}]]}]}]}] 30:T68ba,

### RC Slab Design Overview Reinforced concrete (RC) slabs are essential structural elements that provide flat surfaces in buildings. Their design involves determining dimensions, reinforcement, and ensuring safety under various loads. This cheatsheet covers the design process for one-way and two-way slabs according to the latest codes (e.g., ACI 318, NSCP 2015/2018). #### General Design Steps: 1. **Determine Slab Type:** One-way or Two-way. 2. **Material Properties:** Specify $f_c'$ (concrete compressive strength) and $f_y$ (steel yield strength). 3. **Determine Loads:** Calculate dead loads ($D_L$) and live loads ($L_L$). 4. **Factor Loads:** Apply load factors to get ultimate load ($U$). 5. **Determine Slab Thickness ($h$):** Based on deflection control and strength requirements. 6. **Calculate Moments and Shears:** Use structural analysis (e.g., coefficients, direct design method). 7. **Design Flexural Reinforcement:** Calculate required steel area ($A_s$) for positive and negative moments. 8. **Check Shear Strength:** Ensure concrete can resist shear; design shear reinforcement if needed (rare for slabs). 9. **Detailing:** Specify bar sizes, spacing, and cutoff points. 10. **Deflection Check:** Verify that calculated deflections are within allowable limits. ### Material Properties - **Concrete Compressive Strength ($f_c'$):** Typically 20 MPa to 35 MPa. - **Steel Yield Strength ($f_y$):** Typically 275 MPa to 415 MPa (Grade 40/60). - **Modulus of Elasticity of Concrete ($E_c$):** $E_c = 4700 \sqrt{f_c'}$ (MPa) or $E_c = w_c^{1.5} 0.043 \sqrt{f_c'}$ for normal weight concrete $w_c = 2300 \text{ kg/m}^3$. - **Modulus of Elasticity of Steel ($E_s$):** $200,000 \text{ MPa}$. ### Load Calculations - **Dead Loads ($D_L$):** - Slab self-weight: $h \times 23.5 \text{ kN/m}^3 \text{ (for normal weight concrete)}$ - Finishes, partitions, ceiling, etc. (specified by architect) - **Live Loads ($L_L$):** Based on occupancy (refer to NSCP/ACI). - **Factored Ultimate Load ($U$):** - $U = 1.2 D_L + 1.6 L_L$ (for gravity loads, NSCP/ACI) ### One-Way Slab Design A one-way slab primarily bends in one direction, transferring loads to supporting beams or walls along its longer edges. This occurs when the ratio of the longer span ($L_y$) to the shorter span ($L_x$) is greater than or equal to 2 ($L_y/L_x \ge 2$). Design is done for a 1-meter wide strip. #### 1. Minimum Thickness ($h_{min}$) for Deflection Control (NSCP/ACI Table 7.3.1.1) - **Simply Supported:** $L/20$ - **One End Continuous:** $L/24$ - **Both Ends Continuous:** $L/28$ - **Cantilever:** $L/10$ - *Note:* Values are for $f_y = 415 \text{ MPa}$. For other $f_y$, multiply by $(0.4 + f_y/700)$. - *Note:* $L$ is the clear span for continuous members and span length for simply supported. #### 2. Analysis for Moments and Shears - For uniformly distributed loads on continuous slabs, use NSCP/ACI Moment and Shear Coefficients (Table 6.5.2.1). - Example: $M_u = w_u L_n^2 / C$, where $C$ is the coefficient. - $V_u = w_u L_n / 2$ (at face of support for interior spans) #### 3. Flexural Reinforcement Design - **Required Steel Area ($A_s$):** - $M_u = \phi A_s f_y (d - a/2)$, where $a = (A_s f_y) / (0.85 f_c' b)$. - Solve for $A_s$ iteratively or using quadratic formula. - Use $\phi = 0.90$ for flexure. - **Minimum Reinforcement ($A_{s,min}$):** (NSCP/ACI 7.6.1.1) - For $f_y \le 415 \text{ MPa}$: $A_{s,min} = 0.0018 b h$ - For $f_y > 415 \text{ MPa}$: $A_{s,min} = 0.0018 (415/f_y) b h$ - A