ME 623A

Finite Element Methods

Course Project

Finite Element Analysis of Bimetallic Strip

Project Guide

Prof. Dr N.N Kishore

Group 13

11250Dhrupal R Shah

11552Rachit Tripathi

11606Riya Khurana

A Finite Element Method Analysis for Bimetallic strip

Problem: A bimetallic strip of aluminum and brass is heated and the steady state temperature is non-uniform and is given by100e200xy . Brass strip thickness is given (= 1cm) and Aluminum strip thickness is to be find so that strip just touches a limit switch at a distance of 2mm. Also the maximum von-mises stress is to be determined.

Governing Equation:

The given problem is similar to the general solid mechanics stress strain problem with given boundary conditions. The bimetallic strip has a strain induced in it due to thermal expansion. The governing differential equations for both the case are similar

xxyy

xyyxyx

X 0

Y 0

X = Body force in element in X direction.

Y = Body force in element in Y direction.

x = Stress vector of the element in x plane in x direction

y = Stress vector of the element in y plane in y direction

xy

= Stress vector of the element in x plane in y direction

The strain induced in the body due to thermal effect is defined as follows

(m

o

xx

(

)

th

m o

yy

0

xy

Where

m

(i

j3

k )

o Constant Room Temperature

xx , yy ,xyrespectively.

are strains along x in x plane, along y in y plane, along y in x plane

Virtual Work Formulation

The given problem can be solved using virtual work formulation where we will equate internal work with the external work. The following equation gives the integral from of the virtual work approach.

uT PdS

T dV

eS e

eV e

Where

P= External Force applied

u and are virtual displacement and virtual strain

The stress in material having thermal strains is defined as

e{} [D]([B]{u

} {th })

where

D E

1* 1

0 0 E and are Youngs Modulus and Poissons ratio

(1 2 )

1

00

2

bi

B 0ci

0bj0bkci0c j0bic jbjck

0

ck bk

bi ( y j yk ) / 2 * Ae

ci (xk x j ) / 2 * Ae

bj ( yk yi ) / 2 * Ae

c j (xi xk ) / 2 * Ae

bk ( yi y j ) / 2 * Ae

ck (x j xi ) / 2 * Ae

Where

xi , yi , x j , y j , xk , yki

are the coordinates of nodes of triangular element

The weak form of the equation can be deuced to give the following simple result

{ue }T [B]T [D]([B]{ue } {e S

{ue }T [B]T [D][B]{ue }dS

th })dS 0

{ue }T [B]T [D]{

}dS

{ue }T *

A1 { f e }dS

e Se S

thee S

{ue }T [K e ]{ue } {ue }T { f e }ee[K e ] [B]T [D][B]* Ae

[K e ]{ue } { f e }

{ f e } [B]T [D]{

th}* Ae

With this simplified condition we can proceed further to find the elemental k and then formulate the global matrix, the after applying boundary conditions we can calculate the displacement of the nodes and finally we can compute the von mises stress.

Boundary Conditions

The given problem has only displacement boundary condition. The traction force is zero on all points except the ones having displacement boundary condition. The following conditions represent the boundary condition.

u 0

v 0

on y-axis at x=0

x nx xy ny 0 y ny xy nx 0

y = 5, y = 0, and x = 40 lines

Defining the mesh

We have taken the mesh2D code by Darren Engwirda, available in matlab website. The code takes the points of polygon to form a triangular mesh within it and returns the matrix of nodes which gives X and Y coordinates of the nodes and a triangle matrix which gives the number of nodes which together form a triangle.

The Problem we faced in defining mesh was to have a line after the end of first strip defining the nodes. We first defined a mesh with the dimensions of the brass strip and extracted the top nodes of the mesh. We stored it variable topline(). Then we defined another mesh of dimensions of aluminum strip with the nodes of topline() as the point of the polygon.

Then using the user defined function mergeNode.m we can merge two meshes and form a global mesh with triG and NodeG as node matrix and triangle matrix.

Defining various variable used in program

The element of the mesh is accessed by defining a structure sG. The structure has various variable like [K],[B], {sigma} ,{phim} etc.

The variable nodeG represent coordinates of various nodes in the mesh. The variable triG represent nodes of various triangles in the mesh.The KG defines the global K matrix and fG is global thermal force matrix.

Formulating the Global Matrix

The global matrix are defined by removing the rows and columns of the nodes with known displacements. We can calculate the {f} matrix for other nodes from the strain of nodes. The nodal displacement matrix can be found out by taking the inverse of modified K matrix and multiplying it with the modified {f} matrix. Thus we can find out the displacements of each node.

{ug }[K g ]{ug } {u g }T { f g } [K g ]{u g } { f g }

With modified K and f we can find the global U matrix.

{ug } [k g ]1{ f g }

The inverse of the {Kg} is taken out by using pseudo inverse function. This function takes out inverse even in case of singular matrix.

Solving for trivial problem

For verification of code, we have taken a trivial problem of a Brass strip heated uniformly to 100C with no displacement boundary condition allowing free expansion. The results of displacements along X and along Y were studied and they were calculated manually. The following result were obtained.

Manual Calculation:

Length along X= 0.4 mLength along Y= 0.06 m

brass 20 * 10^(-6) K-1

Tm = 1000C

T0 = 200C

T 800C

Elongation in X= 0.4*20*10(-6)*80 = 7.104*10(-4) m= 0.7104 mm Elongation in Y= 0.06*20*10(-6)*80 = 1.065*10(-4) m= 0.1065 mm Results From Code:The temperature distribution is given below.

The X and Y displacements of the various positions of the node are plotted below.

The given graphs suggest that the strip is expands in both the direction having positive in one and negative in the other.

The above diagram shows the Von-mises stress at each element. As material is elongated freely the stress should be near to zero. The maximum stress found out in this case is 4 *10-4 Pa which is almost negligible as compared to usual values of stress.

The above plot shows the strip being elongated in x direction with no bending present.

Problem of Bending strip :

A Bimetallic Strip of aluminum and brass is uniformly heated to 100C with no displacement boundary condition allowing free expansion with the strip bending on its own. The strip will bend towards the material having lower . Following plot shows the constant temperature.

The X elongation according to code is 0.6399mm and along Y direction is 0.09525 mm.

The following graph shows the X and Y displacement plot the bimetallic strip in uniform temperature lying on its own

The given Plot shows that the strip bends towards the brass side.

The Maximum stresses are observed at the junction of the two materials

Solving Problem for given conditions

The dimensions given in the questions were not suitable and were not giving required result (2mm deflection). So we changed the dimensions so as to get the required result. The given problem is solved with the given dimension of strip as 40cm * 1cm of brass200xy

The Von Mises Stress distribution is given by following plot.

strip. The temperature distribution were chosen as 100e

where x and y are

chosen in m. The results were iterated for the thickness of aluminum strip and the thickness is found out to be 3.5cm.

Temperature distribution plot

The X and Y displacements of the various positions of the node are plotted below with the deflection scaled up by the factor of 10. This helps us in analyzing the given solution.

X- Displacement distribution

Y- Displacement distribution

Maximum X deflection= 0.5mmMaximum Y deflection= 2.0mm

Deflected Strip (scale 1:1)

Iterating aluminum thickness

We iterated the code by changing the aluminum thickness every time and the following table was constructed.

Iteration NumberAluminum Thickness(cm)Y-Deflection of tip(mm)

1.12.528423435996

2.22.309994462993

3.41.916808539135

4.32.081664067357

5.3.51.985361719762

So finally we took the value of thickness to be 3.5 cm

222Calculating the Von- Mises Stress

'

(x

y

3*xy

x y

With the given displacements, we can found out the stress on the element with the given formula. The von mises stress was calculated and 2d plot was created.

The above stress distribution is for the case when both X,Y displacements at the fixed edge of the bimetallic strip are constrained. The above distribution is for very fine mesh (h.data variable in code is set to be 0.003).

We have considered another case where only X displacements at the fixed edge is constrained. The stress distribution is shown below.

The given plot suggests that the stress in aluminum is comparatively small as compared to that of brass. The stress is maximum at points near support. The stress at the lower boundary of the strip is significantly larger then that at inside.

Conclusions

The thickness of the aluminum strip is found out to be 3.5 cm. The maximum von mises stress is found out to be 391MPa With the stress distribution shown the maximum stress occurs at the point of support with relatively very small stress at other places. The Bimetallic strip deflects in other direction when heated to uniform constant temperature and it is not attached to any given boundary as the of aluminum is more than that of brass. The of aluminum is more than that of brass, still the strip bends in opposite direction i.e towards the aluminum side due to the dominant effect of given temperature distribution. The stresses at the wall are reduced when the Y constraints are removed fromthe fixed edge boundary condition while keeping the X constraints (as shown in t stress distribution plots )