Reinforcement Detailing of Beams from ETABS output
AIM:- To run the analysis and design the RCC moment resisting frame for G+4 building:-
a) 3 span continuous beam along grid A
b) 7 span continuous beam along grid 3
To provide details of longitudinal and shear reinforcement for the two continuous beams. To sketch the elevation details or draft it using AutoCAD software.
To provide reasons for failure of the middle span along Grid A and the possible ways to resolve the issue.
To 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.
INTRODUCTION:- All the steps are going to be mentioned along with step by step procedure
PROCEDURE:
- Download the file and open the ETABS software.
- Check the model
- Model has been checked
- Click on concrete frame design
- Choose it as total longitudinal reinforcing
-
Longitudinal Reinforcement :
* the top as well as the bottom reinforcement shall consist of at least two bars throughout the member length
Analysis needs to be done
*Concrete design is done for this model
-
3 span continuous beam along grid A
At the top:
Maximum at end support 2046 mm^2
Minimum at end support 938 mm ^2
Ast for 32 diameter = 804 .24
Ast for 25 diameter = 490
Ast for 20 diameter= 314
Ast for 16 diameter = 201
Ast for 12 diameter = 113
Maximum at the mid supports is 511 mm^2
-
** 2 bars of diameter 32mm and 1 bars of 25 mm satisfies the overall Ast
2 x T32 = 1608 mm ^2
T25 = 490 mm^2
2 X T32 + T25 = 2098 mm ^2
(or)
** 3 bar of diameter 32 mm satidfies the overall Ast
3 T32 = 2412 mm ^2
-
At the bottom :
Maximum at the end support is 1945 mm ^2
Maximum at the mid support is 789 mm ^2
** 2 bars of diameter 32 mm and 1 bar of 25 mm satisfies the overall Ast
2 x T32 = 1608 mm ^2
T25 = 490 mm^2
2 X T32 + T25 = 2098 mm ^2
- Grid A
Grid 3
7 span continuous beam along grid 3
At the top :
**Maximum at end supports is 1940 mm ^2
**Maximum at Mid supports is 485 mm ^2
** 2 bars of diameter 32 mm and 1 bar of 25 mm satisfies the overall Ast
2 x T32 = 1608 mm ^2
T25 = 490 mm^2
2 X T32 + T25 = 2098 mm ^2
At the bottom :
** Maximum at end supports is 1105 mm^2
** Maximum at mid supports is 1501 mm^2
** 2 bars of diameter 32mm satisfies the overall Ast
2 X T32 = 1608 mm ^2
Shear details :
- From the above ETABS output it has been shown that the limiting value of shear force and developed shear force remains the same. Hence it is not safe
The possible ways to solve this problem can be :-
a) By providing higher grade of concrete
b) By reducing effective depth
From ETABS Design shear force = 209.58 KN
Provide 2 legged 8 mm dia stirrups with 100 mm spacing
Providing 30 mm clear cover, Asv = 2*3.14/4*10^2 = 100 mm^2
Effective depth = 500-60 = 440 mm
Vus = 0.87*fy*Asv*d/Sv = 0.87*415*100*440/100 = 158.862 KN
Since Vus < V ( Design shear force), hence safe.
So provide 2 legged 8 mm dia stirrups with 100 mm spacing.
RESULTS: The folloxing has been illustrated properly along with results:-
a) Analysis and design the RC moment frame for G+4 building
b) Providing details of longitudinal and shear reinforcement for beam along with elevation details
c) Providing reasons for failure of the middle span along grid A
d) Calculating value of maximum shear forces in any of the span