Structural Modelling using Etabs 2018

CHALLENGE - 10

ANSWERS

1. AIM & INTRODUCTION:

• To Model the G+7 Residential building using ETAB analysis.
• The Analysis is based on IS1893 & IS 13920

Given data are:

Pinned support at base, G+7 ie Base +9 story (Ground to mumty)

Column – 400x600mm, 400x900mm

Beam – 400x500mm, 400x600mm

Shear wall – 300mm thick.

Live load – 2.50 KN/M², Floor finish – 1.50 KN/M², Wall load – 5 KN/M,

Water tank – 20KL over mumty ie.. 200 KN over load.

Conceptual design consideration before analysis are:

Re-entrant and cut out check & modification (Compromise)

Vertical irregularity (Soft story)

Planer irregularity (Looks symmetrical)

2% shear wall area with respect to building area (8 m² min)

Stiffness modifier for Column (0.70) and beam (0.35).

The ground floor is Car parking and from Story 1 to Story 7 are same.

MODELLING PROCEDURE ARE :

• Modelling by grid method.
• Define
• Materials (Concrete-M30,Rebar-Fe415,500) and sections (Beam, Column, Shear wall & Slab).
• Add/draw column, beam, shear wall and slab in Model.
• Assign load in beams, slabs.
• Add load patterns.
• Add load weightage in mass source.
• Do load combinations.
• Assign base (foundation) as pin joint.
• Check model for frame joints & connections.
• Assign diaphragm.
• Run Analysis.

The modelled drawing is as follows:

Key points while analyzing the structure are:

1. No slab and wall load at Ground floor.
2. End releases are assigned to secondary beams.
3. No inner wall load at terrace.
4. Compared the static and dynamic forces and revises the factor.

CHECKS AFTER ANALYSIS ARE:

• Modal mass participation for 3 modes dominates at least 65%

• Achieved

• Time period for Mode 1 and mode 2 are differ by 10%

ie 0.998 x 0.90 = 0.898 sec

but it is 0.846 sec which is less than 10%

Hence satisfied.

• Principal modes greater than tortional modes.
• Inter story drift ration is less than 0.004
• 

  Hence satisfied.

  Inter story drift ration is less than 0.004
  • Axial stress in column and shear wall :

  ie 0.40fck x100 = 0.4*30x100 = 1200 KN/M²

  Axial stress is less than 1200 KN/M² in model.

  Achieved

  • Base shear share b/w column and shear wall

  As the shear wall Area is 2x 1.65 = 3.30 m

  2x 1.60 = 3.20 m

  2x 2.40 = 4.80 m

  2 x 1.25 = 2.50 m

  2 x 1.50 = 3.00 m

  2 x 2.55 = 5.10 m

  2 x 2.20 = 4.40 m

  1 x 3.10 = 3.10 m

  Total = 29.40 x 0.30 = 8.82 m²

  Column area = 16 x 0.4x0.90 = 5.76 M²

  2x x0.40x0.60 = 0.48 M²

  Total = 6.24 m²

  Not dual system