RCC numerical: Design of Two Way Slab

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Numerical:

Design a simply supported two way slab over a room 4.8 m x 4.0 m effective, subjected to UDL 5 kN/m2 (inclusive of self wt.) Use M20 and Fe 415. Draw reinforcement detail check for shear may not be given. Take αx = 0.084 and αy = 0.059

Solution:

Given:

Effective dimensions of the two-way slab: $4.8\text{m}\times 4\text{m}4.8\; \backslash ,\; \backslash text\{m\}\; \backslash times\; 4\; \backslash ,\; \backslash text\{m\}$

Load $w=5{\text{kN/m}}^{2}w\; =\; 5\; \backslash ,\; \backslash text\{kN/m\}^2$ (inclusive of self-weight)

Characteristic compressive strength of concrete $f_{ck}=20{\text{N/mm}}^{2}f\_\{ck\}\; =\; 20\; \backslash ,\; \backslash text\{N/mm\}^2$

Yield strength of steel $f_y=415{\text{N/mm}}^{2}f\_y\; =\; 415\; \backslash ,\; \backslash text\{N/mm\}^2$

Coefficient ${\alpha }_x=0.084\backslash alpha\_x\; =\; 0.084$

Coefficient ${\alpha }_y=0.059\backslash alpha\_y\; =\; 0.059$

Design Constants:

For Fe 415, $M_{u\text{lim}}=0.138f_{ck}b{d}^2$



Estimation of Slab Thickness:

Assume MF = 1.4

Therefore,

$d=\frac{\text{span}}{20\times 1.4}d\; =\; \backslash frac\{\backslash text\{span\}\}\{20\; \backslash times\; 1.4\}$

$d=\frac{4000}{20\times 1.4}d\; =\; \backslash frac\{4000\}\{20\; \backslash times\; 1.4\}$

$d=142.86\text{mm}\text{say}150\text{mm}d\; =\; 142.86\; \backslash ,\; \backslash text\{mm\}\; \backslash ,\; \backslash text\{say\}\; \backslash ,\; 150\; \backslash ,\; \backslash text\{mm\}$

Assuming 10 mm Ø main bars, & c = 20 mm,

$D=d+c+\left(\frac{\text{Ø}}{2}\right)D\; =\; d\; +\; c\; +\; \backslash left(\backslash frac\{\backslash text\{Ø\}\}\{2\}\backslash right)$

$D=150+20+\left(\frac{10}{2}\right)D\; =\; 150\; +\; 20\; +\; \backslash left(\backslash frac\{10\}\{2\}\backslash right)$

$D=175\text{mm}D\; =\; 175\; \backslash ,\; \backslash text\{mm\}$

Effective Span:

$L_{xe}=4000\text{mm}L\_\{xe\}\; =\; 4000\; \backslash ,\; \backslash text\{mm\}$

$L_{ye}=4800\text{mm}L\_\{ye\}\; =\; 4800\; \backslash ,\; \backslash text\{mm\}$

Consider 1 m wide strip

Load: $w_{\text{given}}=5{\text{KN/m}}^{2}w\_\{\backslash text\{given\}\}\; =\; 5\; \backslash ,\; \backslash text\{KN/m\}^2$

$w=1\times 1\times 5=5{\text{KN/m}}^{2}w\; =\; 1\; \backslash times\; 1\; \backslash times\; 5\; =\; 5\; \backslash ,\; \backslash text\{KN/m\}^2$

Factored load:

$w_d=1.5\times 5=7.5\text{KN/m}\text{<br>}w\_d\; =\; 1.5\; \backslash times\; 5\; =\; 7.5\; \backslash ,\; \backslash text\{KN/m\}$

Factored Bending Moment:

${\alpha }_x=0.084\backslash alpha\_x\; =\; 0.084$

${\alpha }_y=0.059\backslash alpha\_y\; =\; 0.059$

$M_{xd}={\alpha }_x\cdot w_d\cdot L_{xe}^{2}M\_\{xd\}\; =\; \backslash alpha\_x\; \backslash cdot\; w\_d\; \backslash cdot\; L\_\{xe\}^2$

$M_{xd}=0.084\times 7.5\times 4^{2}M\_\{xd\}\; =\; 0.084\; \backslash times\; 7.5\; \backslash times\; 4^2$

$M_{xd}=10.08\text{KNm}M\_\{xd\}\; =\; 10.08\; \backslash ,\; \backslash text\{KNm\}$

$M_{yd}={\alpha }_y\cdot w_d\cdot L_{xe}^{2}M\_\{yd\}\; =\; \backslash alpha\_y\; \backslash cdot\; w\_d\; \backslash cdot\; L\_\{xe\}^2$

$M_{yd}=0.059\times 7.5\times 4^{2}M\_\{yd\}\; =\; 0.059\; \backslash times\; 7.5\; \backslash times\; 4^2$

$M_{yd}=7.08\text{KNm}\text{<br>}M\_\{yd\}\; =\; 7.08\; \backslash ,\; \backslash text\{KNm\}$

Effective Depth of Slab:

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

$0.138\times 20\times 1000d^2=10.08\times 10^{6}0.138\; \backslash times\; 20\; \backslash times\; 1000\; \backslash ,\; d^2\; =\; 10.08\; \backslash times\; 10^6$

Area and Spacing of Steel:

$A_{stx}=0.5\frac{f_{ck}}{f_y}[1-\sqrt{1-\left(\frac{4.6M_{xd}}{f_{ck}\cdot b\cdot d^2}\right)}]\cdot b\cdot dA\_\{stx\}\; =\; 0.5\; \backslash frac\{f\_\{ck\}\}\{f\_y\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash left(\; \backslash frac\{4.6\; \backslash ,\; M\_\{xd\}\}\{f\_\{ck\}\; \backslash cdot\; b\; \backslash cdot\; d^2\}\; \backslash right)\}\; \backslash right]\; \backslash cdot\; b\; \backslash cdot\; d$

$A_{stx}=\frac{0.5\times 20}{415}[1-\sqrt{1-\left(\frac{4.6\times 10.08\times 10^6}{20\times 1000\times 150\times 150}\right)}]\times 1000\times 150A\_\{stx\}\; =\; \backslash frac\{0.5\; \backslash times\; 20\}\{415\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash left(\; \backslash frac\{4.6\; \backslash times\; 10.08\; \backslash times\; 10^6\}\{20\; \backslash times\; 1000\; \backslash times\; 150\; \backslash times\; 150\}\; \backslash right)\}\; \backslash right]\; \backslash times\; 1000\; \backslash times\; 150$

$A_{stx}=191.278{\text{mm}}^2$

Area and Spacing of Distribution Steel:

Select 8 mm Ø bars, Spacing = min of a, b, c

a) Calculation of spacing $S_{x}S\_x$:

$S_x=\left(\frac{\pi /4\times 8^2\times 1000}{191.27}\right)S\_x\; =\; \backslash left(\backslash frac\{\backslash pi/4\; \backslash times\; 8^2\; \backslash times\; 1000\}\{191.27\}\backslash right)$

$S_x=262.76\text{mm}\text{(say 260 mm)}S\_x\; =\; 262.76\; \backslash ,\; \backslash text\{mm\}\; \backslash ,\; \backslash text\{(say\; 260\; mm)\}$

$S_x=260\text{mm c/c}S\_x\; =\; 260\; \backslash ,\; \backslash text\{mm\; c/c\}$

b) Maximum spacing based on depth:

$3d=3\times 150=450\text{mm}3d\; =\; 3\; \backslash times\; 150\; =\; 450\; \backslash ,\; \backslash text\{mm\}$

c) Given maximum spacing:

$300\text{mm}300\; \backslash ,\; \backslash text\{mm\}$

Therefore, spacing $S_x=260\text{mm c/c}S\_x\; =\; 260\; \backslash ,\; \backslash text\{mm\; c/c\}$.

$d^{\prime }=d-\varnothing =150-8=142\text{mm}d\text{'}\; =\; d\; -\; \backslash varnothing\; =\; 150\; -\; 8\; =\; 142\; \backslash ,\; \backslash text\{mm\}$

$A_{s{t}_y}=\frac{0.5\times f_{ck}}{f_y}[1-\sqrt{1-\left(\frac{4.6\times M_{yd}}{f_{ck}\times b\times d^2}\right)}]\times b\times d^{\prime }A\_\{st\_y\}\; =\; \backslash frac\{0.5\; \backslash times\; f\_\{ck\}\}\{f\_y\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash left(\; \backslash frac\{4.6\; \backslash times\; M\_\{yd\}\}\{f\_\{ck\}\; \backslash times\; b\; \backslash times\; d^2\}\; \backslash right)\}\; \backslash right]\; \backslash times\; b\; \backslash times\; d\text{'}$

$A_{s{t}_y}=\frac{0.5\times 20}{415}[1-\sqrt{1-\left(\frac{4.6\times 7.08\times 10^5}{20\times 1000\times 142\times 142}\right)}]\times 1000\times 142A\_\{st\_y\}\; =\; \backslash frac\{0.5\; \backslash times\; 20\}\{415\}\; \backslash left[\; 1\; -\; \backslash sqrt\{1\; -\; \backslash left(\; \backslash frac\{4.6\; \backslash times\; 7.08\; \backslash times\; 10^5\}\{20\; \backslash times\; 1000\; \backslash times\; 142\; \backslash times\; 142\}\; \backslash right)\}\; \backslash right]\; \backslash times\; 1000\; \backslash times\; 142$

$=141.07{\text{mm}}^2=\; 141.07\; \backslash ,\; \backslash text\{mm\}^2$

$A_{s{t}_{\text{min}}}=\left(\frac{0.12}{100}\right)\times b\times DA\_\{st\_\{\backslash text\{min\}\}\}\; =\; \backslash left(\backslash frac\{0.12\}\{100\}\backslash right)\; \backslash times\; b\; \backslash times\; D$

$=\left(\frac{0.12}{100}\right)\times 1000\times 175=\; \backslash left(\backslash frac\{0.12\}\{100\}\backslash right)\; \backslash times\; 1000\; \backslash times\; 175$

$=210{\text{mm}}^2=\; 210\; \backslash ,\; \backslash text\{mm\}^2$

Since $A_{s{t}_y}A\_\{st\_y\}$ calculated is very low, take $A_{s{t}_x}A\_\{st\_x\}$ and $A_{s{t}_y}A\_\{st\_y\}$ as 210 mm${}^2^2$.

Therefore, spacing of 8 mm diameter bars:

$S_x=S_y=(\frac{\pi }{4}\times \frac{8^2\times 1000}{A_{s{t}_{\text{min}}}})S\_x\; =\; S\_y\; =\; \backslash left(\; \backslash frac\{\backslash pi\}\{4\}\; \backslash times\; \backslash frac\{8^2\; \backslash times\; 1000\}\{A\_\{st\_\{\backslash text\{min\}\}\}\}\; \backslash right)$

$S_x=S_y=\left(\frac{50.26\times 1000}{210}\right)S\_x\; =\; S\_y\; =\; \backslash left(\; \backslash frac\{50.26\; \backslash times\; 1000\}\{210\}\; \backslash right)$

$=239.33\text{mm (say 230 mm c/c, which is less than 3d or 300 mm)}=\; 239.33\; \backslash ,\; \backslash text\{mm\; (say\; 230\; mm\; c/c,\; which\; is\; less\; than\; 3d\; or\; 300\; mm)\}$

Provide 8 mm Ø bars at 230 mm c/c in both directions.

Reinforcement Details: