Question 8.10: A short column has a doubly symmetrical I-section and is fab......

Referring to Fig. S.8.10 and assuming that the flange areas are concentrated at distances d_{1}  and  d_{2} from nn, then Eq.(8.9) becomes

\int_{0}^{d_{1}}\sigma_{x}\mathrm{d}A=\int_{0}^{d_{2}}\sigma_{\text{v}}\ \mathrm{d}A           (8.9)

\sigma_{x}A_{\mathrm{F}}=\sigma_{y}A_{\mathrm{F}}            (i)

and \sigma_{x}=\sigma_{1},\,\sigma_{y}=\sigma_{2}.

But from Eq. (8.12),

{\frac{\mathrm{d}^{2}\text{v}}{\mathrm{d}z^{2}}}={\frac{\sigma_{1}}{E d_{1}}}={\frac{\sigma_{2}}{E_{t}d_{2}}}             (8.12)

\sigma_{1}=E d_{1}\frac{\mathrm{d}^{2}\nu}{\mathrm{d}z^{2}}\mathrm{and}\sigma_{2}=E_{1}d_{2}{\frac{\mathrm{d}^{2}\nu}{\mathrm{d}z^{2}}}

Substitution in Eq.(8.13) then gives

E{\frac{\mathrm{d}^{2}\text{v}}{\mathrm{d}z^{2}}}\int_{0}^{d_{1}}y_{1}\mathrm{d}A-E_{t}{\frac{\mathrm{d}^{2}\text{v}}{\mathrm{d}z^{2}}}\int_{0}^{d^{2}}y_{2}  dA=0           (8.13)

E d_{1}-E_{t}d_{2}=0             (ii)

But d_{1} + d_{2}=d. Then substituting in Eq.(ii) for d_{2} gives

Therefore,

Now I=2A(d^{2}/4)=A d^{2}/2,\,I_{1}=A d_{1}^{2},\,\mathrm{and}\,I_{2}=A d_{2}^{2}. Substituting for d_{1}  and  d_{2} in the expressions for I_{1}  and  I_{2} and then for I, I_{1}  and  I_{2} in Eq. (8.16) and simplifying gives

or

E_r = E\frac{I_1}{I}+ E_t\frac{I_2}{I}           (8.16)