# Flexural and Shear Performance of Prestressed Composite Slabs with Inverted Multi-Ribs
## Appendix A  
In this section, an example of numerical calculations on the flexural behavior of composite JAS is presented. The sectional details and material properties of JAS are shown in Figure 2b and Table 1, respectively.  
**- Input parameters** (Figure A1):  
\[h_{pc}=230\ \text{mm},\,h_{top}=100\ \text{mm},\,d_{p}=290\ \text{mm},\,d^{ \prime}{}_{s}=40\ \text{mm},\,b=1100\ \text{mm},\,b_{w}=150\ \text{mm}\]  
\[f^{\prime}_{c,pc}=36.0\ \text{MPa},\,f^{\prime}_{c,rc}=21.3\ \text{MPa},\,f_{y}=503\ \text{MPa},\,f_{pu}=1936\ \text{MPa},\,f_{se}=0.61f_{pu}\]  
**Figure A1.** Section details of JAS.  
**-Moment-curvature response**  
_Step 1:_ Select \(\varepsilon_{t}=-0.001\)  
_Step 2:_ Assume \(c=100\) mm  
_Step 3:_ Obtain the distribution of strains, as shown in Figure A2.  
\(\varepsilon_{b}=\frac{c-\left(h_{pc}+h_{bp}\right)}{c}\times\varepsilon_{t}= 0.0026\),  
\(\varepsilon^{\prime}_{s}=\frac{c-d^{\prime}_{s}}{c}\times\varepsilon_{t}=-0.00099\), and \(\varepsilon_{ps}=\frac{f_{pc}}{\underline{\varepsilon}_{s}}+\frac{c-d_{p}}{ c}\times\varepsilon_{t}=0.0076\)  
_Step 4:_ Calculate stresses of concrete layers, strands, and compressive reinforcement from constitutive laws of materials shown in Figure 8b.  
_Step 5:_ Calculate forces of concrete, strands, and compressive reinforcement.  
\[C_{pc}=\sum f_{c,pc}b_{i}\cdot\Delta d_{c}=15.1\text{ kN}\]  
\[C_{RC}=\sum f_{c,rc}b_{i}\cdot\Delta d_{c}=-732.4\text{ kN}\]  
\[C_{s}=f^{\prime}_{s}A^{\prime}_{s}=-56.5\text{ kN}\]  
\[T_{ps}=f_{ps}A_{ps}=1258.8\text{ kN}\]  
_Step 6:_ Check force equilibrium.  
\(C_{pc}+C_{RC}+C_{s}+T_{ps}\
eq 0\) (Not satisfied)  
_Step 7:_ Go back to the _step 2_, and update \(c\) until the force equilibrium is satisfied.  
If the neutral axis depth (\(c\)) is assumed to be \(135\) mm, the forces of concrete, strands and compressive reinforcement are calculated, as follows:\[C_{pc}=-119.4\text{ kN},\ C_{RC}=-953.1\text{ kN},\ C_{s}=-56.6\text{ kN},\ T_{ps}=1136.0\text{ kN}\]  
The equilibrium condition is also satisfied, \(C_{pc}+C_{RC}+C_{s}+T_{ps}=6.9\text{ kN}\approx 0\).  
_Step 8_: Calculate curvature (\(\phi\)) and moment (\(M\)).  
\[\phi=\frac{\varepsilon_{t}}{c}=7.4\times 10^{-6}\]  
\[M=\sum f_{c,pc}\cdot b_{i}\cdot d_{c}\cdot\Delta d_{c}+\sum f_{c,rc}\cdot b_{i} \cdot d_{c}\cdot\Delta d_{c}+f_{s}^{\prime}A_{s}^{\prime}d_{s}^{\prime}+f_{ps}A _{ps}d_{p}=276.3\text{ kN}\cdot\text{m}\]  
_Step 9_: Repeat step 1 to 8 with an updated \(\varepsilon_{t}\) until \(\varepsilon_{t}\) reaches the crushing strain of concrete (\(-0.003\)). The moment-curvature response of composite JAS is presented in Figure A3:  
**-Load-deflection response.**  
The loading details of JAS is shown in Figure A4.  
_Step 1_: Select \(P=50\text{ kN}\).  
_Step 2_: Obtain moment and curvature distributions from Figure A3.  
_Step 3:_ Calculate mid-span deflection by numerically integrating curvatures shown in Figure A5, as follows:  
\[\delta=\int_{0}^{0.5I}\phi_{i}xdx=\left(\frac{\phi_{0}x_{0}+\phi_{1}x_{1}}{2} \right)\Delta x_{0}+\left(\frac{\phi_{1}x_{1}+\phi_{2}x_{2}}{2}\right)\Delta x_{ 1}+\cdots+\left(\frac{\phi_{5}x_{5}+\phi_{6}x_{6}}{2}\right)\Delta x_{5}=5.73\; \mathrm{mm}\]  
_Step 4:_ Repeat step 1 to 3 with updating \(P\), then the load-deflection response can be obtained, as shown in Figure 10.