Reinforcement Detailing of Beams from ETABS output
Challenge 7 – Detailed design of concrete buildings -1
Detail design of beams in Etabs 2018.
The ETABS file for a G+4 building is provided. Run the analysis and design the RCC Moment ResistingFrame. The following challenge deals specifically with two continuous beams in the 1st floor.
1) 3 span continuous beam along grid A
2) 7 span continuous beam along grid 3
Address the following issues thereafter.
• Provide details of longitudinal and shear reinforcement for the two continuous beams. Sketch the beam elevation details or draft it using autocad software. A sample of how beam elevation details is prepared in consultancy firms is provided below. Participants can prepare it as per their preference:
• Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved?
• 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. Also, confirm these values from shear force demand calculated by ETABS.Please note that if longitudinal reinforcement provided is more than required (as per ETABS results), the shear force demand will vary from what is provided by ETABS.
Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved?
This failing is due to higher shear stress.As per the above shear force value 3.11mpa aganist the max allowable shear stress 3.1mpa is almost failing but passes numerically.
- Optimizing provided flexural reinforcement
- reducing depth of RC beam which in turn reduces stiffness and flexural requirement
- increasing concrete grade will increase allowable shear stress limit
Calculation:
Grid 3:
At Top:
1) Min Ast at Support requred = 1940MM2
2) Min 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) Min Ast at Support = 970
2) Maximum Ast at Middle = 1498
* 2 Bar dia of 25mm = 980
* 1 dia of 32mm = 800+980=1780
Therefore, to satisfy the support we need to provide 2 dia of 25mm bar and additionally in the middle need to provide 1 dia of 32mm.
lets take 8 dia with 100mm spacing
shear R/f calulation formula use is 1000/S*no of shear leg*A
so shear rainforsment is requred accouring to etabs at the spot is= 1498mm2
1000/100*3*50=1500
here 100mm spacing
3leg stirups
8 dia bar size area =50
shear R/F requred in mid section =1160
1000/100*3*50=1500 becouse we cant give spacing less than 100mm accourding to is 456-2000
Grid A:
At Top:
1) Min Ast at Support = 2045
2) Min Ast at Middle = 477.
* 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 need to provide 1 dia of 25mm.
At Bottom:
1) Maximum Ast at Support = 1927
2) Maximum Ast at Middle = 789
* 2 dia of 32mm = 1606
* 1 dia of 16mm = 200.96
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 16mm.
