Solving <PERSON> beam equation for cantilever beam
Update with solution\ Solution 1\ Problem: Beam clamped at left side, free end on right side, point load pointing downwards. x is defined positive from the clamped end towards the free end.
Create FBD from a section x from the free end gives:\ $M(x) = -P(L-x)$\ $Q(x) = -P$\ \ Differential equation for equilibrium is:\ $-EI\frac{d\alpha}{dx} = M(x) = -P(L-x)$\ Integrating this function results in:\ $EI\alpha(x) = \frac{-Px(2L-x)}{2}+C_{1}$\ Using the boundary condition that the beam is clamped at $x=0$:\ $\alpha(0) = 0 \rightarrow 0 + C_{1} = 0 \rightarrow C_{1} = 0$\ Now using the second differential equation for equilibrium:\ $\frac{d}{dx}\left(\frac{dw}{dx}-\alpha(x)\right)+\frac{p(x)}{\kappa G A} = 0$\ Integration of this equation leads to:\ $\frac{dw}{dx} = \frac{Q(x)}{\kappa A G}+\alpha(x)$\ Substituting in the previous found equation for $\alpha(x)$ gives:\ $\frac{dw}{dx} = -\frac{P}{\kappa A G} - \frac{Px(2L-x)}{2EI}$\ Integrating this once more to find the equation for $w(x)$ leads to:\ $w(x) = -\frac{Px}{\kappa A G} - \frac{Px^2(3L-x)}{6EI} + C_{2}$\ Using the boundary condition $w(0) = 0$ gives:\ $w(0) = 0 = -0 - 0 + C_{2} \rightarrow C_{2} = 0$\ Which gives:\ $w(x) = -\frac{Px}{\kappa A G} - \frac{P}{EI}\left(-\frac{x^3}{6}+\frac{x^2L}{2}\right)$
Solution 2\ Problem: Beam clamped at right side, free end on left side, point load pointing downwards. 402