Problem:
A beam is loaded as shown in Fig. 6.46. Draw the S.F. and B.M. diagrams and find: (i) maximum S.F. (ii) Maximum B.M. (iii) Point of inflexion.
Solution:
At left support, VA = -50.0 kN, and MA = -100.0 kNm
Using the static equilibrium conditions, the algebraic sum of moments at the left end is equated to zero.
Hence, RB = [(MB – MA + VB*(L – a – b) + 0.0 * (0.0) * -2.0 + sum(Pi*xi for i ranges up to number of point loads on beam)]/(L-2.0-2.0)
Hence, RB = 80.0 kN
Also, -(RA + RB) = sum(Pi for i ranges up to number of point loads on beam) + V1 – V2 + 0.5 * (w1 + w2) * 0.0,
or -(RA + RB) = (-90.0 + 0.0 -90.0),
Solving gives, RA = 100.0 kN.
End support reactions are: Vertical Reactions, RA = 100.0 kN, and RB = 80.0 kN
Moment Reactions are MA = -100.0 kNm, and MB = -80.0 kNm
Shear loads near to supports are -50.0 kN, and 40.0 kN
Central span shear and moment nodes are : [ 50. 0. -40.]kN and [-100. 0. 0.]kNm
Shear and Bending Moment diagrams are drawn as shown below.
(i)
Maximum Shear Force is Vmax = 50 kN
(ii)
Maximum positive Bending Moment, Mmax = 0.0 kNm
at x = 0.0 m from the left end.
Minimum Bending Moment (-ve), Mmin = -100.00 kNm, at x = 2.0 m from the left end.
Absolute Maximum B.M. is 100 kN-m at x = 2.0 m.
(iiI)
As the Bending Moment does not go from -ve sign to positive (however, it reaches 0), hence there are no inflexion points.
