Question 26.16: If the thin-walled composite beam of Example 26.15 is subjec......
If the thin-walled composite beam of Example 26.15 is subjected to a bending moment of 0.5 kNm applied in a horizontal plane, calculate the maximum value of direct stress in the beam section.
The second moments of area are, from Ex. 26.15,
\begin{array}{l c r}{{I_{X X}^{\prime}=2.63\times10^{10}\mathrm{Nmm^{2}}}}\\ {{I_{Y Y}^{\prime}=0.83\times10^{10}\mathrm{Nmm^{2}}}}\\ {{I_{X Y}^{\prime}=1.25\times10^{10}\mathrm{Nmm^{2}}}}\end{array}Also M_{X}=0\mathrm{~and~}M_{Y}=0.5\mathrm{~kNm} so that Eq. (26.68) becomes
\sigma_{Z}=E_{Z,i}\Biggl[\left({\frac{M_{Y}I_{X X}^{\prime}-M_{X}I_{X Y}^{\prime}}{I_{X X}^{\prime}I_{Y Y}^{\prime}-I_{X Y}^{\prime 2}}}\right)X+\left({\frac{M_{X}I_{Y Y}^{\prime}-M_{Y}I_{X Y}^{\prime}}{I_{X X}^{\prime}I_{Y Y}^{\prime}-I_{~X Y}^{\prime 2}}}\right)Y\Biggr] (26.68)
\sigma_{z}=E_{z,i}\left(2.12\times10^{-4}\,X-1.01\times10^{-4}\,Y\right) (i)
On the top flange, E_{z,I}=50~{\mathrm{000~N/mm}}^{2}~{\mathrm{and~}}Y=50~{\mathrm{mm}}. Then, from Eq. (i),
\sigma_{z}=10.6\,X-252.5so that at 1 where X=50 mm,
\sigma_{z,1}=277.5\,\mathrm{N/mm^{2}}and at 2 where X=0,
\sigma_{z,2}=-252.5\,\mathrm{N/mm^{2}}In the web E_{z,i}=15\ 000\ \mathrm{N/mm^{2}~a n d}\ X=0. Then
\sigma_{z}=-1.52\,YAt 2,
\sigma_{z,2}=-1.52\times50=-76.0\,\mathrm{N/mm^{2}}The maximum direct stress is therefore 277.5 N/mm² and occurs at point 1.