lecture 1 - УАСГw3.uacg.bg/uaceg_site/acadstaff/userfiles/study_en... · lecture 1 introduction...

Lecture 1IntroductionThe rules of the gamePrint version Lecture Theory of Elasticity and Plasticity of

Dr. D. Dinev, Department of Structural Mechanics, UACEG

1.1

Contents

1 Introduction 11.1 Elasticity and plasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Overview of the course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Course organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Mathematical preliminaries 62.1 Scalars, vectors and tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Index notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Kronecker delta and alternating symbol . . . . . . . . . . . . . . . . . . . . . . 82.4 Coordinate transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.5 Cartesian tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.6 Principal values and directions . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.7 Vector and tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.8 Tensor calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.2

1 Introduction

1.1 Elasticity and plasticity

Introduction

Elasticity and plasticity

• What is the Theory of elasticity (TE)?

– Branch of physics which deals with calculation of the deformation of solid bodies inequilibrium of applied forces

– Theory of elasticity treats explicitly a linear or nonlinear response of structure toloading

• What do we mean by a solid body?

– A solid body can sustain shear

– Body is and remains continuous during the deformation- neglecting its atomic struc-ture, the body consists of continuous material points (we can infinitely ”zoom-in”and still see numerous material points)

• What does the modern TE deal with?

– Lab experiments- strain measurements, photoelasticity, fatigue, material description

– Theory- continuum mechanics, micromechanics, constitutive modeling

– Computation- finite elements, boundary elements, molecular mechanics1.3

1

Introduction


• Which problems does the TE study?

– All problems considering 2- or 3-dimensional formulation1.4

Introduction


• Shell structures1.5

Introduction


• Plate structures1.6

Introduction

2

Elasticity and plasticity• Disc structures (walls)

1.7

Introduction

Mechanics of Materials (MoM)• Makes plausible but unsubstantial assumptions• Most of the assumptions have a physical nature• Deals mostly with ordinary differential equations• Solve the complicated problems by coefficients from tables (i.e. stress concentration fac-

tors)

Elasticity and plasticity• More precise treatment• Makes mathematical assumptions to help solve the equations• Deals mostly with partial differential equations• Allows us to assess the quality of the MoM-assumptions• Uses more advanced mathematical tools- tensors, PDE, numerical solutions

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1.2 Overview of the course

Introduction

Overview of the course• Topics in this class

– Stress and relation with the internal forces

– Deformation and strain

– Equilibrium and compatibility

– Material behavior

– Elasticity problem formulation

– Energy principles

– 2-D formulation

– Finite element method

– Plate analysis

– Shell theory

– Plasticity

Note• A lot of mathematics• Few videos and pictures

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3

1.3 Course organization

Introduction

Course organization

• Lecture notes- posted on a web-site: http://www.uacg.bg/UACEG_site/acadstaff/viewProfile.php?lang=en&perID=151

• Instructor

– Dr. D. Dinev- Room 514, E-mail: [email protected]

• Teaching assistant

– Dr. I. Kerelezova- Room 436

• Office hours

– Instructor: Mon: 10-11; Tue: 13-14

– TA: . . . . . . . . . . . .

Note

• For other time→ by appointment1.10

Introduction

Course organization

• Textbooks

– Elasticity theory, applications, and numerics, Martin H. Sadd, 2nd edition, Elsevier2009

– Energy principles and variational methods in applied mechanics, J. N. Reddy, JohnWiley & Sons 2002

– Fundamental finite element analysis and applications, M. Asghar Bhatti, John Wiley& Sons 2005

– Theories and applications of plate analysis, Rudolph Szilard, John Wiley & Sons2004

– Thin plates and shells, E. Ventsel and T. Krauthammer, Marcel Dekker 20011.11

Introduction

Course organization

• Other references

– Elasticity in engineering mechanics, A. Boresi, K. Chong and J. Lee, John Wiley &Sons, 2011

– Elasticity, J. R. Barber, 2nd edition, Kluwer academic publishers, 2004

– Engineering elasticity, R. T. Fenner, Ellis Horwood Ltd, 1986

– Advanced strength and applied elasticity, A. Ugural and S. Fenster, Prentice hall,2003

– Introduction to finite element method, C.A. Felippa, lecture notes, University of Col-orado at Boulder

– Lecture handouts from different universities around the world1.12

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Introduction

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GradePoints

Course organization

• Grading1.13

Introduction

Course organization

• Grading is based on

– Homework- 20%

– Two mid-term exams- 35%

– Final exam- 45%

• Participation

– Class will be taught with a mixture of lecture and student participation

– Class participation and attendance are expected of all students

– In-class discussions will be more valuable to you if you read the relevant sectionsof the textbook before the class time

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Introduction

Course organization

• Homeworks

– Homework is due at the beginning of the Thursday lectures

– The assigned problems for the HW’s will be announced via web-site

• Late homework policy

– Late homework will not be accepted and graded

• Team work

– You are encouraged to discuss HW and class material with the instructor, the TA’sand your classmates

– However, the submitted individual HW solutions and exams must involve only youreffort

– Otherwise you’ll have terrible performance on the exam since you did not learn tothink for yourself

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2 Mathematical preliminaries

2.1 Scalars, vectors and tensors

Mathematical preliminaries

Scalars, vectors and tensor definitions• Scalar quantities- represent a single magnitude at each pint in space

– Mass density- ρ

– Temperature- T

• Vector quantities- represent variables which are expressible in terms of components in a2-D or 3-D coordinate system

– Displacement- u = ue1 + ve2 +we3

• Matrix quantities- represent variables which require more than three components to quan-tify

– Stress matrix

σ =

σxx σxy σxzσyx σyy σyzσzx σzy σzz

where e1, e2 and e3 are unit basis vectors in the coordinate system

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2.2 Index notation


Index notation• Index notation is a shorthand scheme where a set of numbers is represented by a single

symbol with subscripts

ai =

a1a2a3

, ai j =

a11 a12 a13a21 a22 a23a31 a32 a33

– a1 j → first row– ai1 → first column

• Addition and subtraction

ai±bi =

a1±b1a2±b2a3±b3

ai j±bi j =

a11±b11 a12±b12 a13±b13a21±b21 a22±b22 a23±b23a31±b31 a32±b32 a33±b33

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Index notation• Scalar multiplication

λai =

λa1λa2λa3

, λai j =

λa11 λa12 λa13λa21 λa22 λa23λa31 λa32 λa33

• Outer multiplication (product)

aib j =

a1b1 a1b2 a1b3a2b1 a2b2 a2b3a3b1 a3b2 a3b3

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Index notation

• Commutative, associative and distributive laws

ai +bi = bi +ai

ai jbk = bkai j

ai +(bi + ci) = (ai +bi)+ ci

ai(b jkc`) = (aib jk)c`

ai j(bk + ck) = ai jbk +ai jck

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Index notation

• Summation convention (Einstein’s convention)- if a subscript appears twice in the sameterm, then summation over that subscript from one to three is implied

aii =3

∑i=1

aii = a11 +a22 +a33

ai jb j =3

∑j=1

ai jb j = ai1b1 +ai2b2 +ai3b3

– j- dummy index– subscript which is repeated into the notation (one side of theequation)

– i- free index– subscript which is not repeated into the notation1.20


Index notation- example

• The matrix ai j and vector bi are

ai j =

1 2 00 4 32 1 2

, bi =

240

• Determine the following quantities

– aii = a11 +a22 + . . . = . . . (scalar)- no free index

– ai jai j = a11a11 + a12a12 + a13a13 + . . . = 1× 1 + 2× 2 + . . . = . . . (scalar)- no freeindex

ai jb j = ai1b1 +ai2b2 +ai3b3

=

a11b1 +a12b1 +a13b3. . .. . .

=

. . .. . .. . .

(vector)- one free index

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Index notation- example

• Determine the following quantities

ai ja jk = ai1a1k +ai2a2k +ai3a3k

=

i = 1 a11a1k +a12a2k +a13a3ki = 2 a21a1k + . . .i = 3 a31a1k + . . .

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The first expression gives the components of the 1-st row

a11a1k +a12a2k +a13a3k = k = 1 a11a11 +a12a21 +a13a31 = . . .k = 2 a11a12 +a12a22 +a13a32 = . . .k = 3 a11a13 +a12a23 +a13a33 = . . .

(matrix)- two free indexes• ai jbib j = a11b1b1 +a12b1b2 +a13b1b3 + . . . = . . . (scalar)- no free index

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Index notation- example• Determine the following quantities

– bibi = b1b1 +b2b2 + . . . = . . . (scalar)- no free index

bib j =

b1b jb2b jb3b j

=

b1b1 b1b2 b1b3. . .. . .

= . . .

(matrix)- two free indexes1.23


Index notation- example• Determine the following quantities

– Unsymmetric matrix decomposition

ai j =12(ai j +a ji)︸︷︷︸symmetric

+12(ai j−a ji)︸︷︷︸

antisymmetric

– Symmetric part

12(ai j +a ji) = . . .

– Antisymmetric part

12(ai j−a ji) = . . .

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2.3 Kronecker delta and alternating symbol


Kronecker delta and alternating symbol• Kronecker delta is defined as

δi j ={

1 if i = j0 if i 6= j =

1 0 00 1 00 0 1

• Properties of δi j

δi j = δ ji

δii = 3

δi ja j =

δ11a1 +δ12a2 +δ13a3 = a1. . .. . .

= ai

δi ja jk = aik

δi jai j = aii

δi jδi j = 31.25

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Kronecker delta and alternating symbol• Alternating (permutation) symbol is defined as

εi jk =

+1 if i jk is an even permutation of 1,2,3−1 if i jk is an odd permutation of 1,2,3

0 otherwise=

• Therefore

ε123 = ε231 = ε312 = 1ε321 = ε132 = ε213 =−1ε112 = ε131 = ε222 = . . . = 0

• Matrix determinant

det(ai j) = |ai j|=

∣∣∣∣∣∣a11 a12 a13a21 a22 a23a31 a32 a33

∣∣∣∣∣∣= εi jka1ia2 ja3k = εi jkai1a j2ak3

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2.4 Coordinate transformations


Coordinate transformations• Consider two Cartesian coordinate systems with different orientation and basis vectors

1.27


Coordinate transformations• The basis vectors for the old (unprimed) and the new (primed) coordinate systems are

ei =

e1e2e3

, e′i =

e′1e′2e′3

• Let Ni j denotes the cosine of the angle between x′i-axis and x j-axis

Ni j = e′i · e j = cos(x′i,x j)

• The primed base vectors can be expressed in terms of those in the unprimed by relations

e′1 = N11e1 +N12e2 +N13e3

e′2 = N21e1 +N22e2 +N23e3

e′3 = N31e1 +N32e2 +N33e3

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Coordinate transformations

• In matrix form

e′i = Ni je j

ei = N jie′j

• An arbitrary vector can be written as

v = v1e1 + v2e2 + v3e3 = viei

= v′1e′1 + v′2e′2 + v′3e′3 = v′ie′i

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Coordinate transformations

• Or

v = viN jie′j

• Because v = v′je′j thus

v′j = N jivi

• Similarly

vi = N jiv′j

• These relations constitute the transformation law for the Cartesian components of a vectorunder a change of orthogonal Cartesian coordinate system

1.30

2.5 Cartesian tensors


Cartesian tensors

• General index notation scheme

a′ = a, zero order (scalar)a′i = Nipap, first order (vector)a′i j = NipN jqapq, second order (matrix)

a′i jk = NipN jqNkrapqr, third order

. . .

• A tensor is a generalization of the above mentioned quantities

Example

• The notation v′i = Ni jv j is a relationship between two vectors which are transformed toeach other by a tensor (coordinate transformation). The multiplication of a vector by atensor results another vector (linear mapping).

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Cartesian tensors

• All second order tensors can be presented in matrix form

Ni j =

N11 N12 N13N21 N22 N23N31 N32 N33

• Since Ni j can be presented as a matrix, all matrix operation for 3×3-matrix are valid• The difference between a matrix and a tensor

– We can multiply the three components of a vector vi by any 3×3-matrix

– The resulting three numbers (v′1,v′2v′3) may or may not represent the vector compo-

nents

– If they are the vector components, then the matrix represents the components of atensor Ni j

– If not, then the matrix is just an ordinary old matrix1.32


Cartesian tensors

• The second order tensor can be created by a dyadic (tensor or outer) product of the twovectors v′ and v

N = v′⊗v =

v′1v1 v′1v2 v′1v3v′2v1 v′2v2 v′2v3v′3v1 v′3v2 v′3v3

1.33


Transformation example

• The components of a first and a second order tensor in a particular coordinate frame aregiven by

bi =

142

, ai j =

1 0 30 2 23 2 4

• Determine the components of each tensor in a new coordinates found through a rotation of

60◦ about the x3-axis1.34



11

• The rotation matrix is

Ni j = cos(x′i,x j)

=

cos300◦ cos30◦ cos90◦cos210◦ cos300◦ cos90◦cos90◦ cos90◦ cos0◦

=

12

√3

2 0−√

32

12 0

0 0 1

1.35



• The transformation of the vector bi is

b′i = Ni jb j =

12

√3

2 0−√

32

12 0

0 0 1

1

42

= . . .

• The second order tensor transformation is

a′i j = NipN jpapq =

12

√3

2 0−√

32

12 0

0 0 1

1 0 30 2 23 2 4

12

√3

2 0−√

32

12 0

0 0 1

T

= . . .

1.36

2.6 Principal values and directions


Principal values and directions for symmetric tensor

• The tensor transformation shows that there is a coordinate system in which the componentsof the tensor take on maximum or minimum values• If we choose a particular coordinate system that has been rotated so that the x′3-axis lies

along the vector, then vector will have components

v =

00|v|

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Principal values and directions for symmetric tensor• It is of interest to inquire whether there are certain vectors n that have only their lengths

and not their orientation changed when operated upon by a given tensor A• That is, to seek vectors that are transformed into multiples of themselves• If such vectors exist they must satisfy the equation

A ·n = λn, Ai jn j = λni

• Such vectors n are called eigenvectors of A• The parameter λ is called eigenvalue and characterizes the change in length of the eigen-

vector n• The above equation can be written as

(A−λ I) ·n = 0, (Ai j−λδi j)n j = 0

1.38


Principal values and directions for symmetric tensor• Because this is a homogeneous set of equations for n, a nontrivial solution will not exist

unless the determinant of the matrix (. . .) vanishes

det(A−λ I) = 0, det(Ai j−λδi j) = 0

• Expanding the determinant produces a characteristic equation in terms of λ

−λ3 + IAλ

2− IIAλ + IIIA = 0

1.39


Principal values and directions for symmetric tensor• The IA, IIA and IIIA are called the fundamental invariants of the tensor

IA = tr(A) = Aii = A11 +A22 +A33

IIA =12(tr(A)2− tr(A2)

)=

12(AiiA j j−Ai jAi j)

=∣∣∣∣ A11 A12

A21 A22

∣∣∣∣+ ∣∣∣∣ A22 A23A32 A33

∣∣∣∣+ ∣∣∣∣ A11 A13A31 A33

∣∣∣∣IIIA = det(A) = det(Ai j)

• The roots of the characteristic equation determine the values for λ and each of these maybe back-substituted into (A−λ I) ·n = 0 to solve for the associated principle directions n.

1.40


Example• Determine the invariants and principal values and directions of the following tensor:

A =

3 1 11 0 21 2 0

• The invariants are

IA = . . . , IIA = . . . IIIA = . . .

• The characteristic equation is

−λ3 +3λ

2 +6λ −8 = 0

• The roots are λ1 =−2, λ2 = 1 and λ3 = 41.41

13


Example

• For λ1 =−2 we have (A−λ1I) ·n1 = 0 5 1 11 2 21 2 2

n11n21n31

=

000

• The homogeneous set of equations have linear dependent equations and the solution rep-

resents only the ratio between the solution set• Applying n31 = 1 and solving the first end second equations we get

n1 = . . .

• Similarly for λ2 = 1 and λ3 = 41.42

2.7 Vector and tensor algebra


Vector and tensor algebra

• Scalar product (dot product, inner product)

a ·b = |a||bcosθ

• Magnitude of a vector

|a|= (a ·a)1/2

• Vector product (cross-product)

a×b = det

e1 e2 e13a1 a2 a3b1 b2 b3

• Vector-matrix products

Aa = Ai ja j = a jAi j

aT A = aiAi j = Ai jai

1.43


Vector and tensor algebra

• Matrix-matrix products

AB = Ai jB jk

ABT = Ai jBk j

AT B = A jiB jk

tr(AB) = Ai jB ji

tr(ABT ) = tr(AT B) = Ai jBi j

where ATi j = A ji and tr(A) = Aii = A11 +A22 +A33

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2.8 Tensor calculus


Tensor calculus

• Common tensors used in field equations

a = a(x1,x2,x3) = a(xi) = a(x)− scalarai = ai(x1,x2,x3) = ai(xi) = ai(x)−vectorai j = ai j(x1,x2,x3) = ai j(xi) = ai j(x)− tensor

• Comma notations for partial differentiation

a,i =∂

∂xia

ai, j =∂

∂x jai

ai j,k =∂

∂xkai j

1.45


Tensor calculus- example

• Vector differentiation

ai, j =∂ai

∂x j=

∂a1∂x1

∂a1∂x2

∂a1∂x3

∂a2∂x1

∂a2∂x2

∂a2∂x3

∂a3∂x1

∂a3∂x2

∂a3∂x3

1.46


Tensor calculus

• Directional derivative

– Consider a scalar function φ . Find the derivative of the φ with respect of direction s

dφ

ds=

∂φ

∂xdxds

+∂φ

∂ydyds

+∂φ

∂ zdzds

– The unit vector in the direction of s is

n =dxds

e1 +dyds

e2 +dzds

e3

– The directional derivative can be expressed as a scalar product

dφ

ds= n ·∇φ

1.47

15


Tensor calculus

• Directional derivative

– ∇φ is called the gradient of the scalar function φ and is defined by

∇φ = e1∂φ

∂x+ e2

∂φ

∂y+ e3

∂φ

∂ z

– The symbolic operator ∇ is called del operator (nabla operator) and is defined as

∇ = e1∂

∂x+ e2

∂

∂y+ e3

∂

∂ z

– The operator ∇2 is called Laplacian operator and is defined as

∇2 =

∂ 2

∂x2 +∂ 2

∂y2 +∂ 2

∂ z2

1.48


Tensor calculus• Common differential operations and similarities with multiplications

Name Operation Similarities OrderGradient of a scalar ∇φ ≈ λu vector ↑Gradient of a vector ∇u = ui, jeie j ≈ u⊗v tensor ↑

Divergence of a vector ∇ ·u = ui, j ≈ u ·v dot ↓Curl of a vector ∇×u = εi jkuk, jei ≈ u×v cross→

Laplacian of a vector ∇2u = ∇ ·∇u = ui,kkei

NoteThe ∇-operator is a vector quantity

1.49



• Scalar and vector functions are φ = x2− y2 and u = 2xe1 + 3yze2 + xye3. Calculate thefollowing expressions• Gradient of a scalar

∇φ = . . .

• Laplacian of a scalar

∇2φ = ∇ ·∇φ = . . .

• Divergence of a vector

∇ ·u = . . .

• Gradient of a vector

∇u = . . .

1.50

16



• Curl of a vector

∇×u = det

e1 e2 e3∂

∂x∂

∂y∂

∂ z2x 3yz xy

= . . .

1.51


Tensor calculus

• Divergence (Gauss) theorem ∫S

u ·ndS =∫

V∇ ·udV

where n is the outward normal vector to the surface S1.52


The End

• Welcome and good luck• Any questions, opinions, discussions?

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