Beam Bending Equation Proof: Propped Cantilever Full 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.
Using the above diagram and Macaulay brackets, the beam bending, deflection and slope equations can be found:
$$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega x^3}{6} + A$$ $$\qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega x^4}{24} + Ax + B$$ $$\text{Check the boundary conditions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0$$ $$\qquad \text{Therefore, } B = 0$$ $$\text{At } x = 0, \frac{dv}{dx} = 0$$ $$\qquad \text{Therefore, } A = 0$$ $$M_{A}\text{, }R_{A}\text{ and } R_{B}\text{ are all unknowns} $$ $$\text{At } x = L, v = 0$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{R_{A} L^3}{6} + \frac{\omega L^4}{24}$$ $$\qquad M_{A} = \frac{R_{A}L}{3} - \frac{\omega L^2}{12}$$ $$\text{At } x = L, \text{moment, M } = 0$$ $$\qquad 0 = M_{A} - R_{A}L + \frac{\omega L^2}{2}$$ $$\text{Substitute in the expression for } M_{A} \text{ found above.}$$ $$\qquad 0 = \frac{R_{A}L}{3} - \frac{\omega L^2}{12} - R_{A}L + \frac{\omega L^2}{2}$$ $$\qquad 0 = 4R_{A}L - \omega L^2 - 12R_{A}L + 6\omega L^2$$ $$\qquad 0 = -8R_{A}L + 5\omega L^2 $$ $$\qquad R_{A} = \frac{5\omega L}{8} $$ $$\text{From vertical force equilibrium } R_{B} \text{ can be found}$$ $$\qquad R_{B} = \omega L - R_{A}$$ $$\qquad R_{B} = \frac{3\omega L}{8}$$ $$ M_{A} \text{ can now be found}$$ $$\qquad M_{A} = \frac{5\omega L^2}{24} - \frac{2\omega L^2}{24}$$ $$\qquad M_{A} = \frac{\omega L^2}{8}$$ $$\text{Therefore, }$$ $$\qquad \text{Reactions, }$$ $$ \qquad \qquad R_{A} = \frac{5\omega L}{8}$$ $$ \qquad \qquad R_{B} = \frac{3\omega L}{8}$$ $$\qquad \qquad M_{A} = \frac{\omega L^2}{8}$$ $$\qquad \text{Moment expression, }$$ $$\qquad \qquad M_{x} = M_{A} - R_{A}x + \frac{\omega x^2}{2}$$ $$\qquad \qquad M_{x} = \frac{\omega L^2}{8} - \frac{5\omega xL}{8} + \frac{\omega x^2}{2}$$ $$\qquad \qquad M_{x} = \frac{\omega}{8}((L-4x)(L-x))$$ $$\qquad \text{Slope equation, }$$ $$\qquad \qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega x^3}{6}$$ $$\qquad \qquad EI \frac{dv}{dx} = \frac{\omega xL^2}{8} - \frac{5\omega Lx^2}{16} + \frac{\omega x^3}{6}$$ $$\qquad \qquad \frac{dv}{dx} = \frac{\omega x}{48EI}(6L^2 - 15Lx + 8x^2)$$ $$\qquad \text{Deflection equation, }$$ $$\qquad \qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega x^4}{24}$$ $$\qquad \qquad EI v = \frac{\omega x^2 L^2}{16} - \frac{5\omega Lx^3}{48} + \frac{\omega x^4}{24}$$ $$\qquad \qquad v = \frac{\omega x^2 }{48EI}((3L -2x)(L-x))$$