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**Eccentrically Loaded Columns – Secant Formula**

Consider an eccentrically loaded slender column as shown in Figure 23.1a.

**Fig. 23.1.**

The equation of elastic line becomes,

\[{{{d^2}y} \over {d{x^2}}}=-{{{M_x}} \over {EI}}\] (23.1)

\[\Rightarrow {{{d^2}y} \over {d{x^2}}} + {P \over {EI}}\left( {e + y} \right) = 0\] (23.2)

\[\Rightarrow {{{d^2}y} \over {d{x^2}}} + {k^2}y=-{k^2}e\] (23.3)

where, \[{k^2} = {P / {EI}}\]

The general solution of equation (23.3) is,

\[y=A\cos kx + B\sin kx - e\] (23.4)

Constants* A* and *B* are determined from the boundary conditions as follows,

\[y(x = 0)=0 \Rightarrow A=e\]

\[y(x = l)=0 \Rightarrow B={{1 - \cos kl} \over {\sin kl}}e\]

Substituting A and B in equation (23.4) yields,

\[y=\left( {\cos kx + {{1 - \cos kl} \over {\sin kl}}\sin kx - 1} \right)e\] (23.5)

Lateral deflection at mid-height (y = l/2),

\[{\left. y \right|_{x = l/2}}=\left( {\cos {{kl} \over 2} + {{1 - \cos kl} \over {\sin kl}}\sin {{kl} \over 2} - 1} \right)e\] (23.6)

\[\Rightarrow {\left. y \right|_{x = l/2}}=\left( {\sec {{kl} \over 2} - 1} \right)e\] (23.7)

Writing equation (23.7) in terms of Euler load \[{P_E}={\pi ^2}EI/{l^2}\] ,

\[{\left. y \right|_{x = l/2}}=\left[ {\sec \left( {{\pi\over 2}\sqrt {{P \over{{P_E}}}} } \right) - 1} \right]e\] (23.8)

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