Q. A simply supported beam of span 8 m, carries a udl of 20kN / m upto a distance of 4 m from left support and a concentrated load of 40 kN at a distance of 2 m from right support. Draw SFD and BMD.

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SOLUTION:

Total load 20 * 4 + 40 = 120kN

To find the reactions

Taking moments about A

R{B} * 8 = 40 * 6 + (20 * 4) * 4/2

= 240 + 160 = 400

R{B} = 400/8 = 50kN

R A =Total load – R{B}

= 120 – 50 = 70kN

SFD ( SHEAR FORCE DIAGRAM)

F{A} = 70kN

F{B} = – 50kN

F{C} = – 10kN

F{D} = – 50kN

Max +veSF occurs at A = 70kN and

max – Ve SF occurs at B = -50 KN ( same in DB)

BMD ( BENDING MOMENT DIAGRAM)

M{A} = M{B} = 0

M{C} = 70 * 4 – (20 * 4) * 4/2 = 280 – 160 = 120kNm

M{D} = 50 * 2 = 100kNm

By similar triangles

x/70 = (4 – x)/10

10x = 280 – 70x

80x = 280

x = 280/80 = 3.5m

Max B.M occurs at 3.5 m i. e, at E (where SF changes its sign)

= 70 * 3.5 – (20 * 3.5) 3.5/2 = 122.5kNm

M max =122.5 kNm occurs at E

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