The pin-jointed column shown in Fig. 8.10 carries a compressive load P applied eccentrically at a distance e from the axis of the column. Determine the maximum bending moment in the column.

Question

Accepted Answer

The bending moment at any section of the column is given by

[latex]M=P(e+v)[/latex]

Then, by comparison with Eq. (8.1),

[latex]E I \frac{ d ^{2} v}{ d z^{2}}=-P_{ CR } V[/latex] (8.1)

[latex]E I \frac{ d ^{2} v}{ d z^{2}}=-P(e+v)[/latex]

giving

[latex]\frac{ d ^{2} v}{ d z^{2}}+\mu^{2} v=-\frac{P e}{E I}\left(\mu^{2}=P / E I ight)[/latex] (i)

The solution of Eq. (i) is of standard form and is

[latex]v=A \cos \mu z+B \sin \mu z-e[/latex]

The boundary conditions are [latex]v=0[/latex] when [latex]z=0[/latex] and [latex]( d v / d z)=0[/latex] when [latex]z=L / 2[/latex]. From the first of these, [latex]A=e[/latex], while from the second,

[latex]B=e an \frac{\mu L}{2}[/latex]

The equation for the deflected shape of the column is then

[latex]v=e\left[\frac{\cos \mu(z-L / 2)}{\cos \mu L / 2}-1 ight][/latex]

The maximum value of v occurs at mid-span, where [latex]z=L / 2[/latex]; that is,

[latex]v_{\max }=e\left(\sec \frac{\mu L}{2}-1 ight)[/latex]

The maximum bending moment is given by

[latex]M(\max )=P e+P v_{\max }[/latex]

so that

[latex]M(\max )=P e \sec \frac{\mu L}{2}[/latex]

Question 8.3: The pin-jointed column shown in Fig. 8.10 carries a compress...

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