Deflection by moment-area and conjugate beam methods

Mohr's moment-area theorems and the conjugate-beam method, with support transformations and a stepped-EI cantilever.

Drafted with Aria, reviewed by the AiCanCode.org team. Spotted an error? Use Give Feedback at the bottom of the page.

Why it matters

Aircraft beams are rarely uniform: spars taper towards the tip, and landing-gear legs and fittings change section along their length. Integrating EI·y'' = M is awkward when EI varies, but the moment-area and conjugate-beam methods handle it naturally because they work with the M/EI diagram. They are also the quickest way to get the slope or deflection at one particular point.

Key ideas

The M/EI diagram. Divide the bending moment diagram by the flexural rigidity at each section. Where EI is constant, it is just the BMD scaled; where EI changes, the diagram has steps. Both methods use only the area and first moment of this diagram.

Mohr's first theorem (slope). The change in slope between two points A and B equals the area of the M/EI diagram between them: θ_B − θ_A = ∫ M/(EI) dx. It comes straight from d²y/dx² = M/(EI).

Mohr's second theorem (deviation). The vertical distance of point B on the elastic curve from the tangent drawn at A (the tangential deviation t_B/A) equals the first moment of the M/EI area between A and B taken about B: t_B/A = ∫ (M/EI)·x̄_B dx. Note the moment is taken about the point whose deviation is wanted.

How to use them. Look for a point where the tangent is known:

  • At a fixed end the tangent is horizontal, so for a cantilever the tip deflection is simply t_tip/fixed.
  • In a symmetric simply supported beam the tangent at midspan is horizontal, so the midspan deflection equals t_support/mid.
  • In an unsymmetric beam, draw the tangent at one support, find the other support's deviation from it, divide by the span to get the support slope, then use similar triangles.

Sign: with sagging M positive, a positive area means the slope increases from left to right; for routine problems simply sketch the deflected shape and use magnitudes.

Conjugate beam method. Build an imaginary beam of the same length, loaded by the M/EI diagram as a distributed load. Then:

  • the shear force in the conjugate beam at any section = the slope of the real beam there;
  • the bending moment in the conjugate beam = the deflection of the real beam. The supports are changed so that these statements hold:
  • Fixed end ↔ free end (real: y = 0, θ = 0; conjugate: M = 0, V = 0).
  • Simple end support ↔ simple end support.
  • Internal support ↔ internal hinge, and internal hinge ↔ internal support. A real cantilever fixed at A and free at B therefore becomes a conjugate cantilever free at A and fixed at B. A real determinate beam always gives a stable conjugate beam.

Where they apply. Linear elastic, small deflections, shear deformation neglected. Both methods work for varying EI and, with compatibility conditions, for indeterminate beams (for example finding fixed-end moments).

Standard areas and centroids (from the vertex of a curve with zero slope):

  • Rectangle b × h: area b·h, centroid b/2.
  • Triangle: area b·h/2, centroid b/3 from the tall end.
  • Parabolic spandrel (cantilever UDL BMD): area b·h/3, centroid b/4 from the tall end.
  • Parabolic segment (simply supported UDL, half span): area 2b·h/3, centroid 3b/8 from the peak.

Formulas

θ_B − θ_A = ∫ M/(E·I) dx θ = slope (rad), M = bending moment (N·m), E·I = flexural rigidity (N·m²), x = position (m).

t_B/A = ∫ (M/(E·I))·x̄_B dx t_B/A = deviation of B from the tangent at A (m); x̄_B = distance from the element to B (m).

θ_real = V_conjugate, y_real = M_conjugate Conjugate beam loaded by the M/EI diagram (load intensity in 1/m).

δ = W·L³ / (3·E·I), θ = W·L² / (2·E·I) Cantilever end load; check results with these.

δ = W·L³ / (48·E·I), θ_end = W·L² / (16·E·I) Simply supported, central load.

Worked examples

Example 1 (standard). A simply supported beam of span 6 m carries a central load of 24 kN. E·I = 1.2 × 10⁷ N·m². Find the support slope and the midspan deflection by the moment-area method.

  1. M at midspan = W·L/4 = 24 000 × 6/4 = 36 000 N·m. The BMD is a triangle.
  2. By symmetry the tangent at midspan C is horizontal.
  3. M/EI area from A to C: A = ½ × (L/2) × M/EI = ½ × 3 × 36 000/1.2 × 10⁷ = 4.5 × 10⁻³ rad.
  4. First theorem: θ_A = θ_C − A = 0 − 4.5 × 10⁻³, so the slope at A is 4.5 × 10⁻³ rad (downward towards midspan).
  5. Second theorem, deviation of A from the tangent at C, moment about A: centroid of the triangle from A = (2/3) × 3 = 2 m.
  6. δ_C = t_A/C = 4.5 × 10⁻³ × 2 = 9.0 × 10⁻³ m.
  7. Check: W·L³/(48EI) = 24 000 × 216/(48 × 1.2 × 10⁷) = 9.0 × 10⁻³ m.

Answer: θ_A = 0.0045 rad, δ_mid = 9.0 mm.

Example 2 (GATE level, varying EI). A cantilever of length 3 m is fixed at A and carries P = 10 kN at the free end B. The inner half (from A, 1.5 m) has E·I = 2 × 10⁶ N·m²; the outer half has E·I = 1 × 10⁶ N·m². Find the tip slope and deflection.

  1. Measure x from the tip B: M = P·x (magnitude). The tangent at A is horizontal.
  2. Outer part (0 ≤ x ≤ 1.5 m), EI = 10⁶: M/EI is a triangle from 0 to 10 000 × 1.5/10⁶ = 0.015 m⁻¹. Area A₁ = ½ × 1.5 × 0.015 = 0.01125; centroid from B = (2/3) × 1.5 = 1.0 m.
  3. Inner part (1.5 ≤ x ≤ 3 m), EI = 2 × 10⁶: M/EI rises from 0.0075 to 0.015 m⁻¹. Split into a rectangle A₂ = 0.0075 × 1.5 = 0.01125 (centroid 2.25 m from B) and a triangle A₃ = ½ × 1.5 × 0.0075 = 0.005625 (centroid 1.5 + (2/3) × 1.5 = 2.5 m from B).
  4. Tip slope (first theorem): θ_B = 0.01125 + 0.01125 + 0.005625 = 0.028125 rad.
  5. Tip deflection (second theorem, moments about B): δ_B = 0.01125 × 1.0 + 0.01125 × 2.25 + 0.005625 × 2.5 = 0.01125 + 0.0253125 + 0.0140625 = 0.050625 m.

Answer: θ_B ≈ 0.0281 rad, δ_B ≈ 50.6 mm. For comparison, the whole beam at E·I = 10⁶ would deflect P·L³/(3EI) = 90 mm, and at 2 × 10⁶ it would deflect 45 mm: stiffening the root half recovers most of the benefit.

Common mistakes

  • Taking the first moment about the wrong point; for t_B/A the moments are taken about B.
  • Swapping the conjugate-beam correspondences. Conjugate shear gives real slope; conjugate moment gives real deflection.
  • Keeping the real supports on the conjugate beam; a fixed end must become free and vice versa.
  • Using the BMD instead of the M/EI diagram when EI varies.
  • Using the triangle centroid (b/3) for a parabolic spandrel (b/4).

For GATE AE

Expect slope or deflection at a point of a cantilever or simply supported beam, beams with a stepped EI, ratio questions comparing two beams, and conceptual questions on the conjugate-beam support rules and correspondences. Practise the standard areas and centroids so that each problem becomes a few multiplications, and check with the standard formulas.

Quick check

  1. What does the shear force in the conjugate beam represent?
  2. What does a real fixed end become in the conjugate beam?
  3. State Mohr's first theorem in one line.
  4. What is the area of a cantilever UDL bending moment diagram of base L and height wL²/2?

Answers: 1. The slope of the real beam. 2. A free end. 3. The change in slope between two points equals the area of the M/EI diagram between them. 4. wL³/6.

Interview questions

All Mechanics of Solids interview questions

Try answering each one aloud before you open it.

  1. 1.What is the moment-area method in the context of beam deflection?Concept

    It uses the M/EI diagram instead of integrating the elastic-curve equation. Mohr's first theorem: the change in slope between two points equals the area of the M/EI diagram between them. Mohr's second theorem: the deviation of point B from the tangent at A equals the first moment of that area about B. Choosing a point with a known tangent (a fixed end, or midspan of a symmetric beam) turns these into slopes and deflections.

  2. 2.Explain the conjugate beam method and its purpose.Concept

    An imaginary beam of the same length is loaded with the M/EI diagram of the real beam. The shear force in the conjugate beam at any section equals the slope of the real beam, and its bending moment equals the real deflection. Supports are transformed so this holds: a fixed end becomes free and vice versa, end simple supports stay simple, and internal supports and hinges swap. It is useful for beams with varying EI.

  3. 3.How does the moment-area method differ from the conjugate beam method?Concept

    Both rest on the same M/EI diagram and give identical results. Moment-area works geometrically with tangents and deviations from a reference tangent, so you must pick a point with a known slope. The conjugate-beam method turns the problem into ordinary statics: real slope equals conjugate shear and real deflection equals conjugate moment, with the support conditions transformed. Conjugate beam is often more systematic for overhangs and internal hinges.

  4. 4.Why is the moment-area method preferred for certain types of beams?Application

    The moment-area method is often preferred for statically determinate beams because it provides a straightforward way to calculate deflections using geometric properties of the bending moment diagram. It is particularly useful when the beam has simple loading conditions and supports, making the calculations more intuitive and less complex.

  5. 5.What happens if the boundary conditions are incorrectly applied in the conjugate beam method?Application

    If the boundary conditions are incorrectly applied in the conjugate beam method, the calculated deflections and slopes will be inaccurate. This is because the boundary conditions determine how the conjugate beam is supported and loaded, directly affecting the shear force and bending moment distributions, which are used to find the deflections of the original beam.

  6. 6.In what scenarios would the conjugate beam method be more advantageous than the moment-area method?Application

    It helps when there is no convenient point with a known tangent, for example overhanging beams, beams with internal hinges, or unsymmetric loading, because you simply solve the conjugate beam by statics. It handles varying EI as easily as moment-area does, since both use the M/EI diagram. For a cantilever or a symmetric simply supported beam, moment-area is usually just as quick.

  7. 7.Using the moment-area method, find the midspan deflection of a simply supported beam of span 6 m carrying a UDL of 10 kN/m, in terms of EI.Numerical

    M_max = w·L²/8 = 10 × 36/8 = 45 kN·m, and the tangent at midspan is horizontal by symmetry. The half-span BMD is a parabolic segment: area = (2/3) × 45 × 3 = 90 kN·m², with centroid 5/8 × 3 = 1.875 m from the support. Deflection = moment of the M/EI area about the support = 90 × 1.875/EI = 168.75/EI (kN·m³), which matches 5wL⁴/(384EI).

  8. 8.Using the conjugate beam method, determine the slope at the free end of a cantilever of length 4 m with a point load of 20 kN at the free end.Numerical

    The real BMD is a triangle, zero at the tip and P·L = 80 kN·m at the fixed end. In the conjugate beam the real fixed end becomes free and the real free end becomes fixed, and the beam is loaded by this M/EI triangle. The conjugate shear at the real tip equals the total load: ½ × 4 × 80/EI = 160/EI (kN·m², i.e. radians when EI is in kN·m²), which is P·L²/(2EI). The tip deflection is the conjugate moment there, 160/EI × (2/3 × 4) = 426.7/EI = P·L³/(3EI).

  9. 9.What are the limitations of the moment-area method?Application

    It assumes linear elastic behaviour, small slopes and negligible shear deformation. It needs a reference point with a known tangent, which is awkward for unsymmetric or overhanging beams, and bookkeeping of areas and centroids becomes error-prone for complex M/EI diagrams. It can be used for indeterminate beams, but only together with compatibility conditions, and it gives deflection at chosen points rather than a full equation of the elastic curve.

  10. 10.How does the choice of method (moment-area vs. conjugate beam) affect the accuracy of deflection calculations?Application

    Both methods can provide accurate results if applied correctly, but the choice depends on the complexity of the problem. The moment-area method is straightforward for simple, determinate beams, while the conjugate beam method is better suited for complex or indeterminate beams. Errors in either method typically arise from incorrect application of boundary conditions or assumptions about material properties and beam geometry.

Finished this topic? Mark it so your progress, study plan and readiness keep up.

Stuck on something here?

PreviousDeflection of beams: double integration and Macaulay's methodNext Stress transformation and Mohr's circle