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

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

M_x=P y+Q L-Q x

Now,

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

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

Let \lambda^2=P / E I \text {, therefore } Q/E I=\lambda^2 Q / P . Therefore,

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

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

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

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

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

y(0)=0 \quad \text { or } \quad A_2=\frac{Q L}{P}

y_1(0)=0 \quad \text { or } \quad \frac{ d y}{ d x}=0

or        A_1 \lambda=-\frac{Q}{P}

Therefore,

A_1=-\frac{Q}{P \lambda}

Putting values of A_1 \text { and } A_2 in Eq. (1), we get

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

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

\frac{Q L}{P} \cos \lambda L=\frac{Q}{P \lambda} \sin \lambda L

or      \tan \lambda L=\lambda L           (3)

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

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

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