Q. A beam ABCD is supported at A and D such that AB = 1.5 BC = 2.5 m and AD = 5m . It carries point loads of 30 kN and 20 kN at B and C respectively. It also carries a udl of 10 kN/m including self weight on entire span. Draw SFD and BMD.

by

SOLUTION:


Total load= 10 * 5 + 30 + 20 = 100kN
To find reactions:
R{D} * 5 = (10 * 5) * 5/2 + 30 * 1.5 + 20 * 4
= 125 + 45 + 80 = 250
R{D} = 250/5 = 50kN
R{A} = 100 – 50 = 50kN
SFD (SHEAR FORCE DIAGRAM)
F{A} = 50kN
FB(R) = 5kN ; FB(L) = 35kN
FC(R) = 40kN ; FC(L) = – 20kN ;
F{D} = 50kN
To find x:
x/5 = (2.5 – x)/20
20x = 12.5 – 5x
25x = 12.5
x = 12.5/25 = 1/2 = 0.5m
To find BM:
M{A} = M{D} = 0
M{B} = 50 * 1.5 – 10 * 1.5 * 1.5/2 = 63.75kNm
M{C} = 50 * 1 – (10 * 1) * 1/2 =45 kNm
M max =M 2 =50 * 2 – (10 * 2) 2 /2 =30 * 0.5=65 kNm
Max B.M occurs where S.F changes its sign from + ve to – ve i.e., at 2 m from left support = 65 kNm.

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