# Indeterminate Beams – Macaulay Bracket Method

In the previous section we showed how simple statics could be used to solve statically determinate beams, i.e. beams with the same number of equilibrium equations as unknowns. However, a lot of beams are not statically determinate, they have more unknowns than equilibrium equations and therefore cannot be solved using simple statics. For these statically indeterminate structures it is necessary to examine the stiffness and displacements of the beam in order to find a solution. One method for doing this is to use the Macaulay Bracket Method.

#### Macaulay Brackets

Macaulay brackets are special expressions that are denoted using square brackets. When the equation within the bracket results in a negative value, the whole bracket is set to a value of zero and ignored, when the equation within the bracket is positive the equation value is used as normal.

$$\text{Using a Macaulay bracket with a simple equation inside: } [x-a]$$ $$\text{When } x < a \rightarrow [x-a] = 0$$ $$\text{When } x > a \rightarrow [x-a] = (x-a) $$

Macaulay brackets are useful for turning expressions within an equation “on” or “off” depending on a given value. This is very useful for beam analysis, where you may only want to allow for the contribution of a certain load after a given position on a beam.

In the previous example we established equations for bending moment along a simply supported beam with a point load positioned as position, a, along the beam:

$$\text{Before the point load}$$ $$M_{x} = M_{a} – R_{a}x$$ $$\text{After the point load}$$ $$M_{x} = M_{a} – R_{a}x + P(x-a) $$

This can be expressed in one equation using Macaulay brackets:

$$M_{x} = M_{a} – R_{a}x + P[x-a] $$ $$\text{When } x < a \rightarrow [x-a] = 0$$ $$M_{x} = M_{a} – R_{a}x + \cancel{P[x-a]} $$ $$\text{When } x > a \rightarrow [x-a] = (x-a) $$ $$M_{x} = M_{a} – R_{a}x + P(x-a)$$

#### Macaulay Brackets and Euler-Bernoulli Beam Bending Theory

Euler-Bernoulli beam bending theory, discussed in a separate article, is based on the following identity:

$$EI \frac{d^{2}v}{dx^{2}} = M $$

Where \(E\) is the Young’s modulus of the beam, \(I\) is the Second Moment of Area of a beam, \(M\) is the applied to the beam, \(v\) is the vertical deflection of the beam and \(x\) is the distance along a beam. \(M\) is the moment at the x-distance along the beam that is being examined and so will typically be a different value at each location.

In the above equation \(M\) can be expressed as a “moment function” i.e. a mathematical expression for the moment distribution over a beam. For the previous example of a simply supported beam with a point load positioned at position, a, the following Euler-Bernoulli formula can be developed using Macaulay brackets:

$$EI \frac{d^{2}v}{dx^{2}} = M_{a} – R_{a}x + P[x-a] $$

By integrating the expression and applying suitable boundary conditions for deflection, \(v\) and slope, \(\frac{dv}{dx} \), it is possible to derive expressions for deflection and slope throughout the beam. This is shown in the examples below. For more information on the Macaulay method, please refer to this link.

#### Macaulay Brackets Example - Finding Slope and Deflection

Using the above diagram and Macaulay brackets, the beam bending, deflection and slope equations can be found:

$$\text{From vertical equilibrium}$$ $$ \qquad R_{A} + R_{B}= P$$ $$\text{From moment equilibrium}$$ $$ \qquad R_{A} = Pb/L$$ $$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x} = - R_{A}x + P[x-a]$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{x} = - R_{A}x + P[x-a]$$ $$\qquad EI \frac{dv}{dx} = - \frac{R_{A}x^2}{2} + \frac{P}{2}[x-a]^2 + A$$ $$\qquad EI v = - \frac{R_{A}x^3}{6} + \frac{P}{6}[x-a]^3 + Ax + B$$ $$\text{Check the boundary condtions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0$$ $$\qquad \text{Therefore, } B = 0$$ $$\text{At } x = L, v = 0$$ $$ \qquad \text{Therefore, } $$ $$\qquad 0 = - \frac{R_{A}L^3}{6} + \frac{P}{6}[L-a]^3 + AL$$ $$\qquad \text{Using the solutions, } R_{A} = Pb/L \text{ and } L-a = b$$ $$\qquad A = \frac{PbL}{6} - \frac{Pb^3}{6L}$$ $$\qquad A = \frac{Pb}{6L}(L^2 - b^2)$$ $$\text{Therefore,}$$ $$\qquad \text{Reactions,}$$ $$ \qquad\qquad R_{A} = Pb/L$$ $$ \qquad\qquad R_{B} = P - R_{A}$$ $$\qquad \text{Moment expression,}$$ $$\qquad\qquad M_{x} = - \frac{Pbx}{L} + P[x-a]$$ $$\qquad \text{Slope equation,}$$ $$\qquad\qquad EI \frac{dv}{dx} = - \frac{Pbx^2}{2L} + \frac{P}{2}[x-a]^2 + \frac{Pb}{6L}(L^2 - b^2)$$ $$\qquad\qquad \frac{dv}{dx} = \frac{P}{12EI}(\frac{-6bx^2}{L} + 6[x-a]^2+ \frac{2b}{L}(L^2 - b^2))$$ $$\qquad\qquad \frac{dv}{dx} = \frac{P}{12EIL}( 2b(L^2 - b^2 - 3x^2)+6L[x-a]^2)$$ $$\qquad \text{Deflection equation,}$$ $$\qquad\qquad EI v = - \frac{Pbx^3}{6L} + \frac{P}{6}[x-a]^3 + \frac{Pbx}{6L}(L^2 - b^2)$$ $$\qquad\qquad v = \frac{P}{6EI}(-\frac{bx^3}{L} + [x-a]^3+\frac{bx}{L}(L^2 - b^2))$$ $$\qquad\qquad v = \frac{P}{6EIL}(bx(L^2 - b^2 -x^2)+L[x-a]^3)$$

While the simply supported beam with a point load could be solved using simple statics and a beam bending equation found, the Macaulay bracket method allows an engineer to also find the slope and deflection behaviour of a beam.

#### Macaulay Brackets Example - Finding Reactions and Beam Equations

Where beams cannot be solved using simple statics the Macaulay bracket method can be used to establish the additional unknown variables by providing additional equations that can be compared to the beams boundary conditions. This is shown in the example below:

Using the above diagram and Macaulay brackets, the reactions, beam bending, deflection and slope equations can be found:

$$\text{From vertical equilibrium}$$ $$ \qquad R_{A} = \frac{\omega L}{2}$$ $$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega x^3}{6} + A$$ $$\qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega x^4}{24} + Ax + B$$ $$\text{Check the boundary condtions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0$$ $$\qquad \text{Therefore, } B = 0$$ $$\text{At } x = 0, \frac{dv}{dx} = 0$$ $$\qquad \text{Therefore, } A = 0$$ $$\text{Substitute in } R_{A} \text{ and solve when } x = L, v = 0$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{\omega L^4}{12} + \frac{\omega L^4}{24}$$ $$\qquad 0 = M_{A} - \frac{\omega L^2}{6} + \frac{\omega L^2}{12}$$ $$\qquad M_{A} = \frac{2 \omega L^2}{12} - \frac{\omega L^2}{12}$$ $$\qquad M_{A} = \frac{\omega L^2}{12}$$ $$\text{Therefore, }$$ $$\qquad \text{Reactions, }$$ $$ \qquad \qquad R_{A} = \frac{\omega L}{2}$$ $$\qquad \qquad M_{A} = \frac{\omega L^2}{12}$$ $$\qquad \text{Moment expression, }$$ $$\qquad \qquad M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad \text{Slope equation, }$$ $$\qquad \qquad EI \frac{dv}{dx} = \frac{\omega xL^2}{12} - \frac{\omega Lx^2}{4} + \frac{\omega x^3}{6}$$ $$\qquad \qquad \frac{dv}{dx} = \frac{\omega x}{12EI}(L^2 - 3Lx + 2x^2)$$ $$\qquad \qquad \frac{dv}{dx} = \frac{\omega x}{12EI}(L-2x)(L-x)$$ $$\qquad \text{Deflection equation, }$$ $$\qquad \qquad EI v = \frac{\omega x^2 L^2}{24} - \frac{\omega Lx^3}{12} + \frac{\omega x^4}{24}$$ $$\qquad \qquad v = \frac{\omega x^2 }{24EI}(L^2 -2Lx + x^2)$$ $$\qquad \qquad v = \frac{\omega x^2 }{24EI}(L-x)^2$$

#### Macaulay Method for Partial UDLs

On first glance a beam with a partial UDL presents an issue for the Macaulay bracket method, an equation for the partial UDL needs to be turned “off” before the UDL, then “on” within the UDL and then “off” again after the UDL.

However, a solution to this problem is possible by adding in an additional “dummy UDL”, this allows the partial UDL to be extended to the end of the beam and the “dummy UDL” then acts to counteract the extended section of UDL. This is shown in the diagram below:

Using the above diagram and Macaulay brackets, the beam bending can be expressed as:

$$\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{\omega}{2}[x-a-b]^2$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{\omega}{2}[x-a-b]^2$$

This can then be solved in the same manner as the previous examples.