module 2 thermal stresses

8/13/2019 Module 2 Thermal Stresses

1/11

UCONN ANSYS Module 2 Page 1

Module 2: Thermal Stresses in a 1D Beam Fixed at Both Ends

Table of Contents Page Number

Problem Description 2

Theory 2

Preprocessor 3Scalar Parameters 3

Real Constants and Material Properties 4

Geometry 6

Meshing 7

Loads 8

Solution 9

General Postprocessor 9

Results 11

Validation 11


2/11


Problem Description

Nomenclature:

L =250mm Length of beam

D =25mm Diameter of beam

T =175 C Uniform temperature of beam

= 25

C Room Temperature

E =205GPa Youngs Modulus of ANSI 1030 Steel at Room Temperature =0.30 Poissons Ratio of Steel= Thermal Expansion (Secant) Coefficient of SteelIn this module we will study the thermal stresses resulting from an elevated temperature on a

round beam fixed at both ends. We will model the beam using one dimensional BEAM 4

elements and ANSI 1030, a low carbon steel. The theory for this analysis is shown below:

Theory

Thermal Stress

When most materials are heated, they tend to expand. ANSI is isotropic thus the expansion is

uniform in all directions. The non-dimensionalized form of this expansion is called thermal

strainwhich is given in the form: (2.1)Where is the temperature where the reference length of the beam is considered and is amaterial property known as the Thermal Expansion Coefficient. This constant is called the

Secant Coefficient inMechanical ANSYS APDL.For our beam, when the material tries to expand

in the x direction, the fixed supports provide reaction forces to keep the beam at the initial

length. These reaction forces result in a net compressive stress on the beam. Using some

definitions of axial stress, we can say:

(2.2)Where E is youngs modulus.

y

T

L

D

x


3/11


Substituting equation 2.1 into equation 2.2, we can derive:

= 359.7 MPa (2.3)The yield strength of ANSI 1030 is 441 MPa, so the beam is stressed within the linear elastic

limits. If we want to design against buckling, we can check that the load applied from the fixedsupports doesnt exceed the critical load for buckling. We will simplify this analysis for the

purposes of this tutorial. For more in depth analysis of buckling, see module 3.

Buckling Considerations

First, we must check to see if the beam is a Euler or Johnson column. The criteria are as follows:

( ) (2.4)

(2.5)

{ (2.6)

Where c = 1 in a conservative evaluation and for a circular beam. Evaluating equation2.6 we find that

. Thus the beam is a Johnson Column. Now that we have classified thebeam, we must check to see that the critical buckling load (

is greater than the load applied

by the fixed ends. In a Johnson Column:

= 19.7kN (2.7)For axial stress:

(2.8)where F is the axial load and for a circular cross section. Thus, in order for no bucklingto occur,

(2.9)Evaluating the right hand side of equation 2.6, we get 17.7kN. Thus, no buckling occurs.


4/11


Preprocessor

Scalar Parameters

First, we will declare some variables in ANSYS that will be

used throughout the remainder of the tutorial.

1. Go to Utility Menu -> Parameters ->Scalar Parameters

2. Under SelectiontypePI=acos(-1). ANSYS has thecapability of solving trigonometric functions. After

the statement has been written, press ENTER.

3. Repeat step two for the following statements:D = 0.025

L = 0.25

The screen should look as shown.

4.

Click Close

These variables are stored and can be accessed at any time.

Real Constants and Material Properties

Element Selection

We will be using BEAM 4 in this tutorial. For more information on BEAM 4, see module 1.1.

1. Go to Main Menu -> Preprocessor -> Element Type -> Add/Edit/Delete2. Click Add3. Select Library of Element Types -> Structural Mass -> Beam -> 3D Elastic 34. Click OK5. Click Close

2

4


5/11


Real Constants

Now we will enter the cross sectional properties of the beam.

1. Go to Real Constants -> Add/Edit/Delete2.

Click Add3. Click OK

4. Under AREAenterPI*D*D/45. Under IZZtypePI*D*D*D*D/646. Click OK7. ClickClose

Material Properties

1. Go to Main Menu -> Material Props -> Material Models2. Go to Structural -> Linear -> Elastic -> Isotropic3. UnderEX enter 205E94. Under PRXYenter 0.35. Click OK

6. Go to Structural -> Thermal Expansion ->Secant Coefficient -> Isotropic

7.

Under ALPXenter 11.7E-68. Click OK9. Go to Define Material Model Behavior -> Material -> Exit

4

5

6

3

4

5

7

8


6/11


Geometry

Keypoints

1. Go to Main Menu -> Preprocessor -> Modeling ->Create -> Keypoints -> on Working Plane2. Enter 0,0,0

3. Click Apply4. Repeat steps 3 and 4 for L,0,05. Click OK6. To get rid of the triadgo to the Command Prompt and enter:

/triad,off

/replot

The resulting graphic should look as follows:

Line

1. Go to Main Menu -> Preprocessor -> Modeling -> Create -> Lines -> Straight Line 2. Enter 1,2this connects a line from keypoint 1 to keypoint 23. Click OK

The resulting graphic should look as follows:

2

5


7/11


Meshing

As we will see later in the results section, linear thermal stresses problems are very accurate in

ANSYS. To get the point across, we will mesh the beam with two elements.

1.

Go to Main Menu -> Preprocessor -> Meshing -> MeshTool2. Go to Mesh Tool -> Size Controls: -> Global -> Set3. Under NDIVenter 24. Click OK5. Click Mesh6. Click Pick All7. Click Close

8. Go to Utility Menu -> PlotCtrls -> Numbering 9. Check NODE Node numbers10.Click OK11.Go to Utility Menu -> Plot -> Nodes

Your mesh should look as follows:

2

3

4

5

6


8/11


Loads

Now we will constrain the ends of the beam and select a uniform temperature across the beam.

Fixed Ends

1. Go to Main Menu -> Preprocessor -> Loads -> Define Loads -> Apply ->Structural -> Displacement -> On Nodes

2. Enter 1,2and click OK3. Under Lab2 DOFs to be constrainedclick ALL DOF4. Under Valueenter 0and click OK

The resulting picture should look as shown below:

Uniform Temperature

1. The default reference temperature inMechanical ANSYS APDL is 0. If we are workingwith metric units, the reference temperature is in . If we are in British units, thereference temperature is in . To change the reference temperature to room temperature,go to the Command Promptand enterMP,REFT,1,25This sets the reference

temperature (REFT) material property (MP) on object 1 (the beam) to You areinstructing ANSYS that, at this reference temperature, the object experiences no thermal

strain.2. Go to Main Menu -> Preprocessor -> Loads -> Define Loads -> Apply -> Structural

-> Temperature -> On Lines

3. Click Pick All4. Under VAL1enter 1755. Click OK6. If an error message appears,

ignore it. ANSYS considers

temperature to be a degree of

freedom at each node. Thus,

temperature definitions will betranslated to the nodes. This is a

more important consideration

when a non-uniform temperature

distribution is defined, as an

interpolation algorithm would be

applied across the nodes.

4

5


9/11


Solution

1. Go to Main Menu -> Solution2. In the Command Prompttypesolveand press ENTER in your keyboard.3. Ignore the warning4. Almost instantly, the problem will be solved. Click Closein the Notemenu.

General Postprocessor

As in module 1, APDL has trouble graphing contour plots of stress, so we will access the

postprocessor results in the list files.

To check for buckling, we will first look at the displacement log file.

1. Go to Main Menu -> Postprocessorfailure to do so will not allow access to the log files2. Go to Utility Menu -> List -> Results -> Nodal Solution -> DOF Solution ->

Displacement vector sum3. Click OK

As you can see from the chart, there are no displacements in the beam. Thus, no buckling has

occurred as expected.

Now we will check the forces applied at each node to get the stress distribution in the beam.Using equation 2.8, we can find the axial stress across the beam.

4. Go to Utility Menu -> List -> Results -> Element Solution ->All Available Force Items

5. Click OK


10/11


The log file should appear as follows:

As we can see, the reaction force is uniform across the beam as expected. Using Equation 2.8,

the axial stress at each location is 359.7 MPa.


11/11


Results

Axial Stress Error

The percent error (%E) in our model max deflection can be defined as:

= 0% (2.10)Due to quadrature, beam element functions arefourth order accurate. Since thermal stress

(equation 2.3) is afirst order function, the stresses derived will be 100% accurate every time.

Validation

module 2 thermal stresses

Documents