Name: Cabrera, Ryan M.ES 230: Problem Set 4Professor: Dr. Publico, Andre S.

Note: Only Problem A has been typeset in LateX since I ran out of time in typesetting theremaining problems; hence, they are just presented in manual handwritten forms. Also, thedetailed manual calculations for Problem A is attached for reference.

Problem A: Naviers equations in rectangular Cartesian coordinates.

1. Write out the basic equations of linear elastostatics for plane strain and plane stressconditons. Assume isotropy and use the Lame constants and .

2. From A.1, derive the Naviers equations for plane strain and plane stress. Write theseresults in indicial notation and show that they are the same if, in the plane stresscondition, the expression 2

+2is replaced by .

Solution:

1. Recall the basic equations of linear elastostatics in Cartesian coordinates; they are asfollows:

the strain-displacement relations, we have

ij =1

2(ui,j + uj,i) (1)

the stress-strain relations (or constitutive equations) in terms of the Lame con-stants and , we have

ij = kkij + 2ij (2)

where ij is the Kronecker delta

from the equations of motion,

ij,j + fi = ai

we obtain the equations of equilibrium

ij,j + fi = 0 (3)

when

ai = 0

if vi = 0

if vi 6= vi(t)

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(a) For the plane strain condition (e.g. very long along the x3- direction), we havethe following assumptions:

u1 = u1(x1, x2) (4a)

u2 = u2(x1, x2) (4b)

u3 = 0 (4c)

Using the above assumptions, Eq. (4), for plane strain condition, the strain-displacement relations Eq. (1) become:

11 = u1,1 (5a)

22 = u2,2 (5b)

12 = 21 =1

2(u1,2 + u2,1) (5c)

i3 = 3i = 0, for i = 1, 2, 3 (5d)

It should be noted from the above strain-displacement relations for planestrain, the nonvanishing strains are 11, 22, and 12 = 21, while the vanishingstrains are i3 = 3i = 0, for (i = 1, 2, 3). Using these results, and noting theproperties of the Kronecker delta, ij, i.e.

ij =

{1 , if i = j0 , if i 6= j

the stress-strain relations Eq. (2) become, for plane strain, as follows:

11 = (11 + 22) + 211 (6a)

22 = (11 + 22) + 222 (6b)

33 = (11 + 22) (6c)

12 = 212 = 21 (6d)

i3 = 3i = 0, for i = 1, 2 (6e)

Next, to obtain the equations of equilibrium for plane strain, we note thefollowing nonvanishing stresses, e.g. 11, 22, 33, and 12 = 21; while thevanishing stresses are i3 = 3i = 0, for (i = 1, 2). Then we write out Eqs.(3) and the following are obtained:

11,1 + 12,2 + f1 = 0 (7a)

21,1 + 22,2 + f2 = 0 (7b)

f3 = 0 (7c)

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In addition to the basic equations of linear elastostatics, the following com-patibility equation for plane strain is obtained:

211x22

+222x21

= 2212x1x2

(8)

(b) For the plane stress condition (e.g. very short along the x3- direction, like thatof a thin plate), the following assumptions are made:

ij = ij(x1, x2), for i, j = 1, 2 (9a)

i3 = 3i = 0, for i, j = 1, 2, 3 (9b)

Since the assumptions are with the stresses ij, we then start our analysis withthe constitutive equations, i.e. the stress-strain relations Eq. (2), written outfor plane stress. Again, noting the properties of the Kronecker delta and theassumptions on the stresses Eq. (9), we obtained the following stress-strainrelations for plane stress:

11 = (11 + 22 + 33) + 211 (10a)

22 = (11 + 22 + 33) + 222 (10b)

12 = 212 = 21 (10c)

i3 = 3i = 0, for i = 1, 2, 3 (10d)

It should be noted that the nonvanishing stresses 11 and 22 contain 33which is related to 33 = 0; in particular, we have

33 = (11 + 22 + 33) + 233 = 0

which can be manipulated to obtain

33 = + 2

(11 + 22) (11)

then substituting this 33 into the equations for 11 and 22, i.e. Eqs. (10),we obtain the modified Eqs. (10) as follows:

11 =2

+ 2(11 + 22) + 211 (12a)

22 =2

+ 2(11 + 22) + 222 (12b)

12 = 212 = 21 (12c)

i3 = 3i = 0, for i = 1, 2, 3 (12d)

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wherein 11 and 22 are now in terms of 11 and 22 only.

Next, to obtain the strain-displacement equations, we note the vanishingstresses i.e. i3 = 3i = 0, for (i = 1, 2, 3), to obtain the vanishing strainsi3 = 3i = 0, for (i = 1, 2), since we already know that 33 is nonvanishingas given by Eq. (11). Thus, the strain-displacement Eqs. (1) for plane stressare obtained as follows:

11 = u1,1 (13a)

22 = u2,2 (13b)

33 = + 2

(u1,1 + u2,2) (13c)

12 = 21 =1

2(u1,2 + u2,1) (13d)

i3 = 3i = 0, for i = 1, 2 (13e)

Next, to obtain the equations of equilibrium, we note the vanishing stressesare the same i3 = 3i = 0, for (i = 1, 2, 3), then write out equations ofequilibrium Eqs. (3) and obtained the following:

11,1 + 12,2 + f1 = 0 (14a)

21,1 + 22,2 + f2 = 0 (14b)

f3 = 0 (14c)

which is actually the same to that of plane strain condition.

In addition, we also observed that compatibility equation for plane stress211x22

+222x21

= 2212x1x2

(15)

is the same to that of plane strain condition.

2. Using the results from A.1, the Naviers equations for plane strain and plane stressconditions can be obtained. This is done by converting the stress-strain equations(constitutive equations) into the stress-displacement equations through the strain-displacement equations; then, the stresses (now in terms of the displacements) areinserted into the equations of equilibrium. Thus obtaining the the equations of equi-librium in terms of displacements ui.

(a) For the plane strain condition, recall the following:

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the strain-displacement equations Eqs. (5)

11 = u1,1

22 = u2,2

12 = 21 =1

2(u1,2 + u2,1)

i3 = 3i = 0, for i = 1, 2, 3

the stress-strain equations Eqs. (6)

11 = (11 + 22) + 211

22 = (11 + 22) + 222

33 = (11 + 22)

12 = 212 = 21

i3 = 3i = 0, for i = 1, 2

Now, we can substitute these strain-displacement equations into the stress-strain equations to obtain the stress-displacement equations

11 = (u1,1 + u2,2) + 21,1 (18a)

22 = (u1,1 + u2,2) + 22,2 (18b)

33 = (u1,1 + u2,2) (18c)

12 = 21 = (u1,2 + u2,1) (18d)

i3 = 3i = 0, for i = 1, 2 (18e)

Now, recall the equations of equilibrium for plane strain Eq. (7)

11,1 + 12,2 + f1 = 0

21,1 + 22,2 + f2 = 0

f3 = 0

if we substitute the stress-displacement equations for plane strain, i.e. Eqs.(7), say for example, the first of these equilibrium equations, we have:

11,1 + 12,2 + f1 = 0

x1[(u1,1 + u2,2) + 2u1,1] +

x2[(u1,2 + u2,1)] + f1 = 0

(u1,11 + u2,21) + 2u1,11 + (u1,22 + u2,12) + f1 = 0

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Now, the trick here is letting 2u1,11 = u1,11 + u1,11 and noting that theorder by which differentiation is performed is immaterial, in particular, u21,2 =u2,21. Working out algebraically, then combining like terms, yields

(u1,11 + u1,22) + (+ )(u1,11 + u2,21) + f1 = 0

or in indicial notation, we have

ui,jj + (+ )uj,ji + fi = 0, for i, j = 1, 2 (21)

which is now the Naviers equations for plane strain condition.

(b) For the plane stress condition, recall the following:

the strain-displacement equations Eqs. (13)

11 = u1,1

22 = u2,2

33 = + 2

(u1,1 + u2,2)

12 = 21 =1

2(u1,2 + u2,1)

i3 = 3i = 0, for i = 1, 2

the modified stress-strain equations Eqs. (12), wherein 11 and 22 are interms of 11 and 22, we have

11 =2

+ 2(11 + 22) + 211

22 =2

+ 2(11 + 22) + 222

12 = 212 = 21

i3 = 3i = 0, for i = 1, 2, 3

Now, writing these stress-strain equations in terms of the displacements, wehave

11 =2

+ 2(u1,1 + u2,2) + 2u1,1 (24a)

22 =2

+ 2(u1,1 + u2,2) + 2u2,2 (24b)

12 = 21 = (u1,2 + u2,1) (24c)

i3 = 3i = 0, for i = 1, 2, 3 (24d)

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Now, substituting these stress-displacement equations Eqs. (24) into theequations of equilibrium for plane stress, i.e. Eqs. (14), and observing thesame procedure as what we did for the plane strain case, we obtain the fol-lowing:

ui,jj +

(2

+ 2+

)uj,ji + fi = 0, for i, j = 1, 2 (25)

now if we let

=2

+ 2(26)

then we have

ui,jj + (+ )uj,ji + fi = 0, for i, j = 1, 2 (27)

which is now the Naviers equations for plane stress. It should be noted thatEqs. (27) is the same as the Naviers equations for plane strain, i.e. Eqs.(21), if we let given by Eq. (26).

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