The ideal column has a weight w (force/length) and rests in the horizontal position when it is subjected to the axial load P. Determine the maximum moment in the column at midspan. E I is constant. Hint: Establish the differential equation for deflection, Eq. 13-1, with the origin at the mid span. The general solution is v=C_{1} \sin k x+C_{2} \cos k x+(w /(2 P)) x^{2}-(w L /(2 P)) x-\left(w E I / P^{2}\right) where k^{2}=P / E I.
Moment Functions: FBD ( b ).
\curvearrowleft +\Sigma M_{o}=0 ; \quad w x\left(\frac{x}{2}\right)-M(x)-\left(\frac{w L}{2}\right) x-P v=0 \\M(x)=\frac{w}{2}\left(x^{2}-L x\right)-P v \quad[1]
Differential Equation of The Elastic Curve:
The solution of the above differential equation is of the form
v=C_{1} \sin \left(\sqrt{\frac{P}{E I}} x\right)+C_{2} \cos \left(\sqrt{\frac{P}{E I}} x\right)+\frac{w}{2 P} x^{2}-\frac{w L}{2 P} x-\frac{w E I}{P^{2}}\quad[2]and
\frac{d v}{d x}=C_{1} \sqrt{\frac{P}{E I}} \cos \left(\sqrt{\frac{P}{E I}} x\right)-C_{2} \sqrt{\frac{P}{E I}} \sin \left(\sqrt{\frac{P}{E I}} x\right)+\frac{w}{P} x-\frac{w L}{2 P}\quad[3]The integration constants can be determined from the boundary conditions.
Boundary Condition:
At x=0, v=0 . From Eq. [2],
At x=\frac{L}{2}, \frac{d v}{d x}=0 . From Eq. [3],
0=C_{1} \sqrt{\frac{P}{E I}} \cos \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)-\frac{w E I}{P^{2}} \sqrt{\frac{P}{E I}} \sin \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)+\frac{w}{P}\left(\frac{L}{2}\right)-\frac{w L}{2 P} \\C_{1}=\frac{w E I}{P^{2}} \tan \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)Elastic Curve:
v=\frac{w}{P}\left[\frac{E I}{P} \tan \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right) \sin \left(\sqrt{\frac{P}{E I}} x\right)+\frac{E I}{P} \cos \left(\sqrt{\frac{P}{E I}} x\right)+\frac{x^{2}}{2}-\frac{L}{2} x-\frac{E I}{P}\right]However, v=v_{\max } at x=\frac{L}{2}. Then,
v_{\max } =\frac{w}{P}\left[\frac{E I}{P} \tan \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right) \sin \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)+\frac{E I}{P} \cos \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)-\frac{L^{2}}{8}-\frac{E I}{P}\right] \\=\frac{w E I}{P^{2}}\left[\sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)-\frac{P L^{2}}{8 E I}-1\right]Maximum Moment: The maximum moment occurs at x=\frac{L}{2}. From, Eq.[1],
M_{\max } =\frac{w}{2}\left[\frac{L^{2}}{4}-L\left(\frac{L}{2}\right)\right]-P v_{\max } \\=-\frac{w L^{2}}{8}-P\left\{\frac{w E I}{P^{2}}\left[\sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)-\frac{P L^{2}}{8 E I}-1\right]\right\} \\=-\frac{w E I}{P}\left[\sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2}\right)-1\right]