Engr Help

Beam Deflection by Superposition

In a Nut Shell: Since the differential equation (shown below) governing beam deflection

is linear, solutions may be combined for different loading , M(x), and boundary conditions

by superimposing individual solutions for each loading and for each boundary condition.

d²y/dx² = M(x) / EI

Strategy: Use a table of solutions for deflection, y(x), and slope, dy(x)/dx, under various

types of loading and boundary conditions of beams. Then superimpose them to obtain the

overall resulting deflection and slope for the total loads applied to the beam.

To illustrate, consider the case of a cantilever beam of length, L, and flexural rigidity, EI,

subjected to the combination of a uniformly distributed load, q, (lb/ft) and also a concentrated

load, P, at the end of the beam, AB, as shown in the top figure below.

Instead of the combined loading (top figure) replace it with beam (1) with only the

concentrated load at the end plus beam (2) with only the distributed load q. From

tables

y₁(x) = − P x²/6EI ( 3L − x) and y₂(x) = (q/24EI) ( x⁴ + 4Lx³ − 6L²x²)

The equation o the elastic curve is then

y_total(x) = − P x²/6EI ( 3L − x) + (q/24EI) ( x⁴ + 4Lx³ − 6L²x²)

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