COLUMN BUCKLING EXAMPLE — I-SECTION POSTS SUPPORTING 110 kg
Two vertical I-section posts support a structure carrying a centered mass
$m = 110\,\text{kg}$. Column height $L = 1.0\,\text{m}$. Each post uses the section in the sketch:
overall depth $h = 119\,\text{mm}$, flange width $b_f = 76\,\text{mm}$, flange thickness $t_f = 7.6\,\text{mm}$,
web thickness $t_w = 4\,\text{mm}$. Assume steel with $E = 200\,\text{GPa}$ and $S_y = 345\,\text{MPa}$.
End restraints are approximated as Fixed–Pinned $\Rightarrow K = 0.7$.
Frame with two I-section posts; centered mass $110\,\text{kg}$; column height $1\,\text{m}$.
Given Data
Inputs
| INPUT SUMMARY |
|---|
| Mass carried $[m]$ | 110 | kg |
| Column height $[L]$ | 1.0 | m |
| Section dims $[h \times b_f \times t_f \times t_w]$ | 119 × 76 × 7.6 × 4 | mm |
| Elastic modulus $[E]$ | 200 | GPa |
| Yield strength $[S_y]$ | 345 | MPa |
| End condition | Fixed–Pinned $\Rightarrow K = 0.7$ (assumed) |
Section Properties (about centroid)
-
Area
$$A = 2\,b_f t_f + t_w\,(h - 2 t_f)
= 2(76)(7.6) + 4\,(119 - 2\cdot7.6)
= \mathbf{1{,}570.4\ \text{mm}^2}.$$
-
Second moments
$$d = \frac{h}{2} - \frac{t_f}{2}, \qquad
I_x = 2\!\left(\frac{b_f t_f^3}{12} + b_f t_f\, d^2\right)
+ \frac{t_w (h - 2 t_f)^3}{12},$$
$$I_y = 2\!\left(\frac{t_f b_f^3}{12}\right)
+ \frac{(h - 2 t_f) t_w^3}{12},$$
$$I_x \approx 3.962\times 10^6\ \text{mm}^4,\qquad
I_y \approx 5.566\times 10^5\ \text{mm}^4.$$
-
Radii of gyration
$$r_x = \sqrt{\frac{I_x}{A}} \approx 50.23\ \text{mm},\qquad
r_y = \sqrt{\frac{I_y}{A}} \approx 18.83\ \text{mm},$$
governing axis: $r_{\min} = r_y$.
-
Slenderness
$$S = \frac{K L}{r_{\min}} = \frac{0.7 \cdot 1000}{18.83}
\approx \mathbf{37.18}.$$
-
Johnson/Euler boundary (steel)
$$S_{cr} = \sqrt{\frac{2\pi^2 E}{S_y}}
= \sqrt{\frac{2\pi^2 \cdot 200{,}000}{345}}
\approx \mathbf{106.97}.$$
Since $S < S_{cr}$, use Johnson parabolic.
Buckling Checks (Per Column)
-
Johnson parabolic stress
$$\sigma_J = S_y \left[ 1 - \frac{S_y}{4\pi^2 E}\!
\left(\frac{K L}{r_{\min}}\right)^2 \right]
= 345 \left[ 1 - \frac{345}{4\pi^2 \cdot 200{,}000}
\cdot (37.18)^2 \right]
\approx \mathbf{324.16\ \text{MPa}},$$
$$P_J = \sigma_J\,A \approx 324.16 \times 1570.4
\approx \mathbf{509.06\ \text{kN}}.$$
-
Euler elastic load (reference)
$$P_E = \frac{\pi^2 E\, I_{\min}}{(K L)^2}
= \frac{\pi^2 \cdot 200{,}000 \cdot 5.566\times 10^5}
{(0.7 \cdot 1000)^2}
\approx \mathbf{2242.17\ \text{kN}}.$$
-
Yield (squash) load (reference)
$$P_y = S_y\,A = 345 \times 1570.4
\approx \mathbf{541.79\ \text{kN}}.$$
-
Governing capacity
$$P_{cr} = \min(P_J, P_E, P_y) = \mathbf{509.06\ \text{kN}} \quad(\text{Johnson}).$$
Results & Safety
| RESULTS (PER COLUMN) |
|---|
|
$$W = m g = 110 \times 9.80665 = 1.079\ \text{kN (total)}$$
$$W_{\text{per column}} = \mathbf{0.539\ \text{kN}}$$
(applied reaction, shared by two identical supports)
|
|
$$P_{cr} = \mathbf{509.06\ \text{kN}}$$
(critical load, Johnson parabolic governs)
|
|
$$P_E = 2242.17\ \text{kN}$$
(Euler elastic load, not governing)
|
|
$$P_y = 541.79\ \text{kN}$$
(yield load, not governing)
|
|
$$F_a=\frac{P_{cr}}{n_d}=\mathbf{101.81\ \text{kN}}$$
(allowable load, design factor $n_d=5$ → service check)
|
|
$$\text{fos}=\frac{P_{cr}}{W_{\text{per column}}}\approx \mathbf{944}$$
(factor of safety, very high)
|
The large margin reflects a stout section and a short, well-restrained column ($K{=}0.7$).
If your base or head connections are more flexible, use a larger $K$ (e.g., pinned–pinned $K{\approx}1.0$) and recheck.
Calculator Cross-Check
Comparison with Column Buckling Calculator
| RESULTS COMPARISON |
|---|
| Quantity | Formula Result | Calculator Result* |
|---|
| $r$ (weak axis) | $18.83$ mm | $18.83$ mm |
| $S$ | $37.18$ | $37.18$ |
| $P_J$ (Johnson) | $509.06$ kN | $509.06$ kN |
| $P_E$ (Euler) | $2242.17$ kN | $2242.17$ kN |
| Governing $P_{cr}$ | $509.06$ kN | $509.06$ kN |
* Using identical inputs in the
Column Buckling Calculator.
FAQ
- What if the load is not centered or the posts are not identical?
- Use the eccentric column (Secant) formula and a frame analysis for reactions. Then check each column with its own $P$ and $K$.
- Which axis controls buckling here?
- The weak axis (about $y$) because $r_y < r_x$. Always use $r_{\min}$ for column slenderness.
- How should I choose $K$?
- Follow your design standard (AISC/EC3). Fixed–Pinned ($K{\approx}0.7$) suits a stiff base plate and pin-like beam connection; Pinned–Pinned is $K{\approx}1.0$ (more conservative).
Related