Question 8.8: A long strut AB of length L is of uniform section throughout...

Assuming the column to be fundamental one (i.e., pinned at both ends) with eccentric load at both ends, we show the schematic structural diagram of the column in Figure 8.21(a).
In Figure 8.21(a), we represent the eccentricity at ends A and B as ε and 2ε, respectively. In Figure 8.21(b), we represent the free-body diagram of the entire column while in Figure 8.21(c), we show the free-body diagram of the column of a section cut out at a distance x (0 ≤ x ≤ L) from end A. In Figure 8.21(c), axial force N_x , shear force V_x and bending moment M_x are shown in their positive senses.

Now, from Figure 8.21(b), for equilibrium \text { (note } \Sigma F_x=0 \text { and } \Sigma F_y=0 have automatically been satisfied), we must have:

\sum M_z=0 \Rightarrow Q L=P \varepsilon

or          Q=\frac{P \varepsilon}{L}             (1)

Again from Figure 8.21(c):

\sum M_z=0 \Rightarrow M_x=P \varepsilon+P y+Q x

where y is the depression of the column at a distance x from end A [as shown in Figure 8.21(b) above].
Therefore, from flexure equation:

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

or          \frac{ d ^2 y}{ d x^2}=\left\lgroup -\frac{P}{E I} \right\rgroup y-\left(\frac{Q}{E I}\right) x-\left(\frac{P}{E I}\right) \varepsilon

or          \frac{ d ^2 y}{ d x^2}+\lambda^2 y=-\lambda^2 \varepsilon-\lambda^2 \frac{Q}{P} x \quad \text { (where } \lambda^2=P(E I)

or          \frac{ d ^2 y}{ d x^2}+\lambda^2 y=-\lambda^2\left\lgroup \frac{Q}{P} x+\varepsilon \right\rgroup               (2)

Clearly, the above is a second-order non-homogeneous linear differential equation and will have complementary function and particular integral as its solution components. Therefore,

y=\left(A_1 \cos \lambda x+A_2 \sin \lambda x\right)-\left\lgroup \frac{Q}{P} x+\varepsilon \right\rgroup                (3)

where A_1 \text { and } A_2 are the two arbitrary constants to be determined from the geometric boundary conditions of the problem. Obviously, we have at x = 0, y = 0 and at x = L, y = 0. Thus, from Eq. (3), we get

A_1-\varepsilon=0 \Rightarrow A_1=\varepsilon            (4)

and          \varepsilon \cos \lambda L+A_2 \sin \lambda L-\frac{Q L}{P}-\varepsilon=0

But from Eq. (1), QL/P = ε. Thus, the above expression becomes

\varepsilon \cos \lambda L+A_2 \sin \lambda L-2 \varepsilon=0

A_2=\frac{\varepsilon(2-\cos \lambda L)}{\sin \lambda L}               (5)

Now, from Eq. (3), we obtain

y=\varepsilon\left[\cos \lambda x+\left\lgroup \frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup \sin \lambda x-\frac{x}{L}-1\right]            (6)

This is the equation of the elastic line of the column. Now, bending moment at any distance x is

M_x=P_y+P \varepsilon+Q x

or            \left.M_x=P\left[y+\varepsilon+\frac{\varepsilon x}{L}\right] \quad \text { [as } Q L=P \varepsilon \text { by Eq. (1) }\right]

Putting expression of y from Eq. (6) above, we get

M_x=P \varepsilon\left[\cos \lambda x+\left\lgroup \frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup \sin \lambda x\right]              (7)

\text { For } M_x to be maximum, clearly cos λx +[(2 – cos λL)/ sin λL] sin λx is to be maximum, that is, when

\frac{ d }{ d x}\left\{\cos \lambda x+\left\lgroup \frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup \sin \lambda x\right\}=0

or              -\sin \lambda x+\left\lgroup \frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup \cos \lambda x=0

or            \tan \lambda x=\left\lgroup \frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup             (where λ² = P/EI )

Alternative Solution: Let (2 – cos λL)/sin λL = p. Now,

\cos \lambda x+\left\lgroup\frac{2-\cos \lambda L}{\sin \lambda L} \right\rgroup \sin \lambda x=\cos \lambda x+p \sin \lambda x

=\sqrt{1+p^2} \sin (\lambda x+\theta)

where tan θ= 1/p. Therefore, cos λx + p sin λx is maximum, that is, 1 when

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

or          \lambda x=\frac{\pi}{2}-\theta

or            \tan \lambda x=\cot \theta=p=\frac{2-\cos \lambda L}{\sin \lambda L}             (as before)