Reinforcement Detailing of Beams from ETABS output

Q. 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.

Ans:

• Open the downloaded ETABS file.
  • 
• Click on Analysis.
  • 
• Click on Concrete Frame Design
  • 
• Go to concrete frame design --> Select Design combinations
  • 
• Choose the combinations of only for the dead loads, LL & SIDL except earthquake load.
  • 
• Go to concrete frame design --> Select design group --> Select shear reinforcing --> Apply
• 
• Here it is shown that beam 2-3 of grid A is failing.
• Right-click on the failure beam
  • 
• Go to details.
  • it can be shown that the reason for the failure is written below.
  • 
• So to increase the shear capacity the grade of the concrete can be revised by increasing the grade of concrete.

• For defining the sway combination go to define --> load combination --> add new combo --> rename the load combination as Sway -- > add the scale factor as 1.2 for DL , LL, Brick wall load & floor finishes
  • 
• Ok
• Then go to display frame/pier / spandrel --> select load case as sway --> then select component as shear 2-2 --> apply --> ok
  • 

Detailing:-

• The required area of steel for the required beam Grid 3.
  • 
• Detailing of steel provided along Grid 3.
  • 
• The required area of steel for the required beam Grid A
  • 
• Detailing of steel provided along Grid 3.
  • 
• Provisions for shear reinforcement for detailing according to fig 5 of IS 13920 has to be followed.
  • 

2. Shear reinforcement calculations

Grid line A (B38)

• 
• 

From ETABS output, Design shear force, V = 243 KN

Provide 2-legged 8 mm dia stirrups with 100 mm spacing

Providing 25 mm clear cover, Asv = 2*3.14/4*8^2 = 100.53 mm^2

Effective depth = 500-50 = 450 mm

Vus = 0.87*fy*Asv*d/Sv = 0.87*415*100.53*450/100 = 163.34 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 = 415 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 = 500-50 = 450 mm

Vus = 0.87*fy*Asv*d/Sv = 0.87*415*100.53*450/100 = 163.34 KN

Since Vus < V ( Design shear force), hence safe.

So provide 2-legged 8 mm dia stirrups with 100 mm spacing.

3. 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 effective depth

4. Calculate the value of the maximum shear force

• 1) For sway to the right
• 
• As per IS 13920-2016 a new load combination is to be added.
  • Go to --> Define --> load combination --> Add new combo --> Rename the combination name as sway --> then define load for DL,LL,Brick load & floor finish with factor as 1.2
  • 
• Ok --> Analyse
  • Calculation

  • Sway to left:-

    V(u+a) = 70.58 KN, Mu (As) = 160.776 KN and Mu (Bh) = 233.59 KN

    Vu,a = 70.58 - 1.4*(160.774+233.54)/3.9 = -70.998 KN

    Vu,a = 70.58 + 1.4*(160.774+233.59)/3.9 = 212.14 KN

    Sway to right:-

    V(u+a) = 70.58n KN, Mu (Ah) = 154.79 KN, Mu (Bs) = 219.14 KN

    Vu,a = 70.58 - 1.4*(154.79+219.14)/3.9 = -54.06 KN

    Vu,b = 70.58 + 1.4*(154.79+219.14)/3.9 = 195.22 KN

5. Detailing of beams using AutoCAD

For Grid A

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For Grid 3

• 5. Detailing of beams using AutoCAD

  For Grid A

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  For Grid 3

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