Show that for a column having one end fixed and the other end guided as shown in Figure 8.13, the buckling load of the column must satisfy the transcendental or a non-algebraic equation

$\sqrt{E I} \tan \left\lgroup \sqrt{\frac{P L^2}{E I}} \right\rgroup =\sqrt{P L^2}$

where P is the buckling load of the column.

Question

Show that for a column having one end fixed and the other end guided as shown in Figure 8.13, the buckling load of the column must satisfy the transcendental or a non-algebraic equation

[latex] \sqrt{E I} an \left\lgroup \sqrt{\frac{P L^2}{E I}} ight group =\sqrt{P L^2} [/latex]

where P is the buckling load of the column.

Accepted Answer

Let us draw the free-body diagram of the entire column as well as that of a section A–A considered at a distance x from the fixed-end of the column as shown in Figure 8.14. From the free-body diagram shown in Figure 8.14(b), we can write

[latex] M_x=P y+Q L-Q x [/latex]

Now,

[latex] E I \frac{ d ^2 y}{ d x^2}=-M_x=-P y+Q x-Q L [/latex]

or [latex] \frac{ d ^2 y}{ d x^2}+\frac{P}{E I} y-\frac{Q}{E I} x+\frac{Q L}{E I}=0 [/latex]

Let [latex] \lambda^2=P / E I ext {, therefore } Q/E I=\lambda^2 Q / P [/latex]. Therefore,

[latex] \frac{ d ^2 y}{ d x^2}+\lambda^2 y=\lambda^2(L-x) \frac{Q}{P}[/latex]

This is a second-order linear non-homogeneous differential equation whose solution consists of complementary function and particular integral. The solution of which can be written as

[latex] y=A_1 \sin \lambda x+A_2 \cos \lambda x+\frac{Q}{P}(x-L) [/latex] (1)

and [latex] \frac{ d y}{ d x}=y_1=A_1 \lambda \cos \lambda x-\lambda A_2 \sin \lambda x+\frac{Q}{P} [/latex]

The necessary boundary conditions of the column are
1. at x = 0, y = 0 and dy/dx = 0 (as column is fixed here)
2. at x = L, y = 0 (as column is pinned here)
From these boundary conditions, we get

[latex] y(0)=0 \quad ext { or } \quad A_2=\frac{Q L}{P} [/latex]

[latex] y_1(0)=0 \quad ext { or } \quad \frac{ d y}{ d x}=0 [/latex]

or [latex] A_1 \lambda=-\frac{Q}{P} [/latex]

Therefore,

[latex] A_1=-\frac{Q}{P \lambda} [/latex]

Putting values of [latex] A_1 ext { and } A_2 [/latex] in Eq. (1), we get

[latex] y=\frac{Q L}{P} \cos \lambda x-\frac{Q}{P \lambda} \sin \lambda x+\frac{Q}{P}(x-L) [/latex] (2)

Additionally, applying the second boundary condition, y = 0 at x = L in Eq. (2), we get

[latex] \frac{Q L}{P} \cos \lambda L=\frac{Q}{P \lambda} \sin \lambda L [/latex]

or [latex] an \lambda L=\lambda L [/latex] (3)

The critical load P, thus, will satisfy the transcendental equation obtained by putting [latex] \lambda=(P / E I)^{1 / 2} [/latex] in Eq. (3) as

[latex] an \left\lgroup\sqrt{\frac{P L^2}{E I}} ight group=\sqrt{\frac{P L^2}{E I}} [/latex]

or [latex] \sqrt{E I} an \left\lgroup\sqrt{\frac{P L^2}{E I}} ight group=\sqrt{P L^2} [/latex]

Question 8.6: Show that for a column having one end fixed and the other en...

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