Solution

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Solutions to Problems

in

Fundamentals of

Structural Mechanics

An instructor’s manual to accompany the book

Fundamentals of Structural Mechanics

Keith D. Hjelmstad

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Solutions to Problems

in

Fundamentals of

Structural Mechanics

An instructor’s manual to accompany the book

Fundamentals of Structural Mechanics

by Keith D. Hjelmstad

Keith D. Hjelmstad

University of Illinois at Urbana-Champaign

With assistance from

Ghadir Haikal, Dan Turner, and

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Structural Mechanics

An instructor’s manual to accompany the book Fundamentals of Structural Mechanics by Keith D. Hjelmstad

by

Keith D. Hjelmstad

University of Illinois at Urbana-Champaign

With assistance from

Ghadir Haikal, Dan Turner, and Cara (Liverant) Phillips

E

2007 by Keith D. Hjelmstad

All rights reserved. No part of this document may be translated or reproduced in any form

without the written permission of the author.

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Preface

vii

1 Vectors and Tensors

. . . .

1

2 The Geometry of Deformation

. . . .

39

3 The Transmission of Force

. . . .

77

4 Linear Elastic Constitutive Theory

. . . .

95

5 Boundary Value Problems in Elasticity

. . . .

129

6 The Ritz Method of Approximation

. . . .

141

7 The Linear Theory of Beams

. . . .

173

8 The Linear Theory of Plates

. . . .

239

9 Energy Principles and Static Stability

. . . .

263 10 Fundamental Concepts in Static Stability

. . . .

285 11 The Planar Buckling of Beams

. . . .

309 12 Numerical Computation for Nonlinear Problems

. . . .

359

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Preface

A significant part of the book Fundamentals of Structural Mechanics lives in the problems at the back of each chapter. No student can hope to master the material covered in the book without solving at least some of these problems. An instructor who uses the book as a text for a course on structural mechanics may want to use these problems as a source of exam-ples to illuminate points in the text or as a source of homework problems for the student to work out. This manual should relieve some of the burden of preparation of course mate-rials.

This manual provides detailed solutions to almost all of the 302 problems in

Funda-mentals of Structural Mechanics 2e. The style in which the solutions are presented is

simi-lar to the examples presented in the book, with narrative to help get from one step to the next as well as some discussion of the results. The notation agrees with that used in the book.

In solving the problems it became quite apparent that a program like MATHEMATICAT

is indispensable in carrying out some of the more tedious computations. For those prob-lems that we used MATHEMATICATwe have provided a listing of the script. Most of these

are fairly short, but might be useful to someone who is not very familiar with MATHEMATI -CAT.

The problems in Chapter 12, in almost every case, involve modifying the programs presented in the book. When I revised the book for the second edition I converted all fo the programs from FORTRAN to MATLAB. However, I have not found time to change the problem solutions. Therefore, those are still in FORTRAN. I have included the FOR-TRAN versions from the first edition of the book at the back so that you can follow the solutions presented. You can get the FORTRAN source, for the programs in the text as well as the programs in the solutions manual, from me. Make your request by e-mail to: [email protected]. I will send you the files.

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We have tried very hard to get the answers right and to get them into this book with the highest possible fidelity. Alas, I am sure some errors remain. Please forgive those er-rors. I would appreciate learning of the errors that you discover so that I can eliminate them in future editions.

Putting this manual together was hard work, the burden of which I gladly shared with Ghadir Haikal, Cara (Liverant) Phillips, and Dan Turner. Ghadir managed to locate and organize the solutions to quiz problems that I had collected over the years. She then com-pleted Chapters 1 through 5 while I did 6 through 12. The problems from the first edition were available to us as were the problems that Cara had typset from old hand-written ex-amination solutions a few years ago. Dan proofread a large portion of the document. I would also like to (re)acknowledge the contributors to the solutions manual for the first edition: Ertugrul Taciroglu, Jiwon Kim, Eric Williamson, and Quihai (Ken) Zuo. The problems they worked out are still correct! I am pleased to acknowledge the contributions of Ghadir, Dan, Cara, Ertugrul, Jiwon, Eric and Ken to the preparation of this manual.

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Chapter 1 Vectors and Tensors

1. Compute the values of the following expressions

(a) δii

(b) δijδij

(c) Cijδikδjk

(d) δabδbcδcd... δxyδyz(enough terms to exhaust the whole alphabet)

(a) δii. Compute the value by summing on the repeated index i as

δii= δ11+δ22+δ33= 3.

(b) δijδij. The two indices are repeated. Contract first on index j, then on index i.

δijδij= δii= 3.

(c) Cijδikδjk. The three indices are repeated. Contract first on index i, then on index j,

and finally on index k.

Cijδikδjk= Ckjδjk= Ckk= C11+C22+C33.

(d) δabδbcδcd... δxyδyz(enough terms to exhaust the whole alphabet). All of the

indices are repeated except for a and z. Contract first on index b, then on index c, etc. until all indices have been summed

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2. Let two vectors, u and v, have components relative to some basis as u = (5, −2, 1)

and v = (1, 1, 1). Compute the lengths of the vectors and the angle between them. Find the area of the parallelogram defined by u and v.

The length of a vector can be computed as

‖ u ‖= u ⋅ u = u



iui.

Thus, the lengths of the vectors can be compute as

‖ u ‖ = 25+4+1 = 30 ‖ v ‖ = 1+1+1 = 3

The dot product of the vectors is

u ⋅ v = 5−2+1 = 4

cos θ(u, v) = u ⋅ v∕ ‖ u ‖‖ v ‖ = 4∕ 30  , θ = 1.136 rad3

The area of the parallelogram can be computed either of two ways:

A = ‖ (u × v) ‖ = ‖ (−3, −4, 7) ‖ = 9+16+49 = 8.6023

A = ‖ (u × v) ‖ = ‖ u ‖‖ v ‖ sin θ = 30  sin(1.136 rad) = 8.60233

3. The vertices of a triangle are given by the position vectors a, b, and c. The components

of these vectors in a particular basis are a = (0, 0, 0), b = (1, 4, 3), and c = (2, 3, 1). Using a vector approach, compute the area of the triangle. Find the area of the triangle projected onto the plane with normal n = (0, 0, 1). Find the unit normal vector to the triangle. (a) Using a vector approach, compute the area of the triangle. The area of the original triangle can be computed as

A1 = 1₂‖ (c × b) ‖ = 1₂‖ (5, −5, 5) ‖ = 1₂25+25+25 = 4.330

(b) Find the area of the triangle projected onto the plane with normal n = (0, 0, 1). The projected vectors can be found as follows

c′ = c − (c ⋅ n)n = (2, 3, 1) − 1(0, 0, 1) = (2, 3, 0) b′ = b − (b ⋅ n)n = (1, 4, 3) − 3(0, 0, 1) = (1, 4, 0)

The area of the projected triangle can be computed as

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c a x₁ x₂ x₃ b d x1 x2 x3 a b c b′ c′ n A1 A2

(c) Find the unit normal vector to the triangle.

n = ac × ab

‖ ac × ab ‖

where ab is the vector pointing from a to b. Noting that a is located at the origin of the coordinate system, we find

ab = b = (1, 4, 3) ac = c = (2, 3, 1) ac × ab = 2, 3, 1 ×  1, 4, 3 = 5, −5, 5 ; ‖ ac × ab ‖= 5 3

⇒ n = 1 3

 (1, − 1, 1)

4. Let the coordinates of four points a, b, c and d be given by

the following position vectors a=(1, 1, 1), b=(2, 1, 1), c=(1, 2, 2), and d=(1, 1, 3) in the coordinate system shown. Find vectors normal to planes abc and bcd. Find the angle between those vec-tors. Find the area of the triangle abc. Find the volume of the tet-rahedron abcd.

(a) Find vectors normal to planes abc and bcd.

nabc = ab × ac = b − a × c − a = (1, 0, 0) × (0, 1, 1) = (0, −1, 1)

nbcd = bc × bd = c − b × d − b = (−1, 1, 1) × (− 1, 0, 2) = (2, 1, 1)

(b) Find the angle between those vectors.

cos θ = nabc⋅ nbcd

‖ nabc‖‖ nbcd‖

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(c) Find the area of the triangle abc.

Aabc = ‖ nabc‖ = 2

(d) Find the volume of the tetrahedron abcd.

Vabcd = 1₆ab × ac ⋅ ad = 1₆(0, −1, 1) ⋅ (0, 0, 2) =1₃

5. Demonstrate that (u × v) ⋅ w = uivjwkεijkfrom basic operations on the base vectors.

Equality can be demonstrated by simply substituting the component forms of the vectors into the triple scalar product, collecting the scalars to the front and recognizing the definition of the components of the permutation tensor. To wit,

(u × v) ⋅ w = (uiei× vjej) ⋅ wkek = uivjwk



(ei× ej) ⋅ ek



= uivjwkÁijk

6. Show that the triple scalar product is skew-symmetric with respect to changing the

or-der in which the vectors appear in the product. For example, show that (u × v) ⋅ w = − (v × u) ⋅ w

To generalize this notion, any cyclic permutation (e.g., u, v, w → w, u, v) of the order of the vectors leaves the algebraic sign of the product unchanged, while any acyclic permuta-tion (e.g., u, v, w → v, u, w) of the order of the vectors changes the sign. How does this observation relate to swapping rows of a matrix in the computation of the determinant of that matrix?

For the case given the demonstration is straightforward as, by definition of the cross product, the vector u × v = −(v × u), i.e., a vector of the same magnitude, pointing in the opposite direction. The negative sign persists in the triple product. To see the more general case, consider the component form of the triple scalar product

(u × v) ⋅ w = Áijkuivjwk

To swap an entry, for example v and w we would have

(u × w) ⋅ v = Áijkuiwjvk = Áikjuivjwk

Clearly, then, swapping entries in the triple scalar product is the same as swapping indices on the permutation tensor. A cyclic permutation of indices leaves the sign of

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the product unchanged while an acyclic permutation changes the sign. Two successive acyclic permutations change the sign and then change it back and hence must be equiv-alent to a single cyclic permutation. The determinant of a matrix and the triple scalar product are equivalent because

u × v) ⋅ w = det

u v w

where the components of u, v, and w comprise the rows of the matrix. The effects of permutation of rows of the matrix on the determinant are clearly the same as swapping elements of the triple scalar product.

7. Use the observation that ‖ u−v ‖2_{= u−v  ⋅ u−v along with the distributive law}

for the dot product to show that

u ⋅ v ≡ 1₂



‖ u ‖2_{+ ‖ v ‖}2_{− ‖ v−u ‖}2

_

The solution depends upon the computation of the length of u−v

‖ u − v ‖2_{= u−v  ⋅ u−v}

= u ⋅ u − 2u ⋅ v + v ⋅ v = ‖ u ‖2_{− 2u ⋅ v + ‖ v ‖}2

Rearrange to give desired result. Note that ‖ u−v ‖ = ‖ v−u ‖.

8. Prove the Schwarz inequality, |u ⋅ v| ≤ ‖ u ‖ ‖ v ‖. Try to prove this inequality

with-out using the formula u ⋅ v =‖ u ‖‖ v ‖ cos θ(u, v).

Consider any two vectors u ≠ 0 and v ≠ 0 (equality holds if either one of the vectors is zero). The length of any vector is greater than or equal to zero.

0 ≤ ‖ αu+β v ‖2_{= α u+βv ⋅ αu+βv }

= α2_{‖ u ‖}2_{+ 2α βu ⋅ v + β}2_{‖ v ‖}2

Let α =‖ v ‖2_{and β = −u ⋅ v. Making this specialization gives}

0 ≤ ‖ αu+β v ‖2_{= ‖ v ‖}2_{‖ v ‖}2_{‖ u ‖}2_{− 2 ‖ v ‖}2_{u ⋅ v}2_{+ u ⋅ v}2_{‖ v ‖}2

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Since ‖ v ‖2_{≥ 0 we can conclude that}

0 ≤ ‖ v ‖2_{‖ u ‖}2_{− u ⋅ v }2 _{⇒ ‖ v ‖}2_{‖ u ‖}2_{≥ u ⋅ v}2

9. Show that [u  v]T_{= v  u using the definition of the transpose of a tensor and by}

demonstrating that the two tensors give the same result when acting on arbitrary vectors

a and b.

A tensor A has transpose AT_{defined by a ⋅ Ab = b ⋅ A}T_{a for any vectors a and b.}

Let A ≡ u  v and proceed as follows:

a ⋅ u  vb = a ⋅  b ⋅ vu

= a ⋅ ub ⋅ v = b ⋅ va ⋅ u = b ⋅ a ⋅ uv = b ⋅ v  ua

By definition tensor product of vectors Dot product is a scalar

Scalars are commutative Dot product is a scalar

By definition tensor product of vectors

a

= b ⋅ TT_a

Definition of tensor product Scatter scalars

Rename dummy indices Definition of tensor transpose

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By identification with the penultimate line of the derivation we can observe that the components satisfy [TT_]

kj = [T]jk.

11. Show that the tensor TT_{T is symmetric.}

A tensor A is symmetric if u ⋅ Av = v ⋅ Au for any vectors u and v. Let A ≡ TT_T

and proceed as follows:

u ⋅



TT_T

_

_{v = u ⋅ T}T_Tv

= Tu ⋅ Tv

By definition of composition of tensors By definition of transpose of tensor

= TT_{Tu) ⋅ v}

=



TT_T



_{u ⋅ v}

= v ⋅



TT_T

₌1

6TmiTnjTpkÁijkÁmnp

=1₆TmiTnjTpkÁmnpÁijk

Changing the dummy index i to m, j to n, k to p, and vice--versa ⇒ det



TT

_

₌1

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(b) detST = detS detT. Recall that

u × v) ⋅ w = det

u v w

Therefore, we can compute the determinant of a tensor S as det(S) = det

S₁ S2

S₂ = S1× S2) ⋅ S3

where the components of a vector Siare



Si]j= Sij. Thus, we can compute the

deter-minant of S as

detS = S1iS2jS3kÁijk= S1iS2jS3kÁijk

detT = T1iT2jT3kÁijk= T1iT2jT3kÁijk

detS detT = S1iS2jS3kÁijkdetT

Note that

ÁijkdetT = ÁijkT1pT2mT3nÁpmn= TipTjmTknÁpmn

⇒ detS detT = S1iS2jS3kTipTjmTknÁpmn

= (S1iTip) (S2jTjm)(S3kTkn)Ápmn

= [ST]1p[ ST]2m[ST ]3nÁpmn= detST

(c) [ST]T_{= T}T_ST_{. From the definition of the product of two tensors}

ST = SikTkj[ei ej] [ST]T _{= S} ikTkj[ej ei] = SjkTki[ei ej] Similarly, TT_ST _{= [T}T_] ik[ ST]kj[ei ej] = TkiSjk[ei ej] = SjkTki[ei ej] = [ST]T

(d) [ST]−1_{= T}−1_S−1_{. From the definition of the inverse of a tensor, for this}

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[ST]−1_{[ST] = [ T}−1_S−1_{][ST] = I}

But,

[T−1_S−1_{][ST ] = T}−1_[S−1_{S] T = T}−1_{[I] T = T}−1_{T = I}

⇒ [ST]−1_{= T}−1_S−1

13. Consider two Cartesian coordinate systems, one with basis {e₁, e₂, e₃} and the other with basis {g₁, g₂, g₃}. Let Qij≡ gi⋅ ejbe the cosine of the angle between giand ej.

(a) Show that gi= Qijejand ej= Qijgirelate the two sets of base vectors.

(b) We can define a rotation tensor Q such that ei= Qgi. Show that this tensor can be

expressed as Q ≡ Qij[gi gj], that is, Qijare the components of Q with respect to

the basis [gi gj]. Show that the tensor can also be expressed in the form

Q = [ei gi].

(c) We can define a rotation tensor QT_{, such that g}

i= QTei(the reverse rotation from

part (b). Show that this tensor can be expressed as QT_{≡ Q}

ij[ej ei], that is, Qijare

the components of QT_{with respect to the basis [e}

j ei]. Show that the tensor can also

be expressed in the form QT_{= [g} i ei].

(d) Show that QT_{Q = I, which implies that the tensor Q is orthogonal.}

(a) Show that gi= Qijejand ej= Qijgirelate the two sets of base vectors.

Qik = gi⋅ ek = Qijej⋅ ek = Qijδjk = Qik

Qkj = gk⋅ ej = gk⋅ Qijgi = Qijδki = Qkj

(b) We can define a rotation tensor Q such that ei= Qgi. Show that this tensor can be

expressed as Q ≡ Qij[gi gj], that is, Qijare the components of Q with respect to the

basis [gi gj]. Show that the tensor can also be expressed in the form Q = [ei gi].

Qgk = Qij



gi gj



gk= Qij



gj⋅ gk



gi = Qijδjkgi = Qikgi = ek

Qgk = ei gigk= gi⋅ gkei = δikei = ek

(c) We can define a rotation tensor QT_{, such that g}

i= QTei(the reverse rotation from

part (b). Show that this tensor can be expressed as QT_{≡ Q}

ij[ej ei], that is, Qijare

the components of QT_{with respect to the basis [e}

j ei]. Show that the tensor can also

be expressed in the form QT_{= [g} i ei].

QT_e

k = Qij



ej ei



ek= Qijei⋅ ekej = Qijδikej = Qkjej = gk

QT_e

k = gi eiek= ei⋅ ekgi = δikei = ek

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QT_{Q = g}

i ei[ej gj] =



ei⋅ ej



[gi gj] = δij[gi gj] = [gi gi] = I

14. The components of tensors T and S and the components of vectors u and v are

u ~ 1 1 2 v ~ 1 1 1 S ~ 0 --2 1 2 0 --1 --1 1 0 T ~ 12 20 01 0 1 2

Compute the components of the vector Su. Find the cosine of the angle between u and Su. Compute the determinants of T, S, and TS. Compute TijTijand uiTikSkjvj.

(a) Compute the components of the vector Su. = 1 1 2 Su ~ 0 --2 1 2 0 --1 --1 1 0 0 0 0 (b) Find the cosine of the angle between u and Su.

Since Su has zero length, the angle between u and Su cannot be determined. (c) Compute the determinants of T, S, and TS.

det (T) = --9; det (S) = 0; det(TS) = det(T)det(S) = 0

(d) Compute TijTijand uiTikSkjvj. Note that TijTij= tr



T TT



= tr



T2



since T is

symmetric 1 2 0 2 0 1 0 1 2 T2_~ 1₂ 2₀ 0₁ 0 1 2 = 5 2 2 2 5 2 2 2 5 ⇒ TijTij= tr



T2



= 15 Similarly uiTikSkjvj= u ⋅ TSv 0 --2 1 2 0 --1 --1 1 0 TSv ~ 12 20 01 0 1 2 1 1 1 = 1 --2 1 uiTikSkjvj= u ⋅ TSv = 1

15. Verify that, for the particular case given here, the components of the tensor T and the

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2 --1 0 --1 2 --1 0 --1 2 T ~ T−1 _~ 1 4 3 2 1 2 4 2 1 2 3

That this is actually the inverse can be verified by multiplying the components of the tensor and its inverse together to get the identity

1 4 3 2 1 2 4 2 1 2 3 2 --1 0 --1 2 --1 0 --1 2 1 0 0 0 1 0 0 0 1 =

16. Consider two bases: e1, e2, e3 and g1, g2, g3. The basis g1, g2, g3 is given in

terms of the base vectors  e1, e2, e3 as

g₁= 1

3

 e1+e2+e3, g2=_1₆ 2e1−e2−e3, g3=_1₂e2−e3

The components of the tensor T and vector v, relative to the basis  e₁, e₂, e₃ are

T ~ 0 --11 0 --11

--1 1 0

v ~ 12 3

Compute the components of the vector Tv in both bases. Compute the nine values of

TijTjkTkl(i.e., for i, l = 1, 2 ,3). Find the components of the tensor [T+TT]. Compute Tii.

(a) Compute the components of the vector Tv in both bases. In the basis  e₁, e₂, e₃

Tv = = --21 1 1 2 3 0 --1 1 1 0 --1 --1 1 0

Let bi= Tvibe the i--th component of Tv relative to the basis  g1, g2, g3

bi = Tv ⋅ gi = (e1--2e2+e3) ⋅ gi ⇒ b1= 1 3  -- 2₃+ 2 =₃ 0 b₂= 2 6  + 26-- 1 =6 3 6  b3= -- 2 2  -- 1 =₂ -- 3₂

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(b) Compute the nine values of TijTjkTkl(i.e., for i, l = 1, 2 ,3). --2 1 1 1 --2 1 1 1 --2 TijTjk = 0 --1 1 1 0 --1 --1 1 0 0 --1 1 1 0 --1 --1 1 0 =



TijTjk



Tkl= 0 --1 1 1 0 --1 --1 1 0 0 3 --3 --3 0 3 3 --3 0 = --2 1 1 1 --2 1 1 1 --2

(c) Find the components of the tensor [T+TT_].



T + TT

_

₌ 0 --1 1 1 0 --1 --1 1 0 0 0 0 0 0 0 0 0 0 = 0 1 --1 --1 0 1 1 --1 0 + (d) Compute Tii. Tii = T11+ T22+ T33 = 0

17. Consider two bases: e₁, e₂, e₃ and g₁, g₂, g₃, where

g1= e1+e2+e3, g2= e2+e3, g3= e2−e3

Compute Qijfor the given bases. Compute the value of QikQjk. Explain why the identity

QikQjk= δijdoes not hold in this case.

Now consider a vector v = e1+2e2+3e3and a tensor T given as

T = e2 e1−e1 e2 + e3 e1−e1 e3 + e3 e2−e2 e3

Compute the components of the vector Tv in both bases, i.e., find viand v^iso that the

fol-lowing relationship holds Tv = viei= v^igi. Find the cosine of the angle between the

vec-tor v and the vecvec-tor Tv. Find the length of the vecvec-tor Tv.

(a) Compute Qijfor the given bases. Recall that Qij= gi⋅ ejor the ij--th component

of Q is the j--th component of giin the basis  e1, e2, e3

⇒ Qij=

1 1 1

0 1 1

0 1 --1

(b) Compute the value of QikQjk. Explain why the identity QikQjk= δijdoes not hold

in this case. 3 2 0 2 2 0 0 0 2 QijQkj = 1 1 1 0 1 1 0 1 --1 = 1 0 0 1 1 1 1 1 --1

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For the idendity QikQjk= δijto hold, the base vectors  g1, g2, g3 have to be

orthogo-nal and of unit length. It can be observed that g₁and g₂are not orthogonal, which leads to the non--zero off diagonal terms in Q. The diagonal entries in Q are not equal to 1 because the base vectors are not of unit length.

(c) Now consider a vector v = e1+2e2+3e3and a tensor T given as

T = e2 e1−e1 e2 + e3 e1−e1 e3 + e3 e2−e2 e3

Compute the components of the vector Tv in both bases, i.e., find viand v^iso that

the following relationship holds Tv = viei= v^igi.

In the basis e1, e2, e3 Tv = ₌ --5--2 3 1 2 3 0 --1 --1 1 0 --1 1 1 0 Let v^

i= Tvibe the i--th component of Tv relative to the basis  g1, g2, g3

v^ i = Tv ⋅ gi = (--5 e1--2e2+3e3) ⋅ gi ⇒ v^ 1= − 5 − 2 + 3 = − 4 v^ 2= − 2 + 3 = 1 v^ 3= − 2 − 3 = − 5

(d) Find the cosine of the angle between the vector v and the vector Tv.

cos θ(Tv, v) = Tv ⋅ v

‖ Tv ‖‖ v ‖ = −

5(1) − 2(2) + 3(3)

‖ Tv ‖ ‖ v ‖ = 0

(e) Find the length of the vector Tv.

‖ Tv ‖= Tv ⋅ Tv



= 38

18. A general nth-order tensor invariant can be defined as follows

fn(T) ≡ T_{i1 i2}T_{i2 i3}⋅⋅⋅ T_{in i1}

where {i1, i2, . . ., in} are the n indices. For example, when n = 2 we can use {i, j} to give

f₂(T) = TijTji; when n = 3 we can use {i, j, k} to give f3(T) = TijTjkTki. Prove that

fn(T) is invariant with respect to coordinate transformation.

Let the sets of indices { j₁, j₂, . . ., jn} and {k1, k2, . . ., kn} be identical to the set

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the tensor T from one coordinate system to another as T^ij= QimQjnTmn. Observe that

the Qijterms can be rearranged and the identity QikQij= δijcan be used pairwise to

eliminate the Qij. To wit,

fn(T) ≡ T_{i1 i2}T_{i2 i3}⋅⋅⋅ T_{in i1} = Q_{i1 j1}Q_{i2 k2}T^_{j1 k2}Q_{i2 j2}Q_{i3 k3}T^_{j2 k3}⋅⋅⋅ Qin jnQi1 k1T ^ jn k1 = Q_{i1 j1}Q_{i1 k1}T^_{j1 k2}Q_{i2 k2}Q_{i2 j2}T^_{j2 k3}⋅⋅⋅ Qin knQin jnT ^ jn k1 = δ_{j1 k1}T^_{j1 k2}δ_{j2 k2}T^_{j2 k3}⋅⋅⋅ δjn knT ^ jn k1 = T^_{k1 k2}T^_{k2 k3}⋅⋅⋅ T^_{kn k1}

Since the final expression does not depend upon the basis it is invariant.

19. Use the Cayley-Hamilton theorem to prove that for n ≥ 4 all of the invariants fn(T),

defined in Problem 18, can be computed from f1(T), f2(T), and f3(T).

First let n = 4, f4(T) = tr(T4). According to the Cayley--Hamilton theorem

T3_{− I}

TT2+ IITT − IIITT = 0

Operating on both sides with T

T4_{− I}

TT3+ IITT2− IIITT = 0

⇒ T4_{= I}

TT3− IITT2+ IIITT

Taking the trace of both sides leads to

f4



T) = tr



T4) = 2 ITIIIT− (IIT)2

Following a similar procedure we can find that

T5_{= I}

TT4− IITT3+ IIITT2

f5



T) = IIf4(T) = 2 I2TIIIT− II2TIT

By extension, it can be seen that for any n ≥ 4 all of the invariants fn(T) can be

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20. From any tensor T one can compute an associated deviator tensor Tdevwhich has the

property that the deviator tensor has no trace, i.e., trTdev = 0. Such a tensor can be

ob-tained from the original tensor T simply by subtracting α ≡13trT times the identity

from the original tensor, i.e., Tdev= T−αI. Show that trTdev = 0. Show that the

prin-cipal directions of Tdevand T are identical, but that the principal values of Tdevare reduced

by an amount α from those of the tensor T.

(a) Show that trTdev = 0.

trT′ = tr T−pI = trT − 3p = trT − trT = 0

(b) Show that the principal directions of Tdevand T are identical, but that the

princi-pal values of Tdevare reduced by an amount α from those of the tensor T. Let m and n

be a principal value and a principal direction of the stress tensor S. By definition, then

Sn = mn. Now S = S′+pI. Therefore, the eigenvalue problem can be recast as

S′+pIn = mn, ⇒ S′n = m−pn

By definition of the eigenvalue problem we can observe that n is, indeed, a principal direction of the deviatoric stress tensor S′ corresponding to principal value m−p.

23. Show that the determinant of the tensor T can be expressed as follows

detT = 1₃tr



T3

_

₋1

2ITtr



T2



+1₆IT3

where IT= trT = Tiiis the first invariant of T. Use the Cayley-Hamilton theorem.

The Cayley-Hamilton theorem says that

T3_{− I}

TT2+ IITT − IIITI = 0

where IT= trT, IIT, and IIIT= det T are invariants of T. Taking the trace of the

above expression gives tr(T3_{) − I}

Ttr(T2) + IITtr(T) − 3 det T = 0

Note that (trT)2_−tr(T2_{) = 2II}

T(one can show this identity with a component

com-putation). Substituting for IITin the above expression gives

tr(T3_{) − I}

Ttr(T2) +12



I2T−tr(T2)



IT− 3 det T = 0

Finally, rearranging the expression we get det T = 1₃tr(T3_{) −}1

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As an alternate approach we can observe that if the identity is proven in one coor-dinate system then it is proven in all coorcoor-dinate systems. Let us select the principal coordinates. Then tr(T3_{) = m}3

1+m32+m33and tr(T2) = m21+m22+m23. Now we can

com-pute = 1₃



m3 1+m32+m33



−₂1m1+m2+m3



m21+m22+m23



+1₆m1+m2+m33 = m1m2m3 = det T det T = 1₃tr(T3_{) −}1 2ITtr(T2) +1₆I3T

24. A certain state of deformation at a point in a body is described by the tensor T, having

the components relative to a certain basis of

3 --1 0

--1 5 1

0 1 2

T ~

Find the eigenvalues and eigenvectors of T. Show that the invariants of the tensor T are the same in the given basis and in the basis defined by the eigenvectors for the present case.

(a) Find the eigenvalues and eigenvectors of T.The coefficients of the characteristic equation are the invariants of the tensor: IT= 10, IIT= 29 and IIIT= 25 . Thus, the

characteristic equation is

−m3_{+ 10m}2_{− 29m + 25 = 0}

The three roots of the characteristic equation are m1=1.6228, m2=2.7261 and m3=

5.6511. The corresponding principal directions are

n1= 0.2482, 0.3419, −0.9064

n₂_{= 0.9064, 0.2482, 0.3419} n₃= −0.3419, 0.9064, 0.2482

(b) Show that the invariants of the tensor T are the same in the given basis and in the basis defined by the eigenvectors for the present case. The tensor T in the basis defined by eigenvectors is a diagonal matrix and its components are the eigenvalues them-selves. 1.6228 0 2.7261 0 5.6511

T ~

0 0 0 0

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IT = m1+ m2+ m3 = 10

IIT = m1m2+ m1m3+ m2m3 = 29

IIIT = m1m2m3 = 25

They are the same as the invariants computed above.

25. Find the tensor T that has eigenvalues m₁_{=1, m}₂_{=2, and m}₃_{=3 with two of the}

asso-ciated eigenvectors given by

n₁₌ 1

2

 e1+e2, n2=1₃−2e1+2e2+e3

Is the tensor unique (i.e., is there another one with these same eigenproperties)?

Since the eigenvalues are distinct we can find n3by taking the cross product of

the other two eigenvectors

n₃ = n1× n2 = _1₁₈e1−e2+4e3

Based on the spectral decomposition theorem in the text, T can be computed as

T =



3 i=1 mini ni In components we have T ~ 1₂ 11 11 00 0 0 0 + 2 9 4 --4 --2 --4 4 2 --2 2 1 + 3 18 1 --1 4 --1 1 --4 4 --4 16 = ₁₈1 28 --10 4 --10 28 --4 4 --4 52

The tensor T is unique by the spectral decomposition theorem.

26. Find the tensor T that has eigenvalues m₁=1, m2=3, and m3=3, with two of the

asso-ciated eigenvectors given by

n₁= 1

3

 e1+e2+e3, n2=_1₂−e2+e3

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The solution is similar to Problem 25. n₃ _{= n}₁_{× n}₂ ₌ 1 6   2e1−e2−e3 T =



3 i=1 mini ni T ~ 1₃ 11 11 11 1 1 1 + 3 2 0 0 0 0 1 --1 0 --1 1 + 3 6 4 --2 --2 --2 1 1 --2 1 1 = 1₆ 14--4 14--4 --4--4 --4 --4 14

The eigenvectors n₂and n₃are not unique because the eigenvalues are repeated. Any linear combination of n2and n3is an eigenvector. The tensor T, however, is unique.

27. A certain state of deformation at a point in a body is described by the tensor T, having

the components relative to a certain basis of

T ~ 10−2

14 2 14

2 --1 --16

14 --16 5

Let the principal values and principal directions be designated as m and n. Show that n1

= (--1, 2, 2) is a principal direction and find m1.The second principal value is m2=

9 × 10−2_{, find n}

2.Find m3and n3with as little computation as possible.

(a) Let the principal values and principal directions be designated as m and n. Show that n1= (--1, 2, 2) is a principal direction and find m1.Assume that n1is a principal

direction of T, with eigenvalue m₁× 10−2_{. Then}

= 5−m1 14−m1 2 14 2 −1−m1 − 16 14 − 16 10−2 2 −1 2 0 0 0

From the first equation we have−14 − m1 + 4 + 28 = 0, or m1= −18. To

verify, substitute into the second and third equations − 2 + 2−1 + 18 − 32 = 0 −14 − 2(16) + 25+18 = 0

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Therefore n1is a principal direction.

(b) The second principal value is m₂= 9 × 10−2_{, find n} 2. = (5−9) (14−9) 2 14 2 (−1−9) − 16 14 − 16 10−2 b 1 a 0 0 0 From the first and second rows

2a + 14b = −5 10a + 16b = 2

a = 1 b = −0.5

⇒ The corresponding eigenvector is n2=1₃2, 2−1.

(c) Find m₃and n₃with as little computation as possible. The eigenvalue m₃can be found from the first invariant of Lagrangian strain tensor and n₃can be found from the cross product of the two other principal directions.

IT = T11+ T22+ T33 = 18 = m1+ m2+ m3 = −18 + 9 + m3

The third eigenvalue is m₃= 27 × 10−2

n3 = n1× n2 = 1₃−2, 1, −2

28. The equation for balance of angular momentum can be expressed in terms of a tensor T and the base vectors eias ei×  Tei = 0(sum on repeated index implied). What

specif-ic conditions must the components of the tensor T satisfy in order for this equation to be satisfied?

Carry out the computation for the tensor S as follows

ek× Sij[ei ej]ek = Sijek× (ej⋅ ek)ei = Sijδjkek× ei

∴ ek× Sek = Sikek× ei= 0

If k = i then ek× ei= 0 ⇒ S11, S22, S33can be anything. For the rest of the double

summation we have

S₁₂(e₂× e1) + S13(e3× e1) + S21(e1× e2) + S23(e3× e2) +

S₃₁(e₁× e3) + S32(e2× e3) = 0

S₁₂(−e3) + S13(e2) + S21(e3) + S23(−e1) + S31(−e2) + S32(e1) = 0

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n _v

Since e1, e2, e3≠ 0, we must have

S₃₂= S23 S13= S31 S21= S12

29. The tensor R that operates on vectors and reflects them

(as in a mirror) with unit normal n is given by

R ≡ I−2 n  n

Compute the vector that results from [RR]v. Compute the

length of the vector Rv in terms of the length of v. What is the inverse of the tensor R? Compute the eigenvalues and eigenvectors of R.

(a) Compute the vector that results from [RR]v.

RRv = I − 2n  n I − 2n  n v = I − 2n  n v − 2(n ⋅ v)n

= v − 2(n ⋅ v)n − 2(n ⋅ v)n + 4[n  n]n(n ⋅ v) = v − 4(n ⋅ v)n + 4(n ⋅ n)n(n ⋅ v) = v (b) Compute the length of the vector Rv in terms of the length of v.

 Rv 2

= Rv ⋅ Rv = (v − 2(n ⋅ v) n) ⋅ (v − 2(n ⋅ v) n)

v ⋅ v − 4(n ⋅ v)2_{+ 4(n ⋅ v) = v ⋅ v =  v }2

∴  Rv  =  v  (c) What is the inverse of the tensor R?

RRv = I v = v ⇒ R--1 _{= R}

(d) Compute the eigenvalues and eigenvectors of R.

R = I − 2 n  n = I − n  n − n  n

From the spectral decomposition theorem, we have one distinct eigenvalue and a pair of repeated eigenvalues

m₁= 1; n1_{= n}

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30. Let v(x) and u(x) be two vector fields, and T(x) be a tensor field. Compute the

follow-ing expressions in terms of the components (vi, ui, and Tij) of these fields relative to the

basis  e1, e2, e3: divTv, ∇u ⋅ Tv, ∇Tv, u  Tv

div(x  x) = (xixj),jei = [xi,jxj+ xixj,j] ei = [δijxj+ 3 xj] ei = 4x

div div[x  x] = div(4x) = 4(xiei),j⋅ ej = 4 xi,jei⋅ ej = 4 δijδij = 12

(b) div





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div(x div x) = div(3x) = 3div x = 9 div





= div(9x) = 9 div x = 27 (c) ∇



‖



∇ ‖ x ‖2

_

_‖2



 x 2 = xixi ∇ x 2 = (xixi),jej = [xi,jxi+ xixi,j] ej = 2 δijxiej = 2x  2x 2 = 4  x 2 ∇ 2x 2 = 4 ∇  x 2 = 8x (d) div



x  divx  x



[x  x]ij = xixj div[x  x] = (xixj),jei = (xi,jxj+ xixj,j) ei = (δijxj+ xiδjj) ei = 4xiei = 4x

div



x  divx  x



= div[x  4x] = 4div[x  x] = 16x (e) ∇xdivx div x = xi,i = δii = 3 ∇(x div x) = ∇(3x) = 3∇x = 3I (f) ∇



x ⋅ ∇x ⋅ x



∇(x ⋅ x) = (xixi),jej = (xi,jxi+ xixi,j) ej = 2 xjej = 2x ∇(x ⋅ 2x) = 2∇(x ⋅ x) = 4x

32. Let v(x) = x2−x3 e1+ x3−x1 e2+ x1−x2 e3. Evaluate the following

expres-sions: ∇v, ∇x ⋅ v, div x  v, and ∇x × v, where x = xieiis the position vector.

Evaluate the expressions at the point x = e1+2e2+e3.

(a) ∇v

∇v = _--10 1 --1₀ ₁

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x₁ x₂ x₃ n B (b) ∇x ⋅ v x ⋅ v = (x1x2− x1x3) + (x2x3−x2x1) + (x3x1−x3x2) = 0 ∴∇(x ⋅ v) = 0 (c) div x  v x₃x₁--x₂ x₃x₂--x₃ x1x3--x1 x  v ~ _xx₂1__xx₂2_--x--x₃3_ _x₂__x₃_--x₁_ _xx₂1__xx₁1_--x--x₂2_ x₃x₃--x₁ x₁--x₂ div x  v ~ x₂--x₃ x₃--x₁ (d) ∇x × v x₁x₃--x₁− x2x2--x3 x × v ~ x₂x₁--x₂− x3x3--x1 x3x2--x3− x1x1--x2 x₁+x₂ --2 x₁ --2 x₂ ∇ x × v ~ x₂+x₃ --2 x₃ x₁+x₃ --2 x₁ --2 x₃ --2 x₂

(e) Evaluate the expressions at the point x = e1+2e2+e3.

3 --2 --4 ∇ x × v ~ 3 --2 2 --2 --2 --4 − 1 div x  v ~ 1 0

33. Let v(x) be given by the following explicit function v(x) =



x2

The vector field is defined on the spherical region B of unit ra-dius as shown in the sketch. Give an explicit expression for the

unit normal vector field n(x) to the surface of the sphere. Compute the gradient of the vec-tor field v(x). Compute the product [∇v]n, i.e., the gradient of the vecvec-tor field acting on the normal vector. Compute the divergence of the vector field v(x). Compute the integral of div v over the volume of the sphere. Compute the integral of v ⋅ n over the surface of the sphere.

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(a) Give an explicit expression for the unit normal vector field n(x) to the surface of the sphere.

n = x = xiei

(b) Compute the gradient of the vector field v(x). ∇v =∂v_∂xi j[ei ej] ~ x1 2x1 x₂ x2 x₃ x3 2x₂ x1 2x₃

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. [∇v] n ~ x₁ 2x₁ x₂ x₂ x₃ x₃ _2x₂ _x 1 2x₃ x₁ x₂ x₃ = 2 x2 1+ 2 x2x3 2 x2 2+ 2 x1x3 2 x2 3+ 2 x1x2

(d) Compute the divergence of the vector field v(x). div v = tr(∇v) = 2(x1+x2+x3)

(e) Compute the integral of div v over the volume of the sphere. Given the geometry of the problem, it would be best to compute the integral in spherical coordinates using the following transformations (see figure below)

x₁= r cos φ cos θ x₂= r cos φ sin θ x₃= r sin φ

dV = (r dφ )(r cos φ dθ)dr = r2_{cos φ dφ dθ dr} r*_{d θ} dφ θ r *_{= r cos φ} d θ d r r dφ r z x1 φ x₂ x₃



B div v dV = 2



[cos3_φ_cos3_{θ + cos}3_φ_sin3_{θ + sin}3_φ

+ 3 cos2_φ_{sin φ cos θ sin θ] cos φ dθ dφ = 0}

34. Let v(x) be a vector field given by the

follow-ing explicit function

v(x) = x1e1+x2e2 ln



x21+x22



where ln(⋅) indicates the natural logarithm of (⋅). The vector field is defined on the cylindrical

re-gion B of height h and radius R as shown in the sketch. Give an expression for the unit normal vector field n(x) to the for the cylinder (including the ends). Compute the diver-gence of the vector field v(x) and the integral of div v over the volume of the cylinder.

(a) Give an expression for the unit normal vector field n(x) to the for the cylinder (including the ends).

nt= e3for the top surface

nb= − e3for the bottom surface

ns= 1R(x1e1+ x2e2) for the side surface

(b) Compute the divergence of the vector field v(x) and the integral of div v over the volume of the cylinder. Let r2_{≡ x}2

1+ x22. Since v has no components in the direction

of, we will assume that the index range is 1 to 2 instead of 1 to 3 for this problem. Thus, ∂r2 ∂xi= 2xi Note that vi= xi ln r2 ⇒ div v =∂v_∂xi i= δiiln r 2_{+ x} i_r1₂ ∂r 2 ∂xi= δiiln r 2_{+ 2}xixi r2 = 2 ln r2+ 2

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x₁ x₂ x₃ n B g(x) = 0 where a repeated index implies summation from 1 to 2 .

Computing the integral of div v over the volume would be most conveniently done in cylindrical coordinates. However, we can apply the divergence theorem to find the result more easily. Note that v ⋅ n = 0 on the top and bottom surfaces. On the side surface r2_{= R}2_{= const.} v ⋅ n = ln R2_(x 1e1+ x2e2) ⋅ 1R(x1e1+ x2e2) = R ln R2 ⇒



B div v dV =



Ω v ⋅ n dA =



Ωs v ⋅ nsdA = R ln R2As= 2 π R2h ln R2

35. Consider the scalar field g(x) = x ⋅ x2_{. Compute div}



_∇

_

_{div ∇g(x)}

_



_.

g(x) = (xixi)2 ∇g = 2(xixi) ∂_∂x j(xkxk)ej= 2(xixi)[δkjxk+ xkδkj]ej= 4xixixjej div(∇g) = ∂_∂x j(4xixixj) = 4[δijxixj+ xiδijxj+ xixiδjj] = 20 xixi ∇(div(∇g)) = ∂_∂x j(20 xixi)ej= 20[δijxi+ xiδij] ej= 40 xjej div(∇(div(∇g))) = ∂_∂x j(40 xj) = 40δjj= 120

36. Let v(x) be given by the following explicit function v(x) = x2+x3 e1+ x1+x3 e2+ x1+x2 e3

where x is the position vector of any point and has components x₁, x2, x3 relative to the Cartesian coordinate system as

shown. The vector field is defined on the ellipsoidal region B whose surface is described by the equation g(x) = 2x2

1+x22+2 x23−4 = 0. Give an

ex-pression for the unit normal vector field n(x) to the ellipsoid. Compute the gradient of the vector field v(x). Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. Compute the divergence of the vector field v(x).

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(a) Give an expression for the unit normal vector field n(x) to the ellipsoid. n = ∇_{ ∇g }g _{∇g ~} 4 x2 x12 4 x3  ∇g  = (4 x1) 2_{+(2 x} 2)2+(4 x3)2



(b) Compute the gradient of the vector field v(x).

∇v ~

0 1 1

1 0 1

1 1 0

(c) Compute the product [∇v]n, i.e., the gradient of the vector field acting on the normal vector. [∇v]n = 1_{ ∇g } 4x2x1₂ 4x3 2 x2+ 4 x3 4 x1+ 4 x3 4 x1+ 2 x2 0 1 1 1 0 1 1 1 0 = 1_{ ∇g } (d) Compute the divergence of the vector field v(x).

div v = tr(∇v) = 0

37. Evaluate the expression div



∇x ⋅ Ax



, where A is a constant tensor (i.e., it does not depend upon x), and the vector x has components x = xiei. The derivatives are to be taken

with respect to the independent variables xi. Express the results in terms of the components

of A and x. x ⋅ Ax = xiAijxj ∇g =_∂x∂g kek div v = ∂vk ∂xk ∇(x ⋅ Ax) = ∂_∂x k(xiAijxj)ek= (δijAijxj+ xiAijδjk) ek= (Akjxj+ xiAik)ek div(∇(x ⋅ Ax)) = ∂_∂x k(Akjxj+ xiAik) = Akjδjk+ δikAik= Akk+ Akk= 2Akk

38. Let g(x) = e−‖ x ‖ 2_{be a scalar field in three-dimensional space, where ‖ x ‖ is the}

dis-tance from the origin to the point x. Qualitatively describe the behavior of the function (a one- or two-dimensional analogy might be helpful). Compute the gradient ∇g of the field. Where does the gradient of the function go to zero?

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1 x₁ 2 3 x₂ x₃ Let r2_{≡‖ x ‖}2_{= x}

ixibe the radial distance from the origin. Note that



r2

_

_, j = 2rr,j = xi,jxi+ xixi,j = 2δijxi = 2xj ⇒ r,i=x_ri ∇g(x) = ∂g_∂r_∂x∂r iei = e −r2_(−2r)xi r ei = −2e−r2xiei = −2g(x)x

Therefore, the gradient of the given function is ∇g(x) = −2xe−‖ x ‖2_{. The function has}

value g(0) = 1 at the origin and decays exponentially to zero with distance from the origin. The function has a zero gradient at the origin.

39. Consider a tensor field T defined on a tetrahedral

B

div v dV =



Ω

v ⋅ n dA

Substituting with the results obtained above

(37)

41. The Laplacian of a scalar field is a scalar measure of the second derivative of the field,

defined as ∇2_{g(x) ≡ div∇g(x). Write the component (index) form of the Laplacian of}

g in Cartesian coordinates. Compute the Laplacian of the scalar field of Problem 38.

In components, ∇2_{g(x) ≡ div(∇g(x)) = g,}

ii. From Problem 38 note that

g,i= −2g(r)xi where r2≡‖ x ‖2= xixi.

Thus,

g,ii= −2g,ixi+ gxi,i = −2−2gxixi+ 3g =



4r2−6



g(x)

Therefore, the Laplacian is ∇2_{g(x) =}

_

_4r2₋₆

_

_e−‖ x ‖2

42. ComputedivT, where T(x) = x ⋅ x I−2x  x is a tensor field.

Tij = (xkxk) δij− 2 xixj

divT = ∂T_∂xij

j ei =



2xkδjkδij− 2 δijxj− 2 xiδjj



ei

=



2xjδij− 2 xjδij− 6 xi



,kek=



ui,kTijvj+ uiTij,kvj+ uiTijvj,k



ek

(f) ∇(Tv) =



Tijvj



,kei ek =



Tij,kvj+ Tijvj,k



ei ek

Áijkem

44. Use the same reasoning that was used to derive the three-dimensional version of the

divergence theorem to develop (a) a one-dimensional version

f (x) x a x_i−1 xi b n(xi) n(x_i−1) ∆xi

Noting that in one dimension n = 1 for the normal on the right side and n = −1 for the normal on the left side of the segment

div₁_{f (x) ≡ lim} ∆xi→0 1 ∆xif (xi) − f (xi−1)  ≡ df dx

Divide the region [a,b] into segments with endpoints defined by the x values xo, x1, x2, . . ., xn−1, xn

where xo≡ a and xn≡ b. Let us do the following simple computation



n i=1

f (xi) − f (xi−1) = f (x1) − f (a) + f (x2) − f (x1) + . . .

+ f (xn−1) − f (xn−2) + f (b) − f(xn−1) = f (b) − f (a)

Each term appears once with a plus sign and once with a minus sign except for the two end points. Define the “almost divergence” as follows

Di ≡ 1_∆x

if (xi) − f (xi−1)

Now we can write



n i=1 Di∆xi =



n i=1 f (xi) − f (xi−1) = f (b) − f (a) (*)

Solution

Solutions to Problems

in

Fundamentals of

Structural Mechanics

An instructor’s manual to accompany the book

Fundamentals of Structural Mechanics

Keith D. Hjelmstad

Solutions to Problems

in

Fundamentals of

Structural Mechanics

An instructor’s manual to accompany the book

Fundamentals of Structural Mechanics

by Keith D. Hjelmstad

Keith D. Hjelmstad

University of Illinois at Urbana-Champaign

With assistance from

Ghadir Haikal, Dan Turner, and

E

2007 by Keith D. Hjelmstad

All rights reserved. No part of this document may be translated or reproduced in any form

without the written permission of the author.

Contents

Preface

vii

1

Vectors and Tensors

. . . .

1

2

The Geometry of Deformation

. . . .

39

3

The Transmission of Force

. . . .

77

4

Linear Elastic Constitutive Theory

. . . .

95

5

Boundary Value Problems in Elasticity

. . . .

129

6

The Ritz Method of Approximation

. . . .

141

7

The Linear Theory of Beams

. . . .

173

8

The Linear Theory of Plates

. . . .

239

9

Energy Principles and Static Stability

. . . .

263

10 Fundamental Concepts in Static Stability

. . . .

285

11 The Planar Buckling of Beams

. . . .

309

12 Numerical Computation for Nonlinear Problems

. . . .

359

Preface

Chapter 1

Vectors and Tensors













_

_

_

_

_

_

_

_

_