Q. A simply supported beam of span 6 m carries udl of 10kN / m upto a distance of 2 m from left support and a point load of 15 kN is placed at a distance of 2 m from right support. Draw the SFD and BMD and indicate where max shear force and max B.M occurs.

by

SOLUTION:

Total load = 10 * 2 + 15 = 35 kN

To find reactions

Taking moment about A

R_{B} * 6 = 15 * 4 + (10 * 2) * 2/2 = 80kN R_{B} = 80/6 = 13.33kN

R_{A} = 35 – 13.33 = 21.67kN

SFD (SHEAR FORCE DIAGRAM)

F_{A} = 21.67kN

F_{B} = 1.67kN (D to C constant)

F_{C} = – 13.33kN (C to B constant)

F_{B} = – 13.33kN

Max +ve SF occurs at A = 21.67kN and

max – ve SF occurs at B = – 13.33kN

BMD ( BENDING MOMENT DIAGRAM)

M_{A} = M_{B} = 0

M_{C} = 13.33 * 2 = 26.66kNm

M_{D} = 21.67 * 2 – (10 * 2) * 2/2

= 43.34 – 20 = 23.34kNm ..

Max BM occurs at C (where SF changes its sign)

=M C =M max =26.66 kNm

Leave a Comment