RCC Numerical: Design of RCC Column

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Design of RCC column

Numerical:

Design a R.C column to carry an axial working load 400 kN.The effective length of column is 2.5 m. check the column for min eccentricity. Use M20 and Fe 415 grades of concrete and steel. 

Solution:

Given Data:

$P=400\text{kN}P\; =\; 400\; \backslash ,\; \backslash text\{kN\}$

$L_{\text{eff}}=2.5\text{m}=2500\text{mm}L\_\{\backslash text\{eff\}\}\; =\; 2.5\; \backslash ,\; \backslash text\{m\}\; =\; 2500\; \backslash ,\; \backslash text\{mm\}$

$f_{ck}=20{\text{N/mm}}^{2}f\_\{ck\}\; =\; 20\; \backslash ,\; \backslash text\{N/mm\}^2$

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

Step 1: To Find Factored Load

$P_u=1.5\times PP\_u\; =\; 1.5\; \backslash times\; P$

$P_u=1.5\times 400P\_u\; =\; 1.5\; \backslash times\; 400$

$P_u=600\text{kN}P\_u\; =\; 600\; \backslash ,\; \backslash text\{kN\}$

Step 2: Assume 1% of Steel in Column

Area of steel $A_{sc}A\_\{sc\}$:

$A_{sc}=0.01\times A_{g}A\_\{sc\}\; =\; 0.01\; \backslash times\; A\_g$

Area of concrete $A_{c}A\_c$:

$A_c=A_g-A_{sc}A\_c\; =\; A\_g\; -\; A\_\{sc\}$

$A_c=0.99\times A_{g}A\_c\; =\; 0.99\; \backslash times\; A\_g$

Step 3: To Find Gross Area

$A\_g$

Factored load formula:

$P_u=(0.4\times f_{ck}\times A_c)+(0.67\times f_y\times A_{sc})P\_u\; =\; (0.4\; \backslash times\; f\_\{ck\}\; \backslash times\; A\_c)\; +\; (0.67\; \backslash times\; f\_y\; \backslash times\; A\_\{sc\})$

Substitute values:

$600\times 10^3=(0.4\times 20\times 0.99\times A_g)+(0.67\times 415\times 0.01\times A_g)600\; \backslash times\; 10^3\; =\; (0.4\; \backslash times\; 20\; \backslash times\; 0.99\; \backslash times\; A\_g)\; +\; (0.67\; \backslash times\; 415\; \backslash times\; 0.01\; \backslash times\; A\_g)$

Solve for $A_{g}A\_g$:

$600\times 10^3=7.92\times A_g+2.78\times A_{g}600\; \backslash times\; 10^3\; =\; 7.92\; \backslash times\; A\_g\; +\; 2.78\; \backslash times\; A\_g$

$600\times 10^3=10.70\times A_{g}600\; \backslash times\; 10^3\; =\; 10.70\; \backslash times\; A\_g$

$A_g=\frac{600\times 10^3}{10.70}A\_g\; =\; \backslash frac\{600\; \backslash times\; 10^3\}\{10.70\}$

$A_g=56074.77{\text{mm}}^{2}A\_g\; =\; 56074.77\; \backslash ,\; \backslash text\{mm\}^2$

$A_g=56.07\times 10^3{\text{mm}}^{2}A\_g\; =\; 56.07\; \backslash times\; 10^3\; \backslash ,\; \backslash text\{mm\}^2$

Assuming a square column:

$\text{Side}=\sqrt{56.07\times 10^3}\backslash text\{Side\}\; =\; \backslash sqrt\{56.07\; \backslash times\; 10^3\}$

$\text{Side}\approx 236.79\text{mm}\backslash text\{Side\}\; \backslash approx\; 236.79\; \backslash ,\; \backslash text\{mm\}$



Step 4: Check for Minimum Eccentricity

$e_{\text{min}}=\frac{L}{500}+\frac{D}{30}\text{ OR }20\text{mm whichever is greater}e\_\{\backslash text\{min\}\}\; =\; \backslash frac\{L\}\{500\}\; +\; \backslash frac\{D\}\{30\}\; \backslash text\{\; OR\; \}\; 20\; \backslash text\{mm\; whichever\; is\; greater\}$

Given:

• $L=2500\text{mm}L\; =\; 2500\; \backslash ,\; \backslash text\{mm\}$

• $D=240\text{mm}D\; =\; 240\; \backslash ,\; \backslash text\{mm\}$

$e_{\text{min}}=\frac{2500}{500}+\frac{240}{30}e\_\{\backslash text\{min\}\}\; =\; \backslash frac\{2500\}\{500\}\; +\; \backslash frac\{240\}\{30\}$

$e_{\text{min}}=5+8e\_\{\backslash text\{min\}\}\; =\; 5\; +\; 8$

$e_{\text{min}}=13\text{mm OR }20\text{mm whichever is greater}e\_\{\backslash text\{min\}\}\; =\; 13\; \backslash text\{mm\; OR\; \}\; 20\; \backslash text\{mm\; whichever\; is\; greater\}$

$e_{\text{min}}=20\text{mm}e\_\{\backslash text\{min\}\}\; =\; 20\; \backslash text\{mm\}$

But, $e_{\text{min}}e\_\{\backslash text\{min\}\}$ is more than 0.05D.

$0.05D=0.05\times 240=12\text{mm}0.05D\; =\; 0.05\; \backslash times\; 240\; =\; 12\; \backslash ,\; \backslash text\{mm\}$

So, check for minimum eccentricity is not satisfied.

Increase the Dimension

New dimensions: $320\text{mm}\times 320\text{mm}320\; \backslash ,\; \backslash text\{mm\}\; \backslash times\; 320\; \backslash ,\; \backslash text\{mm\}$

$e_{\text{min}}=\frac{2500}{500}+\frac{320}{30}e\_\{\backslash text\{min\}\}\; =\; \backslash frac\{2500\}\{500\}\; +\; \backslash frac\{320\}\{30\}$

$e_{\text{min}}=5+10.67e\_\{\backslash text\{min\}\}\; =\; 5\; +\; 10.67$

$e_{\text{min}}=15.67\text{mm}e\_\{\backslash text\{min\}\}\; =\; 15.67\; \backslash text\{mm\}$

$0.05D=0.05\times 320=16\text{mm}0.05D\; =\; 0.05\; \backslash times\; 320\; =\; 16\; \backslash text\{mm\}$

Check for minimum eccentricity is satisfied.

Revised Size of Column

Revised size of column: $320\text{mm}\times 320\text{mm}320\; \backslash ,\; \backslash text\{mm\}\; \backslash times\; 320\; \backslash ,\; \backslash text\{mm\}$

Step 5: Calculate Area of Steel

$A_{\text{sc}}=0.01A_{\text{g}}A\_\{\backslash text\{sc\}\}\; =\; 0.01\; A\_\{\backslash text\{g\}\}$

$A_{\text{g}}=320\times 320A\_\{\backslash text\{g\}\}\; =\; 320\; \backslash times\; 320$

$A_{\text{sc}}=0.01\times 320\times 320A\_\{\backslash text\{sc\}\}\; =\; 0.01\; \backslash times\; 320\; \backslash times\; 320$

$A_{\text{sc}}=1024{\text{mm}}^{2}A\_\{\backslash text\{sc\}\}\; =\; 1024\; \backslash ,\; \backslash text\{mm\}^2$

Provide 4 bars of 20 mm Ø bar.

Step 6: Lateral Ties

Diameter of ties:

$\text{Diameter of ties}=\frac{1}{4}\times \text{Diameter of longitudinal steel bar}\backslash text\{Diameter\; of\; ties\}\; =\; \backslash frac\{1\}\{4\}\; \backslash times\; \backslash text\{Diameter\; of\; longitudinal\; steel\; bar\}$

$=\frac{1}{4}\times 20=\; \backslash frac\{1\}\{4\}\; \backslash times\; 20$

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

Since the calculated diameter is less than 6 mm, provide 6 mm diameter lateral ties.

Spacing of Lateral Ties

Spacing should not be greater than:

1. Least lateral dimension of column $=320\text{mm}=\; 320\; \backslash ,\; \backslash text\{mm\}$

2. 16 times the diameter of longitudinal steel $=16\times 20=320\text{mm}=\; 16\; \backslash times\; 20\; =\; 320\; \backslash ,\; \backslash text\{mm\}$

3. 300 mm

Provide lateral ties of 6 mm Ø at 300 mm center-to-center.