RCC numerical: Design of Doubly Reinforced Beam

Hi all, lets understand the step by step calculation to design the Doubly Reinforced Beam. This post will provide you a step by step design calculation from start till end to understand the procedure.

Design of Doubly Reinforced Beam

Numerical:

A doubly reinforced beam 300 mm x 500 mm effective is reinforced with 1035 mm2 at 25 mm below top edge and 1840 mm2 above bottom edge. Take M 20 concrete and Steel Fe 415. Find moment of resistance (Mu). Use fsc= 355 N/mm2 and neglect σcc.

Solution:

Given Data:

Breadth ($bb$) = 300 mm

Effective depth ($dd$) = 500 mm

Cover depth ($d^{\prime }d\text{'}$) = 25 mm

Area of compression steel ($A_{\text{sc}}A\_\{\backslash text\{sc\}\}$) = 1035 mm²

Area of tension steel ($A_{\text{st}}A\_\{\backslash text\{st\}\}$) = 1840 mm²

Characteristic compressive strength of concrete ($f_{\text{ck}}f\_\{\backslash text\{ck\}\}$) = 20 N/mm²

Yield strength of steel ($f_{\text{y}}f\_\{\backslash text\{y\}\}$) = 415 N/mm²

Step 1: To find ${\mathrm{<strong>X</strong>}}_{\mathrm{<strong>u</strong>}<strong>,</strong>\text{<strong>max</strong>}}X\_\{u,\; \backslash text\{max\}\}$

For Fe415 steel:

$X_{u,\text{max}}=0.479dX\_\{u,\; \backslash text\{max\}\}\; =\; 0.479\; \backslash ,\; d$

$X_{u,\text{max}}=0.479\times 500X\_\{u,\; \backslash text\{max\}\}\; =\; 0.479\; \backslash times\; 500$

$X_{u,\text{max}}=239.5\text{mm}X\_\{u,\; \backslash text\{max\}\}\; =\; 239.5\; \backslash ,\; \backslash text\{mm\}$

Step 2: To find actual ${}_{}$${\mathrm{<strong>X</strong>}}_{\mathrm{<strong>u</strong>}}X\_u$

For compression steel:

$f_{\text{cc}}={\sigma }_{\text{cc}}=0f\_\{\backslash text\{cc\}\}\; =\; \backslash sigma\_\{\backslash text\{cc\}\}\; =\; 0$

Area of tension steel in compression ($A_{\text{st2}}A\_\{\backslash text\{st2\}\}$):

$A_{\text{st2}}=\frac{(f_{\text{sc}}-f_{\text{cc}})\times A_{\text{sc}}}{0.87f_{\text{y}}}A\_\{\backslash text\{st2\}\}\; =\; \backslash frac\{(f\_\{\backslash text\{sc\}\}\; -\; f\_\{\backslash text\{cc\}\})\; \backslash times\; A\_\{\backslash text\{sc\}\}\}\{0.87\; \backslash ,\; f\_\{\backslash text\{y\}\}\}$

$A_{\text{st2}}=\frac{(355-0)\times 1035}{0.87\times 415}A\_\{\backslash text\{st2\}\}\; =\; \backslash frac\{(355\; -\; 0)\; \backslash times\; 1035\}\{0.87\; \backslash times\; 415\}$

$A_{\text{st2}}=\frac{1035\times 355}{0.87\times 415}A\_\{\backslash text\{st2\}\}\; =\; \backslash frac\{1035\; \backslash times\; 355\}\{0.87\; \backslash times\; 415\}$

$A_{\text{st2}}=1017.656{\text{mm}}^{2}A\_\{\backslash text\{st2\}\}\; =\; 1017.656\; \backslash ,\; \backslash text\{mm\}²$

Area of tension steel in tension ($A_{\text{st1}}A\_\{\backslash text\{st1\}\}$):

$A_{\text{st1}}=A_{\text{st}}-A_{\text{st2}}A\_\{\backslash text\{st1\}\}\; =\; A\_\{\backslash text\{st\}\}\; -\; A\_\{\backslash text\{st2\}\}$

$A_{\text{st1}}=1840-1017.656A\_\{\backslash text\{st1\}\}\; =\; 1840\; -\; 1017.656$

$A_{\text{st1}}=822.344{\text{mm}}^{2}A\_\{\backslash text\{st1\}\}\; =\; 822.344\; \backslash ,\; \backslash text\{mm\}²$

Neutral axis depth ($X_{u}X\_u$):

$X_u=\frac{0.87f_{\text{y}}{A}_{\text{st1}}}{0.36f_{\text{ck}}b}X\_u\; =\; \backslash frac\{0.87\; \backslash ,\; f\_\{\backslash text\{y\}\}\; \backslash ,\; A\_\{\backslash text\{st1\}\}\}\{0.36\; \backslash ,\; f\_\{\backslash text\{ck\}\}\; \backslash ,\; b\}$

$X_u=\frac{0.87\times 415\times 822.344}{0.36\times 20\times 300}X\_u\; =\; \backslash frac\{0.87\; \backslash times\; 415\; \backslash times\; 822.344\}\{0.36\; \backslash times\; 20\; \backslash times\; 300\}$

$X_u=137.457\text{mm}X\_u\; =\; 137.457\; \backslash ,\; \backslash text\{mm\}$

Since , the section is under-reinforced.

Step 3: To find Moment of Resistance ${}_{}$

$M_u=0.87f_{\text{y}}{A}_{\text{st1}}(d-0.42X_u)+(f_{\text{sc}}-f_{\text{cc}})A_{\text{sc}}(d-d^{\prime })M\_u\; =\; 0.87\; \backslash ,\; f\_\{\backslash text\{y\}\}\; \backslash ,\; A\_\{\backslash text\{st1\}\}\; (d\; -\; 0.42\; \backslash ,\; X\_u)\; +\; (f\_\{\backslash text\{sc\}\}\; -\; f\_\{\backslash text\{cc\}\})\; \backslash ,\; A\_\{\backslash text\{sc\}\}\; (d\; -\; d\text{'})$

$M_u=0.87\times 415\times 822.344(500-0.42\times 137.457)+(355-0)\times 1035(500-25)M\_u\; =\; 0.87\; \backslash times\; 415\; \backslash times\; 822.344\; (500\; -\; 0.42\; \backslash times\; 137.457)\; +\; (355\; -\; 0)\; \backslash times\; 1035\; (500\; -\; 25)$

$M_u=305.83\times 10^6\text{Nmm}M\_u\; =\; 305.83\; \backslash times\; 10^6\; \backslash ,\; \backslash text\{Nmm\}$

$M_u=305.83\text{kNm}$