Problem 7.8. Lateral-torsional buckling

Critical bending moment, M_cr [kNm]=

Step 1.  Calculate the section properties.

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Section properties were calculated in the previous Problem.

$$\begin{gathered} A=2\times 100\times 3+150\times 3=1050 {\mathrm{mm}}^2 \\ {I}_z=2\frac{{100}^3\times 3}{12}+\frac{150\times 5^3}{12}=5.003\times {10}^5 {\mathrm{mm}}^4 \\ {i}_{\mathrm{o}}^2=\frac{I_z}{A}+\frac{I_y}{A}=\frac{5.003\times {10}^5}{1050}+\frac{4.356\times {10}^6}{1050}=4624.6 {\mathrm{mm}}^2 \\ {I}_{\omega }=\frac{b_{\mathrm{f}}^3{t}_{\mathrm{f}}}{24}{d}^2=\frac{{100}^3\times 3}{24}{153}^2=2.926\times {10}^9 {\mathrm{mm}}^6 \\ {I}_{\mathrm{t}}=2\frac{100\times 3^3}{3}+\frac{150\times 3^3}{3}=3150 {\mathrm{mm}}^4 \end{gathered}$$

Step 2.  Give the buckling loads of the beam.

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Results of the previous problem are modified by the different buckling length

$$\begin{gathered} N_{y,\mathrm{cr}}=\frac{{\pi }^{2}E{I}_z}{L^2}=\frac{{\pi }^2\times 210\times {10}^3\times 5.003\times {10}^5}{{\left(4.0\times {10}^3\right)}^2}=64.81\times {10}^3 \mathrm{N}=64.81 \mathrm{kN} \\ {N}_{\omega ,\mathrm{cr}}=\frac{1}{i_{\mathrm{o}}^2}\left(\frac{{\pi }^{2}E{I}_{\omega }}{L^2}+G{I}_{\mathrm{t}}\right)=\frac{1}{4624.6}\left(\frac{{\pi }^{2}210\times {10}^3\times 2.926\times {10}^9}{4.0\times {10}^3}+87.5\times {10}^3\times 3150\right)=141.6\times {10}^3 \mathrm{N}=141.6 \mathrm{kN} \end{gathered}$$

Step 3.  Determine the critical bending moment.

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Eq.(7-158) and Table 7.6.

$$\begin{gathered} {\mathit{M}}_{\mathbf{cr}}=1.13N_{y,\mathrm{cr}}\sqrt{i_{\mathrm{pt}}^2\frac{N_{\omega ,\mathrm{cr}}}{N_{y,\mathrm{cr}}}}=1.13\times 64.81\times {10}^3\sqrt{4624.6\frac{141.6}{64.81}}=7.361\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{361}\mathbf{ }\mathbf{kNm} \\ \mathrm{where} i_{\mathrm{pt}}^2=i_{\mathrm{o}}^2 \end{gathered}$$

The cross section is doubly symmetrical (β = 0) and the load acts above the shear center (in the symmetry axis), Δ = 0. 

Step 4.  Check the beam for lateral-torsional buckling.

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The section properties are given below.

See in the previous Problem.

The buckling loads of the beam are 

Results of the previous problem are modified by the different buckling length

The critical bending moment is determined as follows

Eq.(7-158) and Table 7.6.

$$\begin{gathered} {\mathit{M}}_{\mathbf{cr}}=1.13N_{y,\mathrm{cr}}\sqrt{i_{\mathrm{pt}}^2\frac{N_{\omega ,\mathrm{cr}}}{N_{y,\mathrm{cr}}}}=1.13\times 64.81\sqrt{4624.6\frac{141.6}{64.81}}=7.361\times {10}^6 \mathrm{Nmm}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{361}\mathbf{ }\mathbf{kNm} \\ \mathrm{where} i_{\mathrm{pt}}^2=i_{\mathrm{o}}^2 \end{gathered}$$

The cross section is doubly symmetrical (β = 0) and the load acts above the shear center (in the symmetry axis), Δ = 0. 

$M_{\max }=\frac{p{L}^2}{8}=\frac{0.819\times 4.0^2}{8}=0.1638 \mathrm{kNm} < M_{\mathrm{cr}}=7.361 \mathrm{kNm}$

The beam is safe for lateral-torsional buckling.