Q. A beam of length 6 m is loaded as shown in figure. Draw SF and BM diagrams by indicating values at silent point.
SOLUTION:
Total load = 15 + 10 + 10 * 1 = 35kN Taking moments about A
R{B} * 4 + (10 * 1) * 1/2 = 15 * 3 + 10 * 5
R{B} = (95 – 5)/4 = 90/4 = 22.5kN
R{A} = 35 – 22.5 = 12.5kN SFD ( SHEAR FORCE DIAGRAM)
F{c} = 0
F{A} = 2.5 (same in AD)
F{D} = – 12.5 (same in DB)
F{B} = 10kN (same BE)
Max +ve SF occurs in between BE i.e., 10 kN and max-v SF occurs in between DB = – 12.5kN BMD ( BENDING MOMENT DIAGRAM)
M{C} = M{E} = 0
M{B} = – 10 * 1 = – 10kNm
M{A} = – (10 * 1) * 1/2 = – 5kNm
M{D} = 22.5 * 1 – 10 * 2 = 2.5kNm
Max + ve BM occurs at D = 2.5kNm and
Max – ve BM occurs at B = – 10kNm
To find points of contraflexure (a) Find BM upto xx and equate to zero
12.5(a – 1) – (10 * 1)(1/2 + a – 1) = 0
125a – 12.5 – 5 – 10a + 10 = 0
2.5a = 7.5
a = 7.5/2.5 = 3m from left end (b) Find BM upto yy and equate it to zero
22.5(b – 1) – 10b = 0
225b – 22.5 – 10b = 0
12.5b = 22.5
b = 22.5/12.5 = 1.8m
.. Points of contra flexure occurs at a distances of 3 m from left end C and 1.8 m from right end E
