Tube Flaring

Chapter 15: Tube Flaring

15

Tube Flaring



Summary



Introduction



Requested Solutions



FEM Solutions



Input File(s)

240 241

241 245

241

240 MD Demonstration Problems CHAPTER 15

Summary Title

Chapter 15: Tube Flaring

Features

Deformable-deformable contact Large elastic-plastic deformation

Geometry

Axisymmetric • • • • • •

x=r

Tube diameter = 8 inches Tube thickness = 0.3 inches Tube length = 8 inches Tool apex angle = 30° Tool wall thickness = 0.6 inches Tool length = sufficient to mode the process

y

CL

x=r

y

CL

Material properties

Tube: Young’s modulus = 3.0e7 psi, initial yield stress = 3.6e4 psi, yield stress at 0.1 equivalent plastic strain = 1.8e5 psi, Poisson’s ratio = 0.3 Tool: Young’s modulus = 4.0e7 psi, Poisson’s ratio = 0.3, no yielding

Analysis type

Quasi-static analysis

Boundary conditions

The left end of the tube is prevented from moving in the axial direction but is free to move in the radial direction.

Applied loads

An edge load is applied to the right end of the tool (the end with a larger diameter) to push the tool into the steel tube, then released

Element type

4-node axisymmetric elements

Contact properties

Friction between the tool and the tube is ignored in the analysis

FE results

1. Plot of tube tip versus time. 2. Contours of von Mises stress at maximum load on deformed mesh 3. Contours of plastic strain on deformed mesh after load removal 0.5

Radial Displacement Point A (in)

0.4 0.3

A

0.2 x=r

0.1

y

CL

Time (s) 0.0 0.0

0.5

1.0

1.5

2.0

CHAPTER 15 241 Tube Flaring

Introduction A cone-shaped flaring tool is pushed into a cylindrical tube to permanently increase the diameter of the tube end. The goal of the analysis is to determine whether the final shape of the tube, after the entire process, meets the objective. The nonlinear nature of the problem, along with the irreversible characteristics, makes it impossible to know in advance the load required to drive the tool into the tube. As a result, multiple runs through the analysis cycle may be necessary to achieve the final objective of the analysis. This problem demonstrates the use of MD Nastran SOL 400 to analyze a contact problem involving deformabledeformable contact and large elastic-plastic deformations.

Requested Solutions The requested solutions include the curve of the tube diameter at the right end as a function of loads and the deformed shape of the tube and the tool along with the distributions of von Mises stresses and plastic strains.

FEM Solutions A numerical solution has been obtained with MD Nastran’s SOL 400 for the element mesh (shown in Figure 15-1) using axisymmetric elements.

x=r y

Figure 15-1

Finite Element Mesh

There are two contact bodies. One is the tube and one is the tool. The two contact bodies with ID 3 and 4 are identified as selected elements of the tube and the tools respectively as: BCBODY BSURF

3 3 115

2D 109 116

DEFORM 110 117

3

4 4 32

2D 25 33

DEFORM 26 34

4 27 35

0 111 118

112 119

113 120

114

...

and BCBODY BSURF ...

0 28 36

29 37

30 38

31 39

242 MD Demonstration Problems CHAPTER 15

Furthermore, the BCTABLE entries shown below identify that these bodies can touch each other. BCTABLE

BCTABLE

0 SLAVE

3 0 fbsh MASTERS 3 SLAVE 4 0 fbsh MASTERS 3 1 SLAVE 3 0 fbsh MASTERS 3 SLAVE 4 0 fbsh MASTERS 3

0.05 0

2 100. 0 0.8

0.05 0

100. 0 0.8

0.05 0

2 100. 0 0.8

0.05 0

100. 0 0.8

0.

0.

0

0.

0.

0.

0

0.

0.

0.

0

0.

0.

0.

0

0.

Axisymmetric elements are defined with CQUADX along with PLPLANE and PSHLN2 entries: PLPLANE PSHLN2 + + + + + $ Pset: CQUADX CQUADX

1 1 1 1 C3 AXSOLID C4 AXSOLID C5 IPS C6 AXSOLID C8 AXSOLID "pshell.1" will 109 1 110 1

1 L L L Q Q be imported as: "plplane.1" 10 144 145 1 144 146 147 145

+ + + + +

…

The Young’s modulus and Poisson’s ratios for the tube and the tool are defined as: MAT1* * MAT1* *

1 1. 2 1.

3.+7

1.15385+7

.3

4.+7

1.53846+7

.3

The yield stresses along with the hardening are defined respectively as: MATEP 1 Table 36000. TABLES1,1,2,,,,,,,+, +,0.,36000.,0.1,180000.,ENDT,

1

Isotrop Addmean

The NLPARM entry is used to define the nonlinear analysis iteration strategy. There are two load steps: loading and unloading. One hundred (100) uniform time increments are used to solve each load steps. The stiffness matrix will be updated at each iteration (full Newton-Raphson iteration strategy). NLPARM

1

100

PFNT

1

25

U

YES

NLPARM

2

100

PFNT

1

25

U

YES

CHAPTER 15 243 Tube Flaring

The tube diameter at the right end of the tube gradually increases during the analysis as the load increases and reaches the maximum of 0.4316 inches. The final tube radial displacement after unloading is settled at 0.4093 inches. See Figure 15-2 for the curve of tube diameter as a function of time (load). The entire analysis procedure can be repeated with various load levels to achieve the desired final tube diameter. The curve is not smooth at the loading path because of the discrete finite elements. It can be improved by refining the finite element meshes.

0.5

Radial Displacement Point A (in)

0.4 0.3

A

0.2 x=r

0.1

y

CL

Time (s) 0.0 0.0 Figure 15-2

0.5

1.0

1.5

2.0

Curve of Tube Diameter as a Function of Time

The deformed mesh and the distribution of von Mises stress at the time the applied load reaches maximum are shown in Figure 15-3. It can be observed that the stresses are concentrated in two areas: the tip of deflection where the tube made contact with the tool and in the area where the tube is deformed.

244 MD Demonstration Problems CHAPTER 15

x=r

y

Figure 15-3

Deformed Mesh and Distribution of von Mises Stress at Maximum Load

The deformed shape of the tube and the tool along with the distribution of plastic strains at the end of the analysis are shown in Figure 15-4.

x=r

y

Figure 15-4

Deformed Mesh and Distribution of Equivalent Plastic Strains at the End of Analysis

CHAPTER 15 245 Tube Flaring

Input File(s) File nug_15.dat

Description Tube flaring input file.