Deflection of beams: double integration and Macaulay's method

The elastic curve EI·y″ = M, boundary and continuity conditions, double integration, Macaulay's singularity brackets and the standard deflection results, with a cantilever and a two-load beam solved.

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Why it matters

A beam can be strong enough and still be useless if it sags too much: a chassis that flexes misaligns the driveline, a gearbox shaft that deflects ruins gear mesh, and a leaf spring is designed for a specific deflection. Deflection limits are usually set by function, not strength. The double integration method gives the full elastic curve; Macaulay's method makes it practical for beams with several loads.

Key ideas

Elastic curve. Under load the neutral axis bends into the elastic (deflection) curve y(x). For small slopes, its curvature is 1/R ≈ d²y/dx². From the flexure equation M/I = E/R, the governing equation is E·I·d²y/dx² = M(x). E·I is the flexural rigidity.

Sign convention. With y measured upward and sagging moment positive, E·I·y″ = M. A sagging beam then has negative y (downward deflection). Many Indian textbooks measure y downward and get positive deflections; either is fine if used consistently.

Assumptions. Linear elastic material, small deflections and slopes (so y′² ≪ 1), plane sections remain plane, and deflection due to shear neglected (fine for slender beams, span/depth above about 10).

Double integration.

  1. Write M(x) for the beam (one expression per segment if the loading changes).
  2. Integrate once: E·I·y′ = ∫M dx + C₁ (slope).
  3. Integrate again: E·I·y = ∬M dx dx + C₁·x + C₂ (deflection).
  4. Find the constants from boundary and continuity conditions.

Boundary conditions.

  • Simple support (pin or roller): y = 0.
  • Fixed end: y = 0 and y′ = 0.
  • Free end: M = 0 and V = 0 (used only when integrating from the load).
  • With several segments, slope and deflection must be continuous at each junction, which gives two extra equations per junction. This is what makes plain double integration tedious.

Macaulay's method. Write one bending-moment expression valid for the whole beam using singularity brackets ⟨x − a⟩, which mean (x − a) when x > a and zero otherwise. Rules:

  • Measure x from one end (usually the left) and keep every term in the form (x − a).
  • Integrate the brackets as a whole: ∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹/(n + 1). Do not expand them.
  • A point load W at a gives −W·⟨x − a⟩; a couple M₀ at a gives ±M₀·⟨x − a⟩⁰; a UDL starting at a gives −w·⟨x − a⟩²/2. If a UDL stops before the end, extend it to the end and add an equal and opposite UDL over the extension.
  • Only two constants of integration appear, from the support conditions; continuity is automatic.

Maximum deflection. It occurs where y′ = 0 (for a simply supported beam, between the supports and not necessarily under the largest load) or at a free end (cantilever).

Standard results (memorise).

  • Cantilever, end load W: y_max = W·L³/(3EI), θ_max = W·L²/(2EI).
  • Cantilever, UDL w: y_max = w·L⁴/(8EI), θ_max = w·L³/(6EI).
  • Simply supported, central load W: y_max = W·L³/(48EI), θ_end = W·L²/(16EI).
  • Simply supported, UDL w: y_max = 5·w·L⁴/(384EI), θ_end = w·L³/(24EI).
  • Simply supported, load W at a from A and b from B: deflection under the load W·a²·b²/(3EIL).

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), sagging positive.

E·I·dy/dx = ∫M dx + C₁, E·I·y = ∬M dx dx + C₁·x + C₂

⟨x − a⟩ⁿ = (x − a)ⁿ for x > a, 0 for x < a; ∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹ / (n + 1)

y = W·x²·(3L − x) / (6EI)

  • Cantilever with end load W, deflection at distance x from the fixed end.

Standard maxima: W·L³/(3EI), w·L⁴/(8EI), W·L³/(48EI), 5·w·L⁴/(384EI), W·a²·b²/(3EIL)

Worked examples

Example 1 (standard): cantilever with UDL by double integration. Given: cantilever of length L = 2 m, fixed at x = 0, UDL w = 4 kN/m, E = 200 GPa, I = 8 × 10⁶ mm⁴. Find the free-end deflection and slope.

  1. With x from the fixed end, M(x) = −w·(L − x)²/2 (hogging).
  2. E·I·y′ = w·(L − x)³/6 + C₁. At x = 0, y′ = 0, so C₁ = −w·L³/6.
  3. E·I·y = −w·(L − x)⁴/24 − w·L³·x/6 + C₂. At x = 0, y = 0, so C₂ = w·L⁴/24.
  4. At x = L: E·I·y = −w·L⁴/6 + w·L⁴/24 = −w·L⁴/8, and E·I·y′ = −w·L³/6.
  5. E·I = 200 × 10⁹ × 8 × 10⁻⁶ = 1.6 × 10⁶ N·m². y = −4000 × 2⁴ / (8 × 1.6 × 10⁶) = −5.0 × 10⁻³ m; y′ = −4000 × 2³ / (6 × 1.6 × 10⁶) = −3.33 × 10⁻³ rad. Answer: free-end deflection 5.0 mm downward; slope 0.00333 rad (0.19°).

Example 2 (GATE level): Macaulay's method, two point loads. Given: simply supported beam AB, span 6 m, loads 20 kN at 2 m and 10 kN at 4 m from A. E·I = 12 000 kN·m². Find the deflection under the 20 kN load, at mid-span, and the maximum deflection.

  1. Reactions: R_B = (20 × 2 + 10 × 4)/6 = 13.33 kN, R_A = 16.67 kN.
  2. M = 16.67·x − 20·⟨x − 2⟩ − 10·⟨x − 4⟩ (kN·m).
  3. EI·y′ = 8.333·x² − 10·⟨x − 2⟩² − 5·⟨x − 4⟩² + C₁.
  4. EI·y = 2.778·x³ − 3.333·⟨x − 2⟩³ − 1.667·⟨x − 4⟩³ + C₁·x + C₂.
  5. y(0) = 0 gives C₂ = 0. y(6) = 0: 600 − 213.33 − 13.33 + 6·C₁ = 0, so C₁ = −62.22.
  6. At x = 2: EI·y = 22.22 − 124.44 = −102.22, so y = −102.22/12 000 = −8.52 × 10⁻³ m.
  7. At x = 3: EI·y = 75.0 − 3.33 − 186.67 = −115.0, so y = −9.58 × 10⁻³ m.
  8. Maximum: y′ = 0 in 2 < x < 4 gives 8.333x² − 10(x − 2)² − 62.22 = 0, i.e. 1.667x² − 40x + 102.22 = 0, so x = 2.908 m; there y = −9.59 × 10⁻³ m. Answer: 8.52 mm under the 20 kN load, 9.58 mm at mid-span, maximum 9.59 mm at 2.91 m from A.

Common mistakes

  • Expanding Macaulay brackets before integrating, or keeping a bracket that is negative for the section considered.
  • Forgetting to extend a part-span UDL and subtract the extension.
  • Using y = 0 and y′ = 0 at a simple support (only y = 0 holds).
  • Unit inconsistency: kN with N·m² EI, or mm⁴ with m.
  • Assuming maximum deflection always occurs under the load or at mid-span.
  • Sign slips between "y up" and "y down" conventions.

For GATE ME

Expect standard deflection and slope formulas applied to new numbers, ratios (how deflection changes when L doubles or a load moves), deflection at a point with two loads by Macaulay or superposition, and boundary-condition reasoning. Know the standard cases by heart and use Macaulay only when needed.

Quick check

  1. A simply supported beam's span doubles under the same UDL. By what factor does the maximum deflection change?
  2. What boundary conditions apply at a fixed end?
  3. Cantilever, L = 4 m, end load 10 kN, EI = 1050 kN·m². Find the deflection at 2 m from the fixed end.
  4. Simply supported, L = 5 m, 15 kN at 2 m from one end, EI = 150 000 kN·m². Deflection under the load?

Answers: 1. 16 times. 2. y = 0 and dy/dx = 0. 3. 10 × 2² × (12 − 2)/(6 × 1050) = 0.0635 m. 4. 15 × 2² × 3²/(3 × 150 000 × 5) = 0.24 mm.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is the deflection of a beam, and why is it important in engineering mechanics?Concept

    Deflection of a beam refers to the displacement of a point on the neutral axis of the beam from its original position under the action of loads. It is important because excessive deflection can lead to structural failure, discomfort in structures like bridges and floors, and misalignment in machinery. Engineers must ensure that deflection is within acceptable limits to maintain structural integrity and functionality.

  2. 2.Explain the double integration method for calculating beam deflection.Concept

    Starting from E·I·d²y/dx² = M(x), you write the bending moment as a function of x, integrate once to get E·I times the slope and again to get E·I times the deflection. The two constants of integration come from support conditions, such as y = 0 at a simple support or y = 0 and dy/dx = 0 at a fixed end. When the loading changes along the beam, each segment needs its own M(x) and two constants, fixed by continuity of slope and deflection at the junctions, which is why Macaulay's method is used for multi-load beams.

  3. 3.What is Macaulay's method, and how does it differ from the double integration method?Concept

    Macaulay's method is double integration with a single bending-moment expression valid along the whole beam, written using brackets ⟨x − a⟩ that are taken as zero when x < a. The brackets are integrated as a whole without expanding, so only two constants of integration appear and continuity between segments is automatic. Plain double integration needs separate moment equations and two constants for every segment, plus continuity conditions at each load point, which becomes tedious with several loads.

  4. 4.Why is it necessary to consider boundary conditions when calculating beam deflection?Application

    Boundary conditions are essential because they define the constraints and supports of the beam, such as fixed, simply supported, or free ends. These conditions influence the beam's response to loads and are used to solve the constants of integration in deflection equations. Without considering boundary conditions, the calculated deflection would not accurately reflect the beam's behavior in its actual setting.

  5. 5.What happens if a beam is subjected to a load beyond its elastic limit?Application

    If a beam is subjected to a load beyond its elastic limit, it will undergo plastic deformation, meaning it will not return to its original shape after the load is removed. This can lead to permanent deflection, structural damage, or failure. Engineers must design beams to operate within the elastic range to ensure safety and functionality.

  6. 6.How does the moment of inertia affect the deflection of a beam?Application

    The moment of inertia is a measure of a beam's resistance to bending. A higher moment of inertia indicates greater resistance to deflection under a given load. It depends on the beam's cross-sectional shape and size. Engineers can reduce deflection by increasing the moment of inertia, often by altering the beam's geometry.

  7. 7.Calculate the deflection at the center of a simply supported beam with a span of 6 meters, subjected to a uniform load of 5 kN/m. Assume E = 200 GPa and I = 8 × 10⁶ mm⁴.Numerical
    1. Use δ_max = 5·w·L⁴/(384·E·I), with w the load per unit length, w = 5000 N/m (not the total load).
    2. E·I = 200 × 10⁹ × 8 × 10⁻⁶ = 1.6 × 10⁶ N·m².
    3. δ_max = 5 × 5000 × 6⁴ / (384 × 1.6 × 10⁶) = 32.4 × 10⁶ / 614.4 × 10⁶ = 0.0527 m. The mid-span deflection is about 52.7 mm, which is large (about span/114), so this section would normally be too flexible.
  8. 8.Using Macaulay's method, determine the deflection at a point 2 meters from the left end of a cantilever beam of length 4 meters, with a point load of 10 kN at the free end. Assume E = 210 GPa and I = 5 × 10⁶ mm⁴.Numerical
    1. Take x from the fixed end. With the 10 kN load at x = L = 4 m, M(x) = −10(4 − x) kN·m.
    2. EI·y′ = −10(4x − x²/2) + C₁ and EI·y = −10(2x² − x³/6) + C₁x + C₂; the fixed end gives y′(0) = 0 and y(0) = 0, so C₁ = C₂ = 0.
    3. Hence y = −W·x²(3L − x)/(6EI). EI = 210 × 10⁹ × 5 × 10⁻⁶ = 1.05 × 10⁶ N·m².
    4. At x = 2 m: y = −10 000 × 4 × (12 − 2)/(6 × 1.05 × 10⁶) = −0.0635 m. The deflection 2 m from the fixed end is about 63.5 mm downward (the free-end deflection is W·L³/(3EI) = 203 mm, so this beam is very flexible).
  9. 9.What are the limitations of using the double integration method for beam deflection analysis?Application

    The double integration method is limited to beams with simple loading and support conditions. It becomes cumbersome and complex when dealing with multiple loads or discontinuous loading. Additionally, it requires solving for constants of integration using boundary conditions, which can be challenging if the conditions are not straightforward. For more complex scenarios, methods like Macaulay's are preferred.

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