>[!info] Beam deflections by superposition (determinant) >**Goal:** break down complicated loading on a beam into simple loading components and used published deflection formulas to determine beam deflection through linear superposition >- Each component must be in equilibrium >- For piece-wise defined loadings and intermediate supports, the superposed solutions must be written in a piece-wise way >- The components chosen are not unique; there are typically many ways to solve >[!example] The beam is made of a material with an elastic modulus $E$ and has a cross-sectional second area moment $I,$ both of which are constant along the length of the beam. Determine reactions at $A$ and $B.$ >![[Pasted image 20261019163924.png|center|300]] >###### Equilibrium eqns. > >$ > \sum M_{A}=M_{0}+B_{y}(L)-M_{A}=0 >$ >$ > \sum F_{y}=A_{y}+B_{y}=0 >$ >>[!info] **METHOD 1: INTEGRATION** >>Section between A and B: >> >>$ >> \sum M_{C}=M_{C}-M_{A}-A_{y}x=0\to M(x)=M_{A}+A_{y}x >>$ >>Load deflection eqn: >> >>$ >> \theta(x)=\cancel{ \theta(0) }+\frac{1}{EI} \int_{0}^x M(x)dx= \int_{0} ^x (M_{A}+A_{y}x)dx\to \theta(x)= \frac{1}{EI} \left( M_{A}\cdot A_{y} \frac{x^2}{2} \right) >>$ >> >>$ >> v(x)=\cancel{ v(0) }+ \int_{0}^x \theta(x)dx= \int_{0}^x \frac{1}{EI}\left( M_{A}x+A_{y} \frac{x^2}{2} \right)dx=\frac{1}{EI}\left( M_{A} \frac{x^2}{2}+A_{y} \frac{x^3}{6} \right) >>$ >>Apply boundary condition, $v(x=L)=0$ at point B: >> >>$ >> v(L)= \frac{1}{EI}\left( M_{A} \frac{L^2}{2}+A_{y} \frac{L^3}{6} \right)=0 >>$ >> >>$ >> 3M_{A}+A_{y}L=0 >>$ >> >>$ >> M_{A}+A_{y}L=M_{0} >>$ >> >>$ >> A_{y}=-B_{y} >>$ >> >>$ >> \boxed{ M_{A}=-\frac{M_{0}}{2},\quad A_{y}=\frac{3M_{0}}{2L},\quad B_{y}=-\frac{3M_{0}}{2L} } >>$ > >>[!info] **METHOD 2: LINEAR SUPERPOSITION** >> ![[Pasted image 20261019164202.png|center|450]] >>$ >> v_{1}(x)=\frac{M_{0}x^2}{2EI} >>$ >> >>$ >> v_{2}(x)=\frac{B_{y}x^2}{6EI}(3L-x) >>$ >> >>$ >> v(x)=v_{1}(x)+v_{2}(x)=\frac{M_{0}x^2}{2EI}+\frac{B_{y}x^2}{6EI}(3L-x) >>$ >>Apply boundary condition, $v(L)=0$: >> >>$ >> \frac{M_{0}L^2}{2EI}+\frac{B_{y}L^2}{6EI}(3L-L)=0 >>$ >> >>$ >> \boxed{ B_{y}=-\frac{3M_{0}}{2L},\quad A_{y}=\frac{3M_{0}}{2L},\quad M_{A}=-\frac{M_{0}}{2} } >>$