at AB = -10kN
S.F at BC = -30kN
S.F at CD = -50kN
Max B.M
MA = 0kN.m
MB = -10kN.m
MC = -40kN.m
MD = -90kN.m