Principle of superposition with example

Principle of superposition: for a linear elastic structure (small deformations, Hooke's law valid), the total effect (reaction, shear force, bending moment, stress or deflection) produced by several loads acting together is equal to the algebraic sum of the effects produced by each load acting separately.

Explanation

• Each load is applied alone on the original, undeformed structure and its effect is found.
• The individual effects are then added, taking care of signs.
• It works because the equations of equilibrium and the stress–strain relation are linear: doubling a load doubles every effect, and the effect of one load does not change the effect of another.

Example

A simply supported beam of span $L = 6$ m carries a UDL $w = 2$ kN/m over the full span and a point load $P = 6$ kN at mid-span.

        P = 6 kN
 w = 2 kN/m  |
 ↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓
 A▲------- 3 m -------●------- 3 m -------▲B

Case 1 – UDL alone: $M_{mid,1} = \dfrac{wL^2}{8} = \dfrac{2\times6^2}{8} = 9$ kN·m.

Case 2 – point load alone: $M_{mid,2} = \dfrac{PL}{4} = \dfrac{6\times6}{4} = 9$ kN·m.

Superposing: $M_{mid} = 9 + 9 = 18$ kN·m. Similarly the reactions are $R_A = R_B = 6 + 3 = 9$ kN, which equals the sum of the reactions of the two cases ($6$ kN from the UDL and $3$ kN from the point load).

Hence a complicated loading is split into simple standard cases whose results are tabulated, and the answers are added.