MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MATERIALS SCIENCE AND ENGINEERING
CAMBRIDGE, MASSACHUSETTS 02139 3.22 MECHANICAL PROPERTIES OF MATERIALS
PROBLEM SET 4 SOLUTIONS
1. Consider a 500 nm thick aluminum film on a 500 µm thick silicon wafer, 200 mm in diameter. After deposition at 60°C, the wafer is cooled to room temperature. What is the radius of curvature of the film? (ESi = 150 GPa, νSi = 0.17, αSi = 3×10-6 °C-1, EAl = 69 GPa, νAl = 0.33, αAl = 23×10-6 °C-1) The use of the Stoney formula is appropriate for this system because the aluminum film is very thin compared to the relatively thick substrate. The Stoney formula is given by
2
6
s s m f
E h ρ h
= σ ,
where ρ is the radius of curvature, Esis the biaxial modulus of the substrate, hs and hf are the substrate and film thickness, and σ is the stress at the interface. m
The biaxial modulus of the substrate is calculated by 150 GPa
180.7 GPa
1 1-0.17
s s s
E E
= ν = =
− .
The stress at the interface in the aluminum film is given by 1
f
m m f m
f
E E
σ ε ε
= = ν
− .
To find the strain at the interface, we must calculate the mismatch strain due to the thermal expansion coefficient mismatch.
6 -1 6 -1
( ) (3 10 C 23 10 C ) (20 C 60 C) 0.0008
m s f T
ε = α α− ⋅ ∆ = × − D − × − D ⋅ D − D = Solving for the mismatch stress,
69 GPa
0.0008 82.39 MPa
1 1 0.33
f
m m f m
f
E E
σ ε ε
= = ν = =
− −
The radius of curvature of the film-on-substrate system can be calculated from the Stoney formula.
2 6 2
9
(180.7 GPa)(500 10 m) 182.8 m 6 6(0.08239GPa)(500 10 m)
s s m f
E h ρ h
σ
−
−
= = × =
×
The positive curvature (concave) indicates that the film is in tension after cooling to room temperature.
Because the thermal expansion coefficient for aluminum is larger than silicon, the aluminum attempts to contract more than the silicon during the cooling. However, at the interface, the aluminum cannot contract fully due to the constraining silicon, and as a result the aluminum is in tension.
2. A bilyer is composed of a 1 mm thick coating (E=350 GPa,α = ×9 10−6 DC−1) on a 5 mm substrate ( ) and has a width of 5 mm. The bilayer has zero curvature at the initial temperature of 20°C. The bilayer is then heated to 100°C. Calculate the resulting curvature and the maximum tensile and compressive stresses. Where do the maximum internal stresses occur?
70 GPa, 23 106 C E= α = × − D −1
To find the curvature of the bilayer, we can use the equation for the most general bilayer derived in class
2 1 0
1 1 2 1
1 1 2 2
( )( )
2( ) 1 1
2
T T E I E I h
h E a b E a
κ= α α− −
+ + +
b
where the coating properties are denoted by the subscript 1 and the substrate properties by 2. To calculate the curvature, we need the moments of inertia for each layer.
3 3
13 2 1
1
(0.005 m)(0.001 m)
4.1667 10 m
12 12
I =ba = = × −
3 3
11 4 2
2
(0.005 m)(0.005 m)
5.20833 10 m
12 12
I =ba = = × −
Entering the parameters of the bilayer system into the given equation, we find that the curvature of the bilayer is κ =0.25 m−1 .
The maximum stresses for a bilayer were derived in class and exist at the interface of the two layers.
1 1 2 2 1 1
max,1
1
1 1 2 2 2 2
max,2
2
2( )
107 MPa 2
2( )
56.4 MPa 2
E I E I a E ha b
E I E I a E ha b
σ κ
σ κ
+
= + =
+
= − + = −
3. The structure and mechanics of cork: The structure and properties of cork are approximately axisymmetric. In the schematics on the next page, we see a macroscopic view of cork as it exists on a tree, a microscopic look at the arrangement of cork cells, and finally a view of an individual cork cell.
Ignoring the natural variations in cell structure, we can model this cellular material as a linear-elastic solid by
ij
S
ijkl klε = σ
.Axisymmetry reduces the number of independent compliances to five (note that the axis of symmetry is the x1 axis).
11 1111 11 1122 22 1122 33
22 1122 11 2222 22 2233 33
33 1122 11 2233 22 2222 33
23 2222 2233 23
31 1212 31
12 1212 12
( )
2 2
S S S
S S S
S S S
S S
S S
ε σ σ σ
ε σ σ σ
ε σ σ σ
ε σ
ε σ
ε σ
= + +
= + +
= + +
= −
=
=
Measurement of four of the five compliances is straightforward. Simple tensile or compressive tests are done on the cork along the axis of symmetry (x1 axis) and orthogonal to this direction (x2 and x3 axes) to measure the Young’s modulus and Poisson’s ratio in these directions.
1 1111
2 3 2222
12 1 13 1 21 2 31 2 1122
23 2 32 2 2233
1
1 1
E S
E E S
E E E E S
E E S
ν ν ν ν
ν ν
=
= =
= = = = −
= = −
The modulus G23 in the x2, x3-plane is obtained from these measurements by
23 23 2 2222 2233
1G =2(1+ν ) E =2(S −S ).
To determine the fifth compliance value, S1212, which is related to the shear modulus, G12, we rotate the cork through 45° about the x3-axis and cut a cube with one face normal to the x3-axis, and the other two at 45° to the x2-axis and x1-axis. A simple compression test in the new x1 direction then gives a new Young’s modulus,
E′
.a. What is the relationship between the shear modulus, G12, and S1212?
The shear modulus G12 relates the engineering shear strain and shear stress. Using matrix notation, we have
6 66 6
12
S 1
G 6
ε = σ = σ .
To convert between matrix and tensor notation, we have the relation that
66 4 1212
S = S . Thus we find that
12 1212
14 G = S .
b. By rotating the Sijkl through 45° about the x3-axis, obtain an expression for in terms of the compliances in the original orientation.
S′1111
The direction cosines for this change in orientation by rotation about the x3-axis is
1 1 0
2 2
cos 45 sin 45 0
1 1
sin 45 cos 45 0 0
2 2
0 0 1 0 0
ij
−
1
−
= =
D D
D D
A .
The tensor equation for the transformation of a fourth order tensor is Sijkl′ =A A A Ami nj ok plSmnop. To find the S′1111 component, we use the transformation rule.
1111 m1 n1 o1 p1 mnop
S′ =A A A A S
Although the expanded form of the tensor transformation has 81 terms per equation, we note that A and A are the only nonzero direction cosine terms that appear in the tensor
transformation. The resulting expanded form is
11 21
1111 11 11 11 11 1111
11 11 11 21 1112 11 11 21 11 1121 11 21 11 11 1211 21 11 11 11 2111 11 11 21 21 1122 11 21 21 11 1221 11 21 11 21 1212
21 11 11 21 2112 21 11 21 11 2121 21 21 11
S S
S S S S
S S S
S S
′ = +
+ + +
+ + +
+ +
A A A A "
A A A A A A A A A A A A A A A A "
A A A A A A A A A A A A "
A A A A A A A A A A A A11 2211
21 21 21 11 2221 21 21 11 21 2212 21 11 21 21 2122 11 21 21 21 1222 21 21 21 21 2222
S
S S S S
S
+
+ + +
"
A A A A A A A A A A A A A A A A "
A A A A
+
+
From the symmetry of the strain tensor and stress tensor that are related by the compliance tensor, we can simplify the expression. (Note that S1112 and S2212 are zero)
4 3
1111 ( 11) 1111 4( 11) ( 21) 1112
S′ = A S + A A S +2(A11) (2 A21)2S1122+4(A11) (2 A21)2S1212+4(A21) (3 A11)S2221 + A( 21) S4 2222
In tensor notation, the expression for the compliance in the new direction is
1111 1111 1122 1212 2222
1 1 1
4 2 4
S′ = S + S +S + S
c. Use your answers from parts (a) and (b) to find an expression for the shear modulus G12, in terms of experimental parameters E E1, 2,ν12, andE′ .
From part (a) we have
1212 12
1 4S
G = .
From the answer to part (b) we can solve for S1212 and substitute this expression into the equation above.
1111 1111 1122 2222
12
1 1 1 1
4 S 4S 2S 4S G
′
= − − −
Replacing the compliance values with the corresponding experimental values leads to the result (Note that the value of the compliance in the new direction is related to the new Young’s modulus by S1111′ =1E′.)
12
12 1 1 2
12
12 1 2
1 1 1 1 1 1 1
4 4 2 4
1 2
1 4 1
G E E E E
G E E E
ν ν
−
= ′− − −
= − − −
′
.
d. For the isotropic case, what does the equation derived in part (c) become?
For an isotropic material, which is more symmetric than the transversely isotropic symmetry assumed above, the moduli and Poisson’s ratios do not vary with direction. Hence, we can rewrite the answer in part (c) as
1 4 1 2 1
G E E E
ν
= − − − . Simple algebra leads to the result
1 4 1 2 1
1 2 2
2(1 )
G E
G E
G E ν ν
ν
− + −
=
= +
⇒ =
+
Thus, the expression in part (c) correctly reduces to the relationship between the elastic modulus, shear modulus and Poisson’s ratio for the isotropic case.
4. An isotropic sample of material subjected to a compressive stress σ = − is confined so that it cannot z p deform in either the x- or y-directions.
a. Do the stresses occur in the material in the x- and y-directions? If so, how are they related to σ ? z
If the sample were not confined in the x- and y-directions, we would expect the material to expand in the transverse direction upon compressive loading. The fact that the material is constrained in the transverse direction indicates that a stress must evolve to maintain zero strain in the transverse direction. Looking at Hooke’s law for an isotropic material in three dimensions, we find
( )
x x y
E E
σ ν
ε = − σ +σz = . 0
Due to the symmetry of the problem, we know σx =σy. Thus, we can find a relationship between the transverse stress and the applied compressive stress
x y 1 z
σ σ ν σ
= = ν
− .
b. Determine the stiffness E′ =σ εz z in the direction of the applied stress in terms of the isotropic elastic constants and E ν for the material. Is E′ equal to the elastic modulus from uniaxial loading? Why or why not?
Again using Hooke’s law to find the strain in the z-direction, 2 2
( ) 1
1
z z
z x y
E E E
σ ν σ ν
ε σ σ
ν
= − + = − − .
Using this expression for the strain in the z-direction, we can calculate the stiffness in the z- direction.
2 2
2 2
1 1
1 1
z z
z z
E E
E
σ σ
ε σ ν ν
ν ν
′ = = =
− −
− −
The stiffness calculated here is not equal to the elastic modulus in uniaxial loading. In uniaxial loading, the transverse stresses are zero and do not contribute to the uniaxial deformation. However, in this case the transverse stresses do affect the strain in the direction of loading.
c. What happens if the Poisson’s ratio for the material approaches 0.5?
As Poisson’s ratio approaches 0.5, the strain in the z-direction goes to zero for any applied compressive stress. This can be understood without derivation. We know that materials with a Poisson’s ratio of 0.5 are incompressible (εx+εy+εz = ). Because the transverse directions are 0 constrained (zero strain), the strain in the loading direction must be zero to satisfy the
incompressibility condition, regardless of the applied compressive stress.
5. Consider a flat plate of isotropic material that lies in the x-y plane and which is subjected to applied loading in this plane only. Such a plate is under plane stress, so that σz =τyz =τxz =0 MPa.
a. Does the thickness of the plate usually change when the plate is loaded?
For all non-zero Poisson’s ratio materials, there will be some change in the thickness of the plate due to the Poisson effect. By inspecting Hooke’s law for a three dimensional isotropic material
( )
z
z E E x y
σ ν
ε = − σ +σ
we notice that even in a state of plane stress, the stresses in the orthogonal directions contribute to the strain in the out-of-plane direction.
b. Under what conditions does the thickness not change? That is, when is this state of plane stress also a state of plane strain?
Looking at Hooke’s law again, it is clear that if the stresses in the orthogonal directions to the out-of- plane direction are equal and opposite, the out-of-plane strain is necessarily zero, indicating no thickness change.
