Consider an ideal column as in Fig. 13-10d, having one end fixed and the other pinned. Show that the critical load on the column is given by P cr =20.19 EI / L ^2. Hint: Due to the vertical deflection at the top of the column, a constant moment

Question

Consider an ideal column as in Fig. 13-10d, having one end fixed and the other pinned. Show that the critical load on the column is given by [latex]P_{ cr }=20.19 EI / L ^{2}[/latex]. Hint: Due to the vertical deflection at the top of the column, a constant moment [latex]M ^{\prime}[/latex] will be developed at the fixed support and horizontal reactive forces [latex]R ^{\prime}[/latex] will be developed at both supports. Show that [latex]d^{2} v / d x^{2}+(P / E I) v=\left(R^{\prime} / E I ight)(L-x)[/latex] . The solution is of the form [latex]v=C_{1} \sin (\sqrt{P / E I} x)+C_{2} \cos (\sqrt{P / E I} x)+\left(R^{\prime} / P ight)(L-x) .[/latex] After application of the boundary conditions show that [latex] an (\sqrt{P / E I} L)=\sqrt{P / E I} L .[/latex] Solve by trial and error for the smallest nonzero root.

Accepted Answer

Equilibrium. FBD(a).
Moment Functions: FBD(b).

[latex]M(x)=R^{\prime}(L-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}}=R^{\prime}(L-x)-P v \\frac{d^{2} v}{d x^{2}}+\frac{P}{E I} v=\frac{R^{\prime}}{E I}(L-x)[/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{R^{\prime}}{P}(L-x)[/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{R^{\prime}}{P}[/latex]

The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x=0, v=0 . From Eq. [1], [latex]C_{2}=-\frac{R^{\prime} L}{P}[/latex]
At x=0, [latex]\frac{d v}{d x}=0 .[/latex] From Eq. [2], [latex]C_{1}=\frac{R^{\prime}}{P} \sqrt{\frac{E I}{P}}[/latex]
Elastic Curve:

[latex]v =\frac{R^{\prime}}{P} \sqrt{\frac{E I}{P}} \sin \left(\sqrt{\frac{P}{E I}} x ight)-\frac{R^{\prime} L}{P} \cos \left(\sqrt{\frac{P}{E I}} x ight)+\frac{R^{\prime}}{P}(L-x) \=\frac{R^{\prime}}{P}\left[\sqrt{\frac{E I}{P}} \sin \left(\sqrt{\frac{P}{E I}} x ight)-L \cos \left(\sqrt{\frac{P}{E I}} x ight)+(L-x) ight][/latex]

However, v=0 at x=L . Then,

[latex]0=\sqrt{\frac{E I}{P}} \sin \left(\sqrt{\frac{P}{E I}} L ight)-L \cos \left(\sqrt{\frac{P}{E I}} L ight) \ an \left(\sqrt{\frac{P}{E I}} L ight)=\sqrt{\frac{P}{E I}} L[/latex]

By trial and error and choosing the smallest root, we have

[latex]\sqrt{\frac{P}{E I}} L=4.49341[/latex]

Then,

[latex]P_{ cr }=\frac{20.19 E I}{L^{2}}[/latex]

Question 13.45: Consider an ideal column as in Fig. 13-10d, having one end f...

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