Beam Bending Equation Proof: Simply Supported Partial UDL

The equations for beam bending, reactions, slope and deflection will be found using Macaulay Brackets and the values from the diagram below. Macaulay brackets are represented with square brackets ("[" and "]"), when the value within the brackets is negative, then the bracketed expression is given a value of zero. Macaulay brackets are used to turn individual expressions "on" or "off", depending at what point along the beam you are interested. For more information on Macaulay brackets, please follow this link.

A useful tool for calculating bending moments, shear forces, deflection and slope for this beam and load combination is contained on this website at this location.

Figure 1: Beam Diagram

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

$$\text{From vertical equilibrium}$$ $$\qquad R_{A} + R_{B} = \omega b$$ $$\text{From moment equilibrium}$$ $$\qquad R_{B}L = (a+\frac{b}{2}) \omega b$$ $$\qquad R_{B} = (a+\frac{b}{2}) \frac{\omega b}{L}$$ $$\qquad R_{A}L = (c+\frac{b}{2}) \omega b$$ $$\qquad R_{A} = (c+\frac{b}{2}) \frac{\omega b}{L}$$ $$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x}= - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{\omega}{2}[x-a-b]^2$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{x}= - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{\omega}{2}[x-a-b]^2$$ $$\qquad EI \frac{dv}{dx} = - \frac{R_{A}x^2}{2} + \frac{\omega}{6}[x-a]^3- \frac{\omega}{6}[x-a-b]^3 + A$$ $$\qquad EI v = - \frac{R_{A}x^3}{6} + \frac{\omega}{24}[x-a]^4- \frac{\omega}{24}[x-a-b]^4 + Ax + B$$ $$\text{Check the boundary conditions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0$$ $$\qquad \text{Therefore, } B = 0$$ $$\text{At } x = L, v = 0$$ $$\qquad EI v = - \frac{R_{A}L^3}{6} + \frac{\omega}{24}(L-a)^4- \frac{\omega}{24}(L-a-b)^4 + AL$$ $$\qquad \text{Substitute in } R_{A} \text{ and } (L-a-b) = c$$ $$\qquad 0 = - \frac{\omega bL^2}{6} (c+\frac{b}{2}) + \frac{\omega}{24}(L-a)^4- \frac{\omega}{24}c^4 + AL$$ $$\qquad \text{Therefore}$$ $$\qquad A = \frac{\omega bL}{6} (c+\frac{b}{2}) - \frac{\omega}{24L}(L-a)^4+ \frac{\omega}{24L}c^4$$ $$\text{Therefore, }$$ $$\qquad \text{Reactions, }$$ $$\qquad \qquad R_{A} = (c+\frac{b}{2}) \frac{\omega b}{L}$$ $$\qquad \qquad R_{B} = (a+\frac{b}{2}) \frac{\omega b}{L}$$ $$\qquad \text{Moment Expression, }$$ $$\qquad \qquad M_{x}= - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{\omega}{2}[x-a-b]^2$$ $$\qquad \text{Slope Equation, }$$ $$\qquad \qquad EI \frac{dv}{dx} = - \frac{R_{A}x^2}{2} + \frac{\omega}{6}[x-a]^3- \frac{\omega}{6}[x-a-b]^3 + \frac{\omega bL}{6} (c+\frac{b}{2}) - \frac{\omega}{24L}(L-a)^4+ \frac{\omega}{24L}c^4$$ $$\qquad \qquad EI \frac{dv}{dx} = - \frac{\omega b x^2}{2L}(c+\frac{b}{2}) + \frac{\omega}{6}[x-a]^3- \frac{\omega}{6}[x-a-b]^3 + \frac{\omega bL}{6} (c+\frac{b}{2}) - \frac{\omega}{24L}(L-a)^4+ \frac{\omega}{24L}c^4$$ $$\qquad \qquad \frac{dv}{dx} = \frac{\omega}{24EIL}\left(-12bx^2(c+\frac{b}{2}) +4L[x-a]^3-4L[x-a-b]^3+4bL^2(c+\frac{b}{2})-(L-a)^4+c^4 \right)$$ $$\qquad \text{Deflection Equation, }$$ $$\qquad \qquad EI v = - \frac{R_{A}x^3}{6} + \frac{\omega}{24}[x-a]^4- \frac{\omega}{24}[x-a-b]^4 + \frac{\omega bxL}{6} (c+\frac{b}{2}) - \frac{\omega x}{24L}(L-a)^4+ \frac{\omega x}{24L}c^4$$ $$\qquad \qquad EI v = - \frac{\omega bx^3}{6L}(c+\frac{b}{2}) + \frac{\omega}{24}[x-a]^4- \frac{\omega}{24}[x-a-b]^4 + \frac{\omega bxL}{6} (c+\frac{b}{2}) - \frac{\omega x}{24L}(L-a)^4+ \frac{\omega x}{24L}c^4$$ $$\qquad \qquad v = \frac{\omega}{24EIL}\left(- 4bx^3(c+\frac{b}{2}) + L[x-a]^4- L[x-a-b]^4 + 4bxL^2(c+\frac{b}{2}) - x(L-a)^4+ xc^4\right)$$

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