Problem 10.7. Local buckling of a box beam

Buckling load of the cross section, P_cr[kN]=

Step 1.  Determine the buckling loads of the flange and the web assuming hinged supports

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See Table 10.4 1^st row or Eqs.(10-79)-(10-80)

${\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}=\frac{N_{\mathrm{w},\mathrm{cr}}}{t}=\frac{1}{t}\frac{4D{\mathrm{\pi }}^2}{b_{\mathrm{w}}^2}=\frac{1}{5}\frac{4\times 2.4\times {10}^3{\mathrm{\pi }}^2}{{245}^2}=0.316\frac{ \mathrm{kN}}{{\mathrm{mm}}^2}$

${\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}=\frac{N_{\mathrm{f},\mathrm{cr}}}{t}=\frac{1}{t}\frac{4D{\mathrm{\pi }}^2}{b_{\mathrm{f}}^2}=\frac{1}{5}\frac{4\times 2.4\times {10}^3{\mathrm{\pi }}^2}{{145}^2}=0.901\frac{ \mathrm{kN}}{{\mathrm{mm}}^2}$

Web buckling is relevant.

Step 2.  Determine the spring constant.

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Eqs.(10-82)-(10-83)

$$\begin{gathered} k=\frac{2D}{b_{\mathrm{f}}}\frac{1}{\psi }=\frac{2\times 2.4}{0.145}\frac{1}{0.650}=21.51 \mathrm{kN} \\ \mathrm{where} \\ \frac{1}{\psi }=1–\frac{{\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}}{{\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}}=1–\frac{0.316}{0.901}=0.650 \end{gathered}$$

Step 3.  Determine the buckling load of the rotationally restrained web.

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See Table 10.4 4^th row or Eqs.(10-81)

$$\begin{gathered} N_{x,\mathrm{cr}}=\frac{{\mathrm{\pi }}^2}{{\left(b_w\right)}^2}\left[2\sqrt{1+4.139\xi }+2+0.44{\xi }^2\right]D= \\ =\frac{{\mathrm{\pi }}^2}{0.{245}^2}\left[2\sqrt{1+4.139\times 0.18}+2+0.44\times 0.{18}^2\right]\times 2.4=1837 \mathrm{kN}/\mathrm{m} \\ \mathrm{where} \\ \mathrm{\varsigma }=\frac{D}{k{b}_{\mathrm{w}}}=\frac{2.4}{21.51\times 0.245}=0.455 \mathrm{and} \xi =\frac{1}{1+10\varsigma }=0.18 \end{gathered}$$

Step 4.  Give the critical force of the total cross section.

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First hinged connections are assumed. The buckling loads of the flange and the web are

See Table 10.4 1^st row or Eqs.(10-79)-(10-80)

${\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}=\frac{N_{\mathrm{w},\mathrm{cr}}}{t}=\frac{1}{t}\frac{4D{\mathrm{\pi }}^2}{b_{\mathrm{w}}^2}=\frac{1}{5}\frac{4\times 2.4\times {10}^3{\mathrm{\pi }}^2}{{245}^2}=0.316\frac{ \mathrm{kN}}{{\mathrm{mm}}^2}$

${\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}=\frac{N_{\mathrm{f},\mathrm{cr}}}{t}=\frac{1}{t}\frac{4D{\mathrm{\pi }}^2}{b_{\mathrm{f}}^2}=\frac{1}{5}\frac{4\times 2.4\times {10}^3{\mathrm{\pi }}^2}{{145}^2}=0.901\frac{ \mathrm{kN}}{{\mathrm{mm}}^2}$

Web buckling is relevant. Now we calculate the buckling load of the rotationally restrained web. The spring constant represents the restraining effect of flanges must be determined.

Eqs.(10-82)-(10-83)

$$\begin{gathered} k=\frac{2D}{b_{\mathrm{f}}}\frac{1}{\psi }=\frac{2\times 2.4}{0.145}\frac{1}{0.650}=21.51 \mathrm{kN} \\ \mathrm{where} \\ \frac{1}{\psi }=1–\frac{{\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}}{{\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}}=1–\frac{0.316}{0.901}=0.650 \end{gathered}$$

The buckling load of the rotationally restrained web is

See Table 10.4 4^th row or Eqs.(10-81)

The critical force of the total cross section results in:

${\mathit{P}}_{\mathbf{c}\mathbf{r}}=\frac{N_{x,\mathrm{cr}}}{t}A=\frac{1837.5}{5\times {10}^{–3}}\left(150\times 5\times 2+220\times 5\right)\times {10}^{–6}\mathbf{=}\mathbf{1433}\mathbf{ }\mathbf{kN}$