Reinforcement Detailing of Beams from ETABS output
Aim:
To Providing details of longitudinal and shear reinforcement for the two continuous beams using ETABS
Step 1:
1) After downloading the model given in the question, click the Concrete frame design tool in the ribbon tab.
2) Choose Select design combinations
3) Select the appropriate combination from the dialogue box
4) Click ok
Step 2:
1) Once the analysis is done, Click concrete frame icon
2) After clicking the icon, it starts checking the concrete design
Step 3:
1) After the checking is done, go to the plan view
2) On the plan view we have a column and beam detailing
3) For this, Grid 3 and Grid A need to provide reinforcement details.
Step 4:
Grid 3:
At Top:
1) Maximum Ast at Support = 194
2) Maximum Ast at Middle = 485.
* 2 dia of 32mm = 1606
* 1 dia of 25 mm = 490
Therefore, to satisfy the support we need to provide 2 dia of 32mm bar and additionally in the middle need to provide 1 dia of 25mm.
At Bottom:
1) Maximum Ast at Support = 1056
2) Maximum Ast at Middle = 1501
* 3 dia of 25mm = 1470
* 1 dia of 16mm = 200.96
Therefore, to satisfy the support we need to provide 3 dia of 25mm bar and additionally in the middle need to provide 1 dia of 16mm.
Step 5:
1) Reinforcement details along Grid A -
i) Max Ast">AstVu,a at bottom support = 1026 mm2
ii) Max Ast">Ast at bottom middle = 757 mm2
iii) Max Ast">Ast at Top support = 1896 mm2
iv) Max Ast">Ast at Top middle = 474 mm2
Reinforcement Calculation -
1) For Top -
i) Total Ast required at top is = 1896 MM^2
So we can provide
3 bars of diameter 20 mm and 5 bars of 16 mm satisfies the overall Ast
3 x T20 = 942 mm2
5 x T16 = 804 mm2
3 X T20 + 5 x T16 = 1946mm2.......................Hence Satisfy .
2) FOR BOTTOM: -
i) Total Ast required at top is = 1026 MM^2.
So we can provide
2 bars of diameter 20 mm and 2 bars of 16 mm satisfies the overall Ast
2 x T20 = 628 mm2
2 x T16 = 402 mm2
2X T20 + 2 x T16 = 1030 mm2........................Hence Satisfy .
Step 6:
1) Right-Click the Grid A beam
2) Click Details and switch it into shear details
Step 7:
1) As per clause 6.3.3 the design force of equivalent Ve = 242.75 kN
2) Shear force under the linear analysis as per the load combination of UDCons66 = 194.825 kN
i) By providing Higher grade of concrete in that particular section
ii) Or by reducing the Shear force in that beam section
iii) Or by providing higher shear reinforcement which can resist the shear force in that section
iii) Calculate value of the maximum shear force in any one of the spans in both the continuous beams,
as per clause 6.3.3 (b) in IS 13920 – 2016.
As we have already calculated the value of shear force as per IS 13920 - 2016 Clause 6.3.3 (b)
We get the Shear sway value Right and Left support within limit