Q. A beam 8 m long carries an udl of 2kN / m on the entire span. The beam has an overhang of 1 m and 1.5 m after left hand support and right hand support respectively. The spacing between the supports is 5.5 m. Draw the SFD and BMD indicating the values at salient points.
Total load= 8 * 2 = 16kN
To find the reactions
Taking moments about A
R{B} * 5.5 + (2 * 1) * 1/2 = (2 * 7) * 7/2
R{B} = (49 – 1)/5.5 = 8.727kN
R{A} = 16 – 8.727 = 7.273kN SFD(SHEAR FORCE DIAGRAM)
F{D} = F{C} = 0
F{A} (just left of A)= – 2kN
FA (just right ofA )=5.273 kN
FB (just left of B )= – 5.727 kN
FB (just right of B) = 3 kN
Max +veSF occurs at A = 5.273kN and
Max ve SF occurs at B = – 5.727kN BMD ( BENDING MOMENT DIAGRAM)
M{D} = M{C} = 0
M{A} = – (2 * 1) * 1/2 = – 1kNm
M{B} = – (2 * 1.5) * 1.5/2 = – 2.25kNm
x/5.273 = (5.5 – x)/5.727
5.727x = 5.273 * 5.5 – 5.273x
x = (5.273 * 5.5)/11 = 2.6365m
M max =M2.6365
= 7.273 * 2.6365 – 2 * 3.6365 * 3.6365/2
= 5.9511 kNm
Max + ve BM occurs at D = 5.9511kNm and max – ve BM occurs at B = – 2.25kNm
