RCC Numerical - Simply Supported Beam Calculate depth and area of steel

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Simply Supported Beam Calculate depth and area of steel

Numerical:

Calculate depth and area of steel at mid span of a simply supported beam over as clear span 6 m. The beam is carrying all inclusive load 20 KN/m. Assume 300 mm bearings. Use M20 and Fe500

Solution:



Given:

• Length of beam (l): 6 m
• Effective length (Le): 6 + 0.3/2 + 0.3/2 = 6.3 m
• Load (w): 20 kN/m
• Characteristic compressive strength of concrete (fck): 20 N/mm²
• Yield strength of steel (fy): 500 N/mm²

Solution:

• Moment Calculation:

$M=\frac{w\cdot L_e^2}{8}M\; =\; \backslash frac\{w\; \backslash cdot\; L\_e^2\}\{8\}$

$=\frac{20\cdot 6.3^2}{8}=\; \backslash frac\{20\; \backslash cdot\; 6.3^2\}\{8\}$

$=99.225\text{ KNm}=\; 99.225\; \backslash text\{\; KNm\}$

• Factored Moment:

$M_d={\gamma }_f\cdot M$

$M\_d\; =\; \backslash gamma\_f\; \backslash cdot\; M$

$=1.5\cdot 99.225$

$\mathrm{}=\; 1.5\; \backslash cdot\; 99.225$

$=148.8375\text{ KNm}=\; 148.8375\; \backslash text\{\; KNm\}$

• Maximum Moment Capacity:

$M_u\text{ lim}=0.133\cdot f_{ck}\cdot b\cdot d^{2}M\_u\; \backslash text\{\; lim\}\; =\; 0.133\; \backslash cdot\; f\_\{ck\}\; \backslash cdot\; b\; \backslash cdot\; d^2$

• Assume beam width (b):

$b=d/2$

Equating $M_u\text{ lim}$ to $M_d$:

$0.133\cdot f_{ck}\cdot b\cdot d^2=148.83\times 10^6$

${\mathrm{}}^{}0.133\; \backslash cdot\; f\_\{ck\}\; \backslash cdot\; b\; \backslash cdot\; d^2\; =\; 148.83\; \backslash times\; 10^6$

$0.133\cdot 20\cdot \left(d/2\right)\cdot d^2=148.83\times 10^6$

${\mathrm{}}^{}0.133\; \backslash cdot\; 20\; \backslash cdot\; (d/2)\; \backslash cdot\; d^2\; =\; 148.83\; \backslash times\; 10^6$

$d^3\cdot 1.33=148.83\times 10^6$

${\mathrm{}}^{}d^3\; \backslash cdot\; 1.33\; =\; 148.83\; \backslash times\; 10^6$

$d^3=111908270.7$

$\mathrm{}d^3\; =\; 111908270.7$

$d=481.89\text{ mm}d\; =\; 481.89\; \backslash text\{\; mm\}$

Say, $d=490\text{ mm}d\; =\; 490\; \backslash text\{\; mm\}$

Assume effective cover ($d^{\prime }d\text{'}$) = 40 mm:

$D=d+d^{\prime }=490+40=530\text{ mm}D\; =\; d\; +\; d\text{'}\; =\; 490\; +\; 40\; =\; 530\; \backslash text\{\; mm\}$

Therefore:

$b=490/2=245\text{ mm}b\; =\; 490\; /\; 2\; =\; 245\; \backslash text\{\; mm\}$

• Percentage of tensile reinforcement:

$P_{t\text{ lim}}=\frac{\text{Ast}}{b\cdot d}\cdot 100$

$\mathrm{}P\_\{t\; \backslash text\{\; lim\}\}\; =\; \backslash frac\{\backslash text\{Ast\}\}\{b\; \backslash cdot\; d\}\; \backslash cdot\; 100$

$=0.038\cdot f_{ck}$

${}_{}=\; 0.038\; \backslash cdot\; f\_\{ck\}$

$=0.038\cdot 20$

$\mathrm{}=\; 0.038\; \backslash cdot\; 20$

$=0.76\%$

$=\; 0.76\; \backslash \%$

$\text{Ast}=\frac{0.76\cdot b\cdot d}{100}$

$\frac{\mathrm{}}{}\backslash text\{Ast\}\; =\; \backslash frac\{0.76\; \backslash cdot\; b\; \backslash cdot\; d\}\{100\}$

$=\frac{0.76\cdot 245\cdot 490}{100}$

$\frac{\mathrm{}}{}=\; \backslash frac\{0.76\; \backslash cdot\; 245\; \backslash cdot\; 490\}\{100\}$

$=912.38{\text{ mm}}^2=\; 912.38\; \backslash text\{\; mm\}^2$

Therefore:

Assuming 20 mm Ø bars to be used:

No. of bars:

$=\frac{912.38}{(\frac{\pi }{4}\cdot 20^2)}=2.90\to \text{Say 3 Nos}=\; \backslash frac\{912.38\}\{\backslash left(\backslash frac\{\backslash pi\}\{4\}\; \backslash cdot\; 20^2\backslash right)\}\; =\; 2.90\; \backslash rightarrow\; \backslash text\{Say\; 3\; Nos\}$

Provide 3 bars of 20 mm Ø with area = 942.48 mm²