Question 26.17: The I-section beam shown in Fig. P.26.17 carries bending mom......
The I-section beam shown in Fig. P.26.17 carries bending moments M_{X} = 0.2 kNm and M_{Y} = 0.2 kNm. If for the flanges E_{x} = 54,100 N/mm² and for the web E_{x} = 17,700 N/mm², determine the distribution of axial force intensity.
The modified section properties are, from Eqs (26.67),
I_{X X}^{\prime}=\int_{A}E_{Z,i}Y^{2}\;\mathrm{d}A,\quad I_{Y Y}^{\prime}=\int_{A}E_{Z,i}X^{2}\;\mathrm{d}A,\quad I_{X Y}^{\prime}=\int_{A}E_{Z,i}X Y\;\mathrm{d}A. (26.67)
I_{X X}^{\prime}=2\times54100\times100\times1\times50^{2}+17700\times100^{3}/12=2.8\times10^{10} \mathrm{Nmm^{2}}I_{Y Y}^{\prime}=2\times54100\times1\times100^{3}/12=0.9\times10^{10} \mathrm{Nmm^{2}}
I_{X Y}^{\prime}=0 since the section is doubly symmetrical about the X and Y axes.
Eq. (26.68) then reduces to
\sigma_{Z}=E_{Z,i}\left[\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}^{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}^{2}}}\right)Y\right] (26.68)
\sigma_{Z}=E_{Z,i}\left({\frac{M_{Y}X}{I_{\mathrm{YY}}^{\prime}}}+{\frac{M_{X}Y}{I_{XX}^{\prime}}}\right) (i)
Substituting in Eq. (i) for M_{Y},\,I_{Y Y}^{\prime} etc. and noting that E_{\mathrm{Z},i}=54100\mathrm{~N/mm^{2},}
\sigma_{Z}=1.2X+0.39Y (ii)
Then, at 1,
\sigma_{Z,1}=1.2\,(-50)+0.39\,(50)=-40.5\,\mathrm{N/mm^{2}}and at 2,
\sigma_{Z,2}=1.2\left(0\right)+0.39\left(50\right)=19.5\,\mathrm{N/mm^{2}}
The corresponding axial force intensities are
For the web the values of M_{Y},I_{Y Y}^{\prime} etc. will be the same as those for the flanges but E_{Z,i}=17700\,\mathrm{N/mm^{2}}.
Eq. (i) then becomes
\sigma_{Z}(\mathrm{web})=0.39X+0.13Y (iii)
Then
\sigma_{Z,2}=0.13\,(50)=6.5\,\mathrm{N/mm^{2}}=-\sigma_{Z,5}The corresponding axial force intensities are then
N_{2}=6.5\times0.5=3.25\,\mathrm{N/mm}=-N_{5}