Q. A simply supported beam of span 10 m carries a udl of 40kN / m over a length of 5 m from LHS and over a length of 2 m beginning from RHS. In addition it carries a point load of 100 kN at 2 m from the RHS. Draw SF and BM diagrams and state the position and magnitude of max. BM.

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SOLUTION:

Total load = 40 * 5 + 100 + 40 * 2 = 380kN
To find the reactions
Taking moments about A
R{B} * 10 = (40 * 5) * 5/2 + 100 * 8 + (40 * 2)(2/2 + 8)
R{B} = 2020/10 = 202kN
R{A} = 380 – 202 = 178kN
SFD (SHEAR FORCE DIAGRAM)
F{A} = 178kN
F{B} = – 202kN
F{C} = – 22kN
F{D} = – 22kN (same in CD)
Max +veSF occurs at A = 178kN and
max -ve SF occurs at B = – 202kN
BMD (BENDING MOMENT DIAGRAM)
M_{A} = M_{B} = 0
M_{C} = 178 * 5 – (40 * 5) * 5/2 = 390kNm
M_{D} = 202 * 2 – (40 * 2) * 2/2 = 324kNm
To find x (by similar triangles)
x/178 = (5 – x)/22
22x = 890 – 178x
200x = 890
x = 890/200 = 4.45m
M E =M max =M 4.45 =178*4.45-(40*4.45)x 4.45/2 = 396 05 kNm

Max BM occurs at E (where SF changes it sign)
=M max =396.05 kNm

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