Structural analysis is fundamental in building technology to ensure safety and serviceability of structures under loads. Computing the slope and deflection of beams and structural elements allows engineers to predict how much a beam bends and tilts under loading, which is critical for design and assessment. This chapter focuses on techniques to calculate slope and deflection, emphasizing Macaulay’s theorem and the assumptions underlying these calculations in building structures common in Kenya, such as beams in school buildings or office blocks.

3.1 Macaulay's theorem

Macaulay’s theorem is a powerful method used to calculate the deflection and slope of beams subjected to various loading conditions, including point loads and distributed loads. It simplifies the treatment of discontinuous loading by using a single expression incorporating Macaulay brackets, which are zero when the argument is negative and equal to the argument when positive. This method is especially useful for building professionals analyzing beams in slabs and frames where loads are applied at discrete points or varying segments.

3.1.1 Concept and formula of Macaulay's theorem

Macaulay’s theorem expresses the bending moment \(M(x)\) along a beam as a function of the distance \(x\) from a reference point, incorporating the effects of loads that begin at different points along the beam. The key formula for slope \(\theta\) and deflection \(y\) uses the relationship between moment, flexural rigidity, and curvature:

$$ EI \frac{d^2 y}{dx^2} = M(x) $$

where \(E\) is the modulus of elasticity, \(I\) is the moment of inertia, and \(M(x)\) is the bending moment at section \(x\).

Using Macaulay brackets \(\langle x, a \rangle^n\), the bending moment expression can be written as:

The slope and deflection of a simply supported beam carrying a point l (Source: [R. K. Bansal]Strength of materials 4th ed[Engineersdaily.com] (1))

$$ M(x) = \text{sum of terms of the form } P \langle x, a \rangle^m $$

where

\[ \langle x, a \rangle^m = \begin{cases} (x - a)^m & \text{if } x > a \\ 0 & \text{if } x \leq a \end{cases} \]

The slope and deflection are found by integrating the moment-curvature equation twice:

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 $$

$$ EI y = \int (\int M(x) dx ) dx + C_1 x + C_2 $$

where \(C_1\) and \(C_2\) are constants determined by boundary conditions.

Worked Examples

Example 1:
Calculate the slope and deflection at any point \(x\) on a simply supported beam of length 6 m carrying a single point load \(P = 10\) kN at 4 m from the left support. The beam has \(E = 200 \times 10^9\) Pa and \(I = 8 \times 10^{-6}\) m\(^4\).

Given:
\(L = 6\) m, \(P = 10,000\) N, \(a = 4\) m, \(E = 200 \times 10^9\) Pa, \(I = 8 \times 10^{-6}\) m\(^4\)

Bending moment using Macaulay’s notation:

$$ M(x) = P \langle x, a \rangle^1 - R_A x $$

Reaction at \(A\) is:

$$ R_A = \frac{P (L - a)}{L} = \frac{10,000 \times (6 - 4)}{6} = \frac{10,000 \times 2}{6} = 3,333.33 \text{ N} $$

So,

$$ M(x) = 10,000 \langle x - 4 \rangle - 3,333.33 x $$

Integrate to find slope:

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 = \int (10,000 \langle x - 4 \rangle - 3,333.33 x ) dx + C_1 $$

$$ EI \frac{dy}{dx} = 10,000 \frac{\langle x - 4 \rangle^2}{2} - 3,333.33 \frac{x^2}{2} + C_1 $$

Integrate again for deflection:

$$ EI y = 10,000 \frac{\langle x - 4 \rangle^3}{6} - 3,333.33 \frac{x^3}{6} + C_1 x + C_2 $$

Boundary conditions for simply supported beam:

At \(x=0\), \(y=0\):

$$ EI y = 0 = 0 - 0 + 0 + C_2 \Rightarrow C_2 = 0 $$

At \(x=6\), \(y=0\):

$$ 0 = 10,000 \frac{(6-4)^3}{6} - 3,333.33 \frac{6^3}{6} + C_1 \times 6 + 0 $$

Calculate:

$$ 10,000 \frac{2^3}{6} = 10,000 \frac{8}{6} = 13,333.33 $$

$$ 3,333.33 \frac{216}{6} = 3,333.33 \times 36 = 120,000 $$

So,

$$ 0 = 13,333.33 - 120,000 + 6 C_1 $$

$$ 6 C_1 = 106,666.67 \Rightarrow C_1 = 17,777.78 $$

Calculate slope at \(x = 3\) m:

Since \(3 < 4\), \(\langle 3 - 4 \rangle = 0\),

$$ EI \frac{dy}{dx} = 10,000 \times 0 - 3,333.33 \frac{3^2}{2} + 17,777.78 $$

$$ = -3,333.33 \times \frac{9}{2} + 17,777.78 = -15,000 + 17,777.78 = 2,777.78 $$

Calculate slope:

$$ \frac{dy}{dx} = \frac{2,777.78}{E I} = \frac{2,777.78}{200 \times 10^9 \times 8 \times 10^{-6}} = \frac{2,777.78}{1,600,000} = 0.001736 $$

Answer: Slope at 3 m = 0.001736 radians

Example 2:
Find the maximum deflection of the beam from Example 1.

Maximum deflection occurs between supports, usually under the load at \(x=4\) m.

Calculate \(y\) at \(x = 4\) m:

Since \(x = 4\), \(\langle 4 - 4 \rangle = 0\),

$$ EI y = 10,000 \frac{0^3}{6} - 3,333.33 \frac{4^3}{6} + 17,777.78 \times 4 + 0 $$

$$ = 0 - 3,333.33 \times \frac{64}{6} + 71,111.12 $$

$$ = -3,333.33 \times 10.6667 + 71,111.12 = -35,555.56 + 71,111.12 = 35,555.56 $$

Deflection:

$$ y = \frac{35,555.56}{E I} = \frac{35,555.56}{1,600,000} = 0.02222 \text{ m} = 22.22 \text{ mm} $$

Answer: Maximum deflection = 22.22 mm

Example 3:
A cantilever beam 3 m long carries a uniformly distributed load \(w = 2\) kN/m over the entire length. Calculate slope and deflection at the free end. Given \(E = 210 \times 10^9\) Pa, \(I = 6 \times 10^{-6}\) m\(^4\).

Given:
\(L = 3\) m, \(w = 2,000\) N/m, \(E = 210 \times 10^9\) Pa, \(I = 6 \times 10^{-6}\) m\(^4\)

Moment at section \(x\):

$$ M(x) = - \frac{w}{2} (L - x)^2 = - \frac{2,000}{2} (3 - x)^2 = -1,000 (3 - x)^2 $$

Use Macaulay brackets:

$$ M(x) = -1,000 \langle 3 - x \rangle^2 $$

Slope at free end \(x = L = 3\):

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 = -1,000 \int \langle 3 - x \rangle^2 dx + C_1 $$

Change variable \(u = 3 - x\), integration yields:

$$ EI \frac{dy}{dx} = -1,000 \times (- \frac{(3-x)^3}{3}) + C_1 = \frac{1,000}{3} (3 - x)^3 + C_1 $$

At fixed end \(x=0\), slope = 0:

$$ 0 = \frac{1,000}{3} (3 - 0)^3 + C_1 = \frac{1,000}{3} \times 27 + C_1 = 9,000 + C_1 $$

$$ C_1 = -9,000 $$

So,

$$ EI \frac{dy}{dx} = \frac{1,000}{3} (3 - x)^3 - 9,000 $$

At free end \(x=3\):

$$ EI \frac{dy}{dx} = \frac{1,000}{3} \times 0 - 9,000 = -9,000 $$

Slope:

$$ \frac{dy}{dx} = \frac{-9,000}{E I} = \frac{-9,000}{210 \times 10^9 \times 6 \times 10^{-6}} = \frac{-9,000}{1,260,000} = -0.00714 \text{ radians} $$

Deflection:

$$ EI y = \int EI \frac{dy}{dx} dx = \int ( \frac{1,000}{3} (3 - x)^3 - 9,000 ) dx + C_2 $$

$$ EI y = \frac{1,000}{3} \times (- \frac{(3 - x)^4}{4}) - 9,000 x + C_2 = -\frac{1,000}{12} (3 - x)^4 - 9,000 x + C_2 $$

At fixed end \(x=0\), deflection = 0:

$$ 0 = -\frac{1,000}{12} (3)^4 - 0 + C_2 = -\frac{1,000}{12} \times 81 + C_2 = -6,750 + C_2 $$

$$ C_2 = 6,750 $$

At free end \(x=3\):

$$ EI y = -\frac{1,000}{12} (0)^4 - 9,000 \times 3 + 6,750 = 0 - 27,000 + 6,750 = -20,250 $$

Deflection:

$$ y = \frac{-20,250}{E I} = \frac{-20,250}{1,260,000} = -0.01607 \text{ m} = -16.07 \text{ mm} $$

Answer: Slope at free end = -0.00714 radians, Deflection at free end = -16.07 mm

3.1.2 Application of Macaulay’s theorem to multiple loads

For beams with several point loads or distributed loads acting at different positions, Macaulay’s theorem allows combining all moments into a single moment function using brackets. This simplifies integration and avoids piecewise functions.

$$ M(x) = \sum P_i \langle x, a_i \rangle^{m_i} + \sum w_j \langle x, b_j \rangle^{n_j} $$

where \(P_i\) are point loads at \(a_i\), \(w_j\) are distributed loads starting at \(b_j\), and \(m_i, n_j\) are powers depending on load type.

Worked Examples

Example 1:
A simply supported beam 8 m long carries two point loads: 15 kN at 3 m and 10 kN at 6 m from the left support. Calculate the deflection at mid-span. Given \(E = 200 \times 10^9\) Pa, \(I = 10 \times 10^{-6}\) m\(^4\).

Given:
\(L=8\) m, \(P_1=15,000\) N at \(a_1=3\) m, \(P_2=10,000\) N at \(a_2=6\) m, \(E=200 \times 10^9\) Pa, \(I=10 \times 10^{-6}\) m\(^4\)

Calculate reactions:

$$ R_A = \frac{P_1 (L - a_1) + P_2 (L - a_2)}{L} = \frac{15,000 \times 5 + 10,000 \times 2}{8} = \frac{75,000 + 20,000}{8} = 11,875 \text{ N} $$

$$ R_B = P_1 + P_2 - R_A = 25,000 - 11,875 = 13,125 \text{ N} $$

Moment function:

$$ M(x) = R_A x - P_1 \langle x - 3 \rangle^1 - P_2 \langle x - 6 \rangle^1 $$

Integrate twice for deflection:

$$ EI y = \int \int M(x) dx dx + C_1 x + C_2 = \int \int (11,875 x - 15,000 \langle x - 3 \rangle - 10,000 \langle x - 6 \rangle ) dx dx + C_1 x + C_2 $$

First integral:

$$ EI \frac{dy}{dx} = \frac{11,875 x^2}{2} - 15,000 \frac{\langle x - 3 \rangle^2}{2} - 10,000 \frac{\langle x - 6 \rangle^2}{2} + C_1 $$

Second integral:

$$ EI y = \frac{11,875 x^3}{6} - 15,000 \frac{\langle x - 3 \rangle^3}{6} - 10,000 \frac{\langle x - 6 \rangle^3}{6} + C_1 x + C_2 $$

Boundary conditions:

At \(x=0\), \(y=0\) implies \(C_2=0\).

At \(x=8\), \(y=0\):

$$ 0 = \frac{11,875 \times 8^3}{6} - 15,000 \frac{(8-3)^3}{6} - 10,000 \frac{(8-6)^3}{6} + 8 C_1 $$

Calculate terms:

$$ \frac{11,875 \times 512}{6} = 11,875 \times 85.3333 = 1,013,020.83 $$

$$ 15,000 \times \frac{125}{6} = 15,000 \times 20.8333 = 312,500 $$

$$ 10,000 \times \frac{8}{6} = 10,000 \times 1.3333 = 13,333.33 $$

So,

$$ 0 = 1,013,020.83 - 312,500 - 13,333.33 + 8 C_1 = 687,187.5 + 8 C_1 $$

$$ 8 C_1 = -687,187.5 \Rightarrow C_1 = -85,898.44 $$

Calculate deflection at mid-span \(x=4\):

Since \(4 > 3\) and \(4 < 6\),

$$ EI y = \frac{11,875 \times 4^3}{6} - 15,000 \frac{(4-3)^3}{6} - 10,000 \times 0 + (-85,898.44) \times 4 + 0 $$

Calculate terms:

$$ \frac{11,875 \times 64}{6} = 11,875 \times 10.6667 = 126,666.67 $$

$$ 15,000 \times \frac{1}{6} = 2,500 $$

So,

$$ EI y = 126,666.67 - 2,500 + 0 - 343,593.76 = -219,427.09 $$

Deflection:

$$ y = \frac{-219,427.09}{E I} = \frac{-219,427.09}{200 \times 10^9 \times 10 \times 10^{-6}} = \frac{-219,427.09}{2,000,000} = -0.1097 \text{ m} = -109.7 \text{ mm} $$

Answer: Deflection at mid-span = 109.7 mm downward

Example 2:
A continuous beam over two spans of 4 m each carries a uniform load of 5 kN/m on the first span only. Using Macaulay’s theorem, find the slope at the mid-support.

Given:
\(E=210 \times 10^9\) Pa, \(I=9 \times 10^{-6}\) m\(^4\), \(w=5,000\) N/m over first 4 m, beam length \(8\) m.

Moment function for first span \(0 \leq x \leq 4\):

$$ M(x) = \frac{w x^2}{2} - R_A x $$

Reactions are calculated from equilibrium but omitted here for brevity.

The Macaulay expression over the full length:

$$ M(x) = R_A x, w \frac{\langle x \rangle^2}{2} $$

Slope at support \(x=4\) is found by integrating twice and applying boundary conditions.

3.1.3 Integration procedure and constants of integration

Double integration of the bending moment equation introduces two constants, \(C_1\) and \(C_2\), which represent the integration constants for slope and deflection. Determining these constants requires applying the beam's boundary conditions such as zero deflection at supports or zero slope at fixed ends.

Worked Examples

Example 1:
Simply supported beam of length 5 m with a uniform load \(w=3\) kN/m. Find constants of integration assuming \(E=200 \times 10^9\) Pa and \(I=7 \times 10^{-6}\) m\(^4\).

Given:
\(L=5\) m, \(w=3,000\) N/m

Moment:

$$ M(x) = R_A x - \frac{w x^2}{2} $$

Reaction:

$$ R_A = \frac{w L}{2} = \frac{3,000 \times 5}{2} = 7,500 \text{ N} $$

Integrate:

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 = \int (7,500 x - 1,500 x^2) dx + C_1 = 3,750 x^2 - 500 x^3 + C_1 $$

$$ EI y = \int EI \frac{dy}{dx} dx + C_2 = 1,250 x^3 - 125 x^4 + C_1 x + C_2 $$

Boundary conditions:

At \(x=0\), \(y=0\):

$$ 0 = 0 + 0 + 0 + C_2 \Rightarrow C_2 = 0 $$

At \(x=5\), \(y=0\):

$$ 0 = 1,250 \times 125 - 125 \times 625 + 5 C_1 + 0 = 156,250 - 78,125 + 5 C_1 $$

$$ 0 = 78,125 + 5 C_1 \Rightarrow 5 C_1 = -78,125 \Rightarrow C_1 = -15,625 $$

Answer:
\(C_1 = -15,625\), \(C_2 = 0\)

Example 2:
Cantilever beam 4 m long with end moment \(M_0 = 5\) kNm. Find constants of integration for slope and deflection.

Given:
\(L=4\) m, \(M_0=5,000\) Nm

Moment function:

$$ M(x) = -M_0 $$

Integration:

$$ EI \frac{dy}{dx} = -M_0 x + C_1 $$

$$ EI y = -\frac{M_0 x^2}{2} + C_1 x + C_2 $$

Boundary conditions at fixed end \(x=0\):

Slope = 0:

$$ 0 = 0 + C_1 \Rightarrow C_1 = 0 $$

Deflection = 0:

$$ 0 = 0 + 0 + C_2 \Rightarrow C_2 = 0 $$

Answer:
\(C_1 = 0\), \(C_2 = 0\)

3.1.4 Using Macaulay’s theorem for combined loading conditions

Beams in buildings often experience combined loading such as point loads, uniform distributed loads, and moments simultaneously. Macaulay’s theorem allows combining these loads into one moment expression to find slope and deflection efficiently.

Worked Examples

Example 1:
Simply supported beam 7 m long with a uniform load \(w=4\) kN/m and a point load \(P=12\) kN at 5 m from left support. Calculate deflection at \(x=3\) m. Given \(E=210 \times 10^9\) Pa, \(I=12 \times 10^{-6}\) m\(^4\).

Reactions:

$$ R_A = \frac{w L}{2} + P \frac{L - a}{L} = \frac{4,000 \times 7}{2} + 12,000 \times \frac{7 - 5}{7} = 14,000 + 3,429 = 17,429 \text{ N} $$

Moment function:

$$ M(x) = R_A x - \frac{w x^2}{2} - P \langle x - 5 \rangle $$

Integrate twice, apply boundary conditions, and calculate deflection at \(x=3\) m.

Chapter Summary

This chapter introduced Macaulay's theorem as a fundamental tool for analyzing beam deflections involving discontinuous loading conditions. It outlined the key assumptions made when calculating slope and deflection, including material linearity, small deflections, and beam behavior under elastic bending. The derivation of slope and deflection was developed from the basic differential equation of the elastic curve, establishing the relationship between bending moment, curvature, slope, and deflection. Two main methods for calculating slope and deflection were presented: Mohr's theorem, which uses the area-moment method to find rotations and deflections, and Macaulay's method, which applies Macaulay’s notation to simplify integration of bending moment expressions. The chapter emphasized step-by-step procedures to obtain numerical values for slope and deflection at specific points along a beam. Mastery of these methods enables accurate structural analysis essential for safe and efficient design.