# Beam Bending Equation Proof: Fully Fixed Triangular Load

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.

Due to symmetry of the support conditions, the case for a decreasing triangular load can also be found using the equations below through mirroring all inputs and outputs about the vertical axis.

$$R_{A} \text{ and } M_{A} \text{ are unknowns}$$ $$\text{Moment expression using Macaulay Brackets}$$ $$\qquad M_{x} = M_{A} - R_{A}x + \frac{ \omega (b+c)}{b} \cdot \frac{[x-a]}{(b+c)} \cdot \frac{[x-a]}{2} \cdot \frac{[x-a]}{3} - \frac{\omega}{2}[x-a-b]^2 - \left(\frac{ \omega (b+c)}{b}- \omega \right) \cdot \frac{[x-a-b]}{c} \cdot \frac{[x-a-b]}{2} \cdot \frac{[x-a-b]}{3}$$ $$\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega }{6b} [x-a]^3- \frac{\omega}{2}[x-a-b]^2 - \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad EI \frac{d^2v}{dx^2} =M_{A} - R_{A}x + \frac{\omega }{6b} [x-a]^3- \frac{\omega}{2}[x-a-b]^2 - \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega }{24b} [x-a]^4- \frac{\omega}{6}[x-a-b]^3 - \frac{1}{24c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^4 +A$$ $$\qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega }{120b} [x-a]^5- \frac{\omega}{24}[x-a-b]^4 - \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^5 +Ax+B$$ $$\text{Check the boundary conditions to find constants } A \text{ and } B$$ $$\text{At } x = 0, v = 0, \frac{dv}{dx} = 0$$ $$\qquad \text{Therefore, } A=0, B = 0$$ $$\text{At } x = L, \frac{dv}{dx} = 0$$ $$\qquad 0 = M_{A}L - \frac{R_{A}L^2}{2} + \frac{\omega }{24b} (L-a)^4- \frac{\omega}{6}(L-a-b)^3 - \frac{1}{24c}\left(\frac{ \omega (b+c)}{b}- \omega \right) (L-a-b)^4$$ $$\text{Using the identities, } a + b = d \text{ and } b+c=e$$ $$\qquad M_{A} = \frac{R_{A}L}{2} - \frac{\omega e^4}{24bL}+ \frac{\omega c^3}{6L} + \frac{c^3}{24L}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\text{At } x = L, v = 0 \text{ therefore,}$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{R_{A}L^3}{6} + \frac{\omega }{120b} (L-a)^5- \frac{\omega}{24}(L-a-b)^4 - \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) (L-a-b)^5$$ $$\qquad 0 = \frac{M_{A}L^2}{2} - \frac{R_{A}L^3}{6} + \frac{\omega e^5}{120b}- \frac{\omega c^4}{24} - \frac{c^4}{120}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\text{Substitute in expression for } M_{A}$$ $$\qquad 0 = \frac{R_{A}L^3}{4} - \frac{\omega e^4 L}{48b}+ \frac{\omega c^3 L}{12} + \frac{c^3 L}{48}\left(\frac{ \omega (b+c)}{b}- \omega \right) - \frac{R_{A}L^3}{6} + \frac{\omega e^5}{120b}- \frac{\omega c^4}{24} - \frac{c^4}{120}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad \frac{R_{A}L^3}{6} - \frac{R_{A}L^3}{4} = - \frac{\omega e^4 L}{48b}+ \frac{\omega c^3 L}{12} + \frac{\omega e^5}{120b}- \frac{\omega c^4}{24} + \left(\frac{c^3 L}{48} - \frac{c^4}{120}\right) \left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad \frac{L^3}{12}(2R_{A} - 3R_{A}) = \frac{\omega}{240}\left( -\frac{5e^4 L}{b} + 20 c^3 L + \frac{2 e^5}{b} - 10c^4 \right) + \frac{c^3}{240}(5L-2c) \left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad -R_{A} = \frac{\omega}{20L^3}\left( -\frac{5e^4 L}{b} + 20 c^3 L + \frac{2 e^5}{b} - 10c^4 \right) + \frac{c^3}{20L^3}(5L-2c) \left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad R_{A} = -\frac{\omega}{20L^3}\left( -\frac{5e^4 L}{b} + 20 c^3 L + \frac{2 e^5}{b} - 10c^4 \right) - \frac{c^3}{20L^3}(5L-2c) \left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\text{Therefore, }$$ $$\qquad \text{Reactions, }$$ $$\qquad\qquad R_{A} = -\frac{\omega}{20L^3}\left( -\frac{5e^4 L}{b} + 20 c^3 L + \frac{2 e^5}{b} - 10c^4 \right) - \frac{c^3}{20L^3}(5L-2c) \left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad\qquad M_{A} = \frac{R_{A}L}{2} - \frac{\omega e^4}{24bL}+ \frac{\omega c^3}{6L} + \frac{c^3}{24L}\left(\frac{ \omega (b+c)}{b}- \omega \right)$$ $$\qquad \text{For brevity reactions, } R_{A} \text{ and } M_{A} \text{ will not be expanded in the expressions below.}$$ $$\qquad \text{Moment expression, }$$ $$\qquad\qquad M_{x} = M_{A} - R_{A}x + \frac{\omega }{6b} [x-a]^3- \frac{\omega}{2}[x-a-b]^2 - \frac{1}{6c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^3$$ $$\qquad \text{Shear expression, }$$ $$\qquad\qquad V_{x} = - R_{A} + \frac{\omega }{2b} [x-a]^2- \omega [x-a-b] - \frac{1}{2c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^2$$ $$\qquad \text{Slope equation, }$$ $$\qquad\qquad EI \frac{dv}{dx} = M_{A}x - \frac{R_{A}x^2}{2} + \frac{\omega }{24b} [x-a]^4- \frac{\omega}{6}[x-a-b]^3 - \frac{1}{24c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^4$$ $$\qquad \text{Deflection equation, }$$ $$\qquad \qquad EI v = \frac{M_{A}x^2}{2} - \frac{R_{A}x^3}{6} + \frac{\omega }{120b} [x-a]^5- \frac{\omega}{24}[x-a-b]^4 - \frac{1}{120c}\left(\frac{ \omega (b+c)}{b}- \omega \right) [x-a-b]^5$$