Q. The thermo couple sandwich assembly is made of steel and brass plates as shown below. The dimensions of the plates are 30x4x1 (mm). So the total thickness of the assembly is 2 mm. BC’s are shown on the surfaces. All surfaces that do not have BC’s, are considered insulated.

During the initial solve, the temperature field will be determined through the bimetallic strip. In the second solution this temperature field will be used as input to calculate the stresses and displacement resulting from the differential expansion of the two metals. Run FEM to solve for:

1. Temperature distribution2. Deformation and Stresses3. You should prepare the models with both Tetra and Hexa elements. Make sure you show

the plots with both type of elements. Compare the results of both models. Try to minimize the number of elements and achieve reasonable results.

4. Plot the stress , thermal contours and the deformed shape. For the solid elements (Hexa and Tetra), look into following stresses : Normal X, Y & Z, Shear XY, XZ & YZ, and max principle. Which element has compressive stresses and which tensile stresses? For thermal temperature distributions are sufficient.

5. Study the behavior of the thermo-couple by varying the thermal properties as follows: Thermal Expansion A for steel & brass – vary by +/- 50% Convection H – 50, 150, 300 Since the total set is reasonably large, try to configure the set that will produce the maximum stresses and deformation of the thermo-couple. Compare the values with the base run. Which property has more effect on deformation of thermo-couple? A or H?

6. Can you support and verify your FEM analysis by your knowledge of heat transfer and MOM?

The Summary/Result section should also answer the following additional questions on a separate page in order:

1. Which element is deforming more? 2. Which element is going through in compression and which in extension?3. There is a bending/warping/curving phenomenon. Bending to which side? What caused

that? 4. Although the thermocouple with heat source is expanding, but the stresses are compressive

as well as tensile. What causes this?

Fixed EndBottom Surface (Steel)Free Convection Temp=20 C

Heat SourceFixed Temp = 395 C

Steel Plate

Brass Plate

Summary:For Base run model:Temperature distribution: in ᵒC

LocationHex Mesh Tetra Mesh

Steel Brass Steel BrassAt heat location 395 395 395 395At Fix Location 306.9 306.9 306.5 306.9

Parameter Mechanics Hexa Mesh Tetra MeshDeflection (mm) 0.871 0.537

Max. Normal Stress X (Mpa)

195.793 Tensile195.793 Compressive

202.2 Tensile133.6 Compressive

337.7 Tensile513.5 Compressive

Max. Normal Stress Y (Mpa)

NA47.61 Tensile

530.8 Compressive73.46 Tensile

1360 CompressiveMax. Normal Stress Z

(Mpa)NA

90.65 Tensile598.0 Compressive

300.4 Tensile1360 Compressive

Max. Shear Stress X-Y (Mpa)

NA 147.4 244.5

Max. Shear Stress X-Z (Mpa)

NA 296.8 322.6

Max. Shear Stress Y-Z (Mpa)

NA 14.03 81.84

Maximum Principle stress (Mpa)

NA 708.1 1579

For Property change Model:Temperature distribution: in ᵒC

LocationHex Mesh Tetra Mesh

Steel Brass Steel BrassAt heat location 395 395 395 395At Fix Location 247.5 248.1 247.5 248.1

Parameter Mechanics Hexa Mesh Tetra MeshDeflection (mm) NA 2.237 1.366Normal Stress X

(Mpa)NA

297.6 Tensile246.3 Compressive

1021 Tensile1162 compressive

Normal Stress Y (Mpa)

NA44.18 Tensile

507.5 Compressive222.2 Tensile

1291 CompressiveNormal Stress Z

(Mpa)NA

254.9 Tensile739.5 Compressive

918.9 Tensile1291 Compressive

Shear Stress X-Y (Mpa)

NA 170.4 254

Shear Stress X-Z NA 232.9 375.9

(Mpa)Shear Stress Y-Z

(Mpa)NA 39.11 131.3

Maximum Principle stress (Mpa)

NA 752.2 1487

Most dominating variableParameter A constant H variable H constant A variableDisplacement (mm) 0.737 2.65

Conclusion:1. It can be seen from the results that thermal stresses in both model are reasonable compared

to the MOM result for the normal stress in X direction. values are higher as in MOM calculations we are not considering effect of convection for the heat transfer.

2. Thermal distribution given by both the meshes is approximately equal.3. Tetra mesh gives abrupt changes in the results due to course mesh to reduced the errors in

results more fine mesh should be used.4. From the results the prediction can be done that free expansion of the brass strip is pulling

steel strip and exerting tensile force on the same, and expansion of the steel is less than the brass so it resists the expansion of brass and exerts compressive force on the brass. i.e. brass is undergoing compressive stress and steel is undergoing tensile stress.

5. Thermal expansion coefficient of brass is more so it deforms more so appears to be in tension in the cantilever thermostat strip.

6. When we use the combination of the A and H values it can be seen that we get the maximum stress value at A value of steel and Brass are 50% and 150% of the original and H is at maximum i.e. 300.

7. From mechanics we can say that deformation is dependent on thermal expansion coefficient so brass expands more than steel.

8. As the deformation is dependent on the value of A so it can be seen from the last result which we got by change in the value of the variables so we can sy that more dominant variable is A i.e. thermal expansion coefficient.

Answers:1. Quad elements are deforming more and predicting the displacements. Elements associated

with Brass are deforming more.2. In this situation there is mismatch between the coefficient of thermal expansion so brass

expand more as it has higher value of CTE and steels expands less, as these two are connected to each other brass pulls the steel inducing tensile force on the system so steel undergoes tension and steel tries to resist the expansion inducing compressive forces so brass it undergoes compressive force.

3. There is bending phenomenon for the cantilever strip condition, this is due to expansion of the brass and steel. As both expands at different rate so that steel resist the expansion of brass and layers near contact tend to expand less and comes near to the expansion of steel and due to pulling effect through brass contact layers of steel expands more than free end. due to these change in expansion the bending occurs.

4. The difference in thermal expansion causes the tensile and compressive forces in the system due to different material properties and different coefficient of thermal expansion.

Overall strategy for Mechanics:1. Problem contains the phenomenon of thermal expansion.2. As the component is bimetallic strip used in thermo stat, it uses different material with

different coefficient of thermal expansion.3. Due to this different coefficient of thermal expansion causes different expansions in both

metals and as they are fixed together they induce stress in each other, i.e. material with less coefficient of thermal expansion produces compression on the other material and which has the higher coefficient of thermal expansion induces tensile stress on the other.

4. Due to these different stresses induced in the strip it gets the deflection.5. We can get the thermal expansion by using following formula.

α X L X(T2-T1)6. We can calculate the stresses from following formula.

σ= P/A where P we can obtain by comparing expansions and resisting forces (lαt+Pl/AE)= (lαt-Pl/AE)

7. We assume that only conduction is taken into consideration for the both metals and convection at the steel face is neglected only for the calculation ease.

8. We only calculate the normal stress in X direction only & compare them.

Overall strategy for FEM:1. As the problem contains various cases and comparison of the results of different elements in

FEM, so we first generate the geometry of the object in Hyper mesh and mesh it with the different elements.

2. As per the problem the thermal analysis need to be carried out for the situation of cantilever so we create the constraints according to that.

3. We generate the thermal loading conditions and convection conditions for the problem.4. Then we analyze the first condition of the problem.5. For the next problems we change the properties for the both the models and try to compare

the results of the same.6. As there are many combinations are involved in the analysis we consider only the condition

which will give the maximum values of the stress which will be achieved by satisfying the following conditions.

Coefficient of thermal expansion for brass is increased i.e. 150% of the original. Coefficient of thermal expansion for steel is decreased i.e. 50% of the original. Convection at steel surface is maximum i.e. H=300

These condition will increase the expansion of brass and reduce thermal expansion of steel which will create the maximum forces in the strip and induce the maximum stress in them.For this case we compare the maximum principle stress and Normal X stress to study the effect for both the model.

7. In the another condition we keep the one value either A or H constant to study the effect of them, we calculate this effect for quad mesh only and look at Normal stress and can predict based on the observations.

8. Now we can compare the results and get the values required.

Material and Property assignment for Hexa Element model in SI systemFor steel

For brass

Load and boundary condition application

Convection surface & ambient temp node

Convection material Steel (1= steel)

Load Application Thermal

Load Application structure

Deformed Quad

Property assignment for Tetra meshFor steel

For brass

Convection surface and ambient temperature node

Deformed Tetra mesh

Thermal distribution:Hex Mesh Tetra mesh

Displacement:Hex Mesh Tetra Mesh

Maximum Principle Stress:Hex Mesh Tetra Mesh

Normal Stress X direction:Hex Mesh Tetra Mesh

Normal Stress Y direction:Hex mesh: Tetra Mesh

Normal Stress Z direction:

Hex Mesh: Tetra Mesh

Shear Stress X-Y:Hex Mesh Tetra Mesh

Shear Stress X-Z:Hex Mesh Tetra Mesh

Shear Stress Y-Z:Hex Mesh: Tetra Mesh

Property Variation i.e. expected maximum stress condition:Hex MeshSteel

Brass

Tetra Mesh:Steel

Brass

Thermal Distribution:Hex Mesh Tetra Mesh

Deflection:Hex Mesh Tetra Mesh

Maximum Principle StressHex Mesh Tetra Mesh

Normal Stress X DirectionHex Mesh Tetra Mesh

Shear Stress X-Y

Hex Mesh Tetra Mesh

Effect of property on deformation:Keeping A constant and varying H

Keeping H constant and varying A