Question 11.2: A viewing platform in a wild-animal park (Fig. 11-20a) is su......
A viewing platform in a wild-animal park (Fig. 11-20a) is supported by a row of aluminum pipe columns having length L = 3.25 m and outer diameter d = 100 mm. The bases of the columns are set in concrete footings and the tops of the columns are supported laterally by the platform. The columns are being designed to support compressive loads P = 100 kN. Determine the minimum required thickness t of the columns (Fig. 11-20b) if a factor of safety n = 3 is required with respect to Euler buckling. (For the aluminum, use E = 72 GPa for the modulus of elasticity and use σ_{pl} = 480 MPa for the proportional limit.)
Because of the manner in which the columns are constructed, we will model each column as a fixed-pinned column (see Fig. 11-19d). Therefore, the critical load is
P_{cr}=\frac{2.046π^2EI}{L^2} (j)
in which I is the moment of inertia of the tubular cross section:
I=\frac{π}{64}[d^4\ -\ (d\ -\ 2t)^4] (k)
Substituting d = 100 mm (or 0. 1 m), we get
I=\frac{π}{64}[(0.1\ m)^4\ -\ (0.1\ m\ -\ 2t)^4] (l)
in which I is expressed in meters.
Required thickness of the columns. Since the load per column is 100 kN and the factor of safety is 3. each column must be designed for the following critical load:
P_{cr} = nP = 3(100 kN) = 300 kN
Substituting this value for P_{cr} in Eq. (j), and also replacing I with its expression from Eq. (1), we obtain
300,000\ N=\frac{2.046π^2(72×10^9\ Pa)}{(3.25\ m)^2}\left(\frac{π}{64}\right)[(0.1\ m)^4\ -\ (0.1\ m\ -\ 2t)^4]
Note that all terms in this equation are expressed in units of newtons and meters.
After multiplying and dividing, the preceding equation simplifies to
44.40×10^{-6}\ m^4=(0.1\ m)^4\ -\ (0.1\ m\ -\ 2t)^4
or (0.1\ m\ -\ 2t)^4=(0.1\ m)^4\ -\ 44.40×10^{-6}\ m^4=55.60×10^{-6}\ m^4
from which we obtain
0.1 m – 2t = 0.08635 m and t = 0.006825 m
Therefore, the minimum required thickness of the column to meet the specified conditions is
t_{min} = 6.83 mm
Notes: Knowing the diameter and thickness of the column, we can now calculate its moment of inertia, cross-sectional area, and radius of gyration. Using the minimum thickness of 6.83 mm, we obtain
I=\frac{π}{64}[d^4\ -\ (d\ -\ 2t)^4]=2.18×10^{6}\ mm^4\\A=\frac{π}{4}[d^2\ -\ (d\ -\ 2t)^2]=1999\ m^2\quad \quad r=\sqrt{\frac{I}{A}} = 33.0 mm
The slenderness ratio L/r of the column is approximately 98, which is in the customary range for slender columns, and the diameter-to-thickness ratio d/t is approximately 15, which should be adequate to prevent local buckling of the walls of the column.
The critical .stress in the column must be less than the proportional limit of the aluminum if the formula for the critical load (Eq. j) is to be valid. The critical stress is
σ_{cr}=\frac{P_{cr}}{A}=\frac{300\ kN}{1999\ mm^2} = 150 MPa
which is less than the proportional limit (480 MPa). Therefore, our calculation for the critical load using Euler buckling theory is satisfactory.
