STRUCTURAL MECHANIC

Modeling review

Jean-Cyril HERNANDEZ Mathilde MOREL Jennifer PINGEON Florent SAIGHI

26/07/2013

Summary

I. Introduction.....................................................................2

II. Simulation........................................................................2

1. Purpose of the model.......................................................2

2. Engineering Model...........................................................2

3. Software...........................................................................2

4. Analysis Model..................................................................3

5. Results Verification...........................................................4

6. Conclusion........................................................................8

III. Check size of two members of the structure............8

IM comments in red

1

I. Introduction

Refer to the requirements in the course document

The following review deals with a portal frame which has a height of 4 meters. A load of 2 700 N/m is applied on the top of the structure.

The aim is to check if the structural model is correct by using software named Lusas, and by making verifications.

II. Simulation

1. Purpose of the model

• Predict axial forces and moments due to all realistic “in-service “loading, to allow sizing of members • Estimate vertical and lateral deflections.

2. Engineering Model

See Figure 1. Key issues; loading, connections, foundation fixity, any longitudinal and end effects. We only use live loading as per code.

Figure 1 Engineering ModelThe two posts are IPE 200The roofing is made with IPE 240

Details of supports, connections?Material?, member sections used

2

3. Software

LUSAS package used. No verification needed.

4. Analysis Model

Portrayal - See Figure 2 Plane frame model based on internal frame. Dimensions of the model?Elements - BEAM elements including shear deformation. Section properties - from tables Supports - pinned Which nodes?Loading - Checking loadcase. 2700 N/m applied vertically (Global distributed) where?Units N,m. E = 210,0E9 N/m

Validation - see Table 1 and figure 2.

Figure 2: Beam’ numbers

All elements are treated as 2-D beam elements including shear deformation.

n° type span(cm) depth(cm) span/depth Validation if span/depth >5

Beam 1 IPE 200 400 20 20,0 Ok

Beam 2 IPE 240 500 24 20,8 Ok

Beam 3 IPE 240 500 24 20,8 Ok

Beam 4 IPE 200 400 20 20,0 Ok

Beam 5 IPE 240 300 24 12,5 Ok

Beam 6 IPE 240 400 24 16,7 Ok

Beam 7 IPE 240 400 24 16,7 Ok

Table 1 Validation span/depth

3

4

Model Development and Acceptance Criteria Assurance

Second order effects are neglected [Frame to be designed to BS5950]

LSR

Linear elastic material behaviour [Stress not greater than yield in any loadcase. Design to BS 5950]

LSR

Foundation connections modelled as pins [Standard practice. Conservative for estimating deformations and moments. Recommended in BS5950.]

S

All connections other than at foundations assumed to be fully rigid [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]

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 modeling process.

Specify a modern code of practiceWill the conncetions be bolted. Welding more common in this type of structureUse of beam elements?

5. Results Verification

Data errors

The software haven’t data errors when we run it - checked S

5

Figure 2: Nodes’ numbers

6

Initial check against checking load case

Node DX (m) DY (m) THZ RSLT

1 -5,29E-05 -9,02E-05 -3,59E-04 1,05E-04

2 8,12E-16 -2,95E-04 7,05E-19 2,95E-04

6 0 0 1,99E-04 0

10 5,29E-05 -9,02E-05 3,59E-04 1,05E-04

14 0 0 -1,99E-04 0

21 8,12E-16 -3,00E-04 1,04E-18 3,00E-04

Table 2: Deformations for checking loadcase

• Deformations at restrained freedoms are zero. ∆x=∆y = 0.0 at nodes 6 and 14. See Table 2 • Symmetry is perfect on the x axis if we look nodes 1 and 10. They have the same deformation on y axis, and opposite deformation on x axis.

Node Fx (N) Fy (N) MZ (N.m) RSLT

6 20,801 1,35E+04 0 1,35E+04

14 -20,801 1,35E+04 0 1,35E+04

Table 3: Support reactions

• Sum of vertical reactions = 13500 + 13500 = 27000 Applied vertical load = 2700 x (lengths’ roof) = 2700 x 10 = 27000

Element Node Fx (N) Fy (N) Mz (N.m)

17 1 1,09E+04 740,809 -2,22E+03

5 1 -1,35E+04 20,801 -83,204

1 1 -1,62E+04 -3,52E+03 2,30E+03

Table 4: Reactions at node 1

7

Check local equilibrium

Check for moment at node 1 = M1+M5+ M17 = 2,30E+03 – 83,204 – 2,22E+3 = 0The sum is equal to 0, check S

Check the form of the results

Figure 3 shows the deflected shape. Note main deflection under load and spreading at eaves - S

Figure 3 Deflected Shape due to vertical load

Internal Force Actions

Figure 4 Diagram of bending moments

8

Note that the bending moments at the apex are positive around 5E-3 N.m. Concerning the eaves the bending moment is around -3E-3. The magnitude both is the same. So the bending moments at the apex and at the eaves are about equal (but of opposite sign). At the end of the roof slope, the bending moment becomes positive around 2,3E-3 due to the clamping. The central moment is reduced to 1/8 (not clear) thanks to this link, so with an articulation the bending will be higher.comment on low moment in column

Making a model comparison with a lateral load

Figure 5 Deflected shape due to a lateral load and bending moments

Figure 6 Deflected shape due to a lateral load and bending moments with a simple frame

Deflection under load (m)

Bending moment at the centre of the left beam (kN.m)

Main model 0,910 50

Checking model 0,927 50

9

Table 4 Checking model comparison

This looks like a good checking model but you have not explained it clearly.

6. Conclusion

All outcomes are satisfactory at this stage.

III. Check size of two members of the structure

fy= 235 γMO= 1

A(mm²) W(mm3) Nc,rd (N) Mpl (N.m)

IPE 200 2850 221000 669750 51935

IPE 240 3910 367000 918850 86245

Figure 2: Beam’ numbers

Check section

n° Section Ned (N) Nc,rd (N) Ned/Nc,rd Ned/Nc,rd < 1 Med (N.m) Mrd (N.m) Med/Mc,rd Med/Mc,rd < 1

Beam 1 IPE 200 -13500 669750 0,020 Ok 0 51935 0,00 Ok

Beam 2 IPE 240 -16230 918850 0,018 Ok 4996 86245 0,06 Ok

Beam 5 IPE 240 1480 918850 0,002 Ok 0 86245 0,00 Ok

Beam 6 IPE 240 10860 918850 0,012 Ok 2298 86245 0,03 Ok

10

We could see that Ned/Nc,rd and Med/Mc,rd are really low. It’s mean that we use IPE oversized. If we want to optimize our project we have to use little profiles.

Good report

11