Question 16.CA.1: A 2-m-long pin-ended column with a square cross section is t...
A 2-m-long pin-ended column with a square cross section is to be made of wood (Fig. 16.9). Assuming E = 13 GPa, \sigma_{\text{all}} = 12 MPa, and using a factor of safety of 2.5 to calculate Euler’s critical load for buckling, determine the size of the cross section if the column is to safely support (a) a 100-kN load, (b) a 200-kN load.
a. For the 100-kN Load. Use the given factor of safety to obtain
P_{cr} = 2.5(100 kN)= 250 kN L = 2 m E = 13 GPa
Use Euler’s formula, Eq. (16.11a), and solve for I:
P_{cr}=\frac{\pi^2 E I}{L^2} (16.11a)
I=\frac{P_{cr} L^2}{\pi^2 E}=\frac{\left(250 \times 10^3 N\right)(2 m)^2}{\pi^2\left(13 \times 10^9 Pa\right)}=7.794 \times 10^{-6} m^4Recalling that, for a square of side a, I=a^4 / 12, write
\frac{a^4}{12}=7.794 \times 10^{-6} m^4 \quad a=98.3 mm \approx 100 mmCheck the value of the normal stress in the column:
\sigma=\frac{P}{A}=\frac{100 kN}{(0.100 m)^2}=10 MPaBecause σ is smaller than the allowable stress, a 100 × 100-mm cross section is acceptable.
b. For the 200-kN Load. Solve Eq. (16.11a) again for I, but make P_{cr}=2.5(200)=500 kN to obtain
\begin{aligned} I=15.588 \times 10^{-6} m^4\\\frac{a^4}{12}=15.588 \times 10^{-6} \quad a=116.95 mm\end{aligned}The value of the normal stress is
\sigma=\frac{P}{A}=\frac{200 kN}{(0.11695 m)^2}=14.62 MPaBecause this is larger than the allowable stress, the dimension obtained is not acceptable, and the cross section must be selected on the basis of its resistance to compression.
\begin{aligned}&A=\frac{P}{\sigma_{\text {all}}}=\frac{200 kN}{12 MPa}=16.67 \times 10^{-3} m^2 \\&a^2=16.67 \times 10^{-3} m^2 \quad a=129.1 mm\end{aligned}A 130 × 130-mm cross section is acceptable.