Problem:
A simply supported beam of length 8 m rests on supports 6 m apart, the right hand end is overhanging by 2 m. The beam carries a uniformly distributed load of 1500 N/m over the entire length. Draw S.F. and B.M. diagrams and find the point of contraflexure, if any.
(Chapter- Shear Force and Bending Moment, Book – Strength of Materials by Dr. R. K. Bansal.)
Solution:
moment at the right support,
MB = -1500.0 * 2.0 * 1.0 = -3000.0 Nm
Similarly, Shear Force VB = Vertical sum of all loads on the right overhang = 3000.0 N
At the left support, MA =0
Using the static equilibrium conditions, algebraic sum of moments at the left end is equated to zero.
Hence, RB = [(MB – MA + VB*(L – a – b) + -1500.0 * 6.0 * 3.0 + sum(Pi*xi for i ranges up to number of point loads on beam)]/(L-0.0-2.0)
Hence, RB = 8000.0 N
Also, RA + RB = sum(Pi for i ranges up to number of point loads on beam) + w1 * 8.0,
Solving gives, RA = 4000.0 N.
In between A and B, for 0.0 m <= x <= 6.0 m:
Shear, Vx = VA -1500.0(x -0.0) = 4000.0 -1500.0x
Moment, Mx = VA(x-0.0) + -1500.0(x -0.0)^2/2 = 4000* (x-0.0) -1500.0*(x -0.0)^2/2
for 6.0 m <= x <= 8.0 m:
Shear, Vx = 3000.0 -1500.0 * (x – 6.0)
Moment, Mx = -3000.0 + 3000.0*(x – 6.0) -1500.0*(x – 6.0)^2/2
Maximum Bending Moment, Mmax = 5333.333 Nm = 5.33 kN-m.
at x = 2.667 m from the left end.
Point of Contraflexure is found by putting Mx = 0,
4000* (x) -1500.0*(x)^2/2 = 0, solving gives x = 5.333 m
