Problem 7.7. Flexural-torsional buckling

Problem a)

Buckling load in x-y plane, N_y,cr[kN]=

Buckling load in x-z plane, N_z,cr [kN]=

Torsional buckling load, N_ω,cr [kN]=

Problem b)

Lowest buckling load, N_cr,1[kN]=

Second lowest buckling load, N_cr,2 [kN]=

Third lowest buckling load, N_cr,3 [kN]=

Problem a)

Step 1.  Calculate the section properties.

Check section propertiesHide section properties

$$\begin{gathered} A=2\times 100\times 3+150\times 3=1050 {\mathrm{mm}}^2 \\ {I}_y=\frac{100\times {156}^3}{12}–\frac{97\times {150}^3}{12}=4.356\times {10}^6 {\mathrm{mm}}^4 \\ {I}_z=2\frac{{100}^3\times 3}{12}+\frac{150\times 5^3}{12}=5.003\times {10}^5 {\mathrm{mm}}^4 \\ {i}_{\mathrm{o}}^2=\frac{I_z}{A}+\frac{I_y}{A}=\frac{5.003\times {10}^5}{1050}+\frac{4.356\times {10}^6}{1050}=4624.6 {\mathrm{mm}}^2 \end{gathered}$$

Square of the radius of gyration is given by Equation (7-123)

$I_{\omega }=\frac{b_{\mathrm{f}}^3{t}_{\mathrm{f}}}{24}{d}^2=\frac{{100}^3\times 3}{24}{153}^2=2.926\times {10}^9 {\mathrm{mm}}^6$

See Table 3.12.

$I_{\mathrm{t}}=2\frac{100\times 3^3}{3}+\frac{150\times 3^3}{3}=3150 {\mathrm{mm}}^4$

Step 2.  Determine buckling loads for flexural-torsional buckling.

Check buckling loadsHide buckling loads

The cross section is doubly symmetrical, thus the buckling loads of the cantilever are

Equation (7-133)

$$\begin{gathered} {\mathit{N}}_{\mathit{z}\mathbf{,}\mathbf{c}\mathbf{r}}=\frac{{\pi }^{2}E{I}_y}{4L^2}=\frac{{\pi }^2\times 210\times {10}^3\times 4.356\times {10}^6}{4\times 1.5\times {10}^3}=1003.0\times {10}^3 \mathrm{N}\mathbf{=}\mathbf{1003}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{kN} \\ {\mathit{N}}_{\mathit{y}\mathbf{,}\mathbf{cr}}=\frac{{\pi }^{2}E{I}_z}{4L^2}=\frac{{\pi }^2\times 210\times {10}^3\times 5.003\times {10}^5}{4\times 1.5\times {10}^3}=115.2\times {10}^3 \mathrm{N}\mathbf{=}\mathbf{115}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{kN} \\ {\mathit{N}}_{\mathit{\omega }\mathbf{,}\mathbf{cr}}=\frac{1}{i_{\mathrm{o}}^2}\left(\frac{{\pi }^{2}E{I}_{\omega }}{4L^2}+G{I}_{\mathrm{t}}\right)=\frac{1}{4624.6}\left(\frac{{\pi }^{2}210\times {10}^3\times 2.926\times {10}^9}{4\times 1.5\times {10}^3}+87.5\times {10}^3\times 3150\right)=205.3\times {10}^3 \mathrm{N}\mathbf{=}\mathbf{205}\mathbf{.}\mathbf{3}\mathbf{ }\mathbf{kN} \end{gathered}$$

Problem b)

Step 1.  Calculate the section properties.

Check section propertiesHide section properties

$$\begin{gathered} A=2\times 100\times 3+150\times 3=1050 {\mathrm{mm}}^2 \\ {I}_y=\frac{100\times {156}^3}{12}–\frac{97\times {150}^3}{12}=4.356\times {10}^6 {\mathrm{mm}}^4 \\ {y}_{\mathrm{c}}=\frac{3\times 100\times 50\times 2+150\times 3\times 1.5}{A}=29.21 \mathrm{mm} \\ {I}_z=2\frac{3\times {100}^3}{12}+2\times 3\times 100(50–29.21{)}^2+\frac{150\times 3^3}{12}+3\times 150(29.21–1.5{)}^2=1.105\times {10}^6 {\mathrm{mm}}^4 \\ {I}_{\mathrm{t}}=2\frac{100\times 3^3}{3}+\frac{150\times 3^3}{3}=3150 {\mathrm{mm}}^4 \\ {I}_{\omega }=\frac{b_{\mathrm{f}}^3{t}_{\mathrm{f}}{d}^2}{12}\frac{3b_{\mathrm{f}}{t}_{\mathrm{f}}+2d{t}_{\mathrm{w}}}{6b_{\mathrm{f}}{t}_{\mathrm{f}}+d{t}_{\mathrm{w}}}=\frac{{100}^3\times 3\times {153}^2}{12}\frac{3\times 100\times 3+2\times 153\times 3}{6\times 100\times 3+153\times 3}=4.71\times {10}^9 {\mathrm{mm}}^6 \\ e=\frac{3b_{\mathrm{f}}^2{t}_{\mathrm{f}}}{6b_{\mathrm{f}}{t}_{\mathrm{f}}+d{t}_{\mathrm{w}}}=\frac{3\times {100}^2\times 3}{6\times 100\times 3+153\times 3}=39.84 \mathrm{mm} \end{gathered}$$

See Table 3.12.

$$\begin{gathered} i_{\mathrm{o}}^2=\frac{I_y}{A}+\frac{I_z}{A}==\frac{1.105\times {10}^6}{1050}+\frac{4.356\times {10}^6}{1050}=5200.7 {\mathrm{mm}}^2 \\ {y}_{\mathrm{sc}}=y_{\mathrm{c}}+e–1.5=29.21+39.84–1.5=67.55 \mathrm{mm} \\ {i}_{\mathrm{pt}}^2=\frac{I_y}{A}+\frac{I_z}{A}+y_{\mathrm{sc}}^2==\frac{1.105\times {10}^6}{1050}+\frac{4.356\times {10}^6}{1050}+67.{55}^2=9764.4 {\mathrm{mm}}^2 \end{gathered}$$

Equations (7-123) and (7-139)

Step 2.  Determine buckling loads for flexural-torsional buckling.

Check buckling loadsHide buckling loads

The displacement are assumed in the following forms:

$v_{\mathrm{o}}=A\sin \frac{\mathrm{\pi }}{L}x, w_{\mathrm{o}}=B\sin \frac{\mathrm{\pi }}{L}x, {\psi }_{\mathrm{o}}=C\sin \frac{\mathrm{\pi }}{L}x$

Substituting the displacement functions the governing equations of the beam result in

Equations (7-140)-(7-143)

$$\begin{gathered} \left[\begin{array}{ccc}{N}_{y,\mathrm{cr}}–N& & \\ & N_{z,\mathrm{cr}}–N& N{y}_{\mathrm{sc}}\\ & N{y}_{\mathrm{sc}}& \left(N_{\omega ,\mathrm{cr}}–N\right)i_{\mathrm{pt}}^2\end{array}\right]\left\{\begin{array}{c}A\\ B\\ C\end{array}\right\}=0, \mathrm{where} \\ {N}_{z,\mathrm{cr}}=\frac{{\pi }^{2}E{I}_y}{4L^2}=\frac{{\pi }^2\times 210\times {10}^3\times 4.356\times {10}^6}{4\times 1.5\times {10}^3}=1003.0\times {10}^3 \mathrm{N}=1003.0 \mathrm{kN} \\ {N}_{y,\mathrm{cr}}=\frac{{\pi }^{2}E{I}_z}{4L^2}=\frac{{\pi }^2\times 210\times {10}^3\times 1.105\times {10}^6}{4\times 1.5\times {10}^3}=254.5\times {10}^3 \mathrm{N}=254.5 \mathrm{kN} \\ {N}_{\omega ,\mathrm{cr}}=\frac{1}{i_{\mathrm{o}}^2}\left(\frac{{\pi }^{2}E{I}_{\omega }}{4L^2}+G{I}_{\mathrm{t}}\right)= \\ =\frac{1}{5200.7}\left(\frac{{\pi }^{2}210\times {10}^3\times 4.71\times {10}^9}{4\times 1.5\times {10}^3}+87.5\times {10}^3\times 3150\right)=261.5\times {10}^3 \mathrm{N}=261.5 \mathrm{kN} \end{gathered}$$

The buckling loads as the eigenvalues of the above matrix result in

Equation (7-146)

$$\begin{gathered} {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{2}}=N_{y,\mathrm{cr}}\mathbf{=}\mathbf{254}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{kN} \\ \left(N_{\mathrm{cr}}–N_{z,\mathrm{cr}}\right)\left(N_{\mathrm{cr}}–N_{\mathrm{\omega },\mathrm{cr}}\right)i_{\mathrm{pt}}^2\mathit{–}{N}_{\mathit{cr}}^{\mathit{2}}{y}_{\mathit{sc}}^{\mathit{2}}=0 \to N_{\mathrm{cr},2}, N_{\mathrm{cr},3} \\ {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{1}}\mathbf{=}\mathbf{229}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{kN} \\ {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{3}}\mathbf{=}\mathbf{2144}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{kN} \end{gathered}$$

Problem a)

The section properties are the following

$$\begin{gathered} A=2\times 100\times 3+150\times 3=1050 {\mathrm{mm}}^2 \\ {I}_y=\frac{100\times {156}^3}{12}–\frac{97\times {150}^3}{12}=4.356\times {10}^6 {\mathrm{mm}}^4 \\ {I}_z=2\frac{{100}^3\times 3}{12}+\frac{150\times 5^3}{12}=5.003\times {10}^5 {\mathrm{mm}}^4 \\ {i}_{\mathrm{o}}^2=\frac{I_z}{A}+\frac{I_y}{A}=\frac{5.003\times {10}^5}{1050}+\frac{4.356\times {10}^6}{1050}=4624.6 {\mathrm{mm}}^2 \end{gathered}$$

Square of the radius of gyration is given by Equation (7-123)

See Table 3.12.

The cross section is doubly symmetrical, thus the buckling loads of the cantilever are

Equation (7-133)

Problem b)

The section properties are calculated below:

$$\begin{gathered} A=2\times 100\times 3+150\times 3=1050 {\mathrm{mm}}^2 \\ {I}_y=\frac{100\times {156}^3}{12}–\frac{97\times {150}^3}{12}=4.356\times {10}^6 {\mathrm{mm}}^4 \\ {y}_{\mathrm{c}}=\frac{3\times 100\times 50\times 2+150\times 3\times 1.5}{A}=29.21 \mathrm{mm} \\ {I}_z=2\frac{3\times {100}^3}{12}+2\times 3\times 100(50–29.21{)}^2+\frac{150\times 3^3}{12}+3\times 150(29.21–1.5{)}^2=1.105\times {10}^6 {\mathrm{mm}}^4 \\ {I}_{\mathrm{t}}=2\frac{100\times 3^3}{3}+\frac{150\times 3^3}{3}=3150 {\mathrm{mm}}^4 \\ {I}_{\omega }=\frac{b_{\mathrm{f}}^3{t}_{\mathrm{f}}{d}^2}{12}\frac{3b_{\mathrm{f}}{t}_{\mathrm{f}}+2d{t}_{\mathrm{w}}}{6b_{\mathrm{f}}{t}_{\mathrm{f}}+d{t}_{\mathrm{w}}}=\frac{{100}^3\times 3\times {153}^2}{12}\frac{3\times 100\times 3+2\times 153\times 3}{6\times 100\times 3+153\times 3}=4.71\times {10}^9 {\mathrm{mm}}^6 \\ e=\frac{3b_{\mathrm{f}}^2{t}_{\mathrm{f}}}{6b_{\mathrm{f}}{t}_{\mathrm{f}}+d{t}_{\mathrm{w}}}=\frac{3\times {100}^2\times 3}{6\times 100\times 3+153\times 3}=39.84 \mathrm{mm} \end{gathered}$$

See Table 3.12.

Equations (7-123) and (7-139)

The displacement are assumed in the following forms:

$v_{\mathrm{o}}=A\sin \frac{\mathrm{\pi }}{L}x, w_{\mathrm{o}}=B\sin \frac{\mathrm{\pi }}{L}x, {\psi }_{\mathrm{o}}=C\sin \frac{\mathrm{\pi }}{L}x$

Substituting the displacement functions the governing equations of the beam result in

Equations (7-140)-(7-143)

The buckling loads as the eigenvalues of the above matrix result in

Equation (7-146)

$$\begin{gathered} {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{2}}=N_{y,\mathrm{cr}}\mathbf{=}\mathbf{254}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{kN} \\ \left(N_{\mathrm{cr}}–N_{z,\mathrm{cr}}\right)\left(N_{\mathrm{cr}}–N_{\mathrm{\omega },\mathrm{cr}}\right)i_{\mathrm{pt}}^2\mathit{–}{N}_{\mathit{cr}}^{\mathit{2}}{y}_{\mathit{sc}}^{\mathit{2}}=0 \to N_{\mathrm{cr},2}, N_{\mathrm{cr},3} \\ {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{1}}\mathbf{=}\mathbf{229}\mathbf{.}\mathbf{7}\mathbf{ }\mathbf{kN} \\ {\mathit{N}}_{\mathbf{cr}\mathbf{,}\mathbf{3}}\mathbf{=}\mathbf{2144}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{kN} \end{gathered}$$