RCC Numerical: Design of One way Slab

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RCC Numerical: Design of One way Slab

Numerical:

A one way slab is to be designed for an effective span 3.3 m . The super imposed load including finishing is 4 KN/m2 .Taking modification factor 1.2.Design the slab. Sketch c/s of slab showing reinforcement details. Use concrete M20 and steel Fe 415 .

Solution:



Given:

• Span (l): 3.3 m
• Live Load (L.L.) and Floor Finish (F.F.): 4 KN/m²
• Modification Factor (M.F.): 1.2

1. Design Constants:

• Yield strength of steel (fy): 415 N/mm²
• Characteristic compressive strength of concrete (fck): 20 N/mm²
• Maximum depth of neutral axis (Xu max): 0.48d
• Ultimate moment of resistance (Mu lim): 0.138 fck b d²

2. Estimation of Slab Thickness:

$d=\frac{\text{span}}{20\times \text{M.F.}}d\; =\; \backslash frac\{\backslash text\{span\}\}\{20\; \backslash times\; \backslash text\{M.F.\}\}$

$d=\frac{3300}{20\times 1.2}d\; =\; \backslash frac\{3300\}\{20\; \backslash times\; 1.2\}$

$d=137.5\text{ mm}d\; =\; 137.5\; \backslash text\{\; mm\}$

Assuming 10 mm Ø main bars and nominal cover as 20 mm:

$D=d+d_c+\frac{\varphi }{2}D\; =\; d\; +\; d\_c\; +\; \backslash frac\{\backslash phi\}\{2\}$

$D=137.5+20+\frac{10}{2}D\; =\; 137.5\; +\; 20\; +\; \backslash frac\{10\}\{2\}$

$D=162.5\text{ mm}D\; =\; 162.5\; \backslash text\{\; mm\}$

Say $D=165\text{ mm}D\; =\; 165\; \backslash text\{\; mm\}$

Therefore:

$d_{avail}=D-d_c-\frac{\varphi }{2}d\_\{avail\}\; =\; D\; -\; d\_c\; -\; \backslash frac\{\backslash phi\}\{2\}$

$d_{avail}=165-20-\frac{10}{2}d\_\{avail\}\; =\; 165\; -\; 20\; -\; \backslash frac\{10\}\{2\}$

$d_{avail}=140\text{ mm}d\_\{avail\}\; =\; 140\; \backslash text\{\; mm\}$

3. Effective Span:

$L_e=3.3\text{ m (given)}\text{<br>}\text{<br>}\text{<br>}L\_e\; =\; 3.3\; \backslash text\{\; m\; (given)\}$

4. Load Calculations (considering 1 m wide strip):

$\text{Self weight of slab}=1\times 1\times 0.165\times 25=4.125\text{ KN/m}\backslash text\{Self\; weight\; of\; slab\}\; =\; 1\; \backslash times\; 1\; \backslash times\; 0.165\; \backslash times\; 25\; =\; 4.125\; \backslash text\{\; KN/m\}$

$\text{Total load (w)}=8.125\text{ KN/m}\backslash text\{Total\; load\; (w)\}\; =\; 8.125\; \backslash text\{\; KN/m\}$

Factored load:

$w_d=1.5\times 8.125w\_d\; =\; 1.5\; \backslash times\; 8.125$

$w_d=12.1875\text{ KN/m}\text{<br>}\text{<br>}w\_d\; =\; 12.1875\; \backslash text\{\; KN/m\}$

5. Factored (Design) Maximum Bending Moment:

$M_d=\left(\frac{w_d\times L_e^2}{8}\right)M\_d\; =\; \backslash left(\backslash frac\{w\_d\; \backslash times\; L\_e^2\}\{8\}\backslash right)$

$M_d=\left(\frac{12.1875\times 3.3^2}{8}\right)M\_d\; =\; \backslash left(\backslash frac\{12.1875\; \backslash times\; 3.3^2\}\{8\}\backslash right)$

$M_d=16.59\text{ KNm}M\_d\; =\; 16.59\; \backslash text\{\; KNm\}$

6. Required Overall Depth and Effective Depth:

Equating $M_u\text{lim}M\_u\; \backslash text\{lim\}$ to $M_{d}M\_d$:

$0.138f_{ck}b{d}^2=M_{d}0.138\; f\_\{ck\}\; b\; d^2\; =\; M\_d$

$0.138\times 20\times 1000\times d^2=16.59\times 10^{6}0.138\; \backslash times\; 20\; \backslash times\; 1000\; \backslash times\; d^2\; =\; 16.59\; \backslash times\; 10^6$

$d=77.53\text{ mm}d\; =\; 77.53\; \backslash text\{\; mm\}$

$d_{avail}=140\text{ mm}>d_{required}d\_\{avail\}\; =\; 140\; \backslash text\{\; mm\}\; >\; d\_\{required\}$:

$\text{Hence, OK.}\backslash text\{Hence,\; OK.\}$

Provide $D=165\text{ mm}D\; =\; 165\; \backslash text\{\; mm\}$ and $d_{avail}=140\text{ mm}d\_\{avail\}\; =\; 140\; \backslash text\{\; mm\}$

7. Area of Main Steel:

$A_{st}=\frac{0.5f_{ck}}{f_y}[1-\sqrt{1-\frac{4.6M_d}{f_{ck}b{d}^2}}]bdA\_\{st\}\; =\; \backslash frac\{0.5\; f\_\{ck\}\}\{f\_y\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash frac\{4.6\; M\_d\}\{f\_\{ck\}\; b\; d^2\}\}\; \backslash right]\; b\; d$

$A_{st}=\frac{0.5\times 20}{415}[1-\sqrt{1-\frac{4.6\times 16.59\times 10^6}{20\times 1000\times 140^2}}]\times 1000\times 140A\_\{st\}\; =\; \backslash frac\{0.5\; \backslash times\; 20\}\{415\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash frac\{4.6\; \backslash times\; 16.59\; \backslash times\; 10^6\}\{20\; \backslash times\; 1000\; \backslash times\; 140^2\}\}\; \backslash right]\; \backslash times\; 1000\; \backslash times\; 140$

$A_{st}=346.13{\text{ mm}}^2$

Spacing of Main Reinforcement

a) Calculating Spacing ($S_{10}S\_\{10\}$)

$S_{10}=\frac{1000\times (\frac{\pi }{4}\times 10^2)}{346.13}S\_\{10\}\; =\; \backslash frac\{1000\; \backslash times\; \backslash left(\backslash frac\{\backslash pi\}\{4\}\; \backslash times\; 10^2\backslash right)\}\{346.13\}$

$S_{10}=\frac{1000\times 78.54}{346.13}S\_\{10\}\; =\; \backslash frac\{1000\; \backslash times\; 78.54\}\{346.13\}$

$S_{10}=226.91\text{mm}S\_\{10\}\; =\; 226.91\; \backslash ,\; \backslash text\{mm\}$

Spacing = 225 mm c/c (rounded to nearest standard size)

b) 3d Calculation

$3d=3\times 140=420\text{mm}3d\; =\; 3\; \backslash times\; 140\; =\; 420\; \backslash ,\; \backslash text\{mm\}$

c) Minimum Spacing

$\text{Minimum Spacing}=300\text{mm}\backslash text\{Minimum\; Spacing\}\; =\; 300\; \backslash ,\; \backslash text\{mm\}$

Ast provided

$\text{Ast provided}=\frac{1000\times (\frac{\pi }{4}\times 10^2)}{225}\backslash text\{Ast\; provided\}\; =\; \backslash frac\{1000\; \backslash times\; \backslash left(\backslash frac\{\backslash pi\}\{4\}\; \backslash times\; 10^2\backslash right)\}\{225\}$

$\text{Ast provided}=\frac{1000\times 78.54}{225}\backslash text\{Ast\; provided\}\; =\; \backslash frac\{1000\; \backslash times\; 78.54\}\{225\}$

$\text{Ast provided}=349.06{\text{mm}}^2$

8. Area and Spacing of Distribution Steel:

Step 1: Calculate the Minimum Area of Distribution Steel (Astmin)

${\text{Ast}}_{\text{min}}=0.15\%\times b\times D\backslash text\{Ast\}\_\{\backslash text\{min\}\}\; =\; 0.15\; \backslash \%\; \backslash times\; b\; \backslash times\; D$

$=\frac{0.15}{100}\times 1000\text{mm}\times 165\text{mm}=\; \backslash frac\{0.15\}\{100\}\; \backslash times\; 1000\; \backslash ,\; \backslash text\{mm\}\; \backslash times\; 165\; \backslash ,\; \backslash text\{mm\}$

$=247.5{\text{mm}}^2=\; 247.5\; \backslash ,\; \backslash text\{mm\}^2$

Step 2: Determine the Spacing of 6 mm Diameter Mild Steel Distribution Bars

$S_d=\left(\frac{\text{Area of one bar}}{{\text{Ast}}_{\text{min}}}\right)\times 1000S\_d\; =\; \backslash left(\; \backslash frac\{\backslash text\{Area\; of\; one\; bar\}\}\{\backslash text\{Ast\}\_\{\backslash text\{min\}\}\}\; \backslash right)\; \backslash times\; 1000$

$\text{Area of one bar}=\pi \times {\left(\frac{6}{2}\right)}^2\backslash text\{Area\; of\; one\; bar\}\; =\; \backslash pi\; \backslash times\; \backslash left(\backslash frac\{6\}\{2\}\backslash right)^2$

$=\pi \times 3^2=\; \backslash pi\; \backslash times\; 3^2$

$=28.27{\text{mm}}^2=\; 28.27\; \backslash ,\; \backslash text\{mm\}^2$

Calculate Spacing (S)

$S_d=\left(\frac{28.27{\text{mm}}^2}{247.5{\text{mm}}^2}\right)\times 1000\text{mm}S\_d\; =\; \backslash left(\; \backslash frac\{28.27\; \backslash ,\; \backslash text\{mm\}^2\}\{247.5\; \backslash ,\; \backslash text\{mm\}^2\}\; \backslash right)\; \backslash times\; 1000\; \backslash ,\; \backslash text\{mm\}$

$=114.22\text{mm}=\; 114.22\; \backslash ,\; \backslash text\{mm\}$

$\approx 110\text{mm}\backslash approx\; 110\; \backslash ,\; \backslash text\{mm\}$

Step 3: Check Maximum Allowable Spacing

$S_{\text{max}}=5\times \text{Depth}S\_\{\backslash text\{max\}\}\; =\; 5\; \backslash times\; \backslash text\{Depth\}$

$=5\times 140\text{mm}=\; 5\; \backslash times\; 140\; \backslash ,\; \backslash text\{mm\}$

$=700\text{mm}=\; 700\; \backslash ,\; \backslash text\{mm\}$

Given Maximum Spacing

$=450\text{mm}=\; 450\; \backslash ,\; \backslash text\{mm\}$

Therefore, the minimum spacing calculated is 114.22 mm, approximated to 110 mm.

Conclusion:

Provide 6 mm diameter distribution bars at 110 mm center-to-center.

9. Reinforcement Details: