3.22 Mechanical Behavior of materials
PS3 Solution
Due: March, 2, 2004 (Tuesday) before class (10:00am)
1. A sample of a material subjected to a compressive stress σ z is confined so that it
cannot deform in the y-direction, but deformation is permitted in the x-direction, as
shown below. Assume that the material is isotropic and exhibits linear-elastic behavior.
Determine the following in terms of σ z and the elastic constants of the material:
(a) The stress that develop in the y-direction.
Hooke’s law for the three-dimensional case is needed as
εx =
[
]
[
]
[
1
1
1
σ x − ν (σ y + σ z ) , ε y = σ y − ν (σ x + σ z ) , ε z = σ z − ν (σ x + σ y )
E
E
E
]
The situation posed requires substituting ε y = 0 and σ x = 0 , and also treating σ z
as a known quantity.
σ y = νσ z
(b) The strain in the z-direction.
εz =
2
1
[σ z − ν (0 + νσ z )] = 1 − ν σ z
E
E
(c) The strain in the x-direction.
εx =
1
[0 − ν (νσ z + σ z )] = − ν (1 + ν ) σ z
E
E
(d) The stiffness E ' = σ z / ε z in the z-direction. Is this apparent modulus equal to the
1
elastic modulus E from a uniaxial test on the material? Why or why not?
σ
E
.
E' = z =
ε z 1 −ν 2
Obviously, this value differs from E, being larger, such as E'=1.10 for a typical value of
ν =0.3. The value E is the ratio of stress to strain only for the uniaxial case, and such
ratios for any other case are determined by behavior according to the three-dimensional
form of Hooke’s Law.
2. An orthotropic material has the compliance matrix:
 S11

 S12
S
S ij =  13
 0

 0
 0

S12
S 22
S13
S 23
0
0
0
0
S 23
S 33
0
0
0
0
S 44
0
0
0
0
0
0
0
S 55
0
0 

0 
0 

0 

0 
S 66 
The reciprocal relation states that:
ν 12 E 2 = ν 21 E1 , where ν 12 = −
ε2
ε1
(a) Prove the reciprocal relation.
(i) apply σ 1 only.
 ε 1   S11
  
 ε 2   S12
ε   S
 3  =  13
ε 4   0
  
ε5   0
ε   0
 6 
S12
S 22
S13
S 23
0
0
0
0
S 23
S 33
0
0
0
0
S 44
0
0
0
0
0
0
0
S 55
0
0  σ 1 
 
0  0 
ε = S11σ 1
0  0 
ε
S
  give 1
, ν 12 = − 2 = − 12
0  0 
ε 2 = S12σ 1
ε1
S11
 
0  0 
S 66  0 
S12 = −ν 12 S11 = −
ν 12
E1
(ii) apply σ 2 only.
2
(A)
 ε 1   S11
  
 ε 2   S12
ε   S
 3  =  13
ε 4   0
  
ε5   0
ε   0
 6 
S12
S 22
S13
S 23
0
0
0
0
S 23
S 33
0
0
0
0
S 44
0
0
0
0
0
0
0
S 55
0
0  0 
 
0  σ 2 
ε = S12σ 2
0  0 
ε
S
  give 1
, ν 21 = − 1 = − 12
0  0 
ε 2 = S 22σ 2
ε2
S 22
 
0  0 
S 66  0 
S12 = −ν 21 S 22 = −
ν 21
(B)
E2
From (A) and (B), ν 12 E 2 = ν 21 E1
(b) A fiber composite is transversely isotropic and has the elastic moduli:
E1 = E 2 = 5 GPa
E3 = 25 GPa
ν 12 = 0.25
ν 13 = 0.33
G13 = 8 GPa
 5 0 4


Calculate the strain tensor for the stress state of σ ij =  0 5 3  MPa.
 4 3 5


S11 = S 22 = 0.2GPa
S 33 = 0.04GPa −1
−1
S12 = −
S13 = −
ν 12
E1
ν 13
E1
=−
0.25
= −0.05GPa −1
5
=−
− 0.33
= −0.066GPa −1 S 66
5
1
= 0.125GPa −1
8
= 2( S11 − S12 ) = 0.5GPa −1
S 44 =
0
0
0  5   4.2 × 10 −4   ε 11 
− 0.05 − 0.066
 ε 1   0 .2
 
  

  
0 .2
0
0
0  5   4.2 × 10 − 4   ε 22 
− 0.066
 ε 2   − 0.05
 ε   − 0.066 − 0.066
0.04
0
0
0  5   − 4.6 × 10 − 4   ε 33 
=
 3=

  = 
0
0
0.125
0
0  3   3.75 × 10 − 4   2ε 23 
ε 4   0
 

ε  

0
0
0
0.125 0  4   5 × 10 − 4   2ε 13 
 5  0
ε   0
  2ε 
0
0
0
0
0.5  0  
0
 6 
  12 
2 .5 
 4 .2 0


ε ij =  0 4.2 1.88  × 10 − 4
 2.5 1.88 − 4.6 


3. (a) How and why would you expect the elastic modulus to vary with melting
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temperature? How would you explain that, although aluminum (Al) and magnesium (Mg)
have very similar melting temperatures (Tm ~650℃), they have very dissimilar elastic
moduli (EAl = 70 GPa, EMg = 45 GPa)?
As melting temperature increases, elastic modulus increases. This is because both the
melting temperature and modulus are indicators of the atomic bond strength of a material.
If more energy is required to break the bonds (melt the material), then more energy will
be required to stretch them (deform the material elastically).
The contrast in moduli of Al and Mg is due to their difference in crystal structure. Al is a
face-centered cubic (FCC) material, whereas Mg is a hexagonal close-packed (HCP)
material in the standard state. Thus, Mg has a lower density ( ρ Mg = 0.6 ρ Al ). Although
the packing factors of FCC and HCP are equal in the close packed planes, the stacking of
these planes is different. (FCC shows ABCABC periodicity whereas HCP shows ABAB
periodicity). This means that these metals will be of different atomic densities orthogonal
to the close-packed planes; HCP will be of lower atomic density along the c-axis, for
example. Since E is a weighed average of all the crystallographic orientations f the
material, the lower density of Mg atoms along some atomic planes will results in its
lower elastic modulus.
(b) Pure Al has an elastic modulus of 70 GPa. An Al alloy commonly used in aero/astro
engineering applications is 7075 T6 Al. The designation “7075” indicates the chemical
composition of the alloy: 1.6% Cu, 2.5%mg, 5.6% Zn, 0.23 Cr, balanced Al. The
designation “T6” indicates that the heat treatment of the alloy: tempered at 650℃.
Despite the addition of so many alloying elements, the elastic modulus of 7075 T6 Al is
not much different from that of the pure Al:E7075 T6=70.7GPa. Explain why alloying has
little effect on E.
Elastic modulus is one of the most stable mechanical properties of a material. Little can
be done to the material via processing to strongly alter its elastic response. Alloying of Al
to create 7075 Al is no exception. A solid solution of approximately 10% impurities does
not significantly alter the equilibrium atomic spacing or the bond strength which are
responsible for the elastic modulus. A notable exception of this is the Al-Li alloys.
Addition of 20%Li to Al makes the alloy lighter and stiffer ( ∆E =+20%) due to
precipitation of a brittle compound in the Al matrix.
(c) Cobalt is magnetostrictive and contracts upon exposure to a magnetic field. What is
the effect of a magnetic field on the elastic modulus of Cobalt?
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In some materials, Young’s modulus can be increased by a factor of 3 due to
magnetostriction. This stems from the change in distance between atoms. (density
increases)
4. (a) As temperature increases, explain how the elastic modulus changes for crystalline
materials and rubbers. Relate each type of behavior to fundamental properties of material.
Crystalline materials experience a decrease in elastic modulus as the temperature
increased. Increasing the temperature causes the atoms to move further apart, due to
thermal expansion. As atomic distances increase, the modulus decreases.
Rubbers, however, experience an increase in elastic modulus as the temperature is
increased. Pulling a rubber in tension causes the chains to align, lowering the entropy in
the system. For rubbers, the entropy governs elastic response. As temperature is increased,
entropy increases. The increase in entropy will therefore increase the resistance of the
material to deformation, increasing the elastic modulus.
(b) Speculate on the relation between the modulus of elasticity of a group of crystalline
solids and their respective melting points. Also relate the modulus of elasticity to the
respective coefficients of thermal expansion.
The modulus of elasticity (E), melting point and coefficient of thermal expansion ( α ) all
depend strongly on the magnitude of atomic bonding forces. Therefore, as E increases,
melting point should increase and α should decrease.
5. Using the figures below for cases of isostress and isostrain, derive the upper and lower
bounds for Young’s modulus for a composite material that were given in class.
(a) Isostrain
(b) Isostress
(a) Isostrain: Lodading is parallel to fibers.
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ε 1 = ε 2 = ε , P = P1 + P2 = E1εA1 + E 2 εA2
P1 + P2
E εA + E 2 εA2
σ
P
=
= 1 1
E= =
ε Aε ( A1 + A2 )ε
( A1 + A2 )ε
V1 = Volume fraction of phase 1, V2 = Volume fraction of phase 2.
E = V1 E1 + V2 E 2 : Upper bound
(b) Isostress : Loading is perpendicular to fibers
σ1 = σ 2 = σ
δ : change in length δ = δ 1 + δ 2 = ε 1l1 + ε 2 l 2
σV σV
ε = ε 1V1 + ε 2V2 = 1 + 2
E1
E=
E2
E1 E 2
σ
σ
: Lower bound
=
=
ε σV1 σV2 V1 E 2 + V2 E1
E1
+
E2
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