Q.Q. A cantilever beam 3 m long carries THREE point loads of 2 kN, 4 kN, and 1kN are acting at 1 m, 2 m, and 3 m respectively from fixed end. Draw SFD and B.M.D.

SOLUTION

TOTAL LOAD =2+4+1=7N

SHEAR FORCE DIAGRAM

FB = 1N
FD = 5N (4+1=5N)
FC = 7N (2+4+1=7N)
MAX SF OCCURS AT FIXED END
A = 7N AND is same in between AC.

BENDING MOMENT DIAGRAM

M{B} = 0
M{D} = – 1 * 1 = – 1 Nm
M{C} = – 1 * 2 – 4 * 1 = – 2 – 4 = – 6Nm
M{A} = – 2 * 1 – 4 * 2 – 1 * 3 = – 2 – 8 – 3 = – 13Nm

Max BM occurs at fixed end A
Mmax = MA = 13 Nm = 13 Nm (hogging)