Q. A simply supported beam of span 4 m carries a point load of 10 kN at its centre. Draw SFD and BMD. Find max Shear force and Вending moment 

SOLUTION:

First To find reactions

As the load is symmetrical, the load shared by the two supports are equal.

Hence R{A} = R{B} = 10/2 = 5kN

(OR)

Taking moments about B
RA × 4 = 10 × 2
RA = (10 × 2)/4 = 5kN
RB = Total load – RA
= 10 – 5 = 5kN

SFD (SHEAR FORCE DIAGRAM)

FA = 5kN
FB = – 5kN
FC =(just left ofC )=+5 kN
FC =(just right ofC )=-5 kN

MaxSF =+5 kN(occurs at A)

BMD (BENDING MOMENT DIAGRAM)
MA = MB = 0
MC = 5×2 = 10 KNm
OR
wl²/4=(10×4)/4 = 10 kNm
Max BM occurs at C = MC=Mmax=10KNm
( where SF changes its sign)