Q.A simply supported beam of span 10 m is subjected to 20 kN, 30 kN, 50 kN and 20 kN loads respectively at 1 m, 3m, 5 m and 8 m from left hand support. Draw the S.F.D and B.M.D indicating the values at salient points.

by

SOLUTION:


Total load = 20 + 30 + 50 + 20 = 120kN
To find reactions:
Taking moments about A
R{B} * 10 = 20 * 1 + 30 * 3 + 50 * 5 + 20 * 8
. .= 20 + 90 + 250 + 160 = 520kN
R{B} = 520/10 = 52kN
R{A} = 120 – 52 = 68kN
To find S.F:
FA = 68kN ;
FB = 52KN
FC = 68kN (just left of C)
FC = 48kN (just right of C)
FD = 48kN (just left of D)
FD = 18kN (just right of D)
FE = 18kN (just left of E)
FE = – 32kN (just right of E)
FF = – 32kN ( just left of F)
FF = – 52kN ( just right of F)
To find B.M:
M{A} = M{B} = 0
M{C} = 68 * 1 = 68kNm
M{D} = 68 * 3 – 20 * 2 = 164 kNm
M{E} = 68 * 5 – 20 * 4 – 30 * 2 = 200kNm
M{F} = 68 * 8 – 20 * 7 – 30 * 5 – 50 * 3 = 104kNm
(or) 52 * 2 = 104 kNm
Max +ve B.M occurs at E and its value = 200 kNm.

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