Question 8.12: A compression member (Fig. P.8.12) is made of circular secti......
A compression member (Fig. P.8.12) is made of circular section tube, diameter d, thickness t. The member is not perfectly straight when unloaded, having a slightly bowed shape that may be represented by the expression
ν=\delta\sin\left({\frac{\pi z}{l}}\right)Show that when the load P is applied, the maximum stress in the member can be expressed as
\sigma_{\mathrm{max}}={\frac{P}{\pi d t}}\biggl[1+{\frac{1}{1-α}}{\frac{4\delta}{d}}\biggr]where
α=P/P_{e},\quad P_{e}=\pi^{2}E I/l^{2}Assume t is small compared with d, so that the following relationships are applicable:
Cross-sectional area of tube = πdt
Second moment of area of tube = πd³t=8
In an identical manner to S.8.6,
E I{\frac{\mathrm{d}^{2}v^{\prime}}{\mathrm{d}z^{2}}}-E I{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}=-P v^{\prime}where \boldsymbol{v^{\prime}} is the total displacement from the horizontal. Thus,
{\frac{\mathrm{d}^{2}v^{\prime}}{\mathrm{d}z^{2}}}+{\frac{P}{E I}}v^{\prime}={\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}or, since
{\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}=-{\frac{\pi^{2}}{l^{2}}}\delta\sin{\frac{\pi}{l}}z\operatorname{and}\mu^{2}={\frac{P}{E I}}
{\frac{\mathrm{d}^{2}v^{\prime}}{\mathrm{d}z^{2}}}+\mu^{2}v^{\prime}=-{\frac{\pi^{2}}{l^{2}}}\delta\sin{\frac{\pi z}{l}} (i)
The solution of Eq. (i) is
\upsilon^{\prime}=A\cos\mu z+B\sin\mu z+{\frac{\pi^{2}\delta}{\pi^{2}-\mu^{2}l^{2}}}\sin{\frac{\pi z}{l}} (ii)
When z = 0 and l, \boldsymbol{v^{\prime}} = 0, hence A=B = 0 and Eq. (ii) becomes
\ v^{\prime}=\frac{\pi^{2}\delta}{\pi^{2}-\mu^{2}l^{2}}\sin\frac{\pi z}{l}The maximum bending moment occurs at the mid-point of the tube so that
M(\operatorname{max})=P v^{\prime}=P{\frac{\pi^{2}\delta}{\pi^{2}-\mu^{2}l^{2}}}={\frac{P\delta}{1-P l^{2}/\pi^{2}E I}}i.e.,
M(\mathrm{max})=\frac{P\delta}{1-P/P_{\mathrm{e}}}=\frac{P\delta}{1-\alpha}The total maximum direct stress due to bending and axial load is then
\sigma(\mathrm{max})=\frac{P}{\pi d t}+\left(\frac{P\delta}{1-\alpha}\right)\frac{d/2}{\pi d^{3}t/8}Hence,
\sigma(\operatorname{max})={\frac{P}{\pi d t}}{\Biggl(}1+{\frac{1}{1-\alpha}}{\frac{4\delta}{d}}{\Biggr)}