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.

Solution:

Procedure:

Steps:

• Download the file and open the ETABS software.
• Check the model and load combinations
• Now run the analysis of the model

• Once the analysis is done
• Start the concrete design checks
• Now got to paln view
• In the same tab click the display design info
• Now select the longitudinal reinforcement

• Design the beam according to the ast required

Grid A

• Beam:38
  • Main Reinforcement at top and bottom - 3nos 20mm dia
  • Additional Reinforcemen of 3 nos 20mm dia at the top in ends
• Beam :34
  • Main Reinforcement at top and bottom - 3nos 20mm dia
  • Additional Reinforcemen of 3 nos 25mm dia at the top and bottom in ends
• Beam :33
  • Main Reinforcement at top and bottom - 3nos 20mm dia
  • Additional Reinforcemen of 3 nos 20mm dia at the top in ends

Grid 3

• Beam:24
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 20mm dia at both ends
    • Bottom: 3 nos 12mm dia at both ends
• Beam:25
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 20mm dia at both ends
    • Bottom: 3 nos 12mm dia at both ends
• Beam:69
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 25mm dia at both ends
    • Mid Span: 3nos 20mm dia
    • Bottom: 3 nos 12mm dia at both ends
• Beam:70
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 25mm dia at both ends
    • Mid Span: 3nos 20mm dia
    • Bottom: 3 nos 12mm dia at both ends
• Beam:73
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 20mm dia at both ends
    • Bottom: 3 nos 12mm dia at both ends
• Beam:14
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 20mm dia at both ends
    • Bottom: 3 nos 12mm dia at both ends
• Beam:16
  • Main Reinforcement at top and bottom - 3nos 16mm dia
  • Additional Reinforcement
    • Top: 3 nos 20mm dia at both ends
    • Bottom: 3 nos 16mm dia at both ends

• Provide shear reinforcement in grid 3 as:- At ends 2L-8ɸ@100mm c/c and for mid span 2L-8ɸ@150mm c/c

GRID A

GRID 3

Reason for failure of the middle span along grid A is if the shear force exceeds the maxximum shear bearing capacity means if the shear force exceeds Гc max.

Calculate the value of maximum shear force

• As per IS 13920-2016, new load combinations has been added for sway combination, the partial factor for dead and live load is considered to be 1.2

Sway to left:-

V(u+a) = 22.633KN, Mu (As) = 326.77 KN and Mu (Bh) = 329.85 KN

Vu,a = 22.633 - 1.4*(326.77+329.85)/3 = -283.787 KN

Vu,a = 22.633 + 1.4*(326.77+329.85)/3 = 329.053 KN

Sway to right:-

V(u+a) = 21.02 KN, Mu (Ah) = 329.85 KN, Mu (Bs) = 326.77 KN

Vu,a = 21.02 - 1.4*(329.85+326.77)/3 = -285.4 KN

Vu,b = 21.02 + 1.4*(329.85+326.77)/3 = 327.44 KN

Result:

• The beams are designed successfully in detail from the Etabs software.
• Provided the details of longitudinal reinforcement and shear reinforcement for the two beams.
• Also calculated the value of maximum shear force in the continuous beams using the Etabs software.