# Beam Bending Equation Proof: Propped Cantilever Triangular Load - Decreasing

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.

Macaulay brackets do not allow loads to be "turned off" once they have been "turned on", i.e. once a load has become active it will continue to the end of the span. To allow for the partial triangular load distribution, the partial triangle must be decomposed into separate loads that remain active once they are "turned on", the partial triangle is then reformed by superimposing each decomposed load. The decomposition of the triangular load into constituent parts is shown in Figure 2 below.

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

$$R_{A} \text{ and } M_{A} \text{ are unknowns}$$ $$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x} =M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2- \frac{ \omega (b+c)}{b} \cdot \frac{[x-a]}{(b+c)} \cdot \frac{[x-a]}{2} \cdot \frac{[x-a]}{3} + \left(\frac{ \omega (b+c)}{b}- \omega \right) \cdot \frac{[x-a-b]}{c} \cdot \frac{[x-a-b]}{2} \cdot \frac{[x-a-b]}{3}$$ $$\qquad M_{x} =M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2 - \frac{\omega }{6b} [x-a]^3 + \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad EI \frac{d^2v}{dx^2} = M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2 - \frac{\omega }{6b} [x-a]^3 + \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega}{6}[x-a]^3 - \frac{\omega }{24b} [x-a]^4 + \frac{1}{24c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^4 +A$$ $$\qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega}{24}[x-a]^4 - \frac{\omega }{120b} [x-a]^5 + \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^5 +Ax+B$$ $$\text{Check the boundary conditions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0, \frac{dv}{dx} = 0$$ $$\qquad \text{Therefore, } A=0, B = 0$$ $$\text{At } x = L, \text{ therefore moment, } M = 0$$ $$\qquad 0 =M_{A} - R_{A}L + \frac{\omega}{2}(L-a)^2 - \frac{\omega }{6b} (L-a)^3 + \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) (L-a-b)^3$$ $$\text{Using the identities, } a + b = d \text{ and } b+c=e$$ $$\qquad 0 =M_{A} - R_{A}L + \frac{\omega e^2}{2} - \frac{\omega e^3}{6b} + \frac{c^3}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad M_{A} = R_{A}L - \frac{\omega e^2}{2}+ \frac{\omega e^3}{6b} - \frac{c^2}{6}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad M_{A} = R_{A}L + \frac{1}{6}\left( \omega e^2\left(\frac{e}{b}-3\right)-c^2\left(\frac{ \omega (b+c)}{b}- \omega \right)\right)$$ $$\text{At } x = L, v = 0$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{R_{A}L^3}{6} + \frac{\omega}{24}(L-a)^4 - \frac{\omega }{120b} (L-a)^5 + \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) (L-a-b)^5$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{R_{A}L^3}{6} + \frac{\omega e^4}{24} - \frac{\omega e^5}{120b} + \frac{c^4}{120}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\text{Substitute in } M_{A}$$ $$\qquad 0 = \frac{R_{A}L^3}{2} + \frac{L^2}{12}\left( \omega e^2\left(\frac{e}{b}-3\right)-c^2\left(\frac{ \omega (b+c)}{b}- \omega \right)\right) - \frac{R_{A}L^3}{6} + \frac{\omega e^4}{24} - \frac{\omega e^5}{120b} + \frac{c^4}{120}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad \frac{R_{A}L^3}{6}-\frac{R_{A}L^3}{2} = \frac{ \omega e^2 L^2}{12}\left(\frac{e}{b}-3\right)-\frac{\omega e^5}{120b}+\frac{\omega e^4}{24}+\left(\frac{c^4}{120}-\frac{c^2L^2}{12}\right)\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad \frac{-L^3 R_{A}}{3} = \frac{ \omega e^2 }{120}\left(\frac{10L^2e}{b}-30L^2-\frac{e^3}{b}+5e^2\right)+\frac{c^2}{120}(c^2-10L^2)\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad R_{A} = \frac{ -\omega e^2 }{40L^3}\left(\frac{10L^2e}{b}-30L^2-\frac{e^3}{b}+5e^2\right)-\frac{c^2}{40L^3}(c^2-10L^2)\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\text{Therefore, }$$ $$\qquad \text{Reactions, }$$ $$\qquad\qquad R_{A} = \frac{ -\omega e^2 }{40L^3}\left(\frac{10L^2e}{b}-30L^2-\frac{e^3}{b}+5e^2\right)-\frac{c^2}{40L^3}(c^2-10L^2)\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad\qquad M_{A} = R_{A}L + \frac{1}{6}\left( \omega e^2\left(\frac{e}{b}-3\right)-c^2\left(\frac{ \omega (b+c)}{b}- \omega \right)\right)$$ $$\qquad \text{For brevity reactions, } R_{A} \text{ and } M_{A} \text{ will not be expanded in the expressions below.}$$ $$\qquad \text{Moment expression, }$$ $$\qquad\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega}{2}[x-a]^2 - \frac{\omega }{6b} [x-a]^3 + \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad \text{Shear expression, }$$ $$\qquad\qquad V_{x} = - R_{A} + \omega [x-a] - \frac{\omega }{2b} [x-a]^2 + \frac{1}{2c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^2$$ $$\qquad \text{Slope equation, }$$ $$\qquad\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega}{6}[x-a]^3 - \frac{\omega }{24b} [x-a]^4 + \frac{1}{24c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^4$$ $$\qquad \text{Deflection equation, }$$ $$\qquad \qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega}{24}[x-a]^4 - \frac{\omega }{120b} [x-a]^5 + \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^5$$