An eccentrically loaded simply supported column is shown in Figure 8.19. Find the value of the average compressive stress, P/A, above which the maximum bending moment will occur between support points A and B.

Question

Accepted Answer

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

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

Now,

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

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

or [latex] \frac{ d ^2 y}{ d x^2}+\lambda^2 y=\lambda^2\left\lgroup -\frac{Q x}{P} ight group \quad\left( ext { taking } \lambda^2=P / E I ight) [/latex]

Solving this we get,

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

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

[latex] \left.A_2=\frac{P \varepsilon}{P \sin \lambda L} \quad ext { (considering } Q L=P \varepsilon ight) [/latex]

or [latex] A_2=\frac{\varepsilon}{\sin \lambda L} [/latex]

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

This is the equation of the deflection curve.
Now,

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

or

[latex] \begin{aligned} M_x & =\left\lgroup \frac{P \varepsilon}{\sin \lambda L} ight group \sin \lambda x-Q x+Q x \ & =\frac{P \varepsilon}{\sin \lambda L} \cdot \sin \lambda x \end{aligned} [/latex]

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

[latex] \lambda x=\frac{\pi}{2} [/latex]

or [latex] x=\frac{\pi}{2 \lambda} [/latex]

According to the problem,

[latex] \frac{\pi}{2 \lambda} \leq L \Rightarrow \frac{\pi}{2 L} \leq \lambda \quad ext { or } \quad \lambda^2 \geq \frac{\pi^2}{4 L^2} [/latex]

or [latex] \frac{P}{E I} \geq \frac{\pi^2}{4 L^2} [/latex]

or [latex] P \geq \frac{\pi^2 E I}{4 L^2} [/latex]

So, the critical value of load is

[latex] P_{ Cr }=\frac{\pi^2 E I}{4 L^2} [/latex]

The compressive stress is

[latex] \sigma_{ ext {comp }}=\frac{\pi^2 E}{4(L / K)^2} \quad\left( ext { as } A K^2=I ight) [/latex]

Thus, the required stress is

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

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

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