Question 11.6: Find the minimum required thickness tmin for a steel pipe co......

We will use the AISC formulas (Eqs. 11-79 through 11-82) when analyzing this column. Since the column has fixed-free end conditions, the effective length is

n_1=\frac{5}{3}+\frac{3\left(\frac{KL}{r}\right)}{8\left(\frac{KL}{r}\right)_c}\ -\ \frac{\left(\frac{KL}{r}\right)^3}{8\left(\frac{KL}{r}\right)_c^3}\quad \quad \frac{KL}{r}≤\left(\frac{KL}{r}\right)_c                           (11-79)

\frac{σ_{allow}}{σ_y}=\frac{\left(\frac{KL}{r}\right)^2_c}{2n_2\left(\frac{KL}{r}\right)^2}\quad \quad \frac{KL}{r}≤\left(\frac{KL}{r}\right)_c                            (11-82)

KL = 2(3.6 m) = 7.2 m

Also, the critical slenderness ratio (Eq. 11-76) is

\left(\frac{KL}{r}\right)_c=\sqrt{\frac{2π^2E}{σ_y}}=\sqrt{\frac{2π^2(200\ GPa)}{250\ MPa}} = 125.7                    (b)

To determine the required thickness of the column, we will use a trial-and-error method. Let us start by assuming a trial value t = 7.0 mm. Then the moment of inertia of the cross-sectional area is

I=\frac{π}{64}[d^4\ -\ (d\ -\ 2t)^4]=\frac{π}{64}[(160\ mm)^4\ -\ (146\ mm)^4]=9.866×10^6\ mm^4

Also, the cross-sectional area and radius of gyration are

A=\frac{π}{4}[d^4\ -\ (d\ -\ 2t)^2]=\frac{π}{4}[(160\ mm)^2\ -\ (146\ mm)^2] = 3365 mm²

r=\sqrt{\frac{I}{A}}=\sqrt{\frac{9.866×10^6\ mm^4}{3365\ mm^2}} = 54.15 mm

Therefore, the slenderness ratio of the column is

\frac{KL}{r}=\frac{2(3.6\ m)}{54.15\ mm} = 133.0

Since this ratio is larger than the critical slenderness ratio (Eq. b), we obtain the factor of safety and the allowable stress from Eqs. (11-80) and (11-82):

n_2=\frac{23}{12}≈1.92\quad \quad \frac{KL}{r}≤\left(\frac{KL}{r}\right)_c                           (11-80)

n_2 = 1.92

\frac{σ_{allow}}{σ_y}=\frac{\left(\frac{KL}{r}\right)^2_c}{2n_2\left(\frac{KL}{r}\right)^2}=\frac{(125.7)^2}{2(1.92)(133.0)^2} = 0.2326

σ_{allow}=0.2326σ_{y} = 0.2326(250 MPa) = 58.15 MPa

Thus, the allowable axial load is

P_{allow}=σ_{allow}A = (58.15 MPa)(3365 mm²) = 196 kN

Since this load is less than the required load of 240 kN, we must try a larger value of the thickness t.
Performing calculations similar to those above, but for t = 8 mm and t = 9 mm, we get the following results:

t = 7.0 mm          P_{allow} = 196 kN

t = 8.0 mm          P_{allow} = 220 kN

t = 9.0 mm          P_{allow} = 243 kN

By interpolation, we see that r = 8.9 mm corresponds to a load of 240 kN. Therefore, the required thickness of the pipe column is

t_{min} = 8.9 mm