Home / Examples / Coupled Analysis / Thermal-Stress Analysis [Watt/Galileo] / Example 3: Deformation due to the Temperature Gradient #3 - Multiple Materials

Example 3: Deformation due to the Temperature Gradient #3 - Multiple Materials


General

 

Analysis Space

Item

Settings

Analysis Space

3D

Model Unit

mm

 

Analysis Conditions

Select Thermal analysis and Stress analysis.

Item

Settings

Solver

Thermal Analysis [Watt]
Stress Analysis [Galileo]

Thermal-Analysis Type

Steady-State Analysis

Options

N/A *

* [Thermal Load] is selected by default for the thermal-stress coupled analysis.

 

The Step/Thermal Load tab is set as follows.

Tab

Setting Item

Settings

Step/Thermal Load *

Reference Temperature

25 [deg]

* The reached temperatures come from the thermal analysis.

Model

The same as Example 10 of thermal analysis.

Body Attributes and Materials

Body Number/Type

Body Attribute Name

Material Name

0/Solid

SUB

006_Glass_epoxy *

1/Solid

GND

008_Cu *

2/Solid

MAINCHIP

001_Alumina *

3/Solid

SUBCHIP

001_Alumina *

4/Solid

HOLE

008_Cu *

5/Solid

HOLE

008_Cu *

6/Solid

HOLE

008_Cu *

7/Solid

HOLE

008_Cu *

8/Solid

HOLE

008_Cu *

9/Solid

HOLE

008_Cu *

10/Solid

HOLE

008_Cu *

11/Solid

HOLE

008_Cu *

* Available from the material DB

 

The heat sources of MAINCHIP and SUBCHIP are set up as follows.

Body Attribute Name

Tab

Settings

MAINCHIP

Heat Source

0.2 [W]

SUBCHIP

Heat Source

0.1 [W]

Boundary Conditions

The heat transfer coefficient for the natural convection is set on the bottom face of GND as follows:

The radiation from the top and the sides is omitted.

Boundary Condition Name/Topology

Tab

Boundary Condition Type

Settings

Bottom/Face

Thermal

Heat Transfer: Convection

Natural convection *: 1.39 [W/m2/deg5/4]

Room temperature: 25[deg]

 

* The coefficient for the natural convection is calculated as follows.

 

2.51×C×(1/L)^(1/4) = 1.39 [W/m2/deg5/4]

where

C = 0.26

Size of the substrate : 0.06x0.04

L (Typical Length) = (0.06 x 0.04 x 2) / (0.06 + 0.04) = 0.048

Results

The temperature distribution as a result of Watt is shown below.

 

The next figure shows the vectors of displacement as a result of Galileo following Watt.

The substrate corners which shows the low temperature are bent downward.