ETABS Modelling

ETABS MODELLING

AUTHOR: VALENTINOS NEOPHYTOU BEng (Hons), MSc

March 2013

ETABS MODELING ACCORDING TO EUROCODES

Valentinos Neophytou BEng (Hons), MSc Page: 2 ETABS MANUAL

Step by step procedure and methodology of how you developing a modelusing ETABS

Step 1: Specify Material Properties for Concrete

1. Poisson ratio is equal to v = 0 (cracked concrete) and v = 0.2 (un-cracked concrete) as (EN1992-1-1,cl.3.1.3)

Table 1: Concrete properties (EN 1992, Table 3.1)

Property Data for concrete

C16/20 (N/mm2)

C20/25 (N/mm2)

C25/30 (N/mm2)

C30/37 (N/mm2)

Mass per unit Volume 2,5E-09 2,5E-09 2,5E-09 2,5E-09

Weight per unit volume 2,5E-05 2,5E-05 2,5E-05 2,5E-05

Modulus of Elasticity 29000 30000 31000 33000

Poisson’s Ratio (cracked concrete) 0 0 0 0

Coeff. of thermal expansion 10E-06 10E-06 10E-06 10E-06

Charact. ConcCyl. Strength, fck 16 20 25 30

Bending Reinf. Yield stress, fyk 500 500 500 500

Shear Reinf. Yield stress, fyk 500 500 500 500

Figure 1: Concrete properties



Step 2: Add frame section for columns

Figure 2: Section properties of concrete columns



Step 3: Add frame section for beams

Figure 3: Effective width of beams (EN1992-1-1,cl.5.3.2.1)

Interior beam

Internal beam supporting an internal and an external slab

Exterior beam supporting cantilever

External beam no cantilever

For practice use beff 1,2 = 0.2lo



Figure 4: Section properties of concrete beams

Notes:

1. Property modification factors are used to reduce moment and torsion stiffness due to crack section. Torsional stiffness of the cracked section should be set equal to 10% of the torsional stiffness of the un-cracked section.

2. Unless a more accurate analysis of the cracked elements is performed, the elastic flexural and shear stiffness properties of concrete and masonry elements may be taken to be equal to one-half of the corresponding stiffness of the un-cracked elements (EN1998-1-1,cl. 4.3.1(7)).

3. These modification factor only affect the analysis properties, they do not affect the design properties.

Column (Line element)

Beam (Line element)

Slab (Shell element) Wall (Shell element)

I22=I33=0.5 I22=I33=0.5 m11=m12=m22=0.5 m11= m12=m22=0.5 It=0.1 It=0.1 It=0.1 It=0.1



Step 4: Add Slabs & Walls

Figure 5: Section properties of concrete slab

Figure 6: Section properties of concrete wall



Step 5: Define Response Spectrum function according to EC8

1. Peak ground acceleration agR=0,25g, 2. Type C or D for building within category of importance I and II, 3. Define two response spectrum cases if the factor q is different in each direction, 4. Modify the existing values of elastic response spectrum case in order to change it into

the design response spectrum.

Figure 7: Response Spectrum to EC8



Figure 8: Design spectrum for elastic analysis data

PERIOD ACCELERATION g = 9.81 m/sec2 T Sd(T) β = 0.2 -‐

0.0000 0.0767 Soil Type = C -‐ 0.0667 0.1150 q = 1.50 -‐

0.1333 0.1533 αgR = 0.10 -‐ 0.2000 0.1917 S = 1.15 -‐ 0.6000 0.1917 TB = 0.20 sec

0.8333 0.1380 TC = 0.60 sec

1.0667 0.1078 TD = 2.00 sec

1.3000 0.0885 T = 0.50 sec 1.5333 0.0750

1.7667 0.0651

Data for soil type -‐ Type Spectrum 1 2.0000 0.0575

index Soil Type S TB TC TD

3.3333 0.0200

1 A 1 0.15 0.4 2 4.6667 0.0200

2 B 1.2 0.15 0.5 2

6.0000 0.0200 3 C 1.15 0.2 0.6 2 7.3333 0.0200 4 D 1.35 0.2 0.8 2 8.6667 0.0200

5 E 1.4 0.15 0.5 2

10.0000 0.0200



Step 6: Define Load Case

Figure 8: Dead/Live/Wind

Step 5: Define Equivalent Static Analysis

Equivalent static analysis can be used if the following case can be met:

1. Ground acceleration: Check seismic zonation map from National Annex

2. Spectrum type 1: 5.5Hz<M (High seismicity areas)

3. Ground type: Normally type B or C can be used (see EN 1998,table 3.1)

4. Lower bound factor for the horizontal design spectrum: 0.2 (EN 1998-1-1,cl.3.2.2.5(4)P)

5. Behavior factor q: See table

6. Correction factor λ (EN1998-1-1,cl.4.3.3.2.2(1Ρ)) λ=0.85 if T1≤2TC and more than 2 storey λ=1.0 in all other case



7. Regular in elevation

8. Regular in elevation and irregular in plan

9. Fundamental period: T1≤4T_c

T1≤2,0s

Table 1: Equivalent Static Force Case

Load case name Direction and Eccentricity % Eccentricity EQXA X Dir + Eccen. Y 0.05 EQYA X Dir – Eccen. Y 0.05 EQXB Y Dir + Eccen. X 0.05 EQYB Y Dir – Eccen. X 0.05



Step 6: Define Load Combination for Equivalent lateral force analysis

Ultimate limit state (ULS)

Static case

COMBO 1. 1.35DL + 1.5LL COMBO 2. 1.35DL + 1.5WINDX + 1.5 (0.7LL + 0.5 SNOW) COMBO 3. 1.35DL + 1.5WINDY + 1.5 (0.7LL + 0.5 SNOW) COMBO 4. 1.35DL + 1.5LL + 1.5 (0.7WINDX + 0.5 SNOW) COMBO 5. 1.35DL + 1.5LL + 1.5 (0.7WINDY + 0.5 SNOW) COMBO 6. 1.35DL + 1.5LL + 1.5 (0.7SNOW + 0.5WINDX) COMBO 7. 1.35DL + 1.5LL + 1.5 (0.7SNOW + 0.5WINDY) COMBO 8. 1.35DL + 1.5SNOW + 1.5 (0.7LL+ 0.5WINDX) COMBO 9. 1.35DL + 1.5SNOW + 1.5 (0.7LL+ 0.5WINDY) COMBO 10. 1.35DL + 1.5SNOW + 1.5 (0.7WINDX + 0.5LL) COMBO 11. 1.35DL + 1.5SNOW + 1.5 (0.7WINDY + 0.5LL) COMBO 12. 1.35DL + 1.5WINDX + 0.7*1.5(LL+SNOW) COMBO 13. 1.35DL + 1.5WINDY + 0.7*1.5(LL+SNOW) COMBO 14. 1.35DL + 1.5(LL+SNOW) + 0.7*1.5WINDX COMBO 15. 1.35DL + 1.5(LL+SNOW) + 0.7*1.5WINDY

Seismic case

COMBO 16. DL + 0.3LL + EQXA + 0.3EQYA COMBO 17. DL + 0.3LL + EQXA – 0.3EQYA COMBO 18. DL + 0.3LL - EQXA + 0.3EQYA COMBO 19. DL + 0.3LL - EQXA – 0.3EQYA COMBO 20. DL + 0.3LL + EQYA + 0.3EQXA COMBO 21. DL + 0.3LL + EQYA – 0.3EQXA COMBO 22. DL + 0.3LL - EQYA + 0.3EQXA COMBO 23. DL + 0.3LL - EQYA – 0.3EQXA

COMBO 24. DL + 0.3LL + EQXB + 0.3EQYB COMBO 25. DL + 0.3LL + EQXB – 0.3EQYB COMBO 26. DL + 0.3LL - EQXB + 0.3EQYB COMBO 27. DL + 0.3LL - EQXB – 0.3EQYB COMBO 28. DL + 0.3LL + EQYB + 0.3EQXB COMBO 29. DL + 0.3LL + EQYB – 0.3EQXB COMBO 30. DL + 0.3LL - EQYB + 0.3EQXB COMBO 31. DL + 0.3LL - EQYB – 0.3EQXB

Serviceability limit state (SLS)

COMBO 32. DL + LL



Step 7: Define Response Spectrum case

Modal Response spectrum

1. Independently in X and Y direction, 2. Define design spectrum, 3. Use CQC rule for the combination of different modes (EN1998-1-1,cl.4.3.3.3.2(3)) 4. Use SRS rule for combined the results of modal analysis for both horizontal directions

(EN1998-1-1,cl.4.3.3.5.1(21)). 5. Accidental eccentricity of each storey cause of uncertainties locatin of masses have

been taken into account 5% (EN1998-1-1,cl.4.3.2). 6. Modal Combination: “Complete Quadratic Combination” (CQC) can be used if the Tj ≤ 0,9 Ti (EN1998-1-1,cl.4.3.3.3.2(3)P).

Figure 9: Response Spectrum case Data for EQY& EQX



Step 8: Define Load Combination for modal analysis

Ultimate limit state (ULS)

Static case

COMBO 1. 1.35DL + 1.5LL COMBO 2. 1.35DL + 1.5WINDX + 1.5 (0.7LL + 0.5 SNOW) COMBO 3. 1.35DL + 1.5WINDY + 1.5 (0.7LL + 0.5 SNOW) COMBO 4. 1.35DL + 1.5LL + 1.5 (0.7WINDX + 0.5 SNOW) COMBO 5. 1.35DL + 1.5LL + 1.5 (0.7WINDY + 0.5 SNOW) COMBO 6. 1.35DL + 1.5LL + 1.5 (0.7SNOW + 0.5WINDX) COMBO 7. 1.35DL + 1.5LL + 1.5 (0.7SNOW + 0.5WINDY) COMBO 8. 1.35DL + 1.5SNOW + 1.5 (0.7LL+ 0.5WINDX) COMBO 9. 1.35DL + 1.5SNOW + 1.5 (0.7LL+ 0.5WINDY) COMBO 10. 1.35DL + 1.5SNOW + 1.5 (0.7WINDX + 0.5LL) COMBO 11. 1.35DL + 1.5SNOW + 1.5 (0.7WINDY + 0.5LL) COMBO 12. 1.35DL + 1.5WINDX + 0.7*1.5(LL+SNOW) COMBO 13. 1.35DL + 1.5WINDY + 0.7*1.5(LL+SNOW) COMBO 14. 1.35DL + 1.5(LL+SNOW) + 0.7*1.5WINDX COMBO 15. 1.35DL + 1.5(LL+SNOW) + 0.7*1.5WINDY

Seismic case

COMBO 16. DL + 0.3LL + EQX + 0.3EQY COMBO 17. DL + 0.3LL + EQX – 0.3EQY COMBO 18. DL + 0.3LL - EQX + 0.3EQY COMBO 19. DL + 0.3LL - EQX – 0.3EQY COMBO 20. DL + 0.3LL + EQY + 0.3EQX COMBO 21. DL + 0.3LL + EQY – 0.3EQX COMBO 22. DL + 0.3LL - EQY + 0.3EQX COMBO 23. DL + 0.3LL - EQY – 0.3EQX

Serviceability limit state (SLS)

COMBO 24. DL + LL



G+0.3Q+Ex+0.3Ey

G+0.3Q+Ex-0.3Ey

G+0.3Q-Ex+0.3Ey

G+0.3Q-Ex-0.3Ey

G+0.3Q+Ey+0.3Ex

G+0.3Q+Ey-0.3Ex



G+0.3Q-Ey+0.3Ex G+0.3Q-Ey-0.3Ex

1.35G+1.5Q





Step 9: Meshing of slab

Assign -> Shell Area -> Area Object Mesh Option

Automatic meshing option for slab element only

Notes:

1. The property assignments to meshed area objectets are the same as the original area object.

2. Load and mass assignments on the original area object are appropriately broken up onto the meshed area objects.



Step 10: Meshing/Label of wall

Edit>Mesh shells and click on the

Mesh/Quads/Triangles at Intersections with visible grid lines:

Assign->Shell/Area->Pier Label or Spandrel Label.



Step 11: Define Auto-Line Constraint

Select area element (slab)->Assign->Shell Are-> Auto-Line Constraint

Step 12: Define mass source

Combination of the seismic action with other actions (EN 1998-1-1,cl.3.2.4):

1. Define the category of building (EN 1991,Table 6.1), 2. Define the reduce factor (EN 199, Table A.1.1).

Table 2: Combination of seismic mass

𝑮𝒌,𝒋 + 𝝍𝑬𝒊𝑸𝒌,𝒊 (ΕΝ1998-1-1,Eq. 3.17)

Combination coefficient for variable action is: 𝜓!" = 𝜙 ∙ 𝜓!! (ΕΝ1998-1-1,Eq. 4.2)

Values of φ for calculating 𝝍𝑬𝒊 (CYS NA EN1998-1-1:2004)

Type of Variable

action

Storey φ

Categories A-C1

Roof Storeys with correlated occupancies Independently occupied storeys

1,0 0,8 0,5

Categories A-F1

1.0



Table 3: Values of ψ coefficients

Category Specific Use ψο ψ1 ψ2

A Domestic and residential 0.7 0.5 0.3 B Office 0.7 0.5 0.3 C Areas for Congregation 0.7 0.7 0.6 D Shopping 0.7 0.7 0.6 E Storage 1.0 0.9 0.8 F Traffic < 30 kN vehicle 0.7 0.7 0.6 G Traffic < 160 kN vehicle 0.7 0.5 0.3 H Roofs 0.7 0 0 Snow, altitude < 1000 m 0.5 0.2 0 Wind 0.5 0.2 0

Figure 10: Adding seismic mass to ETABS



Step 13: Define number of modes

Notes:

1. Minimum number of modes to be taken into account (EN1998-1-1,cl.4.3.3.3.1(5)): k ≥ 3.√n

k is the number of modes taken into account.

n is the number of storeys above the foundation or the top of a rigid basement.



Step 14: Define restrains at the base

Select the entire base joints

Step 15: Define diaphragms to slab



Step 16: Checking the model



MODAL ANALYSIS RESULTS

Step 1: Calculate the effective modal mass

Display> Show Tables > Modal information > Building modal information > Table modal participation mass ratios

1. The sum of the effective modal masses for the modes taken into account amounts to at least 90% of the total mass of the structure (EN 1998-1-1,cl.4.3.3.3.1(3)).

2. All modes with effective modal masses greater than 5% of the total mass are taken into account.

Mode 1 (Translation Y - direction)

Mode 2 (Translation X - direction)



Mode 3 (Torsional)

Step 2: Damage limitations



The damage limitation requirements should be verified in terms of the interstorey drift (dr) (EN 1998-1-1,cl.4.4.3.2) using the equation below:

𝑑! ∙ 𝑣 ≤ 𝑎 ∙ ℎ =>𝑑!ℎ ≤

𝑎𝑣

dr: is the difference of the average lateral displacement ds in CM at the top and bottom of storey.

v: is the reduction factor which takes into account the lower return period of the seismic action.

h: is the storey height

Table 4: Damage limitation (EN1998-1-1,cl.4.4.3)

For non-structural elements of brittle material attached to the structure drv≤0.005h

For building having ductile non structural elements drv≤0.0075h

For building having non-structural elements fixed in a way so as not to interfere with structural deformation

drv≤0.010h

Tab;e 5: Reduction factor of limitation to interstorey drift (CYA NA EN1998-1-1,cl.NA.2.15)

Importance class Reduction factor v

I 0.5 II 0.5 III 0.4 IV 0.4

1. Export results from ETABS to ECXEL



2. Sort the Larger value on top

3. Record the value of each storey in the spread sheet below:



Step 3: Second order effects

1. The criterion for taking into account the second order effect is based on the interstorey drift sensitivity coefficient θ, which is define with equation (EN 1998-1-1,cl.4.4.2.2(2)).

𝜃 =𝑃!"! ∙ 𝑑!𝑉!"! ∙ ℎ

hr: is the interstorey drift,

h: is the storey height,

Vtot: is the total seismic storey shear

Ptot: is the total gravity load at and above storey considered in the seismic design situation (G+0.3Q).

Table 6: Consequences of value of P-Δ coefficient θ on the analysis

θ≤0,1 No need to consider P-Δ effects

0,1≤θ≤0,2 P-Δ effects may be taken into account approximately by amplifying the effects of the seismic actions by !

!!!

0,2≤θ≤0,3 P-Δ effects must be accounted for by an analysis including second order effects explicity

θ≥0,3 Not permitted

1. Explore the results from ETABS to EXCEL

Damage limitation (EN1998-1-1,cl.4.4.3)

X-‐direction dr*v<0,005-‐0,01

Y-‐direction dr*v<0,005-‐0,01

OK OKOK OKStorey 1 0,0017 0,0017 3,00 0,50 0,00028 0,00028

Displacement Drift X dr (m)

Displacement Drift Y dr (m)

Heigh of each storey, h

(m)

Reduction factor

v

v*dr X - direction

v*dr/h Y - direction

Storey 2 0,0026 0,0026 3,00 0,50 0,00043 0,00043



2. Select the combo G+0,3Q and record the highest value from each storey

3. Record the heist value for Vtot



4. Record all values on the spread sheet as showing below

Step 4: Structural regularity plan

Second order effects (EN1998-1-1,cl.4.4.2.2)

θ X-‐direction

θ≤0.1

θ Y-‐direction

θ≤0.1OK OKOK OK

Storey 2 709 3,00 220,00 220,00 0,00260 0,00260Storey 1 1426 3,00 334,00 334,00 0,00170 0,00170

Ptot (kN)

Heigh of each storey,

h (m)

Vtot X-direction

(kN)

Vtot Y-direction

(kN)

Displacement Drift X

dr (m)

Displacement Drift Y dr (m)



1. Slenderness ratio of the building λ=Lmax/Lmin<4 2. A “compact shape”: one in which the perimeter lines is always convex, or at least

encloses not more than 5% re-entrant area. 3. The floor diaphragms shall be sufficient stiff in-plane not to affect the distribution of

lateral loads between vertical elements.

Table 7: Criteria for regularity in plan

Lateral torsional rensponse condition: rx> 3.33eox rx> 3.33eoy

Torsionally rigidity condition: rx> Is

Apply forces as follow:

Regularity in plan (cl. 4.2.3.2)Check 1 - slenderness ratio cl.4.2.3.2(5)Slenderness ratio λ=Lmax/Lmin<4 = 2,80

Regularity in plan (cl. 4.2.3.2)Check 2 - structural eccentricity & torsional radius cl.4.2.3.2(6)

Length in longitudinal direction = 56 mLength in trasverse direction = 20 mStifness in X direction Sx=1000/dx

Stifness in Y direction Sy=1000/dy

Torsional stifness Ts=1000/Rz

Torsional radius ry=Ts/Sx

Torsional radius rx=Ts/Sy

Radius of gyration Is=((Lmax²+Lmin²)12)^0,5

Structural eccentricity in x direction eox=Rz(Fx)/Rz(Mz)

Structural eccentricity in y direction eox=Rz(Fy)/Rz(Mz)

Table 1: Criteria for regularity in plan - Torsionally rigity condition

rx (m)

ry (m)

29,5 30,027,1 24,7

Table 2: Criteria for regularity in plan - Lateral torsional respone condition

Storey 2Storey 1

Storey 2Storey 1

Rotation Rz for

Fx=1000kN

Rotation Rz for

Fy=1000kN8,18E-06 8,18E-068,18E-06 8,18E-06

Rotation Rz for

Mx=1000kNm

Eccentricity eox

8,18E-06 1,008,18E-06 1

Eccentricity eoy

166667136054200000

1,22E+081,22E+08

0.3rx (m)

0.3ry (m)

Is (m)

8,9 9,0 17,2 8,1 7,4 17,2

Is<rx Is<ry

OKOK OK

5 6Storey 2Storey 1

Displacement X (mm)

dx

Displacement Y (mm)

dy

7,35 7,14

Rotation Z (radians)

Rz

OK

3,33eox<rx 3,33eoy<ry

1,00E+001,00 OK

OKOKOK

OK

8,18E-068,18E-06

Stifness X (kN/m)

Sx

Stifness Y (kN/m)

Sy

Torsional Stifness

(kNm/radian) Ts

140056



Storeys Load Case Forces STOREY 1

FX1 FX1=1000kN FY1 FΥ1=1000kN MZ1 MZ1=1000kNm

STOREY 2

FX2 FX2=1000kN FY2 FΥ2=1000kN MZ2 MZ2=1000kNm

Repeat this process for all load case in order to obtain the displacement values.



Step 5: Structural type of the building

Table 8: Classification of structural system

Wall system Vertical and lateral load: Wall resist Vb,wall>65%Vbtotal Frame system Vertical and lateral load: Vb,frame>65%Vbtotal Frame-equivalent dual system Vertical and lateral load: Vb,frame>50%Vbtotal Wall-equivalent dual system Vertical and lateral load: Vb,wall>50%Vbtotal

Display >Show Tables> Support/Sprint/Reaction

1. Explore the results from ETABS to EXCEL



From load case tick the worst-case seismic design combination:

COMBO 1. DL + 0.3LL + EQX + 0.3EQY COMBO 2. DL + 0.3LL + EQX – 0.3EQY COMBO 3. DL + 0.3LL - EQX + 0.3EQY COMBO 4. DL + 0.3LL - EQX – 0.3EQY COMBO 5. DL + 0.3LL + EQY + 0.3EQX COMBO 6. DL + 0.3LL + EQY – 0.3EQX COMBO 7. DL + 0.3LL - EQY + 0.3EQX COMBO 8. DL + 0.3LL - EQY – 0.3EQX

2. Select the worst-case design combo

3. Select the nodes for frames only

4. Calculate the sum of the base shear that can be resist by column in X and Y direction



i.e VTOTAL = 1000KN

VFRAMES, X ,Y = 500KN

VTOTAL / VFRAME 500/1000*100= 50%

Therefore the structural system of building is: Wall-equivalent dual system

How to checking base shear

Base shear can be check as follow:

Table 9: Checking the base shear

Direction Lower bound values Upper bound values X direction Fb = %Effective mass(X dir.)*Mass *Sdx

Fb = ∑mass * Sdx

Y direction Fb = %Effective mass(Y dir.)*Mass *Sdv

Fb = ∑mass * Sdy

Note: The base shear should be within those limits

NOTE: REPEAT ALL THIS PROCESS FROM BEGIN WITH THE NEW Q VALUE

Revised the design spectrum input data with the new q (for example if q=1.5 adopt at initial stage and the new q=3 then you have to repeat the process with the new q)



OUTPUT DATA

Step 1: Print data for steel/concrete design

File > Print Tables > Concrete Frame Design



ADDITIONAL NOTES

SHRINKAGE AREAS

Select Area > Edit > Expand/Srink Area



PIN JOINT

Export model to SAFE

File menu > Export > Save Story as SAFE.f2k Text File

Local Axis

Local axis 1 X - direction Local axis 2 Y- direction



Local axis 3 Z - direction Local axis 2 (My) Y- direction Local axis 3 (Mx) X - direction

Education

ETABS Modelling