Question 10.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. 10.9). 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_{\text {cr }}=2.5(100 kN )=250 kN \quad L=2 m \quad E=13 GPa
Use Euler’s formula, Eq. (10.11a), and solve for I:
P_{ cr }=\frac{\pi^2 E I}{L^2} (10.11a)
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 ^4
Recalling 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 mm
Check the value of the normal stress in the column:
\sigma=\frac{P}{A}=\frac{100 kN }{(0.100 m )^2}=10 MPa
Since σ is smaller than the allowable stress, a 100 × 100-mm cross section is acceptable.
b. For the 200-kN Load. Solve Eq. (10.11a) again for I, but make P_{ cr } = 2.5(200) = 500 kN to obtain
\begin{gathered} I=15.588 \times 10^{-6} m ^4 \\ \frac{a^4}{12}=15.588 \times 10^{-6} \quad a=116.95 mm \end{gathered}
The value of the normal stress is
\sigma=\frac{P}{A}=\frac{200 kN }{(0.11695 m )^2}=14.62 MPa
Since 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.