Question 8.7: An eccentrically loaded simply supported column is shown in ...

Let us draw the bent column with the equivalent force system and its free-body diagram (Figure 8.20).

From Figure 8.20(a), we get

QL = Pε (taking moment about A)

and Figure 8.20(b) gives

M_x=P y+Q x

Now,

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

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

or      \frac{ d ^2 y}{ d x^2}+\lambda^2 y=\lambda^2\left\lgroup -\frac{Q x}{P} \right\rgroup \quad\left(\text { taking } \lambda^2=P / E I\right)

Solving this we get,

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

The boundary conditions are
1. at x = 0, y = 0.
2. at x = L, y = 0.
Using condition (1) we get A_1=0 from Eq. (1). Using condition (2) we get from Eq. (1):

\left.A_2=\frac{P \varepsilon}{P \sin \lambda L} \quad \text { (considering } Q L=P \varepsilon\right)

or          A_2=\frac{\varepsilon}{\sin \lambda L}

so,        y=\varepsilon \frac{\sin \lambda x}{\sin \lambda L}-\frac{Q}{P} x           (2)

This is the equation of the deflection curve.
Now,

\begin{aligned} M_x & =P y+Q x \\ & =P\left[\varepsilon \frac{\sin \lambda x}{\sin \lambda L}-\frac{Q}{P} x\right]+Q x \end{aligned}

or

\begin{aligned} M_x & =\left\lgroup \frac{P \varepsilon}{\sin \lambda L} \right\rgroup \sin \lambda x-Q x+Q x \\ & =\frac{P \varepsilon}{\sin \lambda L} \cdot \sin \lambda x \end{aligned}

Now, M_x=\left(M_x\right)_{\max } \text { when } \sin \lambda x=1 \text { or } \sin \lambda x=\sin \pi / 2 . \text { The least value of } \lambda x that satisfies the relation is as follows:

\lambda x=\frac{\pi}{2}

or        x=\frac{\pi}{2 \lambda}

According to the problem,

\frac{\pi}{2 \lambda} \leq L \Rightarrow \frac{\pi}{2 L} \leq \lambda \quad \text { or } \quad \lambda^2 \geq \frac{\pi^2}{4 L^2}

or        \frac{P}{E I} \geq \frac{\pi^2}{4 L^2}

or        P \geq \frac{\pi^2 E I}{4 L^2}

So, the critical value of load is

P_{ Cr }=\frac{\pi^2 E I}{4 L^2}

The compressive stress is

\sigma_{\text {comp }}=\frac{\pi^2 E}{4(L / K)^2} \quad\left(\text { as } A K^2=I\right)

Thus, the required stress is

\frac{\pi^2 E}{4(L / K)^2}