# Determine the deflection of beams due to combined loads

If you read our article deflection of beams with a single load, you’ll know about deflection of a beam.  Now we’re going to investigate how deflection affects a beam with combined loads.

## Calculating deflections on a beam with combined loads.

Deflections from combined loading on a beam can be calculated by treating the load as a combination of simple loading.  This is called the method of superposition.  There’s a limitation on this method. Each independent load should not cause any appreciable change in the original length or shape of the beam.

## Principle of Superposition

Based on the principle of superposition, a given beam and its loadings can be split into simpler beams and loadings. In the image below, the beam with the distributed load and the point load can be split into two beams. Basically, having one beam with the distributed load and the other with the point load.

Generally for a linearly elastic structure, the effect of several loads acting on a member is equal to the summation of the loads acting separately.

## Assumptions for using the principle of superposition.

There are several assumptions when we use the principle of superposition. We should assume 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).

Let’s look at an example to see what that means in practice.  A 2m load cantilever beam is carrying a load of 20 kN at the free end, and 30 kN 1m from the free end.  We’re going to find the slope and deflection at the free end.  Take E = 200 GN/m2 and I = 150 10-6 m4.

Due to the method of superposition, deflection at point B is the combined effect of the deflection due to 20 kN at B and 30 kN at the centre.

Where 1 and y1 be the slope and deflection at point B due to 20 kN load, and 2 and y2 be the slope and deflection at point B due to the 30 kN load at the centre of the beam.

## Calculate the deflection

E = 200 GN/m2 = 200 109 N/m2

I = 150 10-6 m4.

Load at the centre of beam = 30 kN = 30000 N

Load at the free end of beam = 20 kN = 20000 N

Length of beam = 2 m.

Now, y1 = PL33EI = 20103 233 200 109 150 10-6 = 1.775 10-3 m

Deflection at B due to the 30 kN load at the centre = y2 = Px26EI (3l – x) = 30000 126200 109 150 10-6(3 2 – 1)

y2 = 0. 883 10-3m

So, deflection due to both loads = y1 + y2 = (1.775 + 0.883) 10-3 m = 2.6110-3 m

Slope at point B due to the 20 kN load = 1 = Pl22EI = 20000 222 200 109 150 10-6 = 0.133 rad

Slope at B due to the 30 kN load = 2=Px22EI = 30000 122 200 109 150 10-6 = 0.05 rad

Finally, slope at B due to the combined effect of both loads = B = 1 + 2 = 0.133 + 0.05 = 0.1833 rad

We’re going to continue our series on deflection and loaded beams with even more articles on various calculations concerning various loads on a beam so stay tuned.

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