ME 386P-2, Spring 2014 Homework 2 Assigned: January 23, 2014

ME 386P-2, Spring 2014
Homework 2
1.
Assigned:
Due:
January 23, 2014
January 30, 2014
Consider the following as cases of linear elasticity.
(a) How many independent elastic constants does an isotropic material have?
(b) Materials with a cubic crystal structure, such as aluminum and iron, generally have three independent elastic constants. Yet, polycrystalline forms of these materials often exhibit essentially
isotropic elastic behavior. Explain how this is possible.
Solution:
(a) Two (2) independent elastic constants.
(b) For randomly oriented grains, averaging the elastic behavior over a very large number of grains
produces essentially isotropic behavior.
2. Name the stress state described by these two stress tensors, both given in units of MPa, and explain
how these two tensors are related to each other:




0 0 0
0 0 0
 0 0 5 
 0 5 0  .
and
0 5 0
0 0 −5
Solution: These both represent the same state of pure shear stress but with coordinate systems different by a rotation of 45◦ about the x-axis (8 pts.). If the first (left) tensor represents the original coordinate
system and the second (right) represents a new coordinate system, then the following direction cosine
matrix will transform from the original to new coordinate systems,

1
0
0
p
p
=  0 1/ 2 1/ 2 
p
p
0 −1/ 2 1/ 2

ai j
,
orig
such that σnew
= a i k a j l σkl .
ij
3.
Calculate the following for the stress tensor given below in units of MPa.
(a) The stress deviator tensor (deviatoric stress tensor).
(b) The first invariant of the stress deviator, J 1 .
(c) The second invariant of the stress deviator, J 2 .

5 1 3
σ =  1 8 1  MPa
3 1 5

Solution: The Octave program given at the end of this solution shows the calculations that produce the
required solutions. First, here is a description of the solutions.
(a) To calculate the stress deviator, the hydrostatic stress is first calculated to be σH = 6 MPa. The
hydrostatic stress is then subtracted from the diagonal terms of the stress tensor to produce the
deviatoric stress tensor.


−1 1
3
1  MPa
σ= 1 2
3 1 −1
(b) The first invariant of the stress deviator is the trace of its tensor, trσ0 = σ0i i = 0, which is zero. This
proves that this is, indeed, the stress deviator.
(c) The second invariant of the stress deviator is,
J2
¢
1¡
(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2
6
´
1³
=
(σx − σ y )2 + (σ y − σz )2 + (σz − σx )2 + 6 (τ2x y + τ2y z + τ2xz
6
= 14 MPa2
=
This is related to the von Mises effective stress as,
p
σ = 3 J 2 = 6.48 MPa .
Here is the Octave code and output for these calculations.
octave:1> s = [5,1,3;1,8,1;3,1,5]
s =
5
1
3
1
8
1
3
1
5
octave:2> sh = trace(s)/3
sh = 6
octave:3> sdev = s - sh*eye(3)
sdev =
-1
1
3
1
2
1
3
1
-1
octave:4> trace(sdev)
ans = 0
octave:5> [v,l] = eig(sdev)
v =
-7.0711e-01
-8.3681e-17
7.0711e-01
5.0000e-01
-7.0711e-01
5.0000e-01
5.0000e-01
7.0711e-01
5.0000e-01
l =
Diagonal Matrix
-4.00000
0
0
0
0.58579
0
0
0
3.41421
octave:6> sdev1 = l(1,1); sdev2 = l(2,2); sdev3 = l(3,3);
octave:7> [sdev1, sdev2, sdev3]
ans =
2
-4.00000
0.58579
3.41421
octave:8> j2 = ((sdev1 - sdev2)^2 + (sdev2 - sdev3)^2 + (sdev3 sdev1)^2)/6
j2 = 14.000
octave:9> j2alt = ((sdev(1,1) - sdev(2,2))^2 + (sdev(2,2) sdev(3,3))^2 + (sdev(3,3) - sdev(1,1))^2 + 6*(sdev(1,2)^2 +
sdev(2,3)^2 + sdev(1,3)^2))/6
j2alt = 14
4. Our Engineering Library has numerous materials standards available on-line, including the ASTM
(American Society for Testing and Materials) standards for mechanical testing. These standards include
geometries and dimensions of typical specimens and standards for test conditions to use when measuring particular material properties. Locate the designations and titles of appropriate ASTM standards for
the following:
(a) Tensile testing of metallic materials
(b) Flexural strength testing of ceramics
(c) Plane-strain fracture toughness testing of metals
(d) Fracture toughness testing of ceramics
Remember to log out of your session after you are done so that our University license to the standards
web site will be made available for the next user. (You might also check the homework section of the
class web page.)
Solution: These are examples of the most appropriate ASTM standards. Other standards that may also
be appropriate do exist. Thus, this is not an exclusive list.
(a) ASTM E 8–08: Standard Test Methods for Tension Testing of Metallic Materials
(b) ASTM C 1161–13: Standard Test Method for Flexural Strength of Advanced Ceramics at Ambient
Temperature
(c) ASTM E 399–12: Standard Test Method for Linear-Elastic Plane-Strain Fracture Toughness KIc of
Metallic Materials
(d) ASTM C 1421–10b: Standard Test Methods for Determination of Fracture Toughness of Advanced
Ceramics at Ambient Temperature
5. Consider the cylindrical rod shown below. It is subjected to deformation with time, t , such that the
following displacement field results,
ux
= 0.1 t x
uy
= −0.05 t y
uz
= −0.05 t z
as written in component of the displacement vector, u i .
(a) Calculate the components of the resulting small-strain tensor.
(b) Calculate the components of the resulting Eulerian finite-strain tensor.
(c) Plot for 0 ≤ t ≤ 4 the x and y normal strains from both small-strain and finite-strain theories.
(d) Calculate the time, t , at which E xx deviates from ²xx by 1%, and calculate both these strains at that
point.
(e) Describe what is special about this deformation field. (Hint: Consider Poisson’s ratio.)
3
y
x
Solution:
tions.
The small-strain and the Eulerian finite-strain tensors are calculated from the following equa-
²i j
=
Ei j
=
µ
¶
1 ∂u i ∂u j
+
2 ∂x j ∂x i
µ
¶
1 ∂u i ∂u j ∂u k ∂u k
+
+
2 ∂x j ∂x i
∂x i ∂x j
(a) The components of the small-strain tensor are,
²xx
² y y = ²zz
= 0.1 t
= −0.05 t
All other components are zero.
(b) The components of the Eulerian finite-strain tensor are,
E xx
E y y = E zz
= 0.1 t + 0.005 t 2
= −0.05 t + 0.00125 t 2
All other components are zero.
(c) Plotting these values,
0.4
Strain
Small strain theory
Finite strain theory
0.3
εx
0.2
−εy
0.1
0.0
0
1
2
3
4
t parameter
(d) The following calculation,
E x − ²x
= 1% = 0.01
Ex
yields t = 0.2020, for which ²x = 0.0202 ≈ 2% and E x = 0.0204. This indicates that, for this strain
field, less then 1% error occurs when using the small-strain tensor for strains of less then 2%, which
would be the case for most reasonable elastic loadings of engineering structural materials. Note
that a similar solution is arrived at using the condition (E x − ²x )/²x = 0.01. Either approach is
acceptable.
(e) This deformation field conserves volume, i.e. ν = 0.5, when small-strain theory is applied. This can
be proved by showing that ²i i = 0.
4
6. Consider the case of an elastically anisotropic crystal of α-Fe. The elastic stiffness constants of this
crystal are as follows, in units of GPa.
E 〈111〉
E 〈100〉
c 11
c 12
c 44
276
129
237
141
116
The 1, 2 and 3 directions correspond to the edges of the BCC unit cell of iron, and are taken to be the
same as the x, y and z directions, respectively, in defining this problem. The following stress is applied
to this crystal, in terms of this reference frame,


15 15 0
σi j =  15 15 0  MPa.
0
0 30
(a) Calculate and report the principal stress values and the corresponding principal directions. This is
one of the special stress states considered in HW 1. Which is it and why?
(b) Use the elastic constants provided to calculate the strain tensor which results from this stress state.
Report the strain tensor in the same form as that given above for the stress tensor.
(c) Calculate and report the principal strains and the corresponding principal directions.
i. Are the principal directions for stress and strain the same? Comment on your result.
ii. Are the number of independent principal stresses and strains the same? If not, then why?
(d) Calculate and report i. the hydrostatic stress, ii. the stress deviator and iii. the dilatation.
Solution: This solution utilizes the Octave program of calculations. First enter the stress tensor and
stiffness matrix. Note that stress is given in MPa and stiffness in GPa. Everything is put into units of GPa
for these solutions. Then calculate the compliance matrix, which is the inverse of the stiffness matrix.
octave:1> stens = [15, 15, 0; 15, 15, 0; 0, 0, 30]*10^-3
stens =
0.015000
0.015000
0.000000
octave:2>
octave:3>
octave:4>
octave:5>
c =
237
141
141
0
0
0
0.015000
0.015000
0.000000
0.000000
0.000000
0.030000
c = zeros(6,6);
c(1,1) = c(2,2) = c(3,3) = 237;
c(1,2) = c(2,1) = c(1,3) = c(3,1) = c(2,3) = c(3,2) = 141;
c(4,4) = c(5,5) = c(6,6) = 116; c
141
237
141
0
0
0
141
141
237
0
0
0
0
0
0
116
0
0
0
0
0
0
116
0
0
0
0
0
0
116
octave:6> scomp = inv(c)
scomp =
0.0075867
-0.0028300
-0.0028300
0.0000000
0.0000000
0.0000000
-0.0028300
0.0075867
-0.0028300
0.0000000
0.0000000
0.0000000
-0.0028300
-0.0028300
0.0075867
0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
0.0086207
0.0000000
-0.0000000
0.0000000
0.0000000
0.0000000
0.0000000
0.0086207
-0.0000000
5
0.0000000
0.0000000
0.0000000
-0.0000000
-0.0000000
0.0086207
(a) Now calculate the principal values and directions of the stress tensor by solving the eigenvalue problem.
octave:7> [vstress, lstress] = eig(stens)
vstress =
-0.70711
0.70711
0.00000
0.70711
0.70711
0.00000
0.00000
0.00000
1.00000
lstress =
Diagonal Matrix
0.000000
0
0
0
0.030000
0
0
0
0.030000
Note that the eigenvectors form a left-handed coordinate system. We can reverse the second eigenvector
direction to make a right-handed system, because the sign of the direction is not physically important,
i.e.:
vstress =
-0.70711
0.70711
0.00000
-0.70711
-0.70711
0.00000
0.00000
0.00000
1.00000
Alternatively, the places of two principal directions, and corresponding eigenvalues, could be swapped.
For this special case of stress, the eigenvectors are not all unique. The principal direction corresponding
to the σ1 = 0 principal stress is fixed, and the other two could be in any orthogonal directions within
the plane normal to that direction. You were not expected to provide this level of detail, but it is quite
interesting.
p
p
ˆ 2 + y/
ˆ 2 and zˆ0 = z.
ˆ The
This is a case of balanced biaxial tension in the plane x 0 -z 0 , where xˆ 0 = x/
x i0 coordinates are rotated from x i by 45◦ about the z axis.
(b) Calculating the strain tensor requires the following steps. First, put the stress into matrix form.
Then multiply that by the compliance matrix to calculate the strain matrix. Finally, convert the strain
from matrix to tensor form. Do not forget that the matrix form has the shear strains as engineering
values, i.e., shear strains have a factor of two difference between matrix and tensor forms.
octave:8> stressmat = [stens(1,1); stens(2,2); stens(3,3); stens(2,3);
stens(1,3); stens(1,2)]
stressmat =
0.015000
0.015000
0.030000
0.000000
0.000000
0.015000
octave:9> strainmat = scomp*stressmat
strainmat =
-1.3548e-05
-1.3548e-05
1.4270e-04
6
0.0000e+00
0.0000e+00
1.2931e-04
octave:10> straintens = zeros(3,3);
octave:11> straintens(1,1) = strainmat(1); straintens(2,2) =
strainmat(2); straintens(3,3) = strainmat(3);
octave:12> straintens(2,3) = straintens(3,2) = strainmat(4)/2;
octave:13> straintens(1,3) = straintens(3,1) = strainmat(5)/2;
octave:14> straintens(1,2) = straintens(2,1) = strainmat(6)/2;
octave:15> straintens
straintens =
-1.3548e-05
6.4655e-05
0.0000e+00
6.4655e-05
-1.3548e-05
0.0000e+00
0.0000e+00
0.0000e+00
1.4270e-04
(c) The principal strains and directions are computed through the eigenvalue problem, just as done
for the stress tensor.
octave:16> [vstrain, lstrain] = eig(straintens)
vstrain =
-0.70711
0.70711
0.00000
-0.70711
-0.70711
0.00000
0.00000
0.00000
1.00000
lstrain =
Diagonal Matrix
-7.8203e-05
0
0
0
5.1107e-05
0
0
0
1.4270e-04
It is clear that the principal directions are the same for both the stress and strain tensors, when comparing both sets of eigenvectors as right-handed systems. Even if one considers the non-uniqueness of
the principal directions for stress, the principal directions for both stress and strain can be considered
equivalent. This is because the material exhibits orthotropic or higher symmetry. Some answer stating
the practical equivalence of the principal directions is needed for this part.
The stress tensor has two independent principal values, 0 and 30. The strain tensor, however, has
three. The first principal strain is negative, the result of Poisson contraction along that direction, which is
not loaded, i.e., zero principal stress for that direction. The second and third principal directions have the
same principal stresses, but different principal strains. This is because of elastic anisotropy; the elastic
stiffness (and compliance) are different between the second and third principal directions. This results
in different principal strains along those directions, despite equal principal stresses. The third principal
direction is along the 〈001〉 crystallographic direction, for which E 〈001〉 = 129 GP a. The second principal
direction is along the 〈110〉 crystallographic direction, which has an elastic modulus of E 〈110〉 = 214.8 GPa.
This is why the strain along the second principal direction is significantly less than that along the third
principal direction.
(d) These are readily calculated in Octave. The hydrostatic stress is one-third of the trace of the stress
tensor. The stress deviator is calculated as usual, and the dilatation is the trace of the strain tensor.
octave:17> hydro = trace(stens)/3
hydro = 0.020000
7
octave:18> stressdev = stens - eye(3)*hydro
stressdev =
-0.005000
0.015000
0.000000
0.015000
-0.005000
0.000000
0.000000
0.000000
0.010000
octave:19> dilatation = trace(straintens)
dilatation = 1.1561e-04
8

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