Deflection by Integration — the Elastic Curve

How to turn a bending-moment equation into the exact bent shape of a beam — the elastic curve.

Mechanics of MaterialsMechanical Engineering Year 2Free preview
⏱️ About 18 min

A diving board barely flexes under a child but bends dramatically under an adult, even though the load changed only by a factor of three. Why does the same board, same material, same span, move so differently? The answer is the elastic curve — the mathematical shape a beam takes when it bends. Once you can write that curve, you can predict the deflection and slope at every point along a beam before it is ever built. The tool that produces it is surprisingly direct: integrate the bending moment twice, apply the support conditions, and read off the answer.

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The big idea: The deflected shape of a beam, v(x), is governed by the moment-curvature equation EI·v'' = M(x), where E is Young's modulus, I is the second moment of area, and M(x) is the bending-moment equation. Integrating twice introduces two constants of integration, which are fixed by the boundary conditions at the supports (for example, zero deflection and zero slope at a fixed end). The result is the elastic curve v(x), from which slope and maximum deflection follow directly.
🎯 By the end, you'll be able to
  • State the moment-curvature equation EI·v'' = M(x) and define each symbol
  • Apply boundary conditions to solve for the constants of integration
  • Carry out a full double integration for a cantilever and a simply supported beam
  • Reproduce the classic results PL³/(3EI) and 5wL⁴/(384EI)
📎 Helpful to know first

From bending moment to the elastic curve

When a beam bends, every point on its longitudinal axis moves sideways by a small amount v(x), measured perpendicular to the original (undeflected) axis. The graph of v(x) is called the elastic curve. For small deflections in the elastic range, the curvature of this curve is proportional to the bending moment. That single proportionality gives the governing equation of beam deflection, the moment-curvature equation:

\[ E\,I\,\dfrac{d^{2}v}{dx^{2}} = M(x) \]
Moment-curvature equation. E is Young's modulus, I the second moment of area (so EI is the flexural rigidity), and M(x) the bending-moment equation found from a free-body diagram.
🔑 Boundary conditions fix the constants of integration

Integrating EI·v'' = M(x) twice produces v(x) plus two unknown constants. Those constants are not arbitrary — they are pinned down by what the supports physically enforce:

  • Fixed (built-in) end: neither translation nor rotation is allowed, so v = 0 and v' = θ = 0 — two conditions from a single support.
  • Pinned or roller support: translation is prevented but rotation is free, so only v = 0 — one condition per support.
  • Free end: no condition is imposed on v itself (the moment and shear are zero there instead).

Count the conditions: you always need exactly two, and a beam's supports supply them.

A cantilever beam fixed at the left end: the undeflected axis (dashed) and the deflected elastic curve sagging under a downward end load P. The fixed end carries the boundary conditions v = 0 and theta = 0; the free end shows the maximum deflection PL cubed over 3EI. P Fixed end v = 0, θ = 0 (two boundary conditions) v_max = PL³ / (3EI) free end span L
The elastic curve of a tip-loaded cantilever leaves the wall horizontally (zero slope) and reaches its maximum deflection at the free end.

Worked double integration: cantilever with an end load

Consider a cantilever of span L, fixed at the left end (x = 0), carrying a downward point load P at the free end (x = L). Measuring x from the fixed end, the bending moment is largest at the wall and falls to zero at the tip:

M(x) = −P(L − x)

Now integrate twice. Each integration brings in one constant, which we kill with a boundary condition from the fixed end.

📝 Worked example: Derive the elastic curve, the tip slope, and the maximum deflection for a cantilever of span L with a downward end load P at the free end (fixed end at x = 0). Confirm the classic result v_max = PL³/(3EI).
  1. Moment equation (x from fixed end): M(x) = −P(L − x) = P(x − L).
  2. First integration: EI·v' = P(x²/2 − Lx) + C1.
  3. Boundary condition at the fixed end — zero slope: v'(0) = 0 gives C1 = 0. So EI·v' = P(x²/2 − Lx).
  4. Second integration: EI·v = P(x³/6 − Lx²/2) + C2.
  5. Boundary condition at the fixed end — zero deflection: v(0) = 0 gives C2 = 0.
  6. Elastic curve: EI·v = P(x³/6 − Lx²/2) = −P·x²(3L − x)/6.
  7. Slope at the tip (x = L): EI·v'(L) = P(L²/2 − L²) = −PL²/2, so θ_tip = PL²/(2EI).
  8. Deflection at the tip (x = L): EI·v(L) = −P·L²(3L − L)/6 = −PL³/3, so v_max = PL³/(3EI) at the free end.
✓ Elastic curve EI·v = −P·x²(3L − x)/6; tip slope θ = PL²/(2EI); maximum deflection v_max = PL³/(3EI) at the free end.

Worked double integration: simply supported beam under a UDL

The same procedure handles a simply supported beam. Take a span L with a uniformly distributed load w (force per unit length). By symmetry each support carries half the load, so the reactions are wL/2. The bending-moment equation is a parabola in x:

M(x) = (wL/2)·x − w·x²/2

This time the constants come from the two support conditions, v(0) = 0 and v(L) = 0, rather than from a single fixed end.

📝 Worked example: Derive the midspan deflection of a simply supported beam of span L carrying a full-span uniformly distributed load w. Confirm the classic result v_max = 5wL⁴/(384EI).
  1. Reactions by symmetry: R_A = R_B = wL/2. Moment equation: M(x) = wLx/2 − wx²/2.
  2. First integration: EI·v' = wLx²/4 − wx³/6 + C1.
  3. Second integration: EI·v = wLx³/12 − wx⁴/24 + C1·x + C2.
  4. Boundary condition v(0) = 0 gives C2 = 0.
  5. Boundary condition v(L) = 0: wL⁴/12 − wL⁴/24 + C1·L = 0, which gives C1 = −wL³/24.
  6. By symmetry the slope is zero at midspan (x = L/2), so the maximum deflection is there.
  7. Substitute x = L/2: EI·v(L/2) = wL⁴/96 − wL⁴/384 − wL⁴/48 = −5wL⁴/384.
  8. Therefore v_max = 5wL⁴/(384EI), occurring at midspan.
✓ Maximum midspan deflection v_max = 5wL⁴/(384EI) for a simply supported beam under a full-span UDL.
⚠️ Deflection explodes with span — doubling L is catastrophic

Look at the powers of L in the results above. A point load drives deflection as L³; a distributed load drives it as L⁴. Doubling the span of a tip-loaded cantilever multiplies its deflection by 2³ = 8. Doubling the span of a uniformly loaded simply supported beam multiplies its deflection by 2⁴ = 16. This is why long beams are governed by deflection, not stress: a beam that is perfectly safe from a stress viewpoint can sag unacceptably just because the span grew. Span is the single most dangerous variable in beam deflection.

🎮 Interactive: deflection calculator LIVE
Predict first: Choose the 'cantilever, end load' preset. Set L = 4 m, load = 10, section height h = 300 mm. Read off v_max. Now slide L up to 8 m — watch v_max jump by roughly 8×. Reset, then push h from 300 to 600 mm and see deflection drop to one-eighth (I grows with h³).

An interactive beam deflection calculator with a preset selector, span/load/height sliders, and a canvas drawing the exaggerated deflected shape with v_max and its location reported.

Vary span, load, and section height across five boundary-condition presets. Toggle 'compare' to overlay a second preset at the same load and see how supports dominate stiffness.
✏️ Practice: A steel cantilever (E = 200 GPa) has span L = 2.5 m and a rectangular cross-section 60 mm wide × 120 mm deep. A downward end load of 6 kN acts at the free end. Using v_max = PL³/(3EI), find the maximum deflection, in mm.
mm
Solution
  1. I = b·h³/12 = 60 × 120³ / 12 = 8.64 × 10⁶ mm⁴.
  2. Use N and mm: P = 6000 N, L = 2500 mm, E = 200 000 N/mm².
  3. v_max = P·L³ / (3·E·I) = 6000 × 2500³ / (3 × 200 000 × 8.64 × 10⁶).
  4. Numerator = 6000 × 1.5625 × 10¹⁰ = 9.375 × 10¹³. Denominator = 5.184 × 10¹².
  5. v_max = 18.1 mm (downward, at the free end).
✏️ Practice: A simply supported steel beam (E = 200 GPa) spans L = 4 m and carries a full-span uniformly distributed load w = 8 kN/m. The rectangular cross-section is 100 mm wide × 200 mm deep. Find the midspan deflection, in mm.
mm
Solution
  1. I = 100 × 200³ / 12 = 6.667 × 10⁷ mm⁴.
  2. Convert w to N/mm: w = 8 kN/m = 8 N/mm. L = 4000 mm.
  3. v_max = 5·w·L⁴ / (384·E·I) = 5 × 8 × 4000⁴ / (384 × 200 000 × 6.667 × 10⁷).
  4. Numerator = 1.024 × 10¹⁶. Denominator = 5.12 × 10¹⁵.
  5. v_max = 2.00 mm (downward, at midspan).

Check your understanding

1. Integrating EI·v'' = M(x) twice produces the elastic curve plus how many constants of integration?
Two integrations introduce two constants, fixed by the support boundary conditions (e.g. v = 0 and θ = 0 at a fixed end).
2. If the span of a tip-loaded cantilever is doubled with the load and section unchanged, the deflection becomes:
v_max = PL³/(3EI) scales with the cube of span. Doubling L multiplies deflection by 2³ = 8.
3. At a fixed (built-in) end of a beam, the boundary conditions are:
A fixed end prevents both translation and rotation, supplying two conditions: v = 0 and v' = θ = 0.
🔎 Related lessons
Deflection by Superposition Deflection by the Moment-Area Method · soon
✅ Key takeaways
  • Moment-curvature equation: EI·v'' = M(x) governs the elastic curve v(x).
  • Two integrations introduce two constants, fixed by the support boundary conditions.
  • Cantilever, end load: v_max = PL³/(3EI) at the free end; tip slope PL²/(2EI).
  • Simply supported, UDL: v_max = 5wL⁴/(384EI) at midspan.
  • Deflection scales with L³ (point load) or L⁴ (UDL) — span is the dominant variable.
➡️ Integration gives the exact answer but the algebra is heavy for every new load case. The next lesson shows that most real load cases are just sums of a few standard cases whose answers you can look up — that is superposition.
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