Interactive Lecture
Beam Deflection Using Macaulay's Method and Singularity Functions
by Dr Arnaldo Delli Carri
Macaulay's Method, also known as the Macaulay bracket method, determines deflections and slopes in beams carrying any combination of point loads, moments, and distributed loads. Developed by W. H. Macaulay in 1919, it gives a systematic way to handle discontinuous loading within a single integration.
Macaulay's method uses discontinuity functions (bracket functions) to write a single equation for bending moment valid over the entire beam length, even when loads change position.
The Macaulay bracket is defined as:
\[\langle x - a \rangle^n = \begin{cases} (x-a)^n & \text{if } x \geq a \\ 0 & \text{if } x < a \end{cases}\]
where \(a\) is the position where a load or discontinuity occurs.
The fundamental beam relationships are:
\[\frac{d^2M}{dx^2} = -w(x)\]
\[\frac{dM}{dx} = V(x)\]
\[EI\frac{d^2v}{dx^2} = M(x)\]
where \(E\) is the Young's modulus, \(I\) is the second moment of area, and \(v\) is the deflection.
Macaulay's method is most commonly used to calculate the bending moment diagram M(x). Each load contributes a term with a specific exponent. With the sagging-positive convention, downward loads enter with a minus sign and the upward support reactions (such as \(R_A x\)) enter positive:
A UDL that stops short of the beam's end needs a compensating term. For example, a downward UDL of magnitude w starting at position a and ending at position b contributes:
\[M(x) = -\frac{w}{2}\langle x - a \rangle^2 + \frac{w}{2}\langle x - b \rangle^2\]
The first term "turns on" the UDL at a, and the second term "turns it off" at b by adding an equal and opposite UDL.
Keep the bracket notation intact through every integration. Only evaluate a bracket once its argument is positive (i.e. when \(x > a\)).
Explore how different loading configurations affect beam deflection, shear force, and bending moment diagrams. The simulation solves every case with the finite-element stiffness method, then reconstructs the shear and bending-moment diagrams from the support reactions — the same M(x) you would assemble by hand with Macaulay brackets. The model is normalised to span \(L = 1\) and \(EI = 1\).
Macaulay's method applies wherever a beam carries several discrete loads and you need its deflected shape:
| Concept | Equation | Description |
|---|---|---|
| Macaulay bracket definition | \(\langle x - a \rangle^n = \begin{cases}(x-a)^n & \text{if } x \geq a \\ 0 & \text{if } x < a\end{cases}\) | \(a\) = discontinuity position; \(n\) = exponent |
| Concentrated moment contribution to M(x) | \(M_0\langle x - a \rangle^0\) | \(M_0\) = applied moment at position \(a\) |
| Point load contribution to M(x) | \(-P\langle x - a \rangle^1\) | \(P\) = downward point load at position \(a\) |
| UDL contribution to M(x) | \(-\frac{w}{2}\langle x - a \rangle^2\) | \(w\) = downward UDL intensity starting at position \(a\) |
| Beam curvature equation | \(EI\frac{d^2v}{dx^2} = M(x)\) | \(E\) = Young's modulus; \(I\) = second moment of area; \(v\) = deflection |
| Slope equation (first integration) | \(EI\frac{dv}{dx} = \int M \, dx + C_1\) | \(C_1\) = integration constant from boundary conditions |
| Deflection equation (second integration) | \(EIv = \int\int M \, dx^2 + C_1x + C_2\) | \(C_2\) = integration constant from boundary conditions |
A simply supported beam of length L = 6 m carries a point load P = 10 kN at a distance a = 2 m from the left support. The beam has EI = 12000 kN·m². Calculate:
Show Solution
Solution:
First, find reactions: \(R_A = \frac{P(L-a)}{L} = \frac{10(6-2)}{6} = \frac{20}{3} \approx 6.67\) kN
\(R_B = \frac{Pa}{L} = \frac{10 \times 2}{6} = \frac{10}{3} \approx 3.33\) kN
Bending moment equation:
\(M(x) = R_A x - P\langle x - a \rangle^1 = \frac{20}{3}x - 10\langle x - 2 \rangle^1\)
Integrate to get slope:
\(EI\frac{dv}{dx} = \frac{10}{3}x^2 - 5\langle x - 2 \rangle^2 + C_1\)
Integrate again for deflection:
\(EIv = \frac{10}{9}x^3 - \frac{5}{3}\langle x - 2 \rangle^3 + C_1x + C_2\)
Apply boundary conditions: At \(x = 0\), \(v = 0\) → \(C_2 = 0\)
At \(x = 6\) m, \(v = 0\):
\(\frac{10}{9}(6)^3 - \frac{5}{3}(6-2)^3 + 6C_1 = 0\)
\(240 - 106.7 + 6C_1 = 0 \Rightarrow C_1 = -22.22\) kN·m²
Slope at left support (\(x = 0\)): \(EI\theta_A = C_1 = -22.22\) kN·m²
\(\theta_A = -22.22/12000 = -1.85 \times 10^{-3}\) rad
Slope at right support (\(x = 6\)): \(EI\theta_B = \frac{10}{3}(6)^2 - 5(6-2)^2 + C_1 = 120 - 80 - 22.22 = 17.78\) kN·m²
\(\theta_B = 17.78/12000 = +1.48 \times 10^{-3}\) rad
Maximum deflection: set slope = 0 for \(x > 2\) m → \(3x^2 - 36x + 76 = 0 \Rightarrow x \approx 2.73\) m
\(v_{max} = -3.23\) mm (downward)
Answer: (a) \(v_{max} = -3.23\) mm at \(x = 2.73\) m; (b) \(\theta_A = -1.85 \times 10^{-3}\) rad; (c) \(\theta_B = +1.48 \times 10^{-3}\) rad.
A cantilever beam of length L = 4 m is fixed at the left end and carries a uniformly distributed load w = 5 kN/m over its entire length. EI = 8000 kN·m². Find the deflection and slope at the free end.
Show Solution
Solution:
For a cantilever with UDL over entire length, moment from right end:
\(M(x) = -\frac{w(L-x)^2}{2} = -\frac{5(4-x)^2}{2}\)
Or using Macaulay notation from the left (fixing couple \(-\tfrac{wL^2}{2}\), upward reaction \(wL\), downward UDL):
\(M(x) = -\frac{wL^2}{2} + wL\langle x \rangle^1 - \frac{w}{2}\langle x - 0 \rangle^2\)
Check: at \(x = 0\), \(M = -\tfrac{wL^2}{2} = -40\) kN·m (hogging at the fixed end); at \(x = L\), \(M = 0\) (free end).
Integrating twice and applying boundary conditions at x = 0:
\(v(0) = 0\) and \(\frac{dv}{dx}(0) = 0\)
At the free end (x = L = 4 m):
Deflection: \(v(L) = \frac{wL^4}{8EI} = \frac{5 \times 4^4}{8 \times 8000} = 0.020\) m = 20 mm (downward)
Slope: \(\theta(L) = \frac{wL^3}{6EI} = \frac{5 \times 4^3}{6 \times 8000} = 0.00667\) rad
Answer: Free-end deflection = 20 mm downward; free-end slope = \(6.67 \times 10^{-3}\) rad.
A simply supported beam of length L = 8 m carries a uniformly distributed load w = 12 kN/m that starts at x = 2 m and ends at x = 6 m. The beam has EI = 15000 kN·m². Calculate:
Show Solution
Solution:
Part (a): Support Reactions
Total load: \(W = w \times (6-2) = 12 \times 4 = 48\) kN acting at midpoint x = 4 m
Taking moments about A:
\(R_B \times 8 = 48 \times 4\) → \(R_B = 24\) kN
\(R_A = 48 - 24 = 24\) kN
Part (b): Bending Moment Equation
Using Macaulay's method with the partial UDL notation:
\(M(x) = R_A x - \frac{w}{2}\langle x - 2 \rangle^2 + \frac{w}{2}\langle x - 6 \rangle^2\)
\(M(x) = 24x - 6\langle x - 2 \rangle^2 + 6\langle x - 6 \rangle^2\)
Part (c): Maximum Bending Moment
The maximum moment occurs where shear force V = 0.
\(\frac{dM}{dx} = V(x) = 24 - 12\langle x - 2 \rangle^1 + 12\langle x - 6 \rangle^1 = 0\)
For \(2 < x < 6\): \(24 - 12(x-2) = 0\) → \(x = 4\) m
\(M_{max} = M(4) = 24(4) - 6(4-2)^2 = 96 - 24 = 72\) kN·m
The maximum bending moment is 72 kN·m at x = 4 m (the centre of the beam).
Answer: (a) \(R_A = R_B = 24\) kN; (b) \(M(x) = 24x - 6\langle x - 2 \rangle^2 + 6\langle x - 6 \rangle^2\); (c) \(M_{max} = 72\) kN·m at \(x = 4\) m.