Question 11.3: A 3-m long fixed–fixed ended column of square cross section ......
A 3-m long fixed–fixed ended column of square cross section is to be made of wood in which the maximum allowable stress is 16 MPa and with E = 13 GPa. Using a safety factor with respect to buckling of 2.5, determine the required size of the cross section if the column is to safely support centric loads of (1) 100 kN and (2) 200 kN.
Given: Geometry of column, safety factor.
Find: Size of square cross section for given applied loads.
Assume: Hooke’s law applies; the anisotropy in the wood may be neglected.
1. Since we are asked to use a safety factor of 2.5, to support an actual load of 100 kN, we must have P_{cr} = 250 kN. Designating the length of the side of the square cross section as s and with L_e = 0.5L = 1.5 m,
P_{\mathrm{cr}}=\frac{\pi^2 E I}{L_{\mathrm{e}}^2}=\frac{\pi^2\left(13 \times 10^9 \mathrm{~Pa}\right)\left(1 / 12 \mathrm{~s}^4\right)}{(1.5 \mathrm{~m})^2}=250 \mathrm{kN} .
Then s = 8.5 cm. We must also check that the allowable stress criterion is satisfied:
\sigma=\frac{P}{A}=\frac{100 \mathrm{kN}}{(0.085 \mathrm{~m})^2}=13.8 \mathrm{ksi} .
It is, so the size we have selected is sufficient.
2. Now to support an actual load of 200 kN, we must have P_{cr} = 500 kN.
P_{\mathrm{cr}}=\frac{\pi^2 E I}{L_{\mathrm{e}}^2}=\frac{\pi^2\left(13 \times 10^9 \mathrm{~Pa}\right)\left(1 / 12 \mathrm{~s}^4\right)}{(1.5 \mathrm{~m})^2}=500 \mathrm{kN} .
So s = 10.1 cm. Checking the allowable stress,
\sigma=\frac{P}{A}=\frac{200 \mathrm{kN}}{(0.101 \mathrm{~m})^2}=19.5 \mathrm{ksi} .The size that is sufficient to prevent buckling does not meet the strength criterion for this load. We must redesign a larger column that will meet both criteria using
\sigma_{\text {allow }}=16 \mathrm{MPa}=\frac{200 \mathrm{kN}}{s^2} ,
which gives s = 11.2 cm.