The ideal column is subjected to the force F at its midpoint and 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. The general solution

Question

The ideal column is subjected to the force F at its midpoint and 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. The general solution is [latex]v=C_{1} \sin k x+C_{2} \cos k x-c^{2} x / k^{2}[/latex], where [latex]c^{2}=F / 2 E I, k^{2}=P / E I[/latex].

Accepted Answer

Moment Functions: FBD(b).

[latex]\circlearrowright+\Sigma M_{o}=0 ; \quad M(x)+\frac{F}{2} x+P(v)=0 \M(x)=-\frac{F}{2} x-P v[/latex]

Differential Equation of The Elastic Curve:

[latex]E I \frac{d^{2} v}{d x^{2}}=M(x) \E I \frac{d^{2} v}{d x^{2}}=-\frac{F}{2} x-P v \\frac{d^{2} v}{d x^{2}}+\frac{P}{E I} v=-\frac{F}{2 E I} x\quad(1)[/latex]

The solution of the above differential equation is of the form,

[latex]v=C_{1} \sin \left(\sqrt{\frac{P}{E I}} x ight)+C_{2} \cos \left(\sqrt{\frac{P}{E I}} x ight)-\frac{F}{2 P} x \quad(2)[/latex]

and

[latex]\frac{d v}{d x}=C_{1} \sqrt{\frac{P}{E I}} \cos \left(\sqrt{\frac{P}{E I}} x ight)-C_{2} \sqrt{\frac{P}{E I}} \sin \left(\sqrt{\frac{P}{E I}} x ight)-\frac{F}{2 P} \quad(3)[/latex]

The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x=0, v=0 . From Eq. [2], [latex]C_{2}=0[/latex]
At [latex]x=\frac{L}{2}, \frac{d v}{d x}=0 .[/latex] From Eq.[3],

[latex]0=C_{1} \sqrt{\frac{P}{E I}} \cos \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)-\frac{F}{2 P} \C_{1}=\frac{F}{2 P} \sqrt{\frac{E I}{P}} \sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)[/latex]

Elastic Curve:

[latex]v=\frac{F}{2 P} \sqrt{\frac{E I}{P}} \sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight) \sin \left(\sqrt{\frac{P}{E I}} x ight)-\frac{F}{2 P} x \=\frac{F}{2 P}\left[\sqrt{\frac{E I}{P}} \sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight) \sin \left(\sqrt{\frac{P}{E I}} x ight)-x ight][/latex]
However, [latex]v=v_{\max }[/latex] at [latex]x=\frac{L}{2}[/latex]. Then,

[latex]v_{\max } =\frac{F}{2 P}\left[\sqrt{\frac{E I}{P}} \sec \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight) \sin \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)-\frac{L}{2} ight] \=\frac{F}{2 P}\left[\sqrt{\frac{E I}{P}} an \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)-\frac{L}{2} ight][/latex]

Maximum Moment: The maximum moment occurs at [latex]x=\frac{L}{2}[/latex]. From Eq.[1],

[latex]M_{\max } =-\frac{F}{2}\left(\frac{L}{2} ight)-P v_{\max } \=-\frac{F L}{4}-P\left\{\frac{F}{2 P}\left[\sqrt{\frac{E I}{P}} an \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)-\frac{L}{2} ight] ight\} \=-\frac{F}{2} \sqrt{\frac{E I}{P}} an \left(\sqrt{\frac{P}{E I}} \frac{L}{2} ight)[/latex]

Question 13.42: The ideal column is subjected to the force F at its midpoint...

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