Structure Analysis Report with LUSAS · Web viewPredict axial forces and moments due to all realistic “in-service “ loading, to allow sizing of members Estimate vertical and lateral

Structure Analysis Report with LUSAS software

Validation of portal frame design

Rémy LASJAUNIASThomas DI FRENZA

Julien SAUTRE

Table of contents

1. Purpose..........................................................................................................................................3

2. Engineering model..........................................................................................................................3

3. Software.........................................................................................................................................4

4. Analysis Model...............................................................................................................................4

a) Validation...................................................................................................................................4

b) Checking model..........................................................................................................................5

5. Result verifications.........................................................................................................................6

6. Structure verification......................................................................................................................8

Element 1 : COLUMN..........................................................................................................................8

Element 2 : RAFTER............................................................................................................................9

7. Conclusion....................................................................................................................................12

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REMY LASJAUNIAS CMC 5JULIEN SAUTRETHOMAS DI FRENZA

Modelling review for a portal frame

1. Introduction

Refer to the requirements for the project in the project document. Predict axial forces and moments due to all realistic “in-service “ loading, to allow sizing of

members Estimate vertical and lateral deflections.Engineering Model (Sketch) Check if the structure is correctly dimensioned

2. Engineering model

See Figure 1.

Restrict loading to dead load, live loading as per code and wind in plane of portal frames.

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Figure 1 : Portal Frame

Figure 2 : Detail

Show foundation details. This drawing shows a symbol for a pin support. Specify section types, material

3. Software

LUSAS package used. No verification needed.

4. Analysis Model

Plane frame model based on internal frame. Refer to diagramElements : BEAM elements including shear deformation.Section properties : from tables Supports : pinnedLoading : Checking distributed load 15 kN.m applied vertically downwards on rafter elements Units N, m.E = 210 kN/mm2

a) Validation

Table 1: Validation

Model Development and Acceptance Criteria Assurance

All elements are treated as 2-D beam elements includingshear deformation. [span/depth>5] S

Second order effects are neglected [Frame to be designed toEN 1903] LSR

Linear elastic material behaviour [Stress not greater thanyield in any loadcase. Design to EN 1903] LSR

Foundation connections modelled as pins [Standard practice. Conservative for estimating deformations and moments. Recommended in EN 1903.] S

All connections other than at foundations assumed to be fullyrigid [Standard practice. In reality they will be bolted, causing local rotation but this will be counteracted by the neglect of the finite size of the haunches at the joints] Your engineeerng model does not show that they are bolted!

S

Key:

S - Satisfactory,CL - Check Later. This means that the issue needs to be assessed at a later stage in the modelling process.LSR - Later Stage Requirement. This means that action is needed beyond the modelling process.

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b) Checking modelLUSAS model , nodes :

Verification :

Check if the reaction on the node 1 and 5 is okay with an 15 kN horizontal load on the X axis applied on node 2 and 4.

Computation of the force applied to the X and Y axis to compare with LUSAS value :

ΣFx = X5 + X1+ 30 = 0 so, X5 + X1 = - 30 X5 = X1= - 15 kNΣFy = Y5+Y1= 0 So, Y5 = - Y1

ΣM/5 = -15*7 + -15*7 + 12*Y5= 0 So, Y5 = 17.5 kN Hence : X5 = X1 = -15 kN Y5= -Y1= 17.5 kN

Vérification OK

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Figure 3 : Nodes

Figure 4 : Reactions

15 kN 15 kN

12 m

1.45 m

7 m

5. Result verifications

Sum of vertical forces: Applied loads : 15 kN/mFrom model: 15.0000000000 kN/m OK

Supports and symmetry

Table 2 : Deformations for checking loadcase

Data errors - checked S DX and Dy at node 1 and node 5 are all restrained. Pin supports demonstrated. OK Rotations and displacement are equal and opposite to 15 significant digits - symmetry check

OK Check overall equilibrium . Sum of vertical reactions = 92.7699E3 + 92.7699E3 = 185.54E3 =

applied uniform load – S (185.54E3 / (6.18466*2) = 15 kN/m )

Table 3 : Support Reactions (N)

Check local equilibrium . Check for moment at node 2 = M12 + M23 = -163.6E3 + 163.6E3 = 0.0 – S

Check the form of the results.

Deformations. Figure 5 shows the deflected shape. Note main deflection under load and spreading at eaves - S

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Figure 5 : Deflected shape

Internal Force Actions

Figure 6 : Diagram of bending moments

Does this diagram show results that help to verify the model?

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6. Structure verification

Element 1 : COLUMN

EULER BUCKLING of the Column (HEA) :

HEA 340

Ncr,euleur = π²EI / kL² E = 210 000 Mpa

I = 7436 cm4

L = 7 mk = 2

Ncr,euleur = 786325 N

N = 92770 N

λ = N / Ncr

λ = 0.12 > 0.1 NOT OK

HEA

The difference isn’t that much significant according to the k value = 2. In realty this value will be lower than 2. So the check will be OK

COMPRESSION of the Column (HEA) :

HEA 340

NED < NC,Rd

NC,Rd = Afy / γM0 A = 133.5 cm²

fy = 235 MPa

γM0 = 1

NC,Rd = 3137250 N

NEd = 23370 N

NEd = 2.34E+04 < NC,Rd = 3.14E+06 OK

HEA

Check OK

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BENDING of the Column (HEA) :

Check OK

BEAM deflection (HEA) :

To check the displacement of the structure’s column, need to compare it with 2 different cases as followed :

- Column with fixed support on one side and free on the other side.- Column with fixed support on one side and a pin on the other side.

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HEA 340

Med < Mpl,Rd

Mpl,Rd = Wpl x fy Wpl = 755.9 cm3

fy = 235 MPa

Mpl,Rd = 177637 N.m

Med = 163300 N.m

Med = 163300 < Mpl,Rd = 177637 OK

Wpl > 695 cm3

HEA

15 kN 15 kN

The minor axis was Iyy. To check and compare with the model, the axis has been changed to Izz.

Column with fixed support on one side and free on the other side :

With this formula, we’ve calculated the displacement: WL^3/3EI W= 15000 NL = 7 mE = 210E9 N/m²I = 0.2769E-3 m4

WL^3/3EI = (15000N*7^3/(3*210E9*0.2769E-3))=0.029 m

RESULT FROM LUSAS : 0.030 m

Check OK

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Column with fixed support on one side and pin on the other side :

With this formula, we’ve calculated the displacement: WL^3/12EI W= 15000 NL = 7 mE = 210E9 N/m²I = 0.2769E-3 m4

WL^3/12EI = (15000N*7^3/(12*210E9*0.2769E-3))= 7.337E-3 m

RESULT FROM LUSAS : 7.835 E-3 m

Check OK

Displacement of the model :

RESULT FROM LUSAS : 0.026 m

As expected, the displacement of the model lies between the two other cases.

7.337E-3 m < 0.026 m < 0.030 m

Check OK

checking model for internal forces?

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Element 2 : RAFTER

TENSION of the RAFTER (IPE 240)

Check OK

BENDING of the RAFTER (IPE 240)

IPE 240

Med < Mpl,Rd

Mpl,Rd = Wpl x fy Wpl,y= 366.6 cm3

fy = 235 MPa

Mpl,Rd = 86151 N.m

Med = 163300 N.m

Med = 163300 > Mpl,Rd = 86151 NOT OK

Wpl > 695 cm3

IPE

Check Not OK : Need to increase the Wpl,y value so, need to choose a bigger IPE.

So, in the next part, we compute this model with an IPE 330.

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IPE 240

NED < Nt,Rd

Nt,Rd = Afy / γM0 A = 39.1 cm²

fy = 235 MPa

γM0 = 1

Nt,Rd = 918850 N

NEd = 44000 N

NEd = 4.40E+04 < NC,Rd = 9.19E+05 OK

IPE

Checking value for IPE 330 :

Tension IPE 330 :

Check OK with IPE 330

Bending IPE 330 :

Check OK with IPE 330

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IPE 330

NED < Nt,Rd

Nt,Rd = Afy / γM0 A = 62.6 cm²

fy = 235 MPa

γM0 = 1

Nt,Rd = 1471100 N

NEd = 44000 N

NEd = 4.40E+04 < NC,Rd = 1.47E+06 OK

IPE

IPE 330

Med < Mpl,Rd

Mpl,Rd = Wpl x fy Wpl,y= 804.3 cm3

fy = 235 MPa

Mpl,Rd = 189011 N.m

Med = 163300 N.m

Med = 163300 < Mpl,Rd = 189011 OK

Wpl > 695 cm3

IPE

7. Conclusion

The choice of the HEA 340 is OK. Need to change the IPE 240 with an IPE 330.

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