Deflection of beams

A beam can pass every bending-stress check and still fail its job — a shelf that sags visibly, a floor beam that bounces, a crane girder whose tip droops enough to jam the rails it rides on. Strength tells you a beam will not break; deflection tells you whether it stays usably straight. This article picks up where the elastic curve equation $EI\,d^2y/dx^2 = M(x)$ left off and asks how to actually solve it: by double integration, region by region, using boundary and continuity conditions to pin down the constants, and by superposition, adding tabulated standard-case results for beams carrying several loads at once. Both routes lead to the same handful of results — cantilever and simply-supported beams under point or distributed loads — that cover most design checks. We work through both methods on the same numbers used in the shorter companion article, so nothing here contradicts what you have already seen; it only fills in the reasoning and adds a second, richer worked example.

The core insight

Solving the elastic curve equation $EI\,d^2y/dx^2 = M(x)$ by double integration, or by summing tabulated standard-case results (superposition), gives a beam's deflection and slope under any combination of loads.

The core idea

Deflection analysis rests on one governing equation and two ways to use it.

Double integration starts from $EI\,d^2y/dx^2 = M(x)$ and integrates twice:

EI\frac{dy}{dx} = \int M(x)\,dx + C_1, \qquad EI\,y(x) = \iint M(x)\,dx\,dx + C_1x + C_2

where: $E$ is Young's modulus, $I$ is the second moment of area of the cross-section (so $EI$ is the flexural rigidity), $M(x)$ is the bending-moment expression, $y(x)$ is deflection, and $C_1, C_2$ are integration constants fixed by boundary conditions — $y=0$ at a pin or roller, and both $y=0$ and $dy/dx=0$ at a fixed support. If $M(x)$ changes expression partway along the beam (a point load or an internal support does this), the beam must be split into regions, each with its own pair of constants, tied together by requiring $y$ and $dy/dx$ to match at every region boundary.

Superposition sidesteps the algebra for multi-load beams. Because $M(x)$ is linear in the applied loads, so is $y(x)$ : the deflection from several loads acting together equals the algebraic sum of the deflections each load would cause acting alone. This only works because every load sees the same linear elastic beam — nothing changes shape or stiffness between the "separate" and "combined" cases. Four standard results, for uniform $EI$ , cover the bulk of design problems:

\text{cantilever, tip load } P: \quad y_{max} = \frac{PL^3}{3EI}, \qquad \theta_{max} = \frac{PL^2}{2EI}

\text{cantilever, UDL } w: \quad y_{max} = \frac{wL^4}{8EI}, \qquad \theta_{max} = \frac{wL^3}{6EI}

\text{simply supported, midspan point load } P: \quad y_{max} = \frac{PL^3}{48EI}

\text{simply supported, UDL } w: \quad y_{max} = \frac{5wL^4}{384EI}

where: $L$ is span length, $P$ is a concentrated load, $w$ is load per unit length, and every $y_{max}$ occurs where the figure below marks it — at the free end for a cantilever, at midspan for a simply supported beam. Since $EI$ sits in every denominator, a stiffer material or a deeper section directly shrinks deflection for the same load, which is exactly why deflection limits (commonly $L/360$ for floors) sometimes force a larger section than the bending-stress check alone would demand.

Figure

A cantilever beam fixed at the left carrying a point load at the free end, drawn straight and drawn deflected, with the maximum deflection y max and the tip slope theta max both marked at the free end

Common traps

Summing two different $y_{max}$ values from superposition when they occur at different locations — you may only add deflections evaluated at the same point, using the general $y(x)$ expressions if the maxima do not coincide.
Forgetting the region split in double integration when $M(x)$ changes form partway along the beam — one expression for the whole span silently gives a wrong curve.
Sign errors in $M(x)$ carried straight through both integrations, corrupting the final deflection without any warning.
Mixing units of $EI$ — for example $I$ in mm⁴ with $E$ in GPa without converting both to consistent SI base units before substituting.

Derivation ACantilever tip deflection by double integration

The goal is to derive the standard cantilever-with-tip-load result $y_{max} = PL^3/(3EI)$ directly from the governing equation $EI\,d^2y/dx^2 = M(x)$ , so that the tabulated formula used throughout this article is shown to be a consequence of the same double-integration method used for harder, multi-region beams, not a separately memorised fact.

Step 1. Set up coordinates and the moment expression. Measure $x$ from the fixed end, with the beam of length $L$ fixed at $x=0$ and carrying a point load $P$ downward at the free end $x=L$ . Cutting the beam at any section $x$ and considering the free-end side, the internal bending moment is $M(x) = -P(L-x)$ , negative because the load bends the cantilever concave-down in this sign convention.

Step 2. Integrate once for the slope. $EI\,\dfrac{dy}{dx} = \displaystyle\int -P(L-x)\,dx = -P\!\left(Lx - \dfrac{x^2}{2}\right) + C_1$ . The fixed end cannot rotate, so the boundary condition $dy/dx = 0$ at $x=0$ gives $C_1 = 0$ immediately — no free constant survives at a built-in support.

Step 3. Integrate again for the deflection. $EI\,y(x) = \displaystyle\int -P\!\left(Lx-\dfrac{x^2}{2}\right)dx = -P\!\left(\dfrac{Lx^2}{2} - \dfrac{x^3}{6}\right) + C_2$ . The fixed end also cannot deflect, so $y=0$ at $x=0$ gives $C_2 = 0$ as well. Collecting terms, $EI\,y(x) = -\dfrac{Px^2}{2}\!\left(L - \dfrac{x}{3}\right)$ .

Step 4. Evaluate at the free end and take magnitudes. Setting $x=L$ : $EI\,y(L) = -\dfrac{PL^2}{2}\!\left(L-\dfrac{L}{3}\right) = -\dfrac{PL^2}{2}\cdot\dfrac{2L}{3} = -\dfrac{PL^3}{3}$ , so $y_{max} = \dfrac{PL^3}{3EI}$ (magnitude, taking the free end as sagging downward). Substituting $x=L$ into the Step 2 slope expression gives $EI\,\theta_{max} = -\dfrac{PL^2}{2}$ , i.e. $\theta_{max} = \dfrac{PL^2}{2EI}$ — both standard formulas fall out of the same two integrations, with the fixed-end conditions doing all the work of pinning down the constants.

Worked example

A simply supported timber beam, span $L = 4\ \text{m}$ , flexural rigidity $EI = 3\times10^{6}\ \text{N}\,\cdot\,\text{m}^2$ , carries a uniformly distributed load $w = 2\ \text{kN/m}$ over its full length together with a concentrated load $P = 6\ \text{kN}$ at midspan. Find the total midspan deflection, and check it against a serviceability limit of $L/360$ .

Step 1. Deflection from the distributed load alone, using the standard UDL formula: $y_1 = \dfrac{5wL^4}{384EI} = \dfrac{5(2000)(4)^4}{384(3\times10^6)} = \dfrac{5(2000)(256)}{1.152\times10^9} = 2.22\times10^{-3}\ \text{m} = 2.22\ \text{mm}$ .

Step 2. Deflection from the midspan point load alone: $y_2 = \dfrac{PL^3}{48EI} = \dfrac{(6000)(4)^3}{48(3\times10^6)} = \dfrac{384{,}000}{1.44\times10^8} = 2.67\times10^{-3}\ \text{m} = 2.67\ \text{mm}$ .

Step 3. Superpose. Both individual maxima occur at the same location — midspan — so they can be added directly: $y_{total} = y_1 + y_2 = 2.22 + 2.67 = \mathbf{4.89\ mm}$ .

Step 4. Check the deflection limit. $L/360 = 4000\ \text{mm}/360 = 11.1\ \text{mm}$ . Since $4.89\ \text{mm} < 11.1\ \text{mm}$ , the beam passes the serviceability check comfortably — it has roughly $2.3\times$ margin before the deflection limit governs.

Competitive stretch

If the same beam instead used a stiffer section with $EI = 1\times10^{6}\ \text{N}\,\cdot\,\text{m}^2$ (three times smaller), every $y$ above scales up by the same factor $3\times$ (since $y \propto 1/EI$ ), giving $y_{total} \approx 3 \times 4.89 = 14.7\ \text{mm}$ — now exceeding the $11.1\ \text{mm}$ limit even though the bending-stress check might still pass, illustrating why deflection, not just strength, can govern the final section choice.

Ask the tutor: if setting up the region-by-region integration for a beam with a mid-span point load, or matching continuity conditions between regions, feels uncertain, ask the AI tutor to walk through the setup step by step.

Why do we need continuity conditions in double integration, but not for a simple cantilever with a single tip load?

A single cantilever tip load produces one bending-moment expression valid over the entire span, so one pair of constants, fixed by the two conditions at the built-in end, is enough — that is why the derivation above needed no continuity check. The moment changes to a different algebraic expression as soon as a point load, an internal support, or a change in distributed load appears partway along the beam. Each region then needs its own pair of constants, and continuity conditions — matching $y$ and $dy/dx$ at the boundary between regions — supply the extra equations needed to solve for all of them.

Can I always use superposition instead of double integration to save time?

Only when the beam configuration is a sum of cases you already have tabulated formulas for, with the same support conditions and the same $EI$ throughout. If the loading is unusual, if the beam has a support arrangement that is not one of the standard cases, or if you need the deflection at a point other than the tabulated maximum, superposition either fails outright or requires the full general $y(x)$ expression for each case rather than just $y_{max}$ — at that point double integration is the more reliable route, even if slower.

Why does deflection sometimes govern the design when the bending-stress check already passed?

Strength and stiffness are two independent checks on the same beam. The bending-stress check asks whether the material yields or fractures under the peak bending moment, which depends on the section modulus. Deflection depends instead on the full flexural rigidity $EI$ and the load distribution along the whole span, and design codes cap it (commonly around $L/360$ ) for serviceability — visible sag, cracking of finishes, or vibration are unacceptable even in a beam that is structurally safe. A shallow, wide section can satisfy strength with a small section modulus yet still deflect too much, because $I$ , which enters deflection, grows more slowly with width than with depth.

Deflection of beams

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads