SLAB FS-3
Input Data (Figure -2c):\(D.L=0.8\ kN/m^{2}\)
\begin{equation} L.L=0.64\ kN/m^{2}\nonumber \\ \end{equation}
D.L.F = 1.2 L.L.F= 1.6 Columns = 500 mm x 500 mm
\begin{equation} f_{y}=400\ MPa\nonumber \\ \end{equation}
\(f_{c}^{{}^{\prime}}=30\ MPa\)
\begin{equation} E=200,000\ MPa\nonumber \\ \end{equation}\begin{equation} \gamma_{c}=25\ kN/m^{3}\nonumber \\ \end{equation}
Solution:
1- Divide the slab into suitable triangles and select the triangle with the biggest span length “L ” and linear load “W ”.
The triangular section highlighted in figure 2-c proves to be the triangle with the biggest span length “L” of7.3 m
2 - Minimum slab thickness \(\mathbf{H}_{\mathbf{\min}}\)
\begin{equation} H_{\min}=\left(\frac{L}{30}\right)=\left(\frac{7300}{30}\right)=243.3\ mm\nonumber \\ \end{equation}\begin{equation} H_{\text{design}}=290\ mm\nonumber \\ \end{equation}
3- Determine Ultimate Moment \(\mathbf{M}_{\mathbf{U}}\)
\begin{equation} W_{u}=D.L.F\times\left(D.L+\frac{H_{\text{design}}}{1000}\times\gamma_{c}\right)+(L.L.F\times L.L)\nonumber \\ \end{equation}\begin{equation} W_{u}=10.684\ kN/m^{2}\nonumber \\ \end{equation}\begin{equation} M_{u}=\frac{W_{u}L^{2}}{8}=\ 71.17\ kN/m\nonumber \\ \end{equation}
The design moment for this slab section is selected to be\(M_{\text{design}}=100\ kN/m\)
4- Determine the required depth in flexure for Ultimate design Moment (\(\mathbf{M}_{\mathbf{\text{design}}}\mathbf{)}\).
\begin{equation} k=0.765\times 0.375\times\beta\times\left(1-\frac{0.375\times\beta}{2}\right)=0.204\nonumber \\ \end{equation}\begin{equation} d_{\text{flex}}=\sqrt{\frac{100\times 10^{6}}{0.204\times 30\times 1000}}=127.52\ mm\nonumber \\ \end{equation}\begin{equation} h_{\text{flex}}=\ d_{\text{flex}}+d^{{}^{\prime}}=157.52\ mm\nonumber \\ \end{equation}
5- Finding the required depth for one way shear,\(\mathbf{V}_{\mathbf{u(1)}}\).
\begin{equation} L_{p}=\frac{7.3}{2}-\frac{0.5}{2}=3.4\ mm\nonumber \\ \end{equation}\begin{equation} V_{u(1)}=W_{u}\times L_{p}=10.684\times 3.4\times 1=36.326\ kN\nonumber \\ \end{equation}\begin{equation} 36.326=\ 0.75\times\frac{1}{6}\times\sqrt{30}\times 1\times\left(d\right)\nonumber \\ \end{equation}
\(d=53.05\ mm\ \therefore\ h_{\text{one\ way}}=\ 53.05+30=83.05\ mm\)
6- Finding the two way\(\mathbf{\ }\)shear depth to satisfy punching shear requirement.
\begin{equation} {s=\frac{1}{2}\left(4+7.3+7\right)=9.15\ mm\backslash n}{A=\sqrt{s\times(s-a)\times(s-b)\times(s-c)}=13.69\ \text{mm}^{2}\backslash n}{V_{u(2)}=A\times W_{u}=146.27\ kN}\nonumber \\ \end{equation}\begin{equation} r_{\max}=\frac{146.27}{0.75\times 1\times(290-30)\times\left(\frac{2}{6}\times\sqrt{30}\right)}\text{\ \ \ }\nonumber \\ \end{equation}
\(r_{\max}=0.411\ <1\) (Safe for punching shear)
7- Check for the approximate deflection
\begin{equation} {E_{c}=5000\sqrt{f_{c}^{{}^{\prime}}}=27.38\ GPa\ \backslash n}{\ Moment\ of\ Inertia=I=\ \frac{1\times{0.290}^{3}}{12}=2.032\times 10^{-3}\text{mm}^{4}\backslash n}{W_{s}=\left(D.L+\frac{H_{\text{design}}}{1000}\times\gamma_{c}\right)+(L.L)\backslash n}{W_{s}=8.69\ kN/m^{2}\backslash n}{M_{s}=\frac{W_{s}L^{2}}{8}=\ 57.886\ kN/m\backslash n}{\delta_{\text{approx}}=\frac{57.886}{8\times 27.38\times 2.032\times 10^{-3}}\left[\left(\sqrt{{7.3}^{2}+{7.3}^{2}}\right)-2\times 0.5\right]^{2}=11.301\ mm\ \ }\nonumber \\ \end{equation}\begin{equation} \delta_{\text{approx}}<_{\text{code}}=11.301<20.27\ \ OK!\nonumber \\ \end{equation}
8- Finding the Flexural Capacity \(M_{c}\)
\begin{equation} {Qn=\frac{M_{\text{design}}\times 10^{6}}{0.9\times 1000\times\ \left(290-30\right)^{2}}=1.644\backslash n}{\rho=\frac{0.85\times f_{c}^{{}^{\prime}}}{f_{y}}\times\left(1-\sqrt{1-\frac{2.614\times\text{Qn}}{f_{c}^{{}^{\prime}}}}\right)=4.74\times 10^{-3}\backslash n}{A_{s}=\rho\times b\times(H_{\text{design}}-d^{{}^{\prime}})=1233\ \text{mm}^{2}}\nonumber \\ \end{equation}
Diameter of bar = 14 mm, Number of Bars Nb = 9 with spacing of 120 mm.\(\text{As}_{\text{actual}}=\left(\frac{1000}{\text{Spacing}}+1\right)\times A_{b}=1437\ \text{mm}^{2}\)
9- Flexural capacity\(M_{c}=\varnothing_{b}A_{s}f_{y}\left(d-\frac{a}{2}\right)\ \)
\begin{equation} {a=\frac{A_{s}f_{y}}{0.85f_{c}^{{}^{\prime}}b}=\ \frac{1437\times 400}{0.85\times 30\times 1000}=22.54\ mm\ \backslash n}{M_{c}=0.9\times 1437\times 400\left(260-\frac{22.54}{2}\right)\times 10^{-6}\backslash n}{M_{c}=128.67\ kN.m>M_{\text{design}}\ \ (OK!)}\nonumber \\ \end{equation}
10- Reinforcement detailing (figure-6)