Question 8.2: A short column of rectangular cross-section, width b and dep...

For a rectangular section of width b Eq. (8.19) becomes

E \int_{0}^{d_{1}} y_{1} d A-E_{ t } \int_{0}^{d_{2}} y_{2} d A=0  (8.19)

Integrating and substituting the limits gives

E d_{1}^{2}-E_{ t } d_{2}^{2}=0  (i)

Substituting in Eq. (i) for d_{2}\left(=d-d_{1}\right) we obtain

Solving this quadratic in d_{1} using the formula and simplifying gives

d_{1}=\frac{d}{1+\sqrt{\left(E / E_{ t }\right)}}  (ii)

It follows that

d_{2}=\frac{d}{1+\sqrt{\left(E_{ t } / E\right)}}  (iii)

Now, in Eq. (8.16)

E_{ r }=E \frac{I_{1}}{I}+E_{ t } \frac{I_{2}}{I}  (8.16)

Substituting for d_{1} and d_{2} in these expressions from Eqs. (ii) and (iii) respectively and then in Eq. (8.16) gives

E_{r}=\frac{4 E E_{ t }}{\left(\sqrt{E_{ t }}+\sqrt{E}\right)^{2}}  (iv)

Note that in the case of elastic failure occurring across the complete cross-section of the column then, in Eq. (iv), E=E_{ t } and E_{ r }=E_{ t }. If no elastic failure occurs then E_{ t }=E and E_{ r }=E.