Step-3:Determine Ultimate Moment \(\mathbf{M}_{\mathbf{U}}\) ( figure 1-a)
Ultimate Moment = \(M_{u}=\frac{W_{u}L^{2}}{8}\)
Step-4: Determine the required depth in flexure for Ultimate design Moment (\(\mathbf{M}_{\mathbf{\text{Ud}}}\mathbf{)}\).
\begin{equation} d_{\text{flex}}=\sqrt{\frac{M_{\text{Udes}}\times 10^{6}}{k\times f_{c}^{{}^{\prime}}\times 1000}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)\nonumber \\ \end{equation}
Where;\(k=0.765\times 0.375\times\beta\times\left(1-\frac{0.375\times\beta}{2}\right)\)And,
\(\beta=0.85\ \ for\ \ \mathbf{f}_{\mathbf{c}}^{\mathbf{{}^{\prime}}}\mathbf{\ \leq 30\ MPa}\)
\(\beta=0.85-0.008\left(\mathbf{f}_{\mathbf{c}}^{\mathbf{{}^{\prime}}}-30\right)\geq 0.65\ \ \ for\ \ \mathbf{f}_{\mathbf{c}}^{\mathbf{{}^{\prime}}}\mathbf{\ >30\ MPa}\)
Step-5: Finding the required depth for one way shear,\(\mathbf{V}_{\mathbf{u(1)}}\).
\(V_{c}>V_{u(1)}\) (2)
Where,\(V_{c}=\ \varnothing_{s}\times\frac{1}{6}\times\sqrt{f_{c}^{{}^{\prime}}}\times 1000\times(H_{\text{design}}-d^{\prime}\)) (3)(ACI 318-14 code for shear calculation)And,
\begin{equation} V_{u(1)}=W_{u}\times L_{p}\nonumber \\ \end{equation}\begin{equation} L_{p}=\frac{L}{2}-\frac{\text{Column\ width}}{2}\nonumber \\ \end{equation}