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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 midheight (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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