Q. A beam 9 m long carries an udl of 20 kN/m on the entire span. The beam has an overhang of 1.5m and 1.5 m after left hand support and right hand support respectively. The spacing between the supports is 6 m. Draw the SFD and BMD indicating the values at salient points.
Total load = 20 * 9 = 180kN
To find reactions
As the loading on the beam is symmetrical then
R{A} = R{B} = 180/2 = 90kN SFD ( SHEAR FORCE DIAGRAM)
F{C} = F{D} = 0
F{A} (just left of A)= – 30kN
F{A} (just right of A)=+ 60kN
F{B} (just left of B)= – 60kN ,
FB = 30kN (Just right of B)
Max + ve SF occurs at A = 60kN and
Max -veSF occurs at B = – 60kN BMD( BENDING MOMENT DIAGRAM)
M{C} = M{D} = 0
M{B} = M{A} = (20 * 1.5) 1.5/2 = – 22.5kNm
M max =M E =90 * 3 – 20 * 4.5 * 4.5 /2
= 270 – 202.5 = 67.5kNm
Max +ve BM occurs at E = 67.5kNm and
max -ve BM occurs at support A and B = – 22.5kNm. To find point of contra flexure
Find the BM at a distance ‘a’ and equate it to zero
90(a – 1.5) – 20a * a/2 = 0
90a – 135 – 10a² = 0
90a – 10a² = 135
10a ² – 90a + 135 = 0
a = 1.9019m from C
Since the load is symmetrical hence a = b = 1.9019m from either side.
