The thin-walled composite beam section of Example 26.15 carries a vertical shear load of 2kN applied in the plane of the web. Determine the shear flow distribution.

Question

Accepted Answer

From Ex. 26.15 and Fig. S.26.19 the second moments of area are

[latex]\begin{array}{l c r}{{I_{X X}^{\prime}=2.63 imes10^{10} \mathrm{Nmm^{2}}}}\ {{I_{YY}^{\prime}=0.83 imes10^{10} \mathrm{Nmm^{2}}}}\ {{I_{X Y}^{\prime}=1.25 imes10^{10} \mathrm{Nmm^{2}}}}\end{array}[/latex]

In this case [latex]S_{X}=0,S_{Y}=2\,\mathrm{kN}[/latex] so that Eq. (26.69) becomes

[latex]q_{s}=-E_{Z i}\left[\left(\frac{S_{X}I_{X X}^{\prime}-S_{Y}I_{X Y}^{\prime}}{I_{X X}^{\prime}I_{Y Y}^{\prime}-I_{X Y}^{2}} ight)\int_{0}^{s}t_{t}X\,\mathrm{d}s+\left(\frac{S_{Y}I_{Y Y}^{\prime}-S_{X}I_{X Y}^{\prime}}{I_{X}^{\prime}I_{Y Y}^{\prime}-I_{X Y}^{2}} ight)\int_{0}^{s}t_{i}Y\,\mathrm{d}s ight][/latex] (26.69)

[latex]q_{s}=-E_{Z,i}{\biggl(}-4.03 imes10^{-7}\int_{0}^{s}t_{i}X\,\mathrm{d}s+2.68 imes10^{-7}\int_{0}^{s}t_{i}Y\,\mathrm{d}s{\biggr)}[/latex]

On the top flange, [latex]X=50-s_{1},\,Y=50~\mathrm{mm},\,E_{Z,i}=50~000~\mathrm{N/mm}^{2}.[/latex] Eq. (i) then becomes

[latex]q_{12}=+40.3 imes10^{-3}\int_{0}^{s_{1}}\left(50-s_{1} ight)\!\mathrm{d}s_{1}-26.8 imes10^{-3}\int_{0}^{s_{1}}\!\mathrm{d}s_{1}[/latex]

which gives

[latex]q_{12}=-0.02s_{1}^{2}-1.99s_{1}[/latex]

When [latex]s_{1}=50~\mathrm{mm},~q_{2}=-149.8~\mathrm{N/mm}[/latex]

In the web, [latex]X=0,\,Y=50-s_{2},\,E_{Z,i}=\,15\,\,000\,\,\mathrm{N/mm^{2}}.[/latex] Eq. (i) then becomes

[latex]q_{23}=-8.04 imes10^{-3}\int_{0}^{s_{2}}(50-s_{2})\mathrm{d}s_{2}-149.8[/latex]

so that

[latex]q_{23}=0.40s_{2}+4.02 imes10^{-3}s_{2}^{2}-149.8[/latex]

When [latex]s_{2}=50~\mathrm{mm},~q_{23}=-159.5~\mathrm{N/mm}[/latex]

Question 26.19: The thin-walled composite beam section of Example 26.15 carr......

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