Deflection of beams

In short

Deflection of beams is the study of how far and in what shape a beam bends transverse to its axis under load — the slope (dy/dx, in radians) and the deflection (y, in mm) at every section. A beam can be perfectly safe in stress and still be unusable because it sags, cracks the plaster, jams a door or makes a machine tool cut inaccurately, so deflection is the governing *serviceability* check in IS 800:2007 and IS 456:2000. It is examined heavily in university papers and in GATE (double integration, Macaulay's method, moment-area, standard results, indeterminate beams), and used daily by design engineers at L&T, Tata Projects, AFCONS, Godrej & Boyce, Ashok Leyland and machine-tool firms, and inside every FEA package.

Explained

The elastic curve and the governing equation

• The bent centre-line of the beam is the elastic curve; deflection y is measured perpendicular to the original axis, slope θ ≈ dy/dx is its tangent angle.
• From pure bending theory, 1/R = M/EI. For small slopes (dy/dx)² ≪ 1, so 1/R ≈ d²y/dx², giving the flexure differential equation: EI d²y/dx² = M.
• Successive differentiation links all four quantities: EI d³y/dx³ = V (shear), EI d⁴y/dx⁴ = −w (load intensity). So one can start from load and integrate four times, or from M and integrate twice.
• EI is the flexural rigidity (N·mm²): E is material stiffness, I the second moment of area about the bending axis. Deflection ∝ load × span³ or span⁴, and ∝ 1/EI.
• Assumptions: linearly elastic material (Hooke's law), small deflections, plane sections remain plane, bending only (shear deflection neglected), constant EI unless stated, and loads in a plane of symmetry.
• Sign convention: x along the beam from the left, y upward positive, sagging BM positive — then a sagging beam gives negative y. Many Indian textbooks take downward y positive and write EI d²y/dx² = −M; either is correct if used consistently, and magnitudes are what examiners mark.

Method 1 — Double integration (Macaulay's method)

1. Find the support reactions.
2. Write M as a single expression for a general section x, using Macaulay brackets ⟨x − a⟩: the term is included only when x > a, and is zero (not negative) otherwise.
3. Integrate once → EI dy/dx = ∫M dx + C₁; integrate again → EI y = ∫∫M dx + C₁x + C₂.
4. Evaluate C₁, C₂ from boundary conditions: y = 0 at a simple support or hinge; y = 0 and dy/dx = 0 at a fixed end; dy/dx = 0 at mid-span of a symmetrically loaded beam; continuity of y and dy/dx across an internal section.
5. Substitute x to get slope and deflection anywhere; set dy/dx = 0 to locate maximum deflection.
6. Special Macaulay rules: a UDL starting at x = a is written w⟨x−a⟩²/2 and, if it stops at x = b, cancelled by adding an equal upward UDL from b. An applied couple M₀ at x = a appears as M₀⟨x−a⟩⁰ and must be handled carefully (some texts integrate it as M₀⟨x−a⟩).
7. Advantage: one equation covers the whole beam, no separate constants per segment. Limitation: tedious for many loads or for varying I.

Method 2 — Moment-area (Mohr's) theorems

• Draw the M/EI diagram (the BM diagram divided by EI — for constant EI it is just the BMD scaled).
• Mohr's Theorem I: the change in slope between two points A and B equals the *area* of the M/EI diagram between them: θ_B − θ_A = A_{M/EI}.
• Mohr's Theorem II: the intercept (vertical distance) of point B from the tangent drawn at A equals the *first moment of that area about B*: δ = A·x̄.
• Best for cantilevers (tangent at the fixed end is horizontal, so the intercept *is* the deflection) and for beams with sudden changes of section (split the M/EI diagram into parts).
• For a simply supported beam, draw the tangent at one support, get the intercept at the other support, then work out deflections geometrically.

Method 3 — Conjugate beam and energy methods

• Conjugate beam method: load an imaginary beam of the same span with the M/EI diagram as its load. Then the shear in the conjugate beam = actual slope, and the bending moment in the conjugate beam = actual deflection. Supports are swapped: real fixed end ↔ conjugate free end, real free end ↔ conjugate fixed end, real interior support ↔ conjugate internal hinge.
• Castigliano's second theorem: δ = ∂U/∂P and θ = ∂U/∂M₀, where U = ∫M²dx/2EI. If no load acts where deflection is wanted, apply a dummy load P (or couple), differentiate, then set P = 0.
• Unit-load (virtual work) method: δ = ∫(M·m/EI)dx, where M is the real bending moment and m the moment due to a unit load at the point and direction of the required deflection. This is the standard method for trusses, frames and curved members.
• Superposition: for linear elastic beams, deflections from separate loads simply add; combined with standard results this solves most exam problems in a line or two.

Statically indeterminate beams

• Propped cantilever, fixed beams and continuous beams have more reactions than equilibrium equations; deflection gives the extra (compatibility) equations.
• Method of consistent deformation: remove the redundant support, compute the deflection there under the applied load, then compute the deflection due to the unknown redundant R, and set the sum to zero (or to the known settlement).
• Example result: propped cantilever with UDL → wL⁴/8EI = RL³/3EI → R = 3wL/8 at the prop, fixed-end moment = wL²/8.
• Other standard tools: Clapeyron's theorem of three moments for continuous beams, moment distribution and slope-deflection (covered in structural analysis).
• Fixing or continuity cuts deflection sharply: a UDL on a fixed beam deflects wL⁴/384EI, one-fifth of the simply supported 5wL⁴/384EI.

Factors, codes and Indian practice

• Deflection falls with larger I (depth is far more effective than width: I ∝ d³ for a rectangle, so doubling depth cuts deflection to one-eighth), shorter span, stiffer support conditions and higher E.
• E is almost fixed for a material (steel ≈ 200 GPa, aluminium ≈ 70 GPa, concrete E_c = 5000√f_ck MPa per IS 456, timber per IS 883), so designers change the section or the span, not the material, to control sag.
• IS 800:2007, Table 6 gives deflection limits as fractions of span — roughly span/150 (elastic cladding) to span/180 (brittle cladding) for purlins and girts, span/240 to span/300 for simple spans, span/300 to span/360 for floors and roofs, and much tighter limits for crane gantries: span/500 hand-operated, span/750 to span/1000 for electric cranes. The exact value depends on member and load case and must be read from the table.
• IS 456:2000, cl. 23.2 for RC: final deflection ≤ span/250, and deflection after partitions and finishes are built ≤ span/350 or 20 mm, whichever is less. Compliance is usually shown by the basic span/effective-depth ratios of cl. 23.2.1 — cantilever 7, simply supported 20, continuous 26 (for spans up to 10 m) — modified for tension and compression steel and flanged sections.
• RC deflection also includes long-term creep and shrinkage, estimated with an effective moment of inertia (cracked section) — IS 456 Annexure C; steel deflection is checked under unfactored service loads, not factored loads.
• Shear deflection is neglected in slender beams but matters for deep beams (span/depth less than about 8-10) and for sandwich or timber beams; camber is provided in long steel girders to offset dead-load sag.

Key points

• Flexure equation - EI d²y/dx² = M; third and fourth derivatives give shear and load intensity.
• Flexural rigidity EI - the beam's resistance to bending deformation; deflection is inversely proportional to it.
• Double integration - integrate M twice and fix two constants from boundary conditions.
• Macaulay's method - bracket terms ⟨x−a⟩ switch on only beyond x = a, giving one equation for the whole span.
• Mohr I - change of slope = area of the M/EI diagram between the two points.
• Mohr II - tangential intercept = first moment of the M/EI area about the point.
• Conjugate beam - load with M/EI; its shear = real slope, its moment = real deflection, with supports swapped.
• Castigliano II - δ = ∂U/∂P with U = ∫M²dx/2EI; use a dummy load if none exists at the point.
• Unit load method - δ = ∫(Mm/EI)dx; the general method for frames and curved members.
• Superposition - valid only while behaviour is linear elastic and deflections small.
• Boundary conditions - simple support: y = 0; fixed end: y = 0 and dy/dx = 0; symmetry: dy/dx = 0 at centre.
• Maximum deflection - occurs where dy/dx = 0; for an unsymmetrically placed point load it is *not* under the load but in the longer segment.
• Depth dominates - deflection ∝ 1/d³ for a rectangular section; adding depth beats adding width or changing steel grade.
• Indeterminate beams - solved by compatibility; propped cantilever with UDL gives prop reaction 3wL/8.
• Serviceability - deflection is checked at service (unfactored) loads: IS 456 span/250 overall, IS 800 Table 6 for steel.
• Stiffness is not strength - a high-strength steel beam of the same section deflects exactly as much, because E is unchanged.

Formulas and examples

Example 1 - simply supported beam, central point load. Span 4 m, W = 20 kN, steel I-section I = 8,000 cm⁴ = 8 × 10⁷ mm⁴, E = 200 GPa.

• δ = WL³/48EI = (20,000 × 4000³)/(48 × 200,000 × 8 × 10⁷) = 1.28 × 10¹⁵ / 7.68 × 10¹⁴ = 1.67 mm.
• θ_end = WL²/16EI = (20,000 × 4000²)/(16 × 1.6 × 10¹³) = 1.25 × 10⁻³ rad (0.072°).
• Span/deflection = 4000/1.67 = L/2400 — far stiffer than any IS 800 Table 6 limit.

Example 2 - timber cantilever with UDL. L = 3 m, w = 10 kN/m, section 200 mm wide × 300 mm deep, E = 10 GPa.

• I = bd³/12 = 200 × 300³/12 = 4.5 × 10⁸ mm⁴; w = 10 N/mm.
• δ = wL⁴/8EI = (10 × 3000⁴)/(8 × 10,000 × 4.5 × 10⁸) = 8.1 × 10¹⁴ / 3.6 × 10¹³ = 22.5 mm.
• θ_free = wL³/6EI = (10 × 2.7 × 10¹⁰)/(6 × 4.5 × 10¹²) = 0.010 rad.
• Span/δ = 3000/22.5 = L/133 — too flexible; increasing the depth to 400 mm raises I by (4/3)³ = 2.37 and drops δ to 9.5 mm.

Example 3 - unsymmetrical point load (where the maximum really is). Span 6 m, W = 30 kN at 2 m from A, E = 200 GPa, I = 10,000 cm⁴ = 1 × 10⁸ mm⁴, so EI = 2 × 10¹³ N·mm².

• Deflection under the load: δ = Wa²b²/(3EIL) = (30,000 × 2000² × 4000²)/(3 × 2 × 10¹³ × 6000) = 1.92 × 10¹⁸ / 3.6 × 10¹⁷ = 5.33 mm.
• Location of maximum (in the formula b is the *shorter* distance, 2000 mm, and x is measured from the remote support B): x = √((L²−b²)/3) = √((3.6 × 10⁷ − 4 × 10⁶)/3) = √(1.067 × 10⁷) = 3266 mm from B (2734 mm from A, inside the longer segment).
• δ_max = Wb(L²−b²)^1.5/(9√3·L·EI) = (30,000 × 2000 × 1.810 × 10¹¹)/(15.588 × 6000 × 2 × 10¹³) = 1.086 × 10¹⁹ / 1.871 × 10¹⁸ = 5.81 mm — about 9% more than the value under the load, which is why quoting the load-point value as the maximum loses marks.

Example 4 - superposition plus a serviceability check. Simply supported span 5 m carrying a UDL of 12 kN/m and a central point load of 25 kN; E = 200 GPa, I = 1.5 × 10⁸ mm⁴ (EI = 3 × 10¹³).

• From the UDL: 5wL⁴/384EI = (5 × 12 × 6.25 × 10¹⁴)/(384 × 3 × 10¹³) = 3.75 × 10¹⁶ / 1.152 × 10¹⁶ = 3.26 mm.
• From the point load: WL³/48EI = (25,000 × 1.25 × 10¹¹)/(48 × 3 × 10¹³) = 3.125 × 10¹⁵ / 1.44 × 10¹⁵ = 2.17 mm.
• Total δ = 5.43 mm = L/921, acceptable even against a brittle-cladding limit of about L/300 (16.7 mm). If the same beam were fixed at both ends, the two terms become wL⁴/384EI = 0.65 mm and WL³/192EI = 0.54 mm, total 1.19 mm — roughly one-fifth.

Where it is used

• Structural steel and RC design: every beam, purlin, gantry girder and slab in IS 800 / IS 456 design is checked for deflection as a serviceability limit state; steel girders for metro viaducts and ROBs are cambered for dead-load sag.
• Machine tools and automation: lathe and VMC beds, spindle shafts, gantry axes and robot arms are sized by allowable deflection (microns), not stress — core work at HMT, Bharat Fritz Werner, Ace Micromatic.
• Shafts and rotating machinery: shaft deflection sets bearing alignment, gear-tooth contact and whirling speed in pumps, gearboxes and turbines (KSB, Kirloskar, BHEL).
• Automotive and rail: leaf and torsion-bar springs, chassis side-members, axle beams and bogie frames are designed to deflection and stiffness targets at Tata Motors, Ashok Leyland, Mahindra and ICF/RCF.
• Formwork, scaffolding and cranes: IS 14687 falsework guidelines and crane-girder design use deflection limits to keep the system stable and usable.
• Roles: design engineer, structural/proof-checking consultant, CAE and FEA analyst (ANSYS, Abaqus, STAAD.Pro, ETABS), product validation and test engineer.

Interview questions

• Q: Why is deflection checked when the beam is already safe in bending stress? A: Strength and stiffness are separate limits — excessive sag cracks finishes, causes ponding, misaligns machinery and feels unsafe, so IS 800 and IS 456 impose span-based deflection limits at service loads.
• Q: Will a beam of E 550 steel deflect less than the same section in E 250 (IS 2062)? A: No. Deflection depends on E, which is about 200 GPa for all structural steels; high strength lets you carry more load before yielding, not bend less.
• Q: What is the advantage of Macaulay's method over plain double integration? A: The bracket terms switch on only beyond their point of application, so one bending-moment expression and one pair of constants cover the whole beam instead of separate equations per segment.
• Q: State Mohr's two theorems. A: The change of slope between two points equals the area of the M/EI diagram between them; the tangential intercept of one point from the tangent at the other equals the first moment of that area about the point.
• Q: In the conjugate beam method, what do shear and moment represent? A: Shear in the conjugate beam equals the real beam's slope and bending moment equals its deflection, with support types interchanged (real fixed ↔ conjugate free).
• Q: For a simply supported beam with an off-centre point load, is maximum deflection under the load? A: No — it occurs in the longer segment, at x = √((L²−b²)/3) measured from the support remote from the load (b = the shorter distance), and is slightly larger than the deflection under the load.
• Q: Doubling the depth of a rectangular beam does what to deflection? A: I ∝ d³, so I becomes eight times and deflection one-eighth, for the same load and span.
• Q: How do you find the prop reaction of a propped cantilever under UDL? A: Equate the free-end deflection due to the UDL, wL⁴/8EI, to the upward deflection RL³/3EI from the prop, giving R = 3wL/8.
• Q: When would you apply Castigliano's theorem instead of double integration? A: When the deflection is wanted at one point of a frame, curved member or structure with several load types — differentiating the strain energy (with a dummy load if needed) is quicker than integrating the whole elastic curve.
• Q: Why is shear deflection usually ignored, and when is it not? A: For slender beams it is a small fraction of bending deflection, but in deep beams (span/depth below about 8-10), short sandwich panels and timber members it becomes significant and must be added.
• Q: Which load case is used for the deflection check — factored or service? A: Unfactored service loads, because deflection is a serviceability criterion; factored loads are used only for strength checks.

Common mistakes

• Mixing units: using kN and m with I in cm⁴ and E in GPa. Convert everything to N, mm, N/mm² and mm⁴ before substituting.
• Treating Macaulay brackets like ordinary brackets and letting a term go negative instead of dropping it when x < a.
• Forgetting to cancel a partial UDL beyond its end point by superposing an equal upward UDL.
• Quoting the deflection under an off-centre point load as the maximum deflection.
• Applying simply supported formulas to a fixed or continuous beam — the fixed-end case is five times stiffer under UDL.
• Using the gross I for a reinforced concrete beam and ignoring cracking, creep and shrinkage, which can double or triple long-term deflection.
• Losing the sign/direction convention halfway through, so slope and deflection come out with inconsistent signs.

Related topics

Shear force and bending moment diagrams · Bending, torsion and columns · Strain energy and impact loading · Strength of materials: stress and strain · Principal stresses and theories of failure · Structural analysis of indeterminate beams and frames