Reinforcement Detailing of Beams from ETABS output

AIM-Detail design of beams in Etabs 2018.

Minimum longitudinal Reinforcement as per IS 13920:2016 6.2.1

Criterias

• Minimum Reinforcement for shrinkage and temprature
• Reinforcement required for limiting crack width
• ultimate limit state design rebar requirement
• Ductility Rebar Requirement

and Analysis the model

Display Design Info> go for the longitudinal reinforcement

Reinforcement diagram of Grid A:

Max Ast at bottom support = 1925 mm2

Max Ast at bottom middle = 789 mm2

Max Ast at top support = 2045mm2

Max Ast at top middle = 511mm2

• So providing 3 bars of 20mm (942) and 5bars of 16mm (1004) at bottom

Calculation =3.14*.25*3*400+3.14*.25*5*256 = 1946>1925 (Hence safe)

• Provide 4 bars of 20mm (1256) and 4 bars of 16mm (804) at top

Calculation =3.14*.25*4*400+3.14*.25*4*256 =2059 >2045mm2 (Hence safe)

Now Check the beam in shear

by Etabs Design Checking the shear stresses in beam

6.3.2 The minimum diameter of a link shall be 8 mm. (IS 13920:2016)

Maximum shear Reinforcement Required in the Beam is 25.70 Cm/m

Formula For shear sway calculation

Longitudinal Reinforcement and Shear Reinforcement Drawing Beam of grid -A

Span 1-2

• Providing 3 legged 8 mm Diameter bars with spacing of 100mm at distance 2D
• At middle section 3 legged 8mm bars with 175mm C/C spacing provided

Span 2-3

• Providing 3 legged 12 mm Diameter bars with spacing of 100mm at distance 2D

Span 3-4

• Providing 3 legged 8 mm Diameter bars with spacing of 100mm at distance 2D
• At middle section 3 legged 8mm bars with 175mm C/C spacing provided

Calculation =(11*3.14*0.25*3*100)/100

=25.905 >25.70

FROM Etabs Analysis Reinforcement diagram of Grid 3:

Max Ast at bottom support = 1042 mm2

Max Ast at bottom middle = 1501mm2

Max Ast at top support = 1940 mm2

Max Ast at top middle = 485mm2

So providing 2 bars of 12mm (226) and 4 bars of 16mm (804) at bottom

Calculation =3.14*.25*2*144+3.14*.25*4*256 = 1029~1042

Provide 3 bars of 20mm (942) and 5 bars of 16mm (1004) at top

Calculation =3.14*.25*3*400+3.14*.25*5*256 = 1029~1042

Longitudinal Reinforcement details for GRID -3

Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved?

1. In the overstressed beam there is negligile amount of torsion but there is huge amount of shear force there so we have only three option to make the beam understress
2. so as we know that the Area of cross section of the column is directly propotional to the strength of the beam so we should increase the area of the cross section
3. and if can reduce the imposed forces there than we can make the beam understressed
4. other one option is to increase the grade of concrete to provide the strenth to the column