Mechanics of Multi Layers 3

1

MAE598 Advanced Mechanical Analysis of

Microelectronics Packaging

Week 7

Mechanics of Multilayered Structures (Cont.)

Readings: Chapter 2 of Lau, “Thermal Expansivity and Thermal Stress in Multilayered Structures”

Chapter 4 of Lau, “Transient Thermal Stresses in Multilayered Devices” Chapter 5 of Bar-Cohen/Krauss “Thermal Stress Failures in Microelectronic Components”

2

MAE598 - Advanced Mechanical Analysis of Microelectronics Packaging��Week 7: Mechanics of Multilayers (cont.)

l  Topics for Week 7 l  Thermal stresses in isotropic plates: single layers and multilayers

l  Boundary conditions for plates l  Examples of thermal stresses in single layer isotropic plates l  Thermal stresses in isotropic multilayers

l  Layer-wise constitutive relations l  Resultant loads and moments l  Axial load and bending moment coupling

l  Axisymmetric thermal stresses in N layers l  Kinematics under axisymmetric conditions l  Assumptions and simplifications l  Stress and strain calculation l  Numerical example

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�  Recall from the previous lecture that under the Kirchhoff-Love approximation the strains are linear functions of x3 in multilayered plates. This leads to in-plane strains that are continuous and in-plane stresses that “jump.”

�  The equations also show that there is a coupling between axial loads and moments. Please note the correction on the equations w.r.t. the previous lecture. Thermal effects in isotropic materials do not contribute to shear forces or twisting moments


Thermal stresses in multilayered plates (cont.)

x1

x3

o hi, Ei, ni, ai

Strain Stress

N1N2

N6

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

A11 A12 0A12 A22 00 0 A66

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

ε10

ε20

ε60

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥+

B11 B12 0B12 B22 00 0 B66

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

κ1κ 2

κ 6

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥− NT

110

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

M1

M 2

M 6

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥=

B11 B12 0B12 B22 00 0 B66

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

ε10

ε20

ε60

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥+

D11 D12 0D12 D22 00 0 D66

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

κ1κ 2

κ 6

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥− MT

110

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4

�  Note that the coupling is due to the “B” terms. Let’s examine those more closely

�  The coupling is due to material heterogeneity. If all layers were the same, the “B” terms would be zero. Another way to make these terms identically zero is to build a symmetric multilayer, where layers of the same material and thickness are located at the same distance in opposite sides from the center.

MAE602 - Advanced Mechanical Analysis of Microelectronics Packaging�� Week 8: Mechanics of Multilayers (cont.)

Thermal stresses in multilayered plates (cont.)

B11 = B22 =E (i )

1− ν (i )( )2i=1

N

∑ 12

x3(i )( )2 − x3

(i−1)( )2⎡⎣

⎤⎦

B12 =E (i )ν (i )

1− ν (i )( )212i=1

N

∑ x3(i )( )2 − x3

(i−1)( )2⎡⎣

⎤⎦

B66 =E (i ) (1−ν (i ) )1− ν (i )( )2i=1

N

∑ 12

x3(i )( )2 − x3

(i−1)( )2⎡⎣

⎤⎦

5

�  There are simple engineering approximations that can be used to calculate thermal stresses in unsupported multilayer plates. Three basic cases can be identified depending on the geometry of the plate and the curvature

MAE598 - Advanced Mechanical Analysis of Microelectronics Packaging��Week 7: Mechanics of Multilayers (cont.) Thermal Stresses in N Layers

6

�  The procedure given here follows chapter 2 of Lau’s book for the axisymmetric approximation. However, the general equations can be used to deduce procedures for the strip and cylindrical bending if so desired.

�  For axisymmetric bending there is perfect radial symmetry; therefore,

�  In cylindrical polar coordinates, and recalling that for axisymmetric conditions derivatives with respect to θ are identically zero (shear stresses and strains are zero), the most important strain component is along the radial direction:

MAE598 - Advanced Mechanical Analysis of Microelectronics Packaging��Week 7: Mechanics of Multilayers (cont.) Thermal Stresses in N Layers

�

ε11 = ε22; σ11 = σ 22 ⇒ε11 = σ11

E1−ν( ) + αΔT

7

�  Assuming that the curvature is uniform implies that there is no in-plane variations of stress and strain. Then, the curvature can be related to the strains, by assuming that they are linear functions of x3. In this case, taking an origin for x3 on the bottom layer

�  The curvature of the plate is given by the difference in normal strains between the top and bottom layers divided by the total thickness of the plate, which is very similar to what is done in beam theory.


Axisymmetric Thermal Stresses in N Layers

�

ε11 = ε22 = εrr; σ11 = σ 22 = σ rr ⇒εrr = σ rr

E1−ν( ) + αΔT

�

ε11 = εrr = εrr(b ) + x3

hεrr(t ) −εrr

(b )( ) ⇒ κ rr =εrr( t ) −εrr

(b )( )h

8

�  Assuming that temperature in the plate is uniform, the radial stress in the ith layer is

�  Assuming that the temperature is uniform implies that the stresses are the result of the mismatch of the thermal expansion coefficient between layers.

�  Resultant forces per unit perimeter can be calculated for each layer

�  Moments per unit perimeter about x3=0 can also be obtained for each layer


Axisymmetric Thermal Stresses in N Layers (cont.)

�

σ rr(i) = Ei

1−ν i

εrr(b ) + x3

hεrr

(t ) −εrr(b )( ) −α iΔT

⎛ ⎝ ⎜

⎞ ⎠ ⎟

�

Pi = σ rr( i)

x3( i−1)

x3( i )

∫ dx3 ⇒ Pi = Ei

1−ν i

εrr(b ) −α iΔT( ) x3( i) − x3( i−1)( ) +

εrr( t ) −εrr

(b )( ) x3( i)( )2 − x3

( i−1)( )2( )2h

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

�

Mi = x3σ rr( i)

x3( i−1)

x3( i )

∫ dx3 ⇒ Mi = Ei

1−ν i

εrr(b ) −α iΔT( ) x3

(i)( )2 − x3(i−1)( )2( )

2+

εrr( t ) −εrr

(b )( ) x3( i)( )3 − x3

( i−1)( )3( )3h

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟

9

•  If the total applied forces and moments are known, they can be related to the strains by summing over the layers. In particular, for a free plate

•  Where

•  The result is a linear system of equations on and .



�

Pii=1

N

∑ = 0⇒εrr(b ) 2h A1 − B1( ) + B2 − A2[ ] + εrr

( t ) A2 − B2( ) + 2hΔT D1 −C1( ) = 0

�

Mii=1

N

∑ = 0⇒εrr(b ) 3h A2 − B2( ) + 2B3 − 2A3[ ] + 2εrr

(t ) A3 − B3( ) + 3hΔT D2 −C2( ) = 0

�

Ak = Ei

1−ν i

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

N

∑ x3(i)( )k for k = 1,2,3

�

Bk = Ei

1−ν i

⎛

⎝ ⎜

⎞

⎠ ⎟

i=1

N

∑ x3(i−1)( )k for k = 1,2,3

�

Ck = Ei

1−ν i

⎛

⎝ ⎜

⎞

⎠ ⎟ α i

i=1

N

∑ x3( i)( )k for k = 1,2

�

Dk = Ei

1−ν i

⎛

⎝ ⎜

⎞

⎠ ⎟ α i

i=1

N

∑ x3( i−1)( )k for k = 1,2

�

εrr(b )

�

εrr(t )

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•  The solution for the top and bottom strains is straightforward and is given by

•  Where

•  The strains and stresses can now be calculated in each layer. In addition, the deflection d at the center or a disk of radius r can be estimated by

•  The example done in the textbook is worked out using a symbolic math package. Results are shown in the next few slides.



�

εrr(b ) = hΔT

F4 D1 −C1( ) A3 − B3( ) − 3 D2 −C2( ) A2 − B2( )[ ]

�

εrr(t ) = hΔT

F3 D2 −C2( ) 2h A1 − B1( ) + B2 − A2( ) − 2 D1 −C1( ) 3h A2 − B2( ) + 2 B3 − A3( )( )[ ]

�

F = −2 A3 − B3( ) 2h A1 − B1( ) + B2 − A2( ) + A2 − B2( ) 3h A2 − B2( ) + 2 B3 − A3( )( )

�

δ = r2

2κ rr

11

•  The relevant material properties and geometry are given by

•  The appropriate coefficients can be obtained from the formulas given before and expressions for the field variables in each layer can now be obtained



Layer # 1 2 3 4

Material Ceramic Adhesive Aluminum Encapsulation

Thickness [m] 5.334x10-4 1.524x10-4 7.62x10-4 2.54x10-5

α [1/°C] 6.0x10-6 8.0x10-5 2.3x10-5 4.0x10-5

E [Pa] 102x109 0.68x109 68x109 6.8x109

ν 0.23 0.3 0.32 0.3

∆T [°C] -180 -180 -180 -180

12

•  The numerical values of the strains on bottom and top of the multilayer are

•  Stresses as a function of position along the thickness were calculated from these values. Units have been changed to the metric system, but they correspond to the numerical values given in the text (the plot and some of the equations in the text have typos)



εrr(b) = −5.541×10−4; εrr

(t ) = −5.1544 ×10−3

13

•  If the Al layer is replaced by copper (E=110 GPa, ν=0.34, α=1.69x10-5), the numerical values of the strains on bottom and top of the multilayer are

•  Stresses as a function of position along the thickness were recalculated from these values. Note that, for everything else being equal, the stresses in the copper layer are much higher than in the Al layer



εrr(b) = −6.523×10−4; εrr

(t ) = −5.203×10−3

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Mechanics of Multi Layers 3

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