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′d’) = 25 mm

Area of compression steel (AscA_{text{sc}}) = 1035 mm²

Area of tension steel (AstA_{text{st}}) = 1840 mm²

Characteristic compressive strength of concrete (fckf_{text{ck}}) = 20 N/mm²

Yield strength of steel (fyf_{text{y}}) = 415 N/mm²

Step 1: To find Xu,maxX_{u, text{max}}

For Fe415 steel:

Xu,max=0.479 dX_{u, text{max}} = 0.479 , d

Xu,max=0.479×500X_{u, text{max}} = 0.479 times 500

Xu,max=239.5 mmX_{u, text{max}} = 239.5 , text{mm}

Step 2: To find actual XuX_u

For compression steel:

fcc=σcc=0f_{text{cc}} = sigma_{text{cc}} = 0

Area of tension steel in compression (Ast2A_{text{st2}}):

Ast2=(fsc−fcc)×Asc0.87 fyA_{text{st2}} = frac{(f_{text{sc}} – f_{text{cc}}) times A_{text{sc}}}{0.87 , f_{text{y}}}

Ast2=(355−0)×10350.87×415A_{text{st2}} = frac{(355 – 0) times 1035}{0.87 times 415}

Ast2=1035×3550.87×415A_{text{st2}} = frac{1035 times 355}{0.87 times 415}

Ast2=1017.656 mm2A_{text{st2}} = 1017.656 , text{mm}²

Area of tension steel in tension (Ast1A_{text{st1}}):

Ast1=Ast−Ast2A_{text{st1}} = A_{text{st}} – A_{text{st2}}

Ast1=1840−1017.656A_{text{st1}} = 1840 – 1017.656

Ast1=822.344 mm2A_{text{st1}} = 822.344 , text{mm}²

Neutral axis depth (XuX_u):

Xu=0.87 fy Ast10.36 fck bX_u = frac{0.87 , f_{text{y}} , A_{text{st1}}}{0.36 , f_{text{ck}} , b}

Xu=0.87×415×822.3440.36×20×300X_u = frac{0.87 times 415 times 822.344}{0.36 times 20 times 300}

Xu=137.457 mmX_u = 137.457 , text{mm}

Since Xu<Xu,maxX_u < X_{u, text{max}}, the section is under-reinforced.

Step 3: To find Moment of Resistance

Mu=0.87 fy Ast1(d−0.42 Xu)+(fsc−fcc) Asc(d−d′)M_u = 0.87 , f_{text{y}} , A_{text{st1}} (d – 0.42 , X_u) + (f_{text{sc}} – f_{text{cc}}) , A_{text{sc}} (d – d’)

Mu=0.87×415×822.344(500−0.42×137.457)+(355−0)×1035(500−25)M_u = 0.87 times 415 times 822.344 (500 – 0.42 times 137.457) + (355 – 0) times 1035 (500 – 25)

Mu=305.83×106 NmmM_u = 305.83 times 10^6 , text{Nmm}

Mu=305.83 kNm

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