## 3 Assessment
### Assessment of Pilotis Columns
#### 3.2.2 Deformation and Strength Indices from Flexure
For \(\rm I_{pl}=615\ mm\) all columns demonstrate adequate ductility in the order of 2. Empirically, by following the proposed Equation (4a,b):  
\[\rm\mu_{\phi}=0.45\ \varepsilon_{cu,c}/(\varepsilon_{sy}v_{Ed})\ for\ \
u_{Ed}\geq 0.2\ and\ \rm\mu_{\phi}=0.45\ \varepsilon_{cu,c}/\varepsilon_{sy}\times h/\xi d\ for\ \
u_{Ed}<0.2 \tag{4a}\] Table 1.0  
where, term \(\xi\), from Table 1 should be multiplied with 0.9 (as previously explained), \(\varepsilon_{cu,c}\) = 0.0035, \(\varepsilon_{sy}\) = 0.0025 and \(\theta_{y}=\phi_{y}H/6=\varepsilon_{sy}(H/h)/3=0.64\%\). The \(\rm\mu_{\phi}\) value shall be multiplied by 1.5 to account for the contribution of reinforcement pullout to the rotation capacity. Implementing Equation (4a,b) for example for columns C1/3/7/9 (\(\
u_{\rm Ed}=0.13<0.2\)), result in \(\rm\mu_{\phi}\) = 3.16 and \(\rm\mu_{\phi}\) = 1.5 \(\times\) [0.5 \(\times\) (3.16 + 1)] = 3.1 and \(\rm\theta_{u}\) =3.1\(\cdot\)0.64% = 2%. This value of \(\rm\theta_{u}\) [Table 1, column denoted as (b)] is close to that calculated above in procedure (a) when using \(\rm I_{pl}=615\ mm\) from Equation (3a). The value of the plastic part of the chord rotation capacity of concrete members under cyclic loading is given by the following calibrated Equation (5) (from [33]):  
Table 1.1  
where \(\rm\theta_{u}\) =\(\theta_{y}+\theta_{u}^{Pl}\) (\(\rm\theta_{y}=0.64\%\)). The implementation of this expression is demonstrated for corner columns C1/3/7/9: \(\gamma_{el}=1.8\), \(\alpha_{sl}=1\) because slippage of vertical bars from their anchorage or lap-splice at the lower end of the column is physically possible (presence of splices), \(\
u\) = 0.13, \(\omega=\rm\rho_{sl}f_{sy}/f_{c}=0.53\%\)\(\times\) 500/16 = 0.165, and \(\omega^{\prime}\) (even if \(\rm\rho_{s1}=\rho_{s2}=0.5\rho_{l}=0.53\%\)) is taken double of \(\omega\) due to the presence of splices, \(\rm f_{c}\) = 16 MPa, \(\rm f_{sy}\) = 500 MPa, \(\rm L_{V}\) = 1350 mm, \(\rm h\) = 350 mm, \(\rm\rho_{d}\) = 0 (absence of diagonal reinforcement), \(\rm\rho_{wy}=A_{sy}/(s_{h}b)\) = 2 \(\times\) (\(\pi\times 6^{2}/4\))/(150 \(\times\) 250) = 0.15% (ratio of transverse steel parallel to the direction of loading, here y-direction), \(\rm\alpha_{w}\) = 0.14 (confinement effectiveness factor, limited contribution because stirrrups are sparse) and \(\rm f_{yw}\) = 240 MPa. This results in \(\rm\theta_{u}^{pl}\) = 2.33% and \(\rm\theta_{u}\) = 0.64% + 2.33% \(\approx\) 3%. For the peripheral columns the calculations lead to \(\rm\theta_{u}\) = 0.64% + 1.95% = 3.1%. It is important to note that in this procedure (c) the presence of splice l\({}_{\rm o}\) in the vicinity of plastic hinge length deteriorates the plastic rotation when \(\rm l_{0}<l_{\rm ou,min}\) and then the total rotation capacity should be multiplied by \(\rm l_{0}/l_{\rm ou,min}\); term \(\rm l_{\rm ou,min}\) is given by Equation (5a,b):  
Table 1.2 Table 1.3  
Their implementation results to \(\rm a_{l}=0.32\) and \(\rm l_{\rm ou,min}=\)1513 m. The hooked splice may be assumed as sufficient detailing for earthquake resistance (no need to divide with 1.2). Because \(\rm l_{0}=875\) mm \(<l_{\rm ou,min}=\)1513 mm, thus the ultimate rotation for the corner columns is \(\rm\theta_{u}=875/1513\times 3\%=1.7\%\) and for peripheral columns, \(\rm\theta_{u}=875/1513\times 3.1\%=1.8\%\). The calculations of \(\rm\theta_{u}\) deduced by the three procedures (a-c), led to \(\rm\mu_{\theta}\) results higher than 2, however when considering a narrow plastic length, they may receive values close to 1. The demonstration of the three procedures aims at highlighting the easiness or complexity each one encompasses as per its implementation.