# On the Use of the Effective Width for Simply Supported Generally Loaded Box-Girders
## 5 Average Effective Width
### Displacement Equivalence  
To obtain such an average value, the following strategy is followed: (a) for a given load \(p(z)\), the deflection at the midspan (or at another significant point) is computed via the proposed algorithm, through the series expansion Equation (20), i.e.,  
\[v\biggl{(}\frac{\ell}{2}\biggr{)}=\frac{\ell^{4}}{\pi^{4}E}\sum_{n}\frac{1}{n^ {4}I_{n}}\hat{p}_{n}\sin\Bigl{(}\frac{n\pi}{2}\Bigr{)}; \tag{25}\](b) such a the displacement is equated to \(\vartheta\left(\frac{\ell}{2}\right)\),experienced by a beam whose cross-section has flanges of average effective width \(\tilde{b}_{e}\), i.e., average inertia moment \(\tilde{I}\); (c) from the equality \(\vartheta\left(\frac{\ell}{2}\right)=v\left(\frac{\ell}{2}\right)\), the unknown \(\tilde{b}_{e}\) is derived.  
The procedure is explained in detail for the case of uniformly distributed load \(p(z)=p_{0d}=\text{const}\), for which \(\tilde{p}_{n}\) is given by Equation (9a). Since the midspan displacement of a beam of inertia moment \(\tilde{I}\), under the same load, is \(\vartheta\left(\frac{\ell}{2}\right)=\frac{5}{384}\frac{p_{0d}\ell^{4}}{EI}\), by enforcing the equality, the unknown \(\tilde{I}\) is derived as  
\[\tilde{I}=\frac{5}{768}\frac{\pi^{5}}{\sum_{n}\frac{1}{\frac{\left(1-\left(- \right)^{n}\right)}{n^{5}}\sin\left(\frac{n\pi}{2}\right)}} \tag{26}\]  
Such an expression is still too complicated to draw \(\tilde{b}_{e}\), so that another strong simplification is introduced. It is assumed (only to the scope at hand) that the inertia moment of the reduced cross sections is almost equal to that of the two flanges (i.e., by neglecting the web contributions), whose area is concentrated in their centroids. Accordingly, \(\tilde{I}\simeq\frac{1}{2}\tilde{b}_{e}h_{1}(b_{2}-h_{1})^{2}\) and \(I_{n}=\frac{1}{2}b_{en}h_{1}(b_{2}-h_{1})^{2}\) so that the previous equation provides  
\[\frac{\tilde{b}_{e}}{b_{1}}=\frac{5}{768}\frac{\pi^{5}}{\sum_{n}\frac{\left(1- \left(-\right)\right)^{n}}{n^{2}a_{n}}\sin\left(\frac{n\pi}{2}\right)} \tag{27}\]  
in which \(b_{en}=\alpha_{n}b_{1}\) has been introduced. By remembering Equation (4) for \(\alpha_{n}\), it is concluded that \(\frac{\tilde{b}_{e}}{b_{1}}\) depends on the aspect ratio \(\frac{b_{1}}{\ell}\) and (weakly) on the Poisson ratio. By fixing the latter, a plot of \(\bar{\alpha}=\frac{\tilde{b}_{e}}{b_{1}}\) vs. \(\frac{b_{1}}{\ell}\) can be constructed (see Figure 5a), which is ready for use. Indeed, from known \(\tilde{b}_{e}\), the (now not approximated) inertia moment \(\tilde{I}\) can be computed, and stresses and displacements evaluated, without resorting to series expansions. The accuracy of this simplified method is tested in the next section.  
By repeating the procedure for the concentrated load applied at the midspan (Figure 5a), it is found that  
\[\frac{\tilde{b}_{e}}{b_{1}}=\frac{1}{96}\frac{\pi^{4}}{\sum_{n}\frac{1}{n^{4} a_{n}}\sin^{2}\left(\frac{n\pi}{2}\right)} \tag{28}\]  
and for a segment of load of extension \(\Delta\) centered at the midspan (Figure 5b),  
\[\frac{\tilde{b}_{e}}{b_{1}}=\frac{1}{1536}\frac{\pi^{5}\left(8\delta^{3} \Delta-4\ell\Delta^{3}+A^{4}\right)}{\ell^{4}\sum_{n}\frac{1}{n^{2}a_{n}}\sin \left(\frac{4n\pi}{2\ell}\right)\sin^{2}\left(\frac{n\pi}{2}\right)} \tag{29}\]  
Figure 5: Average shear-lag coefficient \(\bar{\alpha}=\frac{\tilde{b}_{e}}{b_{1}}\) vs. the \(\frac{b_{1}}{\ell}\) aspect ratio: (**a**) uniformly distributed load (DL) and concentrated load at midspan (CLc); (**b**) segment of distributed load centered at midspan (SLc), for different extensions \(\frac{\partial}{\ell}=\frac{1}{10},\frac{1}{4},\frac{1}{2}\); by the displacement equivalence (d.e.), by the stresses equivalence (s.e.). Poisson ratio \(\
u=0.15\).