RC Slab Design (One/Two Way)

Cheatsheet Content

### 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 $f_y > 415 \text{ MPa}$: $A_{s,min} = 0.0018 (415/f_y) b h$ - Alternatively, $A_{s,min} = 0.0014 b h$ if $f_y = 275 \text{ MPa}$. - **Maximum Spacing ($s_{max}$):** (NSCP/ACI 7.6.5.2.1) - $s_{max} = 3h$ or $450 \text{ mm}$, whichever is smaller. - **Shrinkage and Temperature Reinforcement:** (NSCP/ACI 24.4.3.2) - Perpendicular to main reinforcement. - For $f_y \le 415 \text{ MPa}$: $A_{s,temp} = 0.0018 b h$ - For $f_y > 415 \text{ MPa}$: $A_{s,temp} = 0.0018 (415/f_y) b h$ - Maximum spacing: $5h$ or $450 \text{ mm}$, whichever is smaller. #### 4. Shear Check - **Nominal Shear Strength of Concrete ($V_c$):** $V_c = 0.17 \lambda \sqrt{f_c'} b_w d$ (MPa) - **Factored Shear Force ($V_u$):** $V_u \le \phi V_c$ (where $\phi = 0.75$). - Shear reinforcement is usually not required for slabs unless $V_u > \phi V_c$. #### Example 1.1: One-Way Slab (Simply Supported) Design a simply supported one-way slab for a clear span of $3.0 \text{ m}$ to carry a service live load of $2.4 \text{ kPa}$ and a superimposed dead load of $1.0 \text{ kPa}$. Use $f_c' = 28 \text{ MPa}$ and $f_y = 415 \text{ MPa}$. Assume normal weight concrete. **1. Determine Slab Thickness ($h$):** - $h_{min} = L/20 = 3000/20 = 150 \text{ mm}$. - Try $h = 175 \text{ mm}$. - Effective depth $d = h - \text{cover} - d_b/2 = 175 - 20 - 10/2 = 150 \text{ mm}$ (assuming 10mm bar). **2. Load Calculations (per meter width):** - Slab self-weight ($D_L$): $0.175 \text{ m} \times 23.5 \text{ kN/m}^3 = 4.11 \text{ kPa}$ - Superimposed $D_L = 1.0 \text{ kPa}$ - Total $D_L = 4.11 + 1.0 = 5.11 \text{ kPa}$ - $L_L = 2.4 \text{ kPa}$ - Factored load $w_u = 1.2 D_L + 1.6 L_L = 1.2(5.11) + 1.6(2.4) = 6.132 + 3.84 = 9.972 \text{ kN/m}$ **3. Analysis for Moments and Shears:** - Maximum factored moment $M_u = w_u L^2 / 8 = 9.972 (3.0)^2 / 8 = 11.2185 \text{ kN-m}$ - Maximum factored shear $V_u = w_u L / 2 = 9.972 (3.0) / 2 = 14.958 \text{ kN}$ **4. Flexural Reinforcement Design:** - $M_u = \phi A_s f_y (d - a/2)$ - Assume $a \approx 0.1d = 0.1(150) = 15 \text{ mm}$ for initial $A_s$. - $A_s = M_u / (\phi f_y (d - a/2)) = 11.2185 \times 10^6 / (0.90 \times 415 \times (150 - 15/2)) = 210.4 \text{ mm}^2$ - Calculate $a = (A_s f_y) / (0.85 f_c' b) = (210.4 \times 415) / (0.85 \times 28 \times 1000) = 3.68 \text{ mm}$ - Recalculate $A_s$ with new $a$: $A_s = 11.2185 \times 10^6 / (0.90 \times 415 \times (150 - 3.68/2)) = 203.4 \text{ mm}^2$ - Check $A_{s,min} = 0.0018 b h = 0.0018 \times 1000 \times 175 = 315 \text{ mm}^2$. - Since $A_s (203.4) A_{s,min}$. - Use 12mm bars ($A_b = 113 \text{ mm}^2$). Spacing $s = (1000 \times 113) / 380 = 297.3 \text{ mm}$. - Provide **12mm bars @ 290mm O.C.** (top reinforcement). - **For Positive Moment ($M_u^+ = 14.196 \text{ kN-m}$):** - $A_s = 14.196 \times 10^6 / (0.90 \times 415 \times (155 - 6.80/2)) = 260.6 \text{ mm}^2$ (using $a$ from neg moment for initial approximation) - $a = (260.6 \times 415) / (0.85 \times 28 \times 1000) = 4.56 \text{ mm}$ - $A_s = 14.196 \times 10^6 / (0.90 \times 415 \times (155 - 4.56/2)) = 258.0 \text{ mm}^2$ - $A_{s,min} = 324 \text{ mm}^2$. Use $A_s = 324 \text{ mm}^2$. - Use 10mm bars ($A_b = 78.5 \text{ mm}^2$). Spacing $s = (1000 \times 78.5) / 324 = 242.2 \text{ mm}$. - Provide **10mm bars @ 240mm O.C.** (bottom reinforcement). **5. Shrinkage and Temperature Reinforcement:** - $A_{s,temp} = 0.0018 b h = 0.0018 \times 1000 \times 180 = 324 \text{ mm}^2$. - Use 10mm bars ($A_b = 78.5 \text{ mm}^2$). Spacing $s = (1000 \times 78.5) / 324 = 242.2 \text{ mm}$. - Provide **10mm bars @ 240mm O.C.** **6. Shear Check:** - Factored shear at face of interior support: $V_u = 1.15 w_u L_n / 2 = 1.15 \times 14.196 \times 4.0 / 2 = 32.65 \text{ kN}$ - $V_c = 0.17 \sqrt{f_c'} b_w d = 0.17 \sqrt{28} \times 1000 \times 155 = 139.7 \text{ kN}$ - $\phi V_c = 0.75 \times 139.7 = 104.8 \text{ kN}$ - $V_u (32.65 \text{ kN}) ### Two-Way Slab Design A two-way slab bends in two orthogonal directions, transferring loads to supports along all four edges. This occurs when the ratio of the longer span ($L_y$) to the shorter span ($L_x$) is less than 2 ($L_y/L_x \phi V_c$, shear reinforcement (stirrups or shear heads) is required. #### Example 2.1: Two-Way Slab (Interior Panel - Direct Design Method) Design an interior panel of a two-way slab with $L_x = 5.0 \text{ m}$ (short span) and $L_y = 6.0 \text{ m}$ (long span). Service live load $L_L = 4.0 \text{ kPa}$, superimposed dead load $D_L = 1.0 \text{ kPa}$. $f_c' = 28 \text{ MPa}$, $f_y = 415 \text{ MPa}$. Assume column size $400 \text{ mm} \times 400 \text{ mm}$. **1. Determine Slab Thickness ($h$):** - $L_y/L_x = 6.0/5.0 = 1.2 A_{s,min}$) - Use 16mm bars ($A_b = 201 \text{ mm}^2$). Number of bars $N = 1800 / 201 = 8.95 \approx 9$ bars. - Spacing $s = (3000 - 2 \times 20) / (9-1) = 370 \text{ mm}$. Or $s = 3000/9 = 333 \text{ mm}$. - Max spacing $s_{max} = 2h = 2(180) = 360 \text{ mm}$ or $450 \text{ mm}$. Use $360 \text{ mm}$. - Provide **9 - 16mm bars in column strip (top)**. (Spacing $\approx 330 \text{ mm}$) - **Column Strip Positive Reinforcement (at mid-span):** $M_u = 42.252 \text{ kN-m}$, $b = 3000 \text{ mm}$, $d_x = 155 \text{ mm}$. - $A_s = 42.252 \times 10^6 / (0.90 \times 415 \times (155 - a/2))$. Iterative solution: $A_s \approx 775 \text{ mm}^2$. - $A_{s,min} = 972 \text{ mm}^2$. Use $A_s = 972 \text{ mm}^2$. - Use 12mm bars ($A_b = 113 \text{ mm}^2$). Number of bars $N = 972 / 113 = 8.6 \approx 9$ bars. - Provide **9 - 12mm bars in column strip (bottom)**. (Spacing $\approx 330 \text{ mm}$) *(Repeat for middle strip and for long span direction using $d_y$)* **6. Shear Check (Punching Shear):** - Column size $c_1 = c_2 = 400 \text{ mm}$. $d = 155 \text{ mm}$. - Critical perimeter $b_o = 4(c_1 + d) = 4(400 + 155) = 2220 \text{ mm}$. - Area enclosed by critical perimeter $A_{cr} = (c_1+d)^2 = (400+155)^2 = 308025 \text{ mm}^2 = 0.308 \text{ m}^2$. - Factored shear $V_u = w_u (L_x L_y - A_{cr}) = 12.676 (5.0 \times 6.0 - 0.308) = 12.676 (30 - 0.308) = 376.5 \text{ kN}$. - For interior column, $\alpha_s = 40$. - $\beta_c = 400/400 = 1$. - $V_c = \min(0.33 \sqrt{f_c'}, (0.17 + 0.33/\beta_c) \sqrt{f_c'}, (0.17 + 0.083 \alpha_s d/b_o) \sqrt{f_c'}) b_o d$ - $0.33 \sqrt{28} = 1.745 \text{ MPa}$ - $(0.17 + 0.33/1) \sqrt{28} = 0.50 \sqrt{28} = 2.646 \text{ MPa}$ - $(0.17 + 0.083 \times 40 \times 155/2220) \sqrt{28} = (0.17 + 0.231) \sqrt{28} = 0.401 \sqrt{28} = 2.122 \text{ MPa}$ - So, $V_c = 1.745 \text{ MPa} \times 2220 \text{ mm} \times 155 \text{ mm} = 601.5 \text{ kN}$. - $\phi V_c = 0.75 \times 601.5 = 451.1 \text{ kN}$. - $V_u (376.5 \text{ kN}) 1.0$, use interpolation or conservative values. - Negative Moment at exterior support (Col Strip): Use $100\%$ (conservative if $\alpha_f L_y/L_x \ge 1.0$) $\rightarrow 1.0 \times 40.8 = 40.8 \text{ kN-m}$ - Positive Moment (Col Strip): $0.75 \times 81.6 = 61.2 \text{ kN-m}$ - Negative Moment at interior support (Col Strip): $0.75 \times 109.8 = 82.35 \text{ kN-m}$ *(Repeat for middle strip and for short span direction using $d_x$)* **5. Flexural Reinforcement Design (Long Span - examples for Column Strip only):** - **Column Strip Negative Reinforcement (Exterior Support):** $M_u = 40.8 \text{ kN-m}$, $b = 2250 \text{ mm}$, $d_y = 135 \text{ mm}$. - $A_s = 40.8 \times 10^6 / (0.90 \times 415 \times (135 - a/2))$. Iterative solution: $A_s \approx 870 \text{ mm}^2$. - $A_{s,min} = 0.0018 b h = 0.0018 \times 2250 \times 170 = 688.5 \text{ mm}^2$. ($A_s > A_{s,min}$) - Use 12mm bars ($A_b = 113 \text{ mm}^2$). Number of bars $N = 870 / 113 = 7.69 \approx 8$ bars. - Max spacing $s_{max} = 2h = 2(170) = 340 \text{ mm}$ or $450 \text{ mm}$. Use $340 \text{ mm}$. - Provide **8 - 12mm bars in column strip (top)**. (Spacing $\approx 280 \text{ mm}$) - **Column Strip Positive Reinforcement (Mid-span):** $M_u = 61.2 \text{ kN-m}$, $b = 2250 \text{ mm}$, $d_y = 135 \text{ mm}$. - $A_s = 61.2 \times 10^6 / (0.90 \times 415 \times (135 - a/2))$. Iterative solution: $A_s \approx 1300 \text{ mm}^2$. - $A_{s,min} = 688.5 \text{ mm}^2$. ($A_s > A_{s,min}$) - Use 16mm bars ($A_b = 201 \text{ mm}^2$). Number of bars $N = 1300 / 201 = 6.46 \approx 7$ bars. - Provide **7 - 16mm bars in column strip (bottom)**. (Spacing $\approx 320 \text{ mm}$) **6. Shear Check (Punching Shear at Exterior Column):** - Column size $c_1 = c_2 = 350 \text{ mm}$. $d = 145 \text{ mm}$. - Critical perimeter $b_o = c_1 + 2d + c_2 + d = 350 + 2(145) + 350 + 145 = 1135 \text{ mm}$ (for exterior corner column). - For exterior column along edge (not corner): $b_o = c_1 + 2d + c_2/2 + d/2 + c_2/2+d/2 = c_1 + 2d + c_2 + d = 350 + 2(145) + 350 + 145 = 1135 \text{ mm}$. (This calculation for $b_o$ might be for a corner column, check NSCP/ACI figures for exact $b_o$ for edge columns). - Let's assume for exterior column, critical perimeter is $c_1 + 2d + c_2 + 2d = 350 + 2(145) + 350 + 2(145) = 1330 \text{ mm}$ for non-corner (3 sides). Or $b_o = c_1+d + 2(c_2/2+d/2) = c_1+d + c_2+d = 350+145+350+145=990 \text{ mm}$ (if beam is very stiff) - Let's use the definition of $b_o$ for a column at an edge such that $b_o = c_1 + 2d + c_2 + d = 350 + 2(145) + 350 + 145 = 1135 \text{ mm}$ (This assumes the beam is not part of the critical section). - Area enclosed by critical perimeter $A_{cr} = (c_1+d)(c_2+d/2) = (350+145)(350+145/2) = 495(422.5) = 209137.5 \text{ mm}^2 = 0.209 \text{ m}^2$. - Factored shear $V_u = w_u (L_x L_y - A_{cr}) = 10.554 (4.5 \times 5.5 - 0.209) = 10.554 (24.75 - 0.209) = 260.8 \text{ kN}$. - For edge column, $\alpha_s = 30$. - $\beta_c = 350/350 = 1$. - $V_c = \min(0.33 \sqrt{f_c'}, (0.17 + 0.33/\beta_c) \sqrt{f_c'}, (0.17 + 0.083 \alpha_s d/b_o) \sqrt{f_c'}) b_o d$ - $0.33 \sqrt{25} = 1.65 \text{ MPa}$ - $(0.17 + 0.33/1) \sqrt{25} = 0.50 \sqrt{25} = 2.50 \text{ MPa}$ - $(0.17 + 0.083 \times 30 \times 145/1135) \sqrt{25} = (0.17 + 0.318) \sqrt{25} = 0.488 \sqrt{25} = 2.44 \text{ MPa}$ - So, $V_c = 1.65 \text{ MPa} \times 1135 \text{ mm} \times 145 \text{ mm} = 271.7 \text{ kN}$. - $\phi V_c = 0.75 \times 271.7 = 203.8 \text{ kN}$. - $V_u (260.8 \text{ kN}) > \phi V_c (203.8 \text{ kN})$, so punching shear is NOT adequate. Shear reinforcement required or increase slab thickness/column size. - *Correction:* The calculation of $b_o$ and $A_{cr}$ for an edge column needs to be precise based on NSCP/ACI figures. If the beam provides significant stiffness, part of the punching shear capacity is taken by the beam. For simplicity in this example, assuming no beam contribution to punching shear, the slab fails. - If we consider the effect of the edge beam's torsion, the column strip moment distribution changes. The DDM tables account for beam stiffness. For punching shear, the beam's contribution to shear strength is usually considered separately or through more complex methods. In a simplified DDM approach, if $V_u > \phi V_c$, either increase $h$, column size, or provide shear reinforcement. - Let's assume we increase $h$ to $200 \text{ mm}$. - $d = 200 - 20 - 10/2 = 175 \text{ mm}$. - $b_o = 350 + 2(175) + 350 + 175 = 1225 \text{ mm}$. - $A_{cr} = (350+175)(350+175/2) = 525(437.5) = 229687.5 \text{ mm}^2 = 0.230 \text{ m}^2$. - New $w_u = 1.2(0.200 \times 23.5 + 0.8) + 1.6(3.0) = 1.2(4.7+0.8) + 4.8 = 1.2(5.5) + 4.8 = 6.6+4.8 = 11.4 \text{ kPa}$. - New $V_u = 11.4 (4.5 \times 5.5 - 0.230) = 11.4 (24.75 - 0.230) = 280.0 \text{ kN}$. - New $V_c = 1.65 \text{ MPa} \times 1225 \text{ mm} \times 175 \text{ mm} = 354.5 \text{ kN}$. - $\phi V_c = 0.75 \times 354.5 = 265.9 \text{ kN}$. - Still $V_u (280.0 \text{ kN}) > \phi V_c (265.9 \text{ kN})$. - This indicates that for this example, either shear reinforcement is needed, or a larger column/more robust beam connection is required, or a more detailed analysis considering beam contribution to shear is necessary. 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