While superposition can be used for many problems in engineering, they are particularly
useful for beam deflections. Most beam configuration and loading can be split
into simpler beams and loading. For example, the beam at the left with a distributed
load and a point moment load can be split into two beams. One with the distributed
load and one with the moment load.
After the beams are simplified, the deflections of the simplified beams
are still needed. To assist engineers,
the deflection of simple beams are commonly listed in handbooks. This eBook has
many basic beams listed in the
Beam Equations appendix. The solutions for these simple beams
can be derived by
integrating the moment equation or
The deflection equation of the complex beam is the addition of the
two simpler beam equations, or
δ(x) = δ1(x)
There are a number of assumptions when using superposition. It is assumed that
the beam undergoes linear deflection, all deflections are small,
elastic material properties, no shear deflection (i.e. no short thick beams,
and normal boundary conditions.
Superposition for Beam Deflection
One of the best way to understand the principle of superposition for beam deflection
is an example. The same beam as above will be used but with numerical values
for the deflection at the beam center.
The beam is split into two simpler beams; beam 1) with a distributed
load and beam 2) with a point moment load. The deflection for beam 1) with a
distributed load over part of the beam is given in the appendix as
For the center deflections, x = 5 ft, L = 10 ft, and a = 6 ft, giving,
deflection for beam 2) at the beam's center with a point moment load is given
Summing the two solutions give
v = v1 + v2 = -51,325/EI
While this example only has two sub-beam sections, there could be many more,
but the process will be the same.