Solid Mechanics Homework Help

1. Positive- definite elastic energy density

It is reasonable to require that the elastic energy density be a positive-definite quadratic form of the strain tensor. That is, 0>w

for any state of strain, except that 0w =

when the strain tensor vanishes. For an isotropic, linearly elastic solid, confirm that this positive-definite requirement is equivalent to require that 0>E

and 1 1/ 2υ− < < .

Answer 1:

For isotropic linear elastic solid the constitutive law is given by

( )( )1 1 1 2ij ij kk ijE Eυ

σ ε ε δυ υ υ

= ++ + −

Therefore the

elastic energy density is

w = 12σ ijεij =

12

E1

+υ

εijεij +Eυ

1+υ

( ) 1−

2υ

( )εkk( )

2"

#

$$

%

&

''

=12

E1+υ

εijεij +Eυ

1+υ( ) 1− 2υ( )ε112 +ε22

2 +ε332 + 2ε11ε22 + 2ε22ε33 + 2ε33ε11( )

"

#$$

%

&''

For a pure shear situation

(only ε21 = ε12 ≠ 0 ):

w = 12

E1+υ

2ε12ε12!

"#

$

%&=

E1+υ

ε 212 > 0 ,

hence 01Eυ>

+

For a hydrostatic stress situation (only

ε11 = ε22 = ε33 ≠ 0 ):

w = 12

E1+υ

εkk2( )+ Eυ

1+υ( ) 1− 2υ( )εkk( )

2"

#$$

%

&''

=12

E1+υ

3ε112 +

Eυ1+υ( ) 1− 2υ( )

9ε112

"

#$$

%

&''

=3E

2 1− 2υ( )ε112 > 0

,

hence

E1− 2υ

> 0

If E<0,

formula 1.1 gives

1/ 2υ >

and formula 1.2 gives

1υ < − , no solution.

We are left with E>0 and 1 1/ 2υ− < < .

Plug this back to w, one may show the

sufficiency by completing squares.

2. The coefficient of thermal expansion (CTE) is a second-rank tensor.

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strain tensor can change. Consequently, the coefficient of thermal expansion (CTE) is a second-rank tensor. For a crystal of cubic symmetry, show that the CTE is the same in all directions.

Answer 2:

Because the strain tensor is symmetric ( thermalij ijTε α= Δ ) we expect the CTE is also

symmetric.

11 12 31

12 22 23

31 23 33

ij

α α α

α α α α

α α α

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

A second rank tensor is the same in all directions means that it is invariant no matter how the coordinate system changes. This indicates that we should prove the CTE for

cubic crystal has the form of ijij αδα = .

As shown in above figure, a crystal of cubic symmetry has a 3 fold symmetry axis, the [111] direction actually. Looking from this axis, the x, y, z axes are interchangeable (120o rotation about this axis) with the materials properties unchanged. Therefore, we

need to have αααα === zzyyxx .

3. Hooke's law for anisotropic, linearly elastic solids


Hooke's law connecting stresses σij and strains εij for a generally anisotropic

solid can be written in one of the following forms

εij = Sijpqσ pq, σ ij =Cijpqεpq

The usual correspondence is adopted, i.e.

The fourth-order tensors S and C are the compliance and stiffness tensors, respectively. The 6 by 6 matrices s and c (s = c-1) are conventional compliance and stiffness matrices. i) Confirm that

ii) Let m = miei be a unit vector. Show that when the solid is under a

uniaxial stress in the direction m, Young's modulus in this direction is given by

iii) For a cubic solid show the Young's modulus in direction m is given by

( )( ) ⎟

⎠

⎞⎜⎝

⎛ ++⎟⎠

⎞⎜⎝

⎛ −−++= 441244121143

42

41 2

1211 sssssmmm

mE

iv) Find the orientations for a cubic solid which attain minimum or maximum Young's modulus.

Answer 3i)

(1) Assume the following strain state:

11 00( 1, 1)ij i j

ε

ε

≠⎧⎪⎨ = ≠ ≠⎪⎩

or correspondingly 1 00( 1)i i

ε

ε

≠⎧⎨

= ≠⎩

Now we calculate the stress 11σ and 1σ

11 11 1111 11pg pqC Cσ ε ε= =


or

1 1 11 1i ic cσ ε ε= = .

Because 11 1σ σ= and 11 1ε ε= , we get 1111C = 11c

Assume the following stress state:

11 00( 1, 1)ij i j

σ

σ

≠⎧⎪⎨ = ≠ ≠⎪⎩

or correspondingly 1 00( 1)i i

σ

σ

≠⎧⎨

= ≠⎩

Now we calculate the strain 11ε and 1ε

11 11 1111 11pg pqS Sε σ σ= =

or

1 1 11 1i is sε σ σ= =

Therefore 1111S = 11s

(2) Assume the following strain state:

( )⎩⎨⎧

≠=

≠=

32or 23 003223

ijijε

εε or correspondingly

( )⎩⎨⎧

≠=

≠=

4 002 234

iiε

εε

Now we calculate the stress 11σ and 1σ

11 11 1123 23 1132 32 1123 232pg pqC C C Cσ ε ε ε ε= = + =

or

1 1 14 4 14 232i ic c cσ ε ε ε= = =

So 1123C = 14c

Then we calculate the stress 23σ and 4σ


23 23 2323 23 2332 32 2323 232pg pqC C C Cσ ε ε ε ε= = + =

or

4 4 44 4 44 232i ic c cσ ε ε ε= = = .

So 2323C = 44c

Assume the following stress state:

23 32

0( 1, 1)ij i jσ σ

σ

=⎧⎪⎨ = ≠ ≠⎪⎩

or correspondingly 4 23

0( 1)i iσ σ

σ

=⎧⎨

= ≠⎩

Now we calculate the strain 11ε or 1ε

11 11 1123 23 1132 32 1123 232pg pqS S S Sε σ σ σ σ= = + =

or

1 1 14 4i is sε σ σ= =

So 11232S = 14s

Then we calculate the strain 23ε or 4ε

23 23 2323 23 2332 32 2323 232pg pqS S S Sε σ σ σ σ= = + =

or

4 4 44 4i is sε σ σ= = .

So 23234S = 44s

Answer 3ii)

Suppose there is a uniaxial stress in m=miei direction denoted by ( )mσ , then

( ) ( )qp

mqii

mppq mmeeee σσσ =⋅⋅=

)(

( )qpijpq

mpqijpqij mmSS σσε ==


Since ( ) ( )qpjiijpq

mjiij

m mmmmSmm σεε ==

Divide both sides by ( )mσ we have

( ) qpjiijpq mmmmSmE

=1

Answer 3iii)

From “Introduction to solid state physics 8th ed” (p79), we know that for cubic symmetric crystal,

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

6

5

4

3

2

1

44

44

44

111212

121112

121211

6

5

4

3

2

1

000000000000000000000000

σ

σ

σ

σ

σ

σ

ε

ε

ε

ε

ε

ε

ss

ssssssssss

Hence we know the elements in ijpqS

11sSiiii =

)(12 jisSiijj ≠=

)(41

44 jisSijij ≠=

and all the other terms in ijpqS are 0.

We use the results from Answer 3ii), we can get

( )( ) ⎟

⎠

⎞⎜⎝

⎛ ++⎟⎠

⎞⎜⎝

⎛ −−++= 441244121143

42

41 2

1211 sssssmmm

mE

Answer 3iv)

This is equivalent to calculate the maximum or minimum value of )( 43

42

41 mmm ++

under the condition of

123

22

21 =++ mmm


Using change of variable 2i in m= and Lagrange multiplier λ we have

( ) ( )2 2 21 2 3 1 2 3 1 2 3, , , 1f n n n n n n n n nλ λ= + + − + + −

( ) 1 2 32 0 1,2,3ii

f n i n n nn

λ∂

= − = = ⇒ = =∂

( )1 2 3 1 2 3 1 2 31 11 03 3

f n n n n n n m m mλ∂

= − + + − = ⇒ = = = ⇒ = = = ±∂

In this case 4 4 41 2 3 1/ 3m m m+ + = .

We need also check the boundary conditions when one of 321 ,, mmm is equal to

1± , and the other two are 0. It turns out that

4 4 41 2 3 1m m m+ + =

Thus we get the maximum value of )( 43

42

41 mmm ++ is 1 and the minimum is 1/3.

Therefore, we can get the maximum or minimum value of E(m) when

(1) One of 321 ,, mmm is equal to +/-1, and the other two is equal to 0. That’s the

<100> directions.

(2) 31

321 ±=== mmm , that’s the <111> directions.

Whether this gives the maximum or minimum of E(m) depends on the sign of

441211 21 sss −− .

If 021

441211 >−− sss , (1) gives the minimum, and (2) gives the maximum.

If 021

441211 <−− sss , (1) gives the maximum, and (2) gives the minimum.

If 021

441211 =−− sss , E(m) is equal in all directions.

4. Invariants of a tensor When the basis changes, the components of a vector change, but the length of


the vector is invariant. Let f be a vector, and if be the components of the vector

for a given basis. The length of the vector is the square root of

.

The index i is dummy. Thus, this combination of the components of a vector is a scalar, which is invariant under any change of basis. For a vector, there is only one independent invariant. Any other invariant of the vector is a function of the length of the vector. This observation can be extended to high-order tensors. By definition, an invariant of a tensor is a scalar formed by a combination of the components of

the tensor. For example, for a symmetric second-rank tensor , we can form

three independent invariants:

kijkijijijii σσσσσσ ,, .

In each case, all indices are dummy, resulting in a scalar. Any other invariant of the tensor is a function of the above three invariants. Actions:

(a) For a nonsymmetric second-rank tensor, give all the independent invariants. Write each invariant using the summation convention, and then write it explicitly in all its terms.

(b) Give all the independent invariants of a third-rank tensor. Use the summation convention.

Answer 4a) An invariant of a tensor is a scalar formed by a combination of the components of the tensor. So the number of indices included in the expression of each invariant should be even and should not exceed 6 (we only have i, j, k three indices, at most each of them can repeat once). For a nonsymmetric second-rank tensor,

kijkij

ijij

ii

σσσ

σσ

σ

will still be the invariants

Moreover, as jiij σσ ≠ , we have additional invariants as

jiijσσ

kijkji σσσ

For the 3 σ case we notice that each σ should have two different indices in order

to be independent of iiσ . In ij jk kiσ σ σ take i for example, it is the 1st index in ijσ


and the 2nd index in kiσ , so do j and k. In kijkji σσσ both j’s are the 1st index and

both i’s are the 2nd index and one k is the 1st index and one k is the 2nd index. If we

examine ji jk ikσ σ σ , we can see it is actually the same situation of kijkji σσσ . We

cannot do any better than this. Therefore we have only 5 invariants for a nonsymmetric second-rank tensor. If we express the stress tensor as

11 12 31

12 22 23

31 32 33

ij

σ σ σ

σ σ σ σ

σ σ σ

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

We get:

11 22 33iiσ σ σ σ= + +

11 22 33 12 13 21 23 31 32

2 2 2 2 2 2 2 2 2ij ijσ σ σ σ σ σ σ σ σ σ σ= + + + + + + + +

11 22 33 12 21 23 32 31 13

2 2 2 2 2 2ij jiσ σ σ σ σ σ σ σ σ σ σ= + + + + +

kijkij σσσ …

kijkji σσσ …

Answer 4b)

Assume the third-order tensor is ijkd . There are two classes for counting the

invariants.

Class (1): )(ijkqijk dd

in which )(ijkq means the permutations of i, j, and k.

Since

q(ijk)= ijk , jki , kij , ikj , kji , or jik (n=1~6)

There are 6 invariants in this class.

Class (2): )()( jkqijp dd

in which

p(ij)= iij , iji , or jii

q(kj)= kkj , kjk , or jkk


There are 9 combinations of p and q. Since for example jkjiijjjkiji dddd = , we have to

subtract the double counted situations and we are left with 6 invariants in this class.

In conclusion the total number of invariants of ijkd is 12.


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