Continuous Beam – General case

Having determined the movements at the supports by, say, moment distribution, it is necessary to calculate the moments in the spans and also the shear forces on the beam. For a uniformly distributed load, the equations for the shears and the maximum span moments can be derived from the following analysis.

Using the sign convention of figure and taking moments about support B

therefore

(3.1)

and

(3.2)

Maximum span moment occurs at zero shear, and distance to zero shear

(3.3)

therefore

(3.4)

The points of contraflexure occur at , that is

where x is the distance from support A. Taking the roots of this equation gives

so that

(3.5)

and

(3.6)

A similar analysis can be applied to beams that do not support a uniformly distributed load. In manual calculations it is usually not considered necessary to calculate the distances a1, a2 and a3 which locate the points of contraflexure and maximum moment – a sketch of the bending moment is often adequate – but if a computer is performing the calculations these distances may as well be determined also.

The continuous beam shown in figure has a constant cross-section and supports a uniformly distributed dead load including its self-weight of Gk = 25kN/m and an imposed load Qk = 10 kN/m.

The critical loading arrangements for the ultimate limit state are shown in figure 3.6 where the heavy line indicates the region of maximum moments, sagging or possible hogging. Table is the moment distribution carried out for the first loading arrangement: similar calculations would be required for each of the remaining load areas. It should be noted that the reduced stiffness of ¾ I/L has been used for the end spans.

The shearing forces, the maximum span bending moments, and their positions along the beam, can be calculated using the formulae previously derived. Thus for the first loading arrangement and span AB, using the sign convention

Shear

Maximum moment, span

Where w = 306/6.0 = 51 kN/m. Therefore

Distance from