Superposition for deflections and its necessity

Use of superposition to find deflection

For a linear elastic beam, the deflection under several loads is the sum of the deflections due to each load acting alone. The loading is split into simple cases for which standard deflection formulas are known, and the results are added algebraically.

Example: A cantilever of length $L$ carries a point load $P$ at the free end and a UDL $w$ over the whole length. Find the free end deflection.

1. Case 1, point load alone: $\delta_1 = \dfrac{PL^3}{3EI}$
2. Case 2, UDL alone: $\delta_2 = \dfrac{wL^4}{8EI}$
3. Total deflection:

$$\delta = \delta_1 + \delta_2 = \frac{PL^3}{3EI} + \frac{wL^4}{8EI}$$

For example, with $P = 10$ kN, $w = 5$ kN/m, $L = 4$ m, $EI = 20000$ kNm$^2$: $\delta = \dfrac{10\times 64}{3\times 20000} + \dfrac{5\times 256}{8\times 20000} = 0.01067 + 0.008 = 0.0187$ m (18.7 mm).

This works only if the material is linearly elastic and deflections are small.

Why deflection must be determined in design

• Serviceability: Excessive deflection causes cracking of plaster and partitions, damage to finishes and fixtures, and discomfort to users. Codes limit it, e.g. $L/250$ for total deflection and $L/350$ for live load.
• Appearance and function: Sagging beams look unsafe and may pond water on roofs or disturb machinery alignment.
• Indeterminate analysis: Deflections (compatibility) are needed to find the redundant forces.
• Vibration control: Stiffness (related to deflection) controls vibration and natural frequency.
• Check of analysis: Computed deflection can be compared with field measurement.