Reinforcement Detailing of Beams from ETABS output
- AIM : Run the analysis and design the RCC Moment Resisting Frame.
PROCEDURE :
STEP 1 : - Open the challenge 7 attachment which are provided to us and download it.
- After downloading,open the challenge in ETABS software.
STEP 2 : - Now check the load combinations which are already created in given modelling file.
STEP 3 : - Before Run analysis we have to check the model.
- Click on Analyse
- Click on Check model
- A structure is verified without any error and Model has been checked.
STEP 4 : - Now we have to start the beam design by selecting the Start Design/Check Option.
- Once you click on Start / Design check you will see Display Concrete Frame Design result box.
- Where you can select whatever design output you want
- So click on Longitudinal Reinforcement.
STEP 5 : - Once you click on Longitudinal reinfocement details will appear.
STEP 6 :
1.Reinforcement details alang Grid A :
i) Max Ast" > AstVu,a at bottom support = 1026mm^2
ii) Max Ast" > Ast at bottom middle = 757mm^2
iii) Max Ast" > Ast at top support =1896mm^2
iv) Max Ast" > Ast at top middle = 474mm^2
Reinforcement Calculation -
1) For Top -
i) Total Ast required at top is =1896mm^2
So we can provide
3 bars of diameter 20mm and 5 bars of 16mm satisfies the overall Ast
3 x T20=942mm^2
5 x T16=804mm^2
3 x T20 + 5 x T16=1946mm^2.............Hence Satisfy.
2) For Bottom -
i) Total Ast required at top is = 1026mm^2
So we can provide
2 bars of diameter 20mm and 2 bars of 16mm satisfies the overall Ast
2 x T20=628mm^2
2 x T16=402mm^2
2 x T20 + 2 x T16=1030mm^2..............Hence Satisfy.
3) For Shear Details- - Right click on and click on Details
- Click on Shear Details
- As per Clause 6.3.3 we have to provide Spacing till 2d from the face of the column is 100.
- It should not be more than that
- So provide 3 legged stirrups of 10mm diameter with 100mm spacing throught the beam length.
STEP 8 : - Now we have to to check for Sway Shear
A) For Right Sway
We know that the formula to calculate :
From ETABS Model results substitute the value -
i) = 92KN
ii) Mas = 176KN-m
iii) Mbh = 295KN-m
iv) Lab = 40725m
Put this value in above equation, we get ,
V u,a = 48 KN
V u,b = 230 KN
B) For Left Sway
i) Follow the same process
ii) We know the formula
From ETABS model results substitute the value -
i) = 74KN
ii)Mah = 23Kn-m
iii) Mbs = 150KN-m
iv) Lab = 4.725m
Put this value in above equation, we get,
V u,a = 190KN
V u,b = 50KN
C) Existing shear capacity in beams as per IS456:2000
i) We know the formula,
Put the values -
i) fy=415N/mm^2
ii)Asv = 151mm^2
iii)spacing =100mm
iv) d = 450mm
Put these values in above equation , we get,
V us = 246KN....................
230KN(shear sway).........(Hence safe)
Results -
1) Beam Design is done successfully with detailed reinforcement.
2)
i)The main reason of failure of middle span along grid A is shear stress is exceeded the limiting value.
ii) This may be because of the higher Shear force acting on that beam and due to provision of weak shear
iii)This issue can be solved by - - By providing higher grade of concrete in that particular section.
- Or by reducing the shear force in that beam section
- Or by providing higher shear reinforcement which can resist the shear force in that section.
3)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
246KN > 230KN................(safe)