Deflection of beams

The elastic curve and EI·y'' = M, boundary conditions, double integration, Macaulay, moment-area and superposition methods, and the standard deflection results.

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Why it matters

A beam can be strong enough and still unusable if it sags too much: floors crack plaster and feel bouncy, machine-tool beds lose accuracy, shafts misalign gears and bearings, and long spans vibrate. Deflection limits (for example span/250 to span/800, from the relevant code) often govern the design more than stress does. Deflection equations are also how statically indeterminate beams are solved.

Key ideas

Elastic curve. Under load the neutral axis bends into the elastic curve y(x). Deflection y is the vertical displacement of the axis; slope θ = dy/dx is the rotation of the cross-section (small, in radians).

Differential equation. From the flexure formula, curvature 1/R = M/(EI); for small slopes 1/R ≈ d²y/dx². So EI d²y/dx² = M(x). Further differentiation gives EI d³y/dx³ = V and EI d⁴y/dx⁴ = −w. Sign convention here: y positive upward and sagging moment positive, so a downward deflection comes out negative.

Assumptions. Linear elastic material, small deflections and slopes, plane sections remain plane, shear deformation neglected (fine for slender beams with span/depth above about 10).

Boundary conditions.

  • Simple support (pin or roller): y = 0.
  • Fixed end: y = 0 and dy/dx = 0.
  • Free end: M = 0 and V = 0 (used when writing M).
  • At a point where the M expression changes, y and slope are continuous.

Methods.

  1. Double integration: write M(x), integrate twice, apply boundary conditions. Simple for one load segment.
  2. Macaulay's method: write one M(x) for the whole beam using brackets ⟨x − a⟩ that are zero when x < a. Integrate the brackets as a whole; only two constants appear.
  3. Moment-area method: change in slope between two points = area of the M/EI diagram between them; tangential deviation = first moment of that area. Fast for cantilevers and symmetric beams.
  4. Superposition: add standard results for each load (valid because the response is linear).
  5. Energy methods (Castigliano's theorem): see the strain energy topic.

Stiffness. Deflection ∝ load × span³/(EI) for point loads and ∝ w × span⁴/(EI) for distributed loads. Doubling the span of a beam under UDL multiplies deflection by 16; doubling the depth of a rectangular section divides it by 8.

Maximum deflection. Occurs where dy/dx = 0 (simply supported beams) or at the free end (cantilevers). For an off-centre point load, the maximum is in the longer segment, not under the load.

Formulas

EI d²y/dx² = M(x)

  • E = Young's modulus (N/mm²), I = second moment of area (mm⁴), y = deflection (mm), x along beam (mm).

Standard results (magnitudes; P in N, w in N/mm, L in mm):

  • Cantilever, end load P: y_max = P L³ / (3EI), θ_max = P L² / (2EI) at the free end.
  • Cantilever, UDL w: y_max = w L⁴ / (8EI), θ_max = w L³ / (6EI).
  • Cantilever, end couple M0: y_max = M0 L² / (2EI), θ = M0 L / (EI).
  • Simply supported, central load P: y_max = P L³ / (48EI) at mid-span, θ_end = P L² / (16EI).
  • Simply supported, UDL w: y_max = 5 w L⁴ / (384EI), θ_end = w L³ / (24EI).
  • Simply supported, load P at a from A and b from B (a < b): under the load y = P a² b² / (3EIL); maximum y_max = P a (L² − a²)^(3/2) / (9√3 EIL) at √((L² − a²)/3) from B.
  • Fixed–fixed, central load: y_max = P L³ / (192EI); fixed–fixed, UDL: y_max = w L⁴ / (384EI).

Worked examples

Example 1 (standard, superposition). A steel cantilever 3 m long (E = 200 GPa, I = 2 × 10⁷ mm⁴) carries a UDL of 4 kN/m over its full length and a point load of 6 kN at the free end. Find the deflection and slope at the free end.

  1. Units: w = 4 N/mm, P = 6000 N, L = 3000 mm, EI = 200 000 × 2 × 10⁷ = 4 × 10¹² N·mm².
  2. UDL: y1 = w L⁴ / (8EI) = 4 × 3000⁴ / (8 × 4 × 10¹²) = 10.125 mm.
  3. Point load: y2 = P L³ / (3EI) = 6000 × 3000³ / (3 × 4 × 10¹²) = 13.5 mm.
  4. Total deflection: 10.125 + 13.5 = 23.63 mm (downward).
  5. Slope: θ = w L³ / (6EI) + P L² / (2EI) = 0.004 50 + 0.006 75 = 0.011 25 rad.

Answer: y = 23.6 mm downward, θ = 0.0113 rad (0.64°) at the free end.

Example 2 (GATE level, Macaulay). A simply supported beam AB of span 6 m carries a point load of 30 kN at 2 m from A. E = 200 GPa, I = 8 × 10⁷ mm⁴. Find the deflection under the load and the maximum deflection.

  1. Reactions: R_A = 30 × 4/6 = 20 kN, R_B = 10 kN.
  2. Macaulay (x from A, in mm, N): EI y'' = 20 000 x − 30 000 ⟨x − 2000⟩.
  3. Integrate: EI y = 20 000 x³/6 − 30 000 ⟨x − 2000⟩³/6 + C1 x + C2.
  4. y(0) = 0 gives C2 = 0. y(6000) = 0: 20 000 × 6000³/6 − 30 000 × 4000³/6 + 6000 C1 = 0, so C1 = −(7.2 × 10¹⁴ − 3.2 × 10¹⁴)/6000 = −6.667 × 10¹⁰.
  5. EI = 1.6 × 10¹³ N·mm². Under the load (x = 2000): EI y = 20 000 × 8 × 10⁹/6 − 6.667 × 10¹⁰ × 2000 = 2.667 × 10¹³ − 1.333 × 10¹⁴ = −1.067 × 10¹⁴, so y = −6.67 mm. Check: P a² b²/(3EIL) = 30 000 × 2000² × 4000² / (3 × 1.6 × 10¹³ × 6000) = 6.67 mm. ✓
  6. Maximum: in the longer segment, at √((6² − 2²)/3) = 3.266 m from B, i.e. 2.734 m from A. y_max = P a (L² − a²)^(3/2) / (9√3 EIL) = 30 000 × 2000 × (3.2 × 10⁷)^1.5 / (9√3 × 1.6 × 10¹³ × 6000) = 7.26 mm.

Answer: 6.67 mm under the load; maximum 7.26 mm at 2.73 m from A (both downward).

Common mistakes

  • Using mixed units in EI (E in GPa with I in mm⁴). Use N/mm² with mm⁴, or Pa with m⁴.
  • Assuming maximum deflection is under an off-centre load.
  • Integrating Macaulay brackets by expanding them; integrate ⟨x − a⟩ as a unit.
  • Forgetting that a fixed end has zero slope as well as zero deflection.
  • Confusing 5wL⁴/384EI (simply supported) with wL⁴/8EI (cantilever).
  • Losing the sign convention and reporting an upward deflection for a downward load.

For GATE PI

Expect deflection and slope from standard cases, superposition of two loads, ratios (cantilever vs simply supported, effect of doubling span or depth), propped cantilever reactions using compatibility, and Macaulay problems with one off-centre load. Memorise the standard table and practise unit handling in N and mm.

Quick check

  1. By what factor does the deflection of a simply supported beam under UDL change if the span is doubled?
  2. Slope at the fixed end of a cantilever?
  3. Same span, same central load: ratio of central deflection of simply supported to fixed–fixed beam?
  4. Cantilever 2 m, end load 500 N, E = 200 GPa, I = 4 × 10⁶ mm⁴: tip deflection?

Answers: 1. 16 times. 2. Zero. 3. 4 (192/48). 4. 1.67 mm.

See it move

All Production animations
Beam Deflection Visualization

Adjust the sliders to see how different parameters affect the deflection of a simply supported beam with a central point load.

Equations used
  • δ = (P·L³) / (48·E·I) — δ maximum deflection, P point load, L length of the beam, E modulus of elasticity, I moment of inertia

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 floors, and misalignment in machinery. Engineers must ensure that deflection is within acceptable limits to maintain structural integrity and functionality.

  2. 2.Explain the difference between bending moment and shear force in the context of beam deflection.Concept

    Bending moment is the internal moment that causes a beam to bend, while shear force is the internal force that causes the beam to shear. Bending moment affects the curvature of the beam, leading to deflection, whereas shear force affects the beam's ability to resist sliding between its sections. Both are critical in analyzing beam deflection, but bending moment is more directly related to the amount of deflection.

  3. 3.How does the material of a beam affect its deflection?Application

    The material of a beam affects its deflection through its modulus of elasticity (E). A material with a higher modulus of elasticity will deflect less under the same load compared to a material with a lower modulus. This is because a higher modulus indicates a stiffer material, which resists deformation more effectively.

  4. 4.Why is the moment of inertia important in calculating beam deflection?Application

    The moment of inertia (I) is a measure of a beam's resistance to bending and is crucial in calculating deflection. A larger moment of inertia means the beam is more resistant to bending, resulting in less deflection. It depends on the beam's cross-sectional shape and size, making it a key factor in beam design to control deflection.

  5. 5.What happens to the deflection of a beam if its length is doubled, assuming all other factors remain constant?Application

    If the length of a beam is doubled, the deflection increases significantly. For a simply supported beam with a central load, deflection is proportional to the cube of the length (L³). Therefore, doubling the length increases the deflection by a factor of eight, assuming all other factors remain constant.

  6. 6.Explain why cantilever beams generally deflect more than simply supported beams.Application

    For the same span, load and EI, an end-loaded cantilever deflects PL³/(3EI) while a simply supported beam with a central load deflects PL³/(48EI), 16 times less. In the cantilever the whole load is carried back to one support, so the bending moment is larger (PL vs PL/4) and the curvature accumulates over the full length from a single fixed point. Supporting both ends halves the effective lever arm and makes the curve symmetric, which sharply reduces deflection.

  7. 7.How does the cross-sectional shape of a beam influence its deflection?Application

    The cross-sectional shape of a beam influences its deflection through the moment of inertia. Shapes with larger moments of inertia, like I-beams, are more resistant to bending and thus have less deflection. The distribution of material away from the neutral axis increases the moment of inertia, reducing deflection.

  8. 8.Calculate the deflection at the centre of a simply supported beam of span 6 m carrying a uniform load of 2 kN/m, with E = 200 GPa and I = 8 × 10⁶ mm⁴.Numerical

    For a full-span UDL, δ = 5wL⁴/(384EI). In N and mm: w = 2 N/mm, L = 6000 mm, E = 200 000 N/mm², I = 8 × 10⁶ mm⁴. δ = 5 × 2 × 6000⁴ / (384 × 200 000 × 8 × 10⁶) = 1.296 × 10¹⁶ / 6.144 × 10¹⁴ = 21.1 mm. That is span/284, which would fail a span/325 or stricter serviceability limit, so a deeper section would be needed.

  9. 9.A cantilever of length 4 m carries a point load of 500 N at its free end. With E = 210 GPa and I = 5 × 10⁶ mm⁴, find the free-end deflection.Numerical

    For a cantilever with an end load, δ = PL³/(3EI). In N and mm: δ = 500 × 4000³ / (3 × 210 000 × 5 × 10⁶) = 3.2 × 10¹³ / 3.15 × 10¹² = 10.2 mm. The maximum is at the free end, and the slope there is PL²/(2EI) = 0.0038 rad.

  10. 10.What are the potential consequences of ignoring beam deflection in structural design?Application

    Ignoring beam deflection in structural design can lead to several issues, such as structural failure due to excessive bending, discomfort or safety hazards in buildings due to floor vibrations, and misalignment in machinery leading to operational inefficiencies. It can also cause aesthetic problems like sagging floors or ceilings, and in severe cases, it may lead to catastrophic failure of the structure.

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