Question 21.P.6: A straight uniform column of length L and bending stiffness ...
A straight uniform column of length L and bending stiffness EI is subjected to uniform lateral loading w/unit length. The end attachments do not restrict rotation of the column ends. The longitudinal compressive force P has eccentricity e from the centroids of the end sections and is placed so as to oppose the bending effect of the lateral loading as shown in Fig. P.21.6. The eccentricity e can be varied and is to be adjusted to the value which, for given values of P and w, will result in the least maximum bending moment on the column. Show that
e = (w / Pμ²) tan² (μL / 4)
where μ² = P/EI. Also deduce the end moment that gives the optimum condition when P tends to zero.
Referring to Fig. S.21.6 the bending moment at any section of the column is given by
M=-P(e+v)+\frac{w L x}{2}-\frac{w x^{2}}{2}
or
M=-P(e+v)-\frac{w}{2}\left(x^{2}-L x\right) (i)
Then, substituting for M in Eq. (13.3) \frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}=\frac{M}{E I}
E I \frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}+P v=-P e-\frac{w}{2}\left(x^{2}-L x\right)
or
\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}+\mu^{2} v=-\mu^{2} e-\frac{w \mu^{2}}{2_{P}}\left(x^{2}-L x\right) (ii)
where μ² = P/EI. The solution of Eq. (ii) is
v=C_{1} \cos \mu x+C_{2} \sin \mu x-e+\frac{w}{2_{P}}\left(L x-x^{2}\right)+\frac{w}{\mu^{2} P} (iii)
When x = 0, v = 0 so that C_{1} = e – (w/μ²P). Also, when x = L/2, dv/dx = 0 which gives
C_{2}=C_{1} \tan \frac{\mu L}{2}=\left(e-\frac{w}{\mu^{2} P}\right) \tan \frac{\mu L}{2}
Eq. (iii) then becomes
v=\left[e-\frac{w}{\mu^{2} P}\right]\left[\frac{\cos \mu(x-L / 2)}{\cos (\mu L / 2)}-1\right]+\frac{w}{2 P}\left(L x-x^{2}\right) (iv)
The maximum bending moment will occur at the mid-point of the column where x = L/2 and v=v_{\max } . From Eq. (iv)
v_{\max }=\left(e-\frac{E I w}{P^{2}}\right)\left(\frac{\sec \mu L}{2}-1\right)+\frac{w L^{2}}{8 P}
Then, from Eq. (i)
M_{\max }=-P e-P v_{\max }-\frac{w L^{2}}{8}
which gives
M_{\max }=-\left(P e-\frac{w}{\mu^{2}}\right) \sec \frac{\mu L}{2}-\frac{w L^{2}}{\mu^{2}} (v)
For the maximum bending moment to be as small as possible the bending moment at the ends of the column must be numerically equal to the bending moment at the mid-point.
Then
-P e-\left(P e-\frac{w}{\mu^{2}}\right) \sec \frac{\mu L}{2}-\frac{w}{\mu^{2}}=0
from which
e=\frac{w}{P \mu^{2}} \tan ^{2} \frac{\mu L}{4} (vi)
From Eq. (vi) the end moment is
P e=\frac{w}{\mu^{2}} \tan ^{2} \frac{\mu L}{4}=\frac{w L^{2}}{16}\left[\frac{\tan (\mu L / 4)}{\mu L / 4}\right]\left[\frac{\tan (\mu L / 4)}{\mu L / 4}\right]
When P→0, tan(μL/4)→(μL/4) and the end moment becomes wL²/16.
