Problem 10.6. Local buckling with elastically restrained edge

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

See 10^th row of Table 10.4.

Step 1.  Determine the spring constant.

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Eqs.(10-77)-(10-78)

$$\begin{gathered} k=\frac{1}{2}\frac{2D}{b_{\mathrm{w}}}\frac{1}{\psi }=\frac{2.4}{0.225}\frac{1}{0.0425}=0.453 \mathrm{kN} \\ \mathrm{where} \\ \frac{1}{\psi }=1–\frac{{\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}}{{\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}}=1–\frac{1792}{1871.6}=0.0425 \end{gathered}$$

Step 2.  Determine the buckling load of the rotationally restrained flange.

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Step 3.  Give the critical force of the total cross section for local flange buckling.

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See 10^th row of Table 10.4.

First the spring constant is determined.

Eqs.(10-77)-(10-78)

$$\begin{gathered} k=\frac{1}{2}\frac{2D}{b_{\mathrm{w}}}\frac{1}{\psi }=\frac{2.4}{0.225}\frac{1}{0.0425}=0.453 \mathrm{kN} \\ \mathrm{where} \\ \frac{1}{\psi }=1–\frac{{\sigma }_{\mathrm{f},\mathrm{cr}}^{\mathrm{ss}}}{{\sigma }_{\mathrm{w},\mathrm{cr}}^{\mathrm{ss}}}=1–\frac{1792}{1871.6}=0.0425 \end{gathered}$$

The buckling load of the rotationally restrained flange is
$$\begin{gathered} N_{x,\mathrm{cr}}=\frac{D}{{\left(b_f/2\right)}^2}\left[\frac{7}{\sqrt{1+4.12\varsigma }}+6(1–\nu )\right]=\frac{2.4}{0.{075}^2}\left[\frac{7}{\sqrt{1+4.12\times 70.56}}+6(1–0.3)\right]=1967 \mathrm{kN}/\mathrm{m} \\ \mathrm{where} \\ \mathrm{\varsigma }=\frac{D}{k(b_f/2)}=\frac{2.4}{0.453\times 0.075}=70.56 \end{gathered}$$

Finally the critical load, P acting on the cross section which causes the buckling of the flange is given. 

${\mathit{P}}_{\mathbf{c}\mathbf{r}}=\frac{N_{x,\mathrm{cr}}}{t}A=\frac{1966.9\times {10}^{–3}}{5}\left(150\times 5\times 2+220\times 5\right)\mathbf{=}\mathbf{1023}\mathbf{ }\mathbf{kN}$