Tmr4190 Finite Element Methods In Structural Analysis

TMR4190 Finite Element Methods in Structural Analysis ABAQUS EXERCISE By: RENY WATAN & XIAO CHEN

Autumn 2009

The units which have been used to build the ABAQUS model for this project is N – cm. Basically, when using the ABAQUS to do the analysis of a 2D frame, the following tasks will be performed: 1. Sketching the two-dimensional (2D) geometry and creating a part representing the frame. 2. Defining the material properties and section properties of the frame We assigned the material under the name “Steel” with the elastic modulus of 209×103 N/mm2 = 209×105 N/cm2 and Poisson’s ratio of 0.30. Totally, there are seven (7) types of section properties which have been used for this jacket frame, as listed below. Type Pipe Pipe Pipe Pipe Pipe Pipe Pipe

Diameter Thickness (mm) (mm) 700.00 15.00 950.00 25.00 1100.00 50.00 1330.00 25.00 2100.00 50.00 3100.00 50.00 6100.00 50.00

3. Assembling the model. 4. Configuring the analysis procedure and output requests. 5. Applying the loads and boundary conditions to the frame.

Load Name H-409 H-365 H-364 H-382 H-413 H-355 H-42

Type (Horizontal /Vertical) Horizontal Horizontal Horizontal Horizontal Horizontal Horizontal Horizontal

Total V-2500 Total

Vertical

Magnitude in (t)

in (N)

409.00 365.00 364.00 382.00 413.00 355.00 42.00

4012290.00 3580650.00 3570840.00 3747420.00 4051530.00 3482550.00 412020.00

2330.00

22857300.00

2500.00

24525000.00

2500.00

24525000.00

6. Meshing the frame. The approximate global element size of 10 has been defined for this model.

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

Name of Section D700-015 D950-025 D1100-050 D1330-025 D2100-050 D3100-050 D6100-050

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7. Creating a job and submitting it for analysis. 8. Viewing the results of the analysis. For this exercise, we define the cases into four (4) cases. Case 1 is the analysis with a complete structure, as assigned in the Exercise Problem. Case 2 and Case 3 is the analysis with one of the braces removed. Case 4 is the analysis with two braces removed. a. Finding the highest loaded element (element with the highest stress), calculating the scale factor of the load to the yield stress, and finding the typical ratio between stresses from axial forces and moments.

Element with the highest stress 10205

highest stress 1.05E+04

yield stress (σy ) 35000

scale factor 3.32

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

Case 1 (Complete 2D Structure)

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Ratio of Axial and Moment Stress Stress at the Top Fibre of the section

Stress at the Bottom Fibre of the section

Axial Stress

Moment Stress

Ratio (axial/moment)

1.025E+04

1.052E+04

10385.00

135.00

76.9259

b. Checking the global force equilibrium The Equilibrium state that the total forces of the structure should be equal to zero, as written: 1. ∑ = 0 2. ∑ = 0 3. ∑ = 0 Applying equilibrium #1, we write: =0 −

+ 2 × (409 + 365 + 364 + 382 + 413 + 355 + 42) = 0 = 2 × (409 + 365 + 364 + 382 + 413 + 355 + 42)

= 45.7146 × 10 Applying equilibrium #2, we write: =0 + +

− 2 × (2500) = 0

= 5000 = 49.050 × 10

……….( )

Applying equilibrium #2, we write: =0 −(

× 75) + 2 × (409 × 28 + 365 × 56 + 364 × 84 + 382 × 112 + 413 × 140 + 355 × 168 + 42 × 184.5) + (2500 × 24.25) + (2500 × 50.75) . = 0 × 75 = 648422 . = 8645.6266 = 84813597.6

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

= 4660

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Substitute

= 84813597.6 N to equation (a), will result = −3645.62667 = −35.7636 × 10

From the ABAQUS, the reaction forces at the constraints (boundary conditions) are Part Instance PART-1-1 PART-1-1

Part Instance PART-1-1 PART-1-1

Node ID

Elements

RF, RF1

16 24

Elem-1 Elem-1

-4.57146E+07 0

Node ID

Elements

RF, RF2

24 16

Elem-1 Elem-1

8.48136E+07 -3.57636E+07

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

c. Control axial load in the 5th floor of the structure.

5

tan

=

.

tan

.

=

. .

To control the axial load in the 5th floor of the structure, we could substitute the SF1 into the equilibrium equation as expressed below: =0 1 × sin

+ 1 × sin + 1 × sin + = (2 × 2500) × 9.81 × 1000

1 × sin

4.89862 × 10 ≈ 4.9050 × 10 =0 1 × cos

− 1 × cos + 1 × cos + × 9.81 × 1000 = 0 1 × cos

1 × cos

+ 2 × (413 + 355 + 42)

+ 1 × cos − 1 × cos − 1 × cos = 2 × (413 + 355 + 42) × 9.81 × 1000 1.5103 × 10 ≈ 1.5892 × 10

We can conclude that the result from Abaqus is nearly coinciding with the hand calculation result.

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

−

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d. The other cases

Element with the highest stress 988

highest stress 1.648E+04

yield stress (σy ) 35000

scale factor 2.124

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

Case 2

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Element with the highest stress

highest stress

1103

2.514E+04

yield stress (σy ) 35000

scale factor 1.392

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

Case 3

8

Element with the highest stress 5356

highest stress 3.126E+04

yield stress (σy ) 35000

scale factor 1.120

The strength reduction is % of Case reduction case 2 36.10% case 3 58.11% case 4 66.31% The relation between the stress from the axial forces and moments becomes Case

Stress at the Top Fibre of the section

Stress at the Bottom Fibre of the section

Axial Stress

Moment Stress

Ratio (axial/moment)

Case 2 Case 3 Case 4

1.642E+04 2.613E+04 3.159E+04

1.646E+04 2.179E+04 2.605E+04

16440.00 23960.00 28820.00

20.00 -2170.00 -2770.00

822.0000 -11.0415 -10.4043

From the table above it can be shown that the axial stresses dominate the stress of the structure.

TMR4190 Finite Element Methods in Structural Analysis – ABAQUS Exercise

Case 4

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