Challenge

Reinforcement Detailing of Beams from ETABS output

AAkshay Thakur2 Feb 2023

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.

ANSWER

BEAM 33 ALONG GRID A FLRXURAL DETAILING

LIKE WISE WE CALCULATED THE RESULTS FOR

BEAM 34 ALONG GRID A

BEAM 38 ALONG GRID A

NOW ALONG GRID 3

BEAM 24

BEAM 25

BEAM 69

BEAM 71

BEAM 72

BEAM 17

BEAM 14

BEAM 16

Shear reinforcement calculations

Grid line A (B38)

From ETABS output, Design shear force, V = 241.51 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.

Grid line 3 ( B 24)

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.

Reasons for failure of the middle span along grid 3

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 a higher grade of concrete

b) By reducing the g effective depth

Detailing of beams using AutoCAD

Grid A

Grid 3

Calculate the 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.

According to the Clause 6.3.3 as shown below

Calculate the value of maximum shear force