Deflection of beams: double integration and Macaulay's method

Why it matters

In a positioning system, a beam that is strong enough can still be useless if it bends too much: a gantry that sags under the carriage, a long robot link whose tip droops, or a sensor arm that shifts its reading. Deflection limits, not stress limits, often set the size of members in machines. The double integration and Macaulay methods give the full deflected shape, so you can find the deflection and slope anywhere, not just at the standard points in a handbook table.

Key ideas

The elastic curve. Under load, the neutral axis of a beam bends into the elastic curve y(x). For small slopes the curvature is d²y/dx², and from bending theory 1/R = M/(E·I). This gives the governing equation E·I·d²y/dx² = M(x).

Assumptions. Linear-elastic, homogeneous material; small deflections and slopes (so curvature ≈ y″); plane sections remain plane; shear deformation neglected (fine for slender beams, span/depth above about 10).

Sign convention used here. x measured from the left end, y positive upward, M positive when sagging. A sagging moment gives positive curvature (concave upward). Downward deflections then come out negative.

Double integration.

1. Write M(x) for the beam (one expression per segment if the loading changes).
2. Integrate once: E·I·dy/dx = ∫M dx + C₁ (slope).
3. Integrate again: E·I·y = ∫∫M dx dx + C₁·x + C₂ (deflection).
4. Find the constants from boundary conditions:
   • simple support or roller: y = 0;
   • fixed end: y = 0 and dy/dx = 0;
   • free end: no displacement condition (M = 0 and V = 0 are already in M(x));
   • axis of symmetry: dy/dx = 0.
5. Maximum deflection occurs where dy/dx = 0 (or at a free end). With n segments there are 2n constants, and you need continuity conditions (equal slope and deflection) at each junction. That is tedious, which is why Macaulay's method exists.

Macaulay's method (singularity functions). Write one moment expression valid for the whole beam, taking x from the left end, using brackets ⟨x − a⟩ that are zero when x < a and equal to (x − a) when x ≥ a. Rules:

• integrate a bracket as a whole: ∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹/(n + 1);
• never expand a bracket;
• a UDL that stops before the right end is handled by continuing it to the end and adding an equal and opposite UDL from where it stops;
• an applied couple M₀ at x = a enters as M₀·⟨x − a⟩⁰. There are then only two constants, found from the support conditions.

Standard results (for checking).

• Cantilever, end load P: δ = P·L³/(3EI), slope P·L²/(2EI).
• Cantilever, full UDL w: δ = w·L⁴/(8EI), slope w·L³/(6EI).
• Simply supported, central load P: δ = P·L³/(48EI), end slope P·L²/(16EI).
• Simply supported, full UDL w: δ = 5·w·L⁴/(384EI), end slope w·L³/(24EI).
• Simply supported, triangular load with peak w: δ_max = 0.00652·w·L⁴/(EI), at 0.519L from the zero end.

Flexural rigidity. EI combines material and section. Deflection is inversely proportional to EI and grows with L³ (point load) or L⁴ (UDL): doubling span multiplies UDL deflection by 16.

Formulas

E·I·d²y/dx² = M(x)

• E: Young's modulus (Pa); I: second moment of area (m⁴); y: deflection (m); x: position (m); M: bending moment (N·m).

θ = dy/dx

• Slope (rad).

∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹ / (n + 1)

• Macaulay bracket integration; ⟨x − a⟩ = 0 for x < a.

δ = P·L³/(3EI), δ = w·L⁴/(8EI) (cantilever) δ = P·L³/(48EI), δ = 5·w·L⁴/(384EI) (simply supported)

• P: point load (N); w: load per unit length (N/m); L: span (m).

Worked examples

Example 1 (standard): conveyor beam by double integration. Given: a simply supported steel beam, span 6 m, carries a UDL of 2 kN/m. E = 200 GPa, I = 8 × 10⁶ mm⁴, so EI = 200 × 10⁹ × 8 × 10⁻⁶ = 1600 kN·m². Find the end slope and mid-span deflection.

1. Reactions: 6 kN each. M(x) = 6x − 2x²/2 = 6x − x² (kN·m, x in m).
2. EI·y″ = 6x − x².
3. EI·y′ = 3x² − x³/3 + C₁.
4. EI·y = x³ − x⁴/12 + C₁·x + C₂.
5. y(0) = 0 gives C₂ = 0. y(6) = 0 gives 216 − 108 + 6C₁ = 0, so C₁ = −18 kN·m².
6. End slope: y′(0) = −18/1600 = −0.01125 rad.
7. Mid-span: EI·y(3) = 27 − 6.75 − 54 = −33.75 kN·m³, so y = −33.75/1600 = −0.0211 m.
8. Check: 5wL⁴/(384EI) = 5 × 2 × 6⁴/(384 × 1600) = 0.0211 m. Answer: mid-span deflection ≈ 21.1 mm downward; end slope ≈ 0.0113 rad (0.64°).

Example 2 (GATE level): two point loads by Macaulay's method. Given: a simply supported beam AB of span 4 m carries 10 kN at 1 m and 5 kN at 3 m from A. EI = 2000 kN·m². Find the deflection under each load and the maximum deflection.

1. Reactions: R_A × 4 = 10 × 3 + 5 × 1, so R_A = 8.75 kN; R_B = 6.25 kN.
2. EI·y″ = 8.75x − 10⟨x − 1⟩ − 5⟨x − 3⟩.
3. EI·y′ = 4.375x² − 5⟨x − 1⟩² − 2.5⟨x − 3⟩² + C₁.
4. EI·y = 1.4583x³ − 1.6667⟨x − 1⟩³ − 0.8333⟨x − 3⟩³ + C₁·x + C₂.
5. y(0) = 0: C₂ = 0. y(4) = 0: 93.333 − 45 − 0.833 + 4C₁ = 0, so C₁ = −11.875.
6. At x = 1 m: EI·y = 1.458 − 11.875 = −10.417, so y = −5.21 mm.
7. At x = 3 m: EI·y = 39.375 − 13.333 − 35.625 = −9.583, so y = −4.79 mm.
8. Maximum: y′ = 0 in 1 < x < 3: 4.375x² − 5(x − 1)² − 11.875 = 0, which simplifies to x² − 16x + 27 = 0, so x = 1.917 m.
9. EI·y(1.917) = −13.78, so y_max = −6.89 mm. Answer: 5.21 mm under the 10 kN load, 4.79 mm under the 5 kN load, maximum 6.89 mm at 1.92 m from A (all downward). Note the maximum is close to mid-span even though the loads are unequal, which is typical.

Common mistakes

• Expanding Macaulay brackets, or including a bracket term at an x where it should be zero.
• Writing M(x) from the right-hand side in Macaulay's method; take x from the left and include only loads to the left.
• Using the wrong boundary conditions: a cantilever has y = 0 and y′ = 0 at the fixed end, not at the free end.
• Mixing units in EI: 200 GPa × mm⁴ gives N·mm²; convert consistently.
• Assuming maximum deflection is always at mid-span or under the largest load; find where y′ = 0.
• Forgetting that a UDL ending before the support must be extended and cancelled.

For GATE ME

Expect standard cantilever and simply supported deflection and slope results, ratios (how δ changes with L, d or E), Macaulay problems with one or two point loads, and boundary-condition reasoning. Practise the four standard formulas until automatic, and use them to check integrated answers.

Quick check

1. Cantilever, L = 3 m, UDL 5 kN/m, EI = 15 000 kN·m². Find the tip deflection.
2. Simply supported, L = 4 m, central 10 kN, EI = 20 000 kN·m². Find δ_max.
3. Which boundary conditions apply at a fixed end?
4. By what factor does the deflection of a simply supported beam under UDL change if the span doubles?
5. In Macaulay's method, what is the value of ⟨x − 2⟩ at x = 1.5?

Answers: 1. 5 × 81/(8 × 15 000) = 3.4 mm. 2. 10 × 64/(48 × 20 000) = 0.67 mm. 3. y = 0 and dy/dx = 0. 4. 16 times. 5. Zero.