Deflection by Superposition
Why solve the differential equation again and again when most loads are just sums of a few standard cases?
A floor beam in a building rarely carries a single tidy load. It carries its own weight (a uniform load), the weight of the slab above (another uniform load), a partition wall somewhere along its length (a line load), and maybe a column transfer point load from above. Solving the differential equation for that messy combination from scratch would be miserable. The good news: because the beam responds linearly, the deflection under all those loads together is simply the <em>sum</em> of the deflections each load would cause on its own. You look up half a dozen standard results and add them. That is superposition, and it is how engineers actually compute deflection in practice.
The principle of superposition
Linear elasticity is a gift. Because stress, strain, and deflection are all directly proportional to the applied load (double the load, double the response), the response to a combination of loads is the plain algebraic sum of the responses to each load on its own. This is the principle of superposition. It lets us decompose a complicated load case into simple pieces, look up each piece in a table, and add the results.
Superposition is valid only when the response is linear in the load. It fails if:
- The material is stressed beyond its proportional limit (non-linear stress-strain).
- The deflections are large enough to change the geometry of the loading (a cable or a very slender beam sagging so far that the load direction relative to the member shifts).
- The support conditions themselves change with load.
For ordinary beams in the elastic range with small deflections — the overwhelming majority of introductory problems — superposition is exact.
A table of standard cases
The four cases below cover most hand calculations. Each was derived once, by integration, in the previous lesson; from now on we simply cite them. (δ is the maximum deflection, θ a support or tip slope, in the directions you would expect from the load.)
- Cantilever, end load P at the free end: δ = PL³/(3EI), θ_tip = PL²/(2EI).
- Cantilever, full-span UDL w: δ = wL⁴/(8EI), θ_tip = wL³/(6EI).
- Simply supported, central point load P: δ = PL³/(48EI).
- Simply supported, full-span UDL w: δ = 5wL⁴/(384EI).
For non-central point loads, asymmetric supports, overhangs, or couples, fuller tables (in any standard mechanics-of-materials reference) give the corresponding expressions. The method is identical: look up, then add.
Notice that every standard-case deflection has EI in the denominator. Flexural rigidity EI is a product: E comes from the material, I from the shape. To halve a deflection you can either double E (switch from timber at ~10 GPa to steel at ~200 GPa — a twenty-fold jump, expensive and often overkill) or double I. For a rectangular section I = bh³/12, so I grows with the cube of the depth. Increasing the section depth by just 26% doubles I. Geometry is almost always the cheaper lever: a deeper beam of the same material beats a stronger material of the same shape. This is why floor joists are tall and thin, not square.
- Flexural rigidity: I = 120 × 240³ / 12 = 1.3824 × 10⁸ mm⁴. EI = 200 000 × 1.3824 × 10⁸ = 2.765 × 10¹³ N·mm².
- Deflection from the UDL alone (w = 6 N/mm, L = 5000 mm): δ_w = 5wL⁴/(384EI) = 5 × 6 × 5000⁴ / (384 × EI) = 1.766 mm.
- Deflection from the point load alone (P = 10 000 N): δ_P = PL³/(48EI) = 10 000 × 5000³ / (48 × EI) = 0.942 mm.
- By superposition the total midspan deflection is the sum: δ_total = δ_w + δ_P.
- δ_total = 1.766 + 0.942 = 2.71 mm (downward).
- Work in kN and m so the result is in metres: w = 10 kN/m, L = 4 m, EI = 2000 kN·m².
- δ = 5 × 10 × 4⁴ / (384 × 2000) = 5 × 10 × 256 / 768 000 = 12 800 / 768 000.
- δ = 0.01667 m = 16.7 mm.
- Superposition: total deflection = sum of each load's individual contribution.
- δ_total = δ_point + δ_udl = 12 + 8 = 20.0 mm.
- v_max is proportional to L⁴.
- Factor = 2⁴ = 16. Doubling the span multiplies a UDL deflection by sixteen.
Check your understanding
- For linear, small-deflection beams, the response to combined loads is the sum of each load's response.
- A short table of standard cases (cantilever/simply-supported × point/UDL) solves most problems.
- Simply supported + UDL + central point load: δ_total = 5wL⁴/(384EI) + PL³/(48EI).
- EI is a product — increasing section depth (I ∝ h³) is usually the cheapest way to cut deflection.
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