Mechanics of materials (solid mechanics), MENG11100, in 94 solvedproblems

Anton Shterenlikht

Mech Eng Dept, The University of Bristol

ABSTRACT

An absolute minimum of solid mechanics knowledge required to study moreadvanced concepts in years 2 and above. The topics include: stress and strain tensors,stress equilibrium, solutions of simple 1D and 2D stress problems, principal values anddirections, tensor rotations, maximum shear orientation and value, Mohr’s circle, linearisotropic elasticity, idea of a solid mechanics problem, uniaxial stress/strain states, stati-cally indeterminate systems, slender beam bending theory, properties of cross sections,plane strain/stress, axisymmetry, torsion, elastic stability and buckling. Materialandstructural nonlinearity, anisotropy, inelasticity, general solution methods to elastic prob-lems are excluded.

22 April 2016

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Table of Contents

1. Abbreviations . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13. Recommended reading . . . . . . . . . . . . . . . . . . . . . . . . 14. One-dimensional stress/strain analysis. . . . . . . . . . . . . . . . . . . . 3

4.1. Fundamental ideas of stress and strain and linear elasticity. . . . . . . . . . . . 34.1.1. Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . 34.1.2. Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . 64.1.3. Young’s modulus . . . . . . . . . . . . . . . . . . . . . . . 74.1.4. Poisson’s ratio . . . . . . . . . . . . . . . . . . . . . . . . 74.1.5. Displacement. . . . . . . . . . . . . . . . . . . . . . . . . 94.1.6. Elastic energy . . . . . . . . . . . . . . . . . . . . . . . . . 9

4.2. Pin-joined frame . . . . . . . . . . . . . . . . . . . . . . . . . 94.3. Statically indeterminate systems. . . . . . . . . . . . . . . . . . . . 114.4. Euler-Bernoulli (slender beam) bending theory. . . . . . . . . . . . . . . 12

4.4.1. Properties of cross sections. . . . . . . . . . . . . . . . . . . . 164.4.2. Plasticity in bending . . . . . . . . . . . . . . . . . . . . . . 18

4.5. Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.5.1. Buckling of columns . . . . . . . . . . . . . . . . . . . . . . 20

5. Three-dimensional stress/strain analysis. . . . . . . . . . . . . . . . . . . 225.1. Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . 225.2. Stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.2.1. Elementary cube of material. . . . . . . . . . . . . . . . . . . . 275.2.2. Conservation of linear momentum. . . . . . . . . . . . . . . . . . 275.2.3. Conservation of angular momentum. . . . . . . . . . . . . . . . . . 28

5.3. Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.3.1. Eigenvalue / vector . . . . . . . . . . . . . . . . . . . . . . . 30

5.4. Strain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315.5. Maximum shear value . . . . . . . . . . . . . . . . . . . . . . . 335.6. Mohr’s diagram . . . . . . . . . . . . . . . . . . . . . . . . . 335.7. Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . 345.8. Solving solid mechanics problems. . . . . . . . . . . . . . . . . . . 38

6. Special cases. . . . . . . . . . . . . . . . . . . . . . . . . . . . 386.1. Two-dimensional stress/strain problems. . . . . . . . . . . . . . . . . . 39

6.1.1. Plane stress . . . . . . . . . . . . . . . . . . . . . . . . . 396.1.2. Plane strain . . . . . . . . . . . . . . . . . . . . . . . . . 406.1.3. Axisymmetric . . . . . . . . . . . . . . . . . . . . . . . . 416.1.4. Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6.2. Application of tensor theory to properties of areas. . . . . . . . . . . . . . 457. Example problems. . . . . . . . . . . . . . . . . . . . . . . . . . 468. Solutions to example problems. . . . . . . . . . . . . . . . . . . . . . 51

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References

Akivis, 2003.M. Akivis and V. V. Goldberg,Tensor Calculus with Applications,World Scientific (2003).

Beer, 2009.F. P. Beer, E. R. Johnston, Jr., J. T. Dew olf, and D. F. Mazurek,Mechanics of Materials,McGraw-Hill, 5 Ed. (2009).

Bonet, 2008.J. Bonet and R. D. Wood,Nonlinear Continuum Mechanics for Finite Element Analysis,Cambridge,2 Ed. (2008).

Bourne, 1992.D.E. Bourne and P.C. Kendall,Vector analysis and cartesian tensors,Chapman & Hall, 3rd ed.(1992).

Brillouin, 1964.L. Brillouin, Tensors in mechanics and elasticity,Academic Press (1964).

Byars, 1963.E. F. Byars and R. D. Snyder,Engineering Mechanics of Deformable Bodies,International TextbookCompany (1963).

Calcote, 1968.L. R. Calcote,Introduction to continuum mechanics,Van Nostrand (1968).

Case, 1999.J. Case, Lord Chilver, and C. T. F. Ross,Strength of Materials and Structures,Arnold, 4ed. (1999).

Chadwick, 1976.P. Chadwick,Continuum Mechanics,George Allen & Unwin (1976).

Chaves, 2013.E. W. V. Chaves,Notes on Continuum Mechanics,Springer (2013).

Chorlton, 1976.F. Chorlton,Vector & tensor methods,Ellis Horwood (1976).

Crandall, 1978.S. H. Crandall, N. C. Dahl, and T. J. Lardner,An Introduction to the Mechanics of Solids,McGraw-Hill, 2 Ed. with SI units (1978).

Eringen, 1967.A. C. Eringen,Mechanics of Continua,John Wiley & Sons (1967).

Fitzgerald, 1982.R. W. Fitzgerald,Mechanics of Materials,Addison-Wesley, 2 Ed. (1982).

Fleisch, 2012.D. Fleisch,A student’s guide to vectors and tensors,Cambridge (2012).

Ford, 1963.H. Ford,Advanced Mechanics of Materials,Longmans (1963).

Fung, 1969.Y. C. Fung,A First Course in Continuum Mechanics,Prentice-Hall (1969).

Gere, 1997.J. M. Gere and S. P. Timoshenko,Mechanics of Materials,PWS Publishing Company, 4 Ed. (1997).

Hodge, 1970.P. G. Hodge,Continuum mechanics,McGraw-Hill (1970).

Hosford, 2005.W. F. Hosford,Mechanical behavior of materials,Cambridge (2005).

Hosford, 2010.W. F. Hosford,Solid mechanics,Cambridge (2010).

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Hunter, 1976.S. C. Hunter,Mechanics of continuous media,Ellis Horwood (1976).

Malvern, 1969.L. E. Malvern,Introduction to the Mechanics of a Continuous Medium,Prentice Hall (1969).

Mase, 1970.G. E. Mase,Theory and problems of continuum mechanics,Schaum’s outline series in engineering,McGraw-Hill (1970).

Reddy, 2013.J. N. Reddy,An Introduction to Continuum Mechanics,Cambridge, 2 Ed. (2013).

Rees, 2000.D. W. A. Rees,Mechanics of Solids and Structures,Imperial College Press (2000).

Scipio, 1967.L. A. Scipio,Principles of continua, with applications,Wiley (1967).

Segel, 1987.L. A. Segel,Mathematics applied to continuum mechanics,Dover (1987).

Sokolnikoff, 1956.I. S. Sokolnikoff,Matemathical Theory of Elasticity,McGraw-Hill, 2 Ed. (1956).

Sokolnikoff, 1951.I. S. Sokolnikoff,Tensor Analysis,John Wiley & Sons (1951).

Spencer, 1980.A. J. M. Spencer,Continuum mechanics,Longman (1980).

Tadmor, 2012.E. B. Tadmor, R. E. Miller, and R. S. Elliot,Continuum Mechanics and Thermodynamics,Cambridge(2012).

Valliappan, 1981.S. Valliappan,Continuum mechanics,Balkema (1981).

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Index

A

acceleration 3,81-82, 84alternating tensor 30, 85-87area

centroid 17axysymmetric 42

geometry 44

B

basis vectors 88orthonormality 90,95-96

bending 12,116cross section (see moments of area)curvature 14deflection 12,116distributed load 13, 15Euler-Bernoulli theory 12fibres 12four-point 15gradient 15moment 18,116neutral line 12, 18plasticity 18pure 15radius of curvature 14second moment of area 14shear force 15slender beam 12thick beams 41

body force (see force)boundary conditions 38, 81-82buckling 20

column 20critical state 20

C

compatibilitydisplacement 57,59

compliance tensormajor symmetry 106

compliancetensor 106

compression modulus (see elasticbulk modulus)

configurationcurrent 31deformed (see current)original 31undeformed (see original)

conservationangular momentum 28

linear momentum 3, 24constitutive model 18,35constraint 41continuum mechanics 32coordinate system 3, 16, 22, 29, 39, 88, 96-97,121

left-handed 84orthonormal 29principal 90-91,96right-handed 84,88rotated 29,91, 98

coordinate transformation 29, 34, 38-39, 85-87,91-92, 96-97, 113, 121

invariant 29coordinate

axes 16axis 6,80

cross section2nd moment 116polar moment 116

D

deformation 31deformation (see strain)deformation gradient 32, 92-93, 96-97deformation

reversible 32displacement 6,9, 31, 94

field 94gradient 7,32, 92-94, 113vector 31,38, 92, 112

E

eigenvalues (see rank 2 tensorprincipal values)

eigenvectors (see rank 2 tensorprincipal vectors)

Einstein’s summation convention 26elastic tensor

major symmetry 105-106minor symmetry 105

elasticanisotropic 37behaviour 35,38bulk modulus 38, 108compliance 36compliance tensor 108-109conservative 36energy density 106inversible 36isotropic 37Lamé constants 37, 41, 43, 107, 110, 115

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linear 36-37lossless 36non-linear 36path independent 36potential 36recoverable 36reversible 36shear modulus 38, 107superposition 38,118tensor 35-37,105-106, 108, 110Young’s modulus 40,109-110, 115

elasticity 6,18energy 9linear 7,115one-dimensional 6Young’s modulus 7

elementary cube of material 27, 74, 91, 102, 119elongation 6,106energy

elastic 9equilibrium

critical 20force 3moment 3,20new 20perturbation 19stable 19-20unstable 19-20

F

force 3,82-83, 88body 26,42, 82distributed 83gravity 80internal 24reaction 54

free body diagram 3, 81-83

G

Green’s theorem 114

I

impulse 3index

dummy 26,75live 75

instability (see stability)

K

kinematics 31Kronecker delta 29, 32, 76, 80, 87, 92, 110Kronecker delta (see also rank 2 tensor

identity)

L

Lagrange multiplier 99linear

displacement 6

M

materialconstitutive model 38elastic 37incompressible 108isotropic 107symmetry 37

matrixdeterminant 85symmetric 86

Mohr’s diagram 33,103-105, 120moment 3moment of area

2nd 121first 121tensor 121

moments of areafirst 16polar 17second 17

momentumangular 45linear 3

motion 31,93

N

Newtons’ lawssecond 3third 4

O

operatordivergence 75,110gradient 74,110

optimisationconstrained 99objective function 99

P

pin-joined frame 9plastic limit 18plastic

flow 18plasticity 18Poisson’s ratio 7-8,38, 40, 106-107, 115

0.5 108position vector

derivative 80

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pressure 114profile

displacement 6force 4

R

rank 2 tensor 26, 29, 35, 971st invariant 31,87-90, 95, 1082nd invariant 31,87-90, 953rd invariant 31,87-89, 91, 95anti-symmetric 30,79biaxial 31characteristic equation 30, 87, 91, 95contraction 87deformation gradient 32deformation gradient (see also deformation gra-

dient)derivative 92determinant 31,84-86determinant (see also 3rd invariant)diagonal components 79-80equi-biaxial 31,101-102equi-triaxial (see hydrostatic)hydrostatic 31identity 29,76, 80, 92inverse 29,32left Cauchy-Green 32matrix 34maximum shear 34, 100-102, 104-105, 116orthogonal 29,84, 87principal directions 88, 113principal directions (see principal vectors)principal values 30-31,33, 87-88, 90-91, 98,

105principal vectors 30,89-90, 95-96, 105, 117rotation 29,38, 76, 84-87, 89, 92, 96-98, 113,

120-121skew-symmetric (see anti-symmetric)strain 92stretch 76symmetric 30-31,33, 79, 86, 89, 105trace 31,87, 108transpose 29,92, 97triaxial 31two-point 32uniaxial 31

rank 3 tensor 85-86rank 3 tensor (see also alternating tensor)rank 4 tensor 105

elastic (see elastic tensor)elastic compliance 109elasticity 36identity 37

rate of change 3

rigid bodyrotation 57

rotation tensor (see rank 2 tensorrotation

S

sectioning method 3shear

maximum value 33pure 95simple (see pure)

shoulder 3solid mechanics 38

problem 39solution 39

stability 18elastic 20Lyapunov 19

statically determinate 11statically indeterminate 11, 58stiffness 6stiffness tensor (see elastic

tensor)strain 6,31-32strain tensor 92strain

axial 7,55axysymmetric 42compatibility 39,110elastic 94equi-biaxial 102finite 94gauge 97gauge rosette 97incomplressible 108incompressible 33,95-96large 94normal 33,97, 102, 108one-dimensional 6plane 41,115principal 33,95, 105, 107, 115proof (see yeild)pure shear 45, 115quasi-static 32shear 33,98, 102, 105, 108small 33,94state 102symmetry 94tensor 35-36,38, 41, 43, 92-95, 104, 107-110,

113tensor 1st invariant 115three-dimensional 40trace 115transverse 7

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two-dimensional 43uniaxial 109-110volumetric 33yeild 58

stress 3,24stress state 39stress tensor 39stress

av erage 88axysymmetric 42biaxial 84,89, 115compressive 10, 27, 84continuum 13discontinuity 53distibution 83equilibrium 38,42, 81-83, 110, 112equilibrium equations 28field 80-84hydrostatic 101matrix 26,84, 114maximum shear 120non-uniform 82-84normal 23,27, 89, 98, 102-103, 108one-dimensional 3plane 39,115pressure 108,114principal 91,98, 103-104, 107, 113, 115, 117profile 18pure shear 45, 115shear 23,27, 83, 89, 98, 102-103, 108sign convention 82state 23,26, 102symmetry 26,29, 105tensile 10,27, 84tensor 23,26, 35-36, 38, 41, 43, 84, 88-89, 98,

103-104, 108-110, 112-113, 115, 120tensor 1st invariant 115trace 108,115traction 38,81-84triaxial 109two-dimensional 43uniaxial 9,58, 81-83, 88, 106-107, 109-110uniform 81vector 4,24, 89, 98, 112yield 18

stretch 32

T

tensorcomponent 106components 36,105derivative 110equality 30function 35-36

index 111indices 85notation 28rank 1 30, 79rank 2 79rank 2 (see rank 2 tensor)rank 3 30, 79rank 4 30, 79

torsion 43,116cross section 44torque 45twist angle 44

U

unit tensor 76unit tensor (see also rank 2 tensor indentity)

V

vector 120acceleration 3cross product 28, 85derivative 110flux 114force 3,80position 3stress 4velocity 3

velocity 3volume 85,96

change 96infinitecimal 80

Y

Young’s modulus 38,55, 107Young’s modulus (see elasticity)

1. Abbreviations

BC - boundary conditions

BTW - by the way

CS - coordinate system

CT - coordinate transformation

iff - if and only if, also↔R2T - rank 2 tensor

wrt - with regards to, with respect to

2. Notation

In this course a mix of tensor and index notations is used. Although some violation of this notationconvention is unavoidable, mostlyitalic is used for scalar variables, e.g.λ . Vectors and tensors of rank 2are shown either inbold or with indices, e.g. the coordinate vector→x can appear asx or asxi , i = 1, 2, 3.Typically capital letters are used for R2T, e.g. T = Tpq. Tensors of rank 4 are typeset inHelvetica, either inbold, when used with other tensors and vectors, or with indices.For example the elasticity (stiffness) ten-sor can appear asC or asC ijkl . Italic is also used to emphasise key terms or concepts.

↔ means if and only if, also iff.

⋅ means inner product or dot product, contraction on the inner index, e.g. the square of the length ofvector →x is x ⋅ x.

: means tensor product, contraction on the 2 inner indices, e.g.A: A = Aij Aij .

⊗ is a dyadic product, e.g.x⊗x = xi x j = Aij .

3. Recommended reading

All books are available from either the Queen’s building library, the Physics library or the Chemistrylibrary. Some books have multiple copies.

The following books are recommended for understanding the idea of a tensor, index notation, theideas of stress, strain and deformation, ideas of principal strain and stress, elasticity, etc.

Easy/medium introductory level:

→ (Fleisch, 2012) a gentle intro to vectors and tensors with lots of practical examples and illustrations.

→ (Gere, 1997) very easy, chapter 7, inferior notation, stress/strain analysis, little or no tensors. Manycopies.

→ (Mase, 1970) easy/medium, chapters 1-3, lots of solved problems and further problems for self-study.

→ (Tadmor, 2012) the beginning is a very gentle introduction to stress and strain, the need for tensors andindex notation.

→ (Fitzgerald, 1982) easy, sections 6.5 to 6.15, no tensors, just stress/strain analysis.

→ (Case, 1999) very easy, beginning of chapter 5, beware - inferior notation, stress/strain analysis, no ten-sors.

→ (Beer, 2009) very easy, sections 6.1 to 6.6, no tensors, stress/strain analysis.

→ (Crandall, 1978) easy and quite thorough, stress/strain, little or no tensors.

→ (Byars, 1963) very easy, chapters 1-2, no tensors

→ (Hosford, 2005) and (Hosford, 2010) easy, chapter 1 - stress/strain, some tensor language. Multiplecopies.

→ (Valliappan, 1981) explains 3D stress and strain in chapters 1 to 3, although he cuts corners and simpli-fies things a bit. He never mentions tensors, all is done with matrices.

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→ (Calcote, 1968) is ok. Sections 2.1, 2.5-2.7 explain tensors and index notation. Section3 explainsstress, strain and elasticity. Can skip more advanced material.

→ (Spencer, 1980) chapters 1-5, index notation, tensors, stress, strain. Skip more advanced material.Beware: some notation is different to mine. Another copy is in physics library.

→ (Bonet, 2008) is an advanced read overall, but some sections are explained quite easily. Try sections2.2, 4.4, 4.5, 5.2 - tensors and stress.

Intermediate level:

→ (Reddy, 2013) a very good book overall.

→ (Rees, 2000) easy, but misleading due to inferior notation, chapter 13, very applied, use with caution.

→ (Ford, 1963) easy/medium, parts 1-2, skip curvilinear coordinates, lots of copies.

→ (Bourne, 1992) chapters 1, 2, 8, skip vector and tensor calculus. only about tensors and notation, a littlebit about stress, no strain. Multiple copies.

→ (Malvern, 1969) a classic book, you only the beginning - stress/strain analysis and tensor basics.

→ (Brillouin, 1964) chapters 1,2 - slightly outdated notation, skip covariant/contravariant and curvilinearsections.

→ (Scipio, 1967) part 1 contains all you need: index notation, tensors, stress, strain. Beware: some nota-tion is different to mine. Another copy is in physics library.

→ (Sokolnikoff, 1956) the beginning is suitable as an introduction to stress, strain, tensors and index nota-tion.

Advanced level:

→ (Segel, 1987) chapters 1-2 - index notation, tensors. Skip the theorems and more advanced material.

→ (Chorlton, 1976) chapters 12-13, tensors/stress

→ (Hunter, 1976) chapter 4 - notation, chapter 5 - strain.

→ (Chaves, 2013) has lots of nice diagrams. Chapter 1 - tensors, chapter 3 - stress. Skip chapter 2 onstrain - too advanced.

→ (Fung, 1969) is a demanding but still a very good book. Read chapters 1-5, skip more advanced mate-rial.

→ (Chadwick, 1976) and (Hodge, 1970) are more advanced, with different notation, but still useful, if onewants to know more.

→ (Eringen, 1967) a classic book on the subject

→ (Akivis, 2003) extra knowledge of tensors and tensor calculus.

→ (Sokolnikoff, 1951) another classic book, the beginning is not too hard, and very relevant.

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4. One-dimensional stress/strain analysis

4.1. Fundamental ideas of stress and strain and linear elasticity

4.1.1. Stress

We consider as known and understood the concepts of force,f, position vector,x, velocity,v = dx/dt = x, accelerationa = dv/dt = x and coordinate system.

We also consider as known and understood the Newton’s laws, e.g. the second law of Newton can bewritten as:

f = ma = mdvdt

(1)

or, if m is not constant, then:

f =d(mv)

dt(2)

or

d(mv) = fdt (3)

meaning that the total force acting on a body causes the body to accelerate, with the acceleration vectorbeing proportional to the force vector, Eqn. (1). Or we can say that the force is proportional to the rate ofchange of thelinear momentum, Eqn. (2). Or we can say that the linear momentum is always conserved,Eqn. (3), i.e. the change in the linear momentum can be caused only by theimpulse, the product of force bytime.

If dv = 0, then the total force isf = 0. We call this force equilibrium.

We introduce a concept of amoment. If forcef acts in point A, then the moment off about someother point B is the product of the force by the distance from B to the line of action off. This distance,h,is typically called ashoulder. The moment,m, is a vector.

fA

BC

h

m = fh (4)

(Ex. prob. 1).

If m = 0 then we have amoment equilibrium.

Consider a pulley lifting massm.

m

I want to know the minimum diameter of the rope, required for lifting the mass.

First we need to introduce the idea of afree body diagram, which is contructed using thesectioningmethod. Inthis method we introduce imaginary cuts to the system to isolate (free) body or bodies of inter-est. We substitute the actions of removed bodies by forces and moments.I imaginary cut the rope at two

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points and substitute the action of missing parts by forces and moments.

1(x)

m

mg

p r

sq

t

u

v

w

I also introduce a coordinate axis, 1 orx, with a positive direction defined.

Finally I need thethird law of Newton, i.e. "action=-reaction"

Since all parts are in static equilibrium, I quickly conclude that:q = t = s = w = 0,p = u = r = v = mg. So the force in the rope ismg along its axis. Since my cuts were in arbitrary positionsalong the rope, I conclude that the force in the rope is everywheremg.

Graphically the forceprofile this can be shown as this.

mg

We can now introduce the concept of astress vector, t. If the cross-section of the rope isA, and theaxial force isf, then I define 1D stress as:

t =fA

(5)

In the one-dimensional class of problems, like the rope, where there is only a single spatial dimension, thestress and force are always along the axis. Hence we can often neglect the vectorial nature of the stressvector, and simply use the term stress.

t =f

A(6)

By convention, tensile stress is positive and compressive stress is negative.

t is really a pressure. The units of stress are those of pressure - force per unit area. In SI the units ofstress are Pascal, abbreviated as Pa. 1Pa = 1 N / 1 m2. 1 Pa is a very low stress in the majority of engi-neering applications.Typically we use MegaPascal, MPa, which can be conveniently defined as

1 MPa = 1 N / 1 mm2

Some typical values of stress or pressure are: bicycle tyre - 0.5 MPa, car tyre - 0.2-0.3 MPa, tensile strengthof nylon - 70 MPa and of polyamide - 110 MPa, steel yield stress - 300-700 MPa. (Theconcepts of thestrength and yield stress will be discussed in detail in Properties of Materials. These are fundamendal prop-erties of engineering materials).

Why are bike tyre pressures higher than in car tyres, even though cars are much heavier than bikes?

Because car tyres are much wider that in bikes. Socar tyres have a much bigger contact area with theroad, which allows to reach the required reaction force.

Back the rope problem. The magnitude of axial stress in the rope ist = mg/A.

What’s so clever about the idea of stress? Stress is a relative measure of force. It is force related tothe area over which it acts. If I imaginary split the rope into several ropes of smaller cross sectional area,

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e.g. 4, then the force in each rope will bef/4, whereas stress remains the same,t, because the cross sectionarea in each "sub-rope" has decreased proportionally.

=

f/4 f/4 f/4 f/4A/4 A/4 A/4 A/4

+ + +

t

fA

t t t t

We can now pose and solve simple solid mechanics (or mechanics of materials) problems on the pul-ley. Let’s assume the rope is of a circular cross section.

Problem A: What is the minimum diameter of rope, used in a simple pulley, for the stress in the ropenot to exceed 100 MPa while lifting a mass of 200 kg?

The stress in the rope ist = mg/A, whereA = π d2/4, d is the diameter. So

mg

t=

π d2

4→ d = √ 4mg

π t(7)

The smallestd will result from the highest possiblet, which is 100 MPa. If we use MPa as the unit ofstress, kg as the unit of mass and m / s2 for the freefall acceleration, then the units for diameter will be mm.Assumingg ≈ 10 m / s2:

d ≈ √ 4 × 200× 10

3. 14× 100≈ 5

So the answer is 5 mm.

Problem B: If the stress in the rope must not exceed 200 MPa, what is the maximum load that a pul-ley with a rope of diameter 3 mm can lift?

This is the opposite problem to problem A. All we need to do is rearrange Eqn. (7).

m =π d2t

4g≈

3. 14× 32 × 200

4 × 10= 141. 3

So the answer is 141.3 kg.

Problem C: What will be the maximum stress in a rope of 10 mm diameter lifting a mass of 2 tons?

Again, I just rearrange Eqn. (7).

t =4mg

π d2=

4 × 2 × 103 × 10

3. 14× 102≈ 255 (8)

The answer is 255 MPa.

The follow-up to problem C isProblem D: Choose an engineering material for the rope, that cancarry the stress of 255 MPa.

This problem is outside the scope of this course. It is addressed fully in the Properties of Materialscourse.

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4.1.2. Strain

A pulley rope will elongateunder the application of load. The higher the load - the greater the elon-gation. In fact, under certain conditions, the elongation is proportional to the force. This is the famous lawof elasticity, first discovered in this form by Robert Hooke in 1660, when he published it as an anagram:"ceiiinosssttuv". In1678 he published the solution to the anagram: "ut tensio, sic vis", which can be trans-lated as "as the extension, so the force".

If the original length of the rope on one side of the pulley wasl1, then after massm was attached tothe pulley, the length of the rope increased tol2. The elongation is∆l = l2 − l1.

∆l

l2

m

l1

Since the stress in the rope ismg, we can say that

∆lk1 = mg

wherek1 is thestiffnessof the rope of lengthl1.

Although the above equation has some use for calculating elongations, it has a major problem -kdepends on the length of the rope. Indeed, if the rope is initially twice as long, under the application of thesame load it’s elongation will be twice as high. It would be good to have stiffness as amaterialproperty,that does not depend on the geometry of the problem.

To address this problem, we introduce the idea ofstrain, which is a relative measure of elongation,similar to the idea of stress as a relative measure of force. By definition, a one-dimensional strain is

e =∆l

l(9)

wherel is the original undeformed length.

Strain is dimensionless.

By convention, tensile strain is positive and compressive strain is negative.

If we fix a coordinate axis with the rope,x, then we can talk ofdisplacement, u, which is by howmuch each point of the rope moved from the origin.We can choose the origin arbitrarily, but let’s take it,for simplicity, at the point where the rope leaves the pulley roller. The displacement profile will look likethis.

∆l

x

u

l1

0

It is of key importance that the displacement profile is linear.

We can now redefinee from Eqn. (9) as

e =u

x(10)

Sinceu is a linear function ofx, e is constant along the length of the rope.We can say that the

-7-

displacementgradientis constant. Straine is really a displacement gradient, or relative displacement.

4.1.3. Young’s modulus

For linearly elastic materials

t = Ee (11)

whereE is theYoung’s modulus, named after Thomas Young (1773-1829), an English scientist.

Sincee is dimensionless,E has the units of pressure.For engineering materials we usually use MPaor GigaPascal, GPa. Sometypical values are: steel - 200 GPa, aluminum alloys - 70 GPa, brass - 100 GPa,concrete - 30 GPa, nylon - 2 GPa, wood - 9 GPa.

With the ideas of strain and elasticity added to the idea of stress, some further pulley problems can beposed and solved.

Problem E: What should be the minimum diameter of a steel wire, used for the pulley rope, of initiallength of 20 m, so that it does not extend more than 10 mm under the application of mass of 500 kg?

Let’s assume the Young’s modulus of 200 GPa for steel. From Eqn. (9) strain is

e =∆l

x=

10

20× 103= 5 × 10−4

From Eqn. (11) stress, in MPa, is

t = Ee= 2 × 105 × 5 × 10−4 = 100

Substituingt into Eqn. (7) we can calculate the diameter:

d = √ 4mg

π t= √ 4 × 500× 10

3. 14× 100≈ 8

Since we use value for stress in MPa, the answer is in mm - 8 mm.

Problem F:

Calculate the maximum extension of a nylon pulley rope of initial length of 3 m and 1 mm diameter,under the application of mass of 2 kg.

First we need to calculate the stress in the rope. Using Eqn. (8):

t =4mg

π d2=

4 × 2 × 10

3. 14× 12≈ 25

Sinced was in mm, the stress in in MPa - t = 25 MPa. Now, let’s calculate strain in the rope using Eqn.(11). Let’s assumeE = 2 GPa for nylon.

e =t

E=

25

2 × 103= 1. 25× 10−2

Now the elongation can be calculated from Eqn. (9). Using mm:

∆l = le = 3 × 103 × 1. 25× 10−2 = 37. 5

4.1.4. Poisson’s ratio

A closer look at the pulley example reveals the first complication with our one-dimensional straintheory developed so far. Ropes made of most ordinary materials will contract intransversedirection whenextended in the axial direction.

-8-

r0

l0

l1

∆l

d1

1

23

d0

So, in addition to the axial strain, we now need to have a definition for the transverse strain. Similar to theaxial strain we take it as a ratio of the change in diameter,∆d = d1 − d0, to the original diameter.

Let’s add another two coordinate axes - 2 in the plane of the paper, and 3 normal to the paper, asshown above.

We now use subscript 11 for axial strain,e11, and 22 and 33 for two transverse strains,e22, e33.

e22 = e33 =∆d

d0(12)

If we take the origin in trasverse direction at the axis of the rope, thenu2 is the transverse displacement ofany point in the rope along axis 2, andu3 is the transverse displacement along axis 3. So the definitions ofthe transverse strains can be rewritten as:

e22 =u2

x2(13)

e33 =u3

x3(14)

The profiles of the transverse displacement and strain, e.g. along 2, are simple:

∆d

2d0

1

23

−∆d

2

∆d

2

u2

e22

Hence, transverse strain is constant along the diameter.

It turns out the degree to which ropes contract in transverse direction is described by a single materialparameter,ν , the Poisson’s ratio, named after Siméon Denis Poisson (1781-1840), a French mathematician.By definition:

ν == −e22

e11= −

e33

e11(15)

The typical values of the Poisson’s ratio for common engineering materials are: steel - 0.33, Alu-minium alloys - 0.33, concrete - 0.1, rubber - 0.49.

The Poisson’s ratio of cork is 0. Why does this make it an ideal material to close bottles?

-9-

4.1.5. Displacement

A careful look at Eqns. (10), (13) and (14), will show that all three strain measures defined so far,axial and transversal, have the same structure.We mentioned before that strain is a gradient of displace-ment, which we now write explicitly using differentiation.

e11 =du1

dx1; e22 =

du2

dx2; e33 =

du3

dx3(16)

From Eqn. (16) displacements can be calculated via integration:

u1 = ∫x1

e11dx1 ; u2 = ∫x2

e22dx2 ; u3 = ∫x3

e33dx3 (17)

So a linear displacement profile means a constant corresponding strain and vice versa.

4.1.6. Elastic energy

When bodies deform internal forces doworkon corresponding displacements. Or we can say thatstresses do work on corresponding strains. Ifx1 is along the axis of the rope in tension (or a column undercompression), then the stress ist and the corresponding strain ise11. The work is an integral of stress overstrain, which for linear elasticity,t = Ee11, is just the area of a triangle.

Et

e11

Hence the storedelastic energyis

H =1

2te11 (18)

Note that the units ofH are the units of stress. In other wordsH is the elastic energy per unit of vol-ume, J / m3 or N / m2. To calculate the elastic energy for the whole body, one has to integrateH over vol-ume. If H is constant everywhere in the body, then the total elastic energy is simplyHV, whereV is thevolume of the body.

Elastic material is just like a convertional spring - when loads are applied, the body is deforming andthe external work is transferred into stored internal elastic energy. When the body is allowed to relax backto its original configuration, the stored elastic energy is trasnferred back into work. A watch spring a sim-ple analogy to an elastic material.

(Ex. probs. 2, 3, 4, 5.)

4.2. Pin-joined frame

Pin-joined frame is an idealised engineering structure in which there are no moments in the joints,and hence no bending moments anywhere. Somevery complex stress/strain fields will exist in the immedi-ate vicinity of the pins. If these are ignored, then the stress state in each bar is one-dimensional, oruniax-ial, tension or compression. These structures are easy to analyse.

Consider a pin-joined frame structure in 2D space, loaded by force F:

-10-

F

From moment equlibrium it follows that forces in the bars are always along bar axes. Thefree body dia-grams for this problem look like this:

x1

F

R2

R2

R1

R2R1 R1

x1

where both reaction forces are found from equilibrium equations. It’s clear that in this problemR2 is ten-sile (positive) and R1 is compressive (negative).

Let’s use superscripts 1 or 2 to denote the bar number. The stresses in bars aret1 = R1/A1 andt2 = R2/A2, whereA1 andA2 are the cross sections of the two bars.

If in each barx1 is along the axis, and if both bars are made of the same material, then

e111 =

R1

EA1

e211 =

R2

EA2

Axial displacements are found by integration.

u11 =

1

EA1 ∫ R1dx1 + C

u21 =

1

EA2 ∫ R2dx1 + D

whereC andD are integration constants found from the boundary conditions - displacements are zero atthe constraints.

For bar 1:u1(x1 = 0) = 0→C = 0.

For bar 2:u2(x1 = 0) = 0→D = 0.

Finally

u1 =R1

EA1x1

-11-

u2 =R2

EA2x1

The maximum displacements are clearly at the point where the force is applied:

u1max =

R1L1

EA1

u2max =

R2L2

EA2

However, if the displacement is substantial, then one also has to consider the rigid body motion:

F

F

In this problem this is only rigid body rotation. There is no rigid body translation.

(Ex. prob. 6.)

Note that solution of problems of this sort relies on the ability to solve for all reaction forces and/orreaction moments only from the equations of force and/or moment equilibrium. Such problems are calledstatically determinate. Problems where it is not possible to calculate reaction forces and/or moments onlyfrom the equations of equilibrium are calledstatically indeterminate.

4.3. Statically indeterminate systems

Consider a force applied to the end point ofn wires:

4

F

1 3 n...2

There are only two useful equation of equilibrium we can use from statics of rigid bodies - sum of theforces in 2 directions (in the plane of drawing), e.g. vertical and horizontal, is equal to zero. However, therearen unknown reaction forces.We need a furthern − 2 equations.

These can only come from some other information found in the problem. In this particular examplewe’d exploit the fact that displacements of the end points of all wires are equal:u1 = u2, u1 = u3, . . .,u1 = un, n − 1 equations altogether. So the problem becomes solvable.

(Ex. prob. 7.)

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4.4. Euler-Bernoulli (slender beam) bending theory

Mathematically a beam is a cylinder of arbitrary cross sectionA, i.e. if A is moved along a straightline, normal toA, then the bounding contour ofA will describe a surface of a beam.

Without the loss of generality let’s assume the beam is bent about axis 3:

3

2

1

A beam is calledslenderif some characteristic dimension of the cross section is much smaller than thelength, e.g. if√ A << L, whereL is the length of the beam.

The specific theory we study in this section, the Euler-Bernoulli theory, has several further assump-tions, which make the solution of a bending problem particularly simple:

• There is aneutral linein the beam, which passes through the same point in any cross section normalto the beam axis. The normal line does not change length.

• The deflections are small. This is required to ensure that the previous assumption is valid.

• Planar sections which are initially normal to the axis of the beam remain planar and normal to theaxis throughout bending;

• Any straight line normal to the neutral axis remains straight throughout the deformation.

• The body force is negligible.

• The deformation is slow.

• Only tensile or compressive deformation is allowed.

Together all above assumption mean that the stress state at any point is described only by the axialstresst, and strain is described by the axial and the two transverse strains,e11, e22, e33. If the bar is imagi-nary split into multiple thinfibres, running along its axis, then the behaviour of each such fibre can bedescribed by one-dimensional theory of Sec. 4.1.

e11 < 0

e11 > 013

2

Usually, in a bending problem, one is interested in findingdeflectionsresulting from the applicationof external loading.We designate deflection of the neutral line asw.

Let’s consider an infinitecimal length of the beam,dx1, separated by two cross sections, in deformedconfiguration.

-13-

dx1

w

α

orig. neutral line

2

13

u1

neutral lineafter deform.

u1

dw

Our assumptions lead to the following expressions for displacements:

u1 = x2 tanα = x2dw

dx1(19)

We use full differentials becausew is a function of onlyx1: w = w(x1).

u2 = w + ∫ e22dx2 (20)

u3 = ∫ e33dx3 (21)

Let’s consider force and momentum equilibrium for a "slice" of thicknessdx1:

q

dx1

2

1

F F+dFM M+dM

t

3

C

whereq is distributed load, i.e. load per length.

Note that in all problems studied so far, axial stresst was constant along the cross section. However,in bending problemst will change alongx2. This means our previous definition,t = f/A, Eqn. (5), is nolonger adequate.We need a new, continuum mechanics, definition of stress, i.e. for stress as a continuouslychanging function.

t = lim∆A→0f

∆A(22)

Exactly how it changes will come out as part of this analysis. Cross section force and momentchange smoothly alongx1, hence the use ofdF anddM. Sum of forces along 2 is zero:

dF = qdx1 → q =dF

dx1

Sum of moments about C is zero:

dM = Fdx1 + qdx1dx1

2

or, neglecting the quadratic term,:

dM = Fdx1 → F =dM

dx1→ q =

d2M

dx21

(23)

-14-

where

M = ∫ tdAx2 (24)

dA is the infinitecimal cross section area,tdA is the infinitecimal force normal to the cross section, andtdAx2 is the moment of this infinitecimal force about the neutral axis. The integral sums all such infiniteci-mal moments to give the total moment in the cross section.

From Eqn. (19):

e11 =du1

dx1= x2

d2w

dx21

From one-dimensional Hooke’s law:

t = Ee11 = Ex2d2w

dx21

(25)

This means thatt is a linear function ofx2. This is a very important result of this bending theory.

xmax2

neutral line

tmax

So that

t = tmax x2

xmax2

(26)

Putting this into (24):

M =tmax

xmax2

∫ x22dA

Note that the integral depends solely on the cross section. It is called thesecond moment of areaand isdenotedI :

I = ∫ x22dA (27)

So that

tmax =Mxmax

2

I(28)

By combining (28), (26) and (25):

d2w

dx21

=M

IE(29)

This is the famous ordinary differential equation (ODE) of 2nd order, linking the curvature with the bend-ing moment, the mechanical properties of material (the Young’s modulus) and the property of the cross sec-tion (the second moment of area).

The second derivative of deflection iscurvature, the inverse of theradius of curvature, ρ:

d2w

dx21

=1

ρ

Combining (29) with (23):

q =d2

dx21

IE

d2w

dx21

(30)

-15-

This is the famous 4th order ODE, suitable for variableI andE. If these are constant along the axis of thebeam, then:

q = IEd4w

dx41

(31)

It is interesting to note that the first 4 derivatives of deflections all have easy mechanical meaning:

dw

dx1= tanα ≈ α − angle, slope, gradient

d2w

dx21

=1

ρ∝M − curvature, bending moment, axial stress

d3w

dx31

∝F − shear force

d4w

dx41

∝q − distributed load

What this means is, if one is able to capture the deflection of the beam, then the stress/strain state can becalculated via numerical differentiation, e.g. graphically.

For uniaxial stress state

e22 = e33 = −ν e11

Sincee11 = Et, and t is positive on one side of the neutral line and negative on the other, thereforee22 ande33 are of different signs either side of the neutral line.For example, an initially square cross section mightdeform after bending like this:

ε11 < 0

31

2ε11 > 0

Above the neutral line the material is in tension, hencee22 = e33 < 0, and the width and the height of thecross section will decrease. Below the neutral line the material is in compression, hencee22 = e33 > 0, andthe width and the height of the cross section will increase.

The Euler-Bernoulli theory becomes exact if there is no shear force acting on the cross section. Oth-erwise, the theory is approximate, and the degree of deviation of this theory from experiment is directlyrelated to the magnitude of the shear force. This special case of zero shear force is calledpure bending, andis usually achieved experimentally with afour-point bendingsetup like this:

-16-

2P P

P

-PF

FL

L L

M

3 1

The added advantage of this geometry is a constant bending moment between the innermost rollers. Thisfact is widely exploited in laboratory materials testing experiments.

Eqn. (28) shows that the maximum stress in a cross section depends onxmax2 and I - two of the prop-

erties of the cross section. This means we need to examine the properties of cross sections in more detail.This is the subject of the next section.

(Ex. prob. 8).

4.4.1. Properties of cross sections

Consider a cross section as an arbitrary 2D area. It can be regular or irregular, with a single or multi-ple boundaries.We use a coordinate system (CS), with arbitrary orientation and origin O.We denote thisCS asx j , where j can be 1 or 2. So the coordinate axes arex1 andx2.

dA

1

2

1′

x1

x2

S2

2′

x2′

O

C

x1′S1

By definition thefirst moments of areain x j are:

i1 = ∫ x2dA (32)

i2 = ∫ x1dA (33)

wheredA is an ifinitecimal element of area located at positionx1, x2, and the integration is done over thewhole area. So the moments depend on CS!

In another CS shifted bySj :

x′ j = x j − Sj (34)

the first moments become:

i1′ = ∫ x2′dA = ∫ (x2 − S2)dA = i1 − S2A (35)

i2′ = ∫ x1′dA = ∫ (x1 − S1)dA = i2 − S1A (36)

-17-

By definitioncentroid, C, is a point, such that if the origin of a CS is put there, then both first moments ofinertia vanish.

From Eqns. (35) and (36), settingi1′ = i2′ = 0, one findsSj , the coordinates ofC in x j :

S1 =i2

A(37)

S2 =i1

A(38)

If cross section had thickness and constant density, then the centroid would coincide with the centreof mass.

By definition, thesecond moments of areaare

I11 = ∫ x22dA (39)

I22 = ∫ x21dA (40)

I12 = ∫ x1x2dA (41)

I r = ∫ r 2dA (42)

wherer 2 = x21 + x2

1. I r is called thepolar second moment of area. Often J is used instead ofI r .

(Ex. prob. 9.)

Although the moments can be calculated wrt any CS, for bending one needs the second moments wrtthe neutral axis. This means that one needs to be able to calculate the second moments wrtC.

(Ex. probs. 10, 11, 12, 13, 14, 15, 16, 17.)

Note that the units of second moments of area are [I ] = L4.

The moments of complex cross sections can be calculated by splitting the original section into easyto integrate areas.A hole, or a cutout, is a negative area. Ingeneral, ifA = ΣN

i=1 Ai , then

i1 = Σ j=Nj=1 i j

1 (43)

where

i11 = ∫A1

x2dA; i21 = ∫A2

x2dA; . . . i N1 = ∫AN

x2dA (44)

And the centroids of each sub-area are

S12 =

i11

A1; S2

2 =i21

A2; . . . SN

2 =i N1

AN(45)

From (45) and (43)

S2 =i1

A= Σ j=N

j=1

Sj2 A j

A=

1

A Σ j=Nj=1 Sj

2 A j (46)

S1 is calculated similarly.

The same applies to the second moments, e.g. forI r

I r = Σ j=Nj=1 I j

r (47)

where

I 1r = ∫A1

r 2dA; I 2r = ∫A2

r 2dA; . . . I Nr = ∫AN

r 2dA (48)

-18-

Care should be taken to refer the second moments of each sub-section to the centroid of the wholesection.

(Ex. prob. 18).

4.4.2. Plasticity in bending

Eqn. (28) says thattmax∝M , so with increasingM stress eventually reachesplastic limit. Theconsti-tutive modelat that point switches from elasticity to some plastic behaviour. The exact description of plas-ticity is beyond the scope of this course. Popular criteria for the onset of plasticity, such as due to vonMises or Tresca have very simple formulation for uniaxial stress state - whent = σY, whereσY is the yieldstress, then plasticity starts.

The stress profile is linear, with zero stress always on the neutral line.With increasing moment,stress increases maintaining a linear profile, until the maximum stress reaches yield value,σY. The diagramshows 5 stress profiles, withM increasing from profile 1 to profile 5.

1, 2, 3, 4, 5

xmax2

neutralline

tmax

plastic elastic

plastic

σY σY

Plastic flowfirst starts wherex2 = xmax2 , i.e. the layer of material furthest from the neutral line. The 2nd

stress profile is drawn at that moment.

As M keeps increasing, plasticity progresses inwards towards the neutral line. Eventually |t| on theother side will reachσY and plastic flow will start there too. The 4th stress profile is drawn at that moment.

Note that this diagram is nothing but qualitative. Many things might happen during plastic flow,including a shift of the neutral line.

Conclusion: if one wants to avoid plasticity in bending, one must ensure that

Mxmax2

I< σY (49)

(Ex. prob. 19).

4.5. Stability

Consider a pendulum under the action of a gravitational force. The pendulum has two equilibriumpositions - one with the pendulum directly below the support (left) and the other with the pendulum directlyabove the support (right):

-19-

α α

mgmg

mgmg

However, only the left configuration isstable. The right configuration isunstable. This can be verified eas-ily by considering a small angularperturbation, or disturbance, α , from the ideal vertical alignment of thependulum. Inthe stable configuration (left) there will be a horizontal force,mgsinα cosα , returning thependulum to the equilibrium position. In the unstable configuration (right) horizontal force of the samemagnitude pushes the pendulum away from the equilibrium position.

The idea of linking the concept of stability to a perturbation can be mathematically formalised as fol-lows.

ε

x

stable

unstable

δ

f (x)

f (x)

f (x)

f (x)

Consider functionf (x) which describes some system. This function has a specific boundary, or initial,condition atx = 0: f (0). If this initial condition is perturbed slightly, so that the value of the function isf (0), then the evolution of the system withx might be different. Thefollowing stability definition is due toLyapunov. f (x) is stable, if

∀ε > 0 ∃δ (ε ) > 0 ∋ if | f (0) − f (0)| <δ , then |f (x) − f (x)| < ε ∀x ≥ 0 (50)

where that maths symbols are∀ - for all, ∃ - there exist(s),∋ - such that.

Consider adding a horizontal supporting elastic spring of stiffnessk to help stabilise the inverted pen-dulum:

l

mgmg

x

Consider that the pendulum length isl . For a small horizontal perturbationx, there is now a horizontalrestoring force, pushing the pendulum back to the equilibrium position. The moment of the weight ismgx.The restoring moment from the spring iskxl. By comparing these moments, one can see that ifkl > mg,

-20-

then the system will be restored to the vertical equilibrium position.We can say that under this conditionthe vertical equilibrium is stable. Ifkl < mg, then the spring will not be able to restore the pendulum tovertical equilibrium. We can say that under this condition the vertical equilibrium becomes unstable. Ifkl = mg, then anew equilibrium position is possible, with the pendulum deviating from the vertical posi-tion. Thisequilibrium is stable.

Conditionkl = mg, which separates stable and unstable regions for the initial equilibrium configura-tion, is called thecritical condition. Inmost practical applications one wants to find a critical condition fora particular equilibrium configuration.

For the inverted pendulum example the equilibrium regions can be summarised as follows.

kl > mg; stable

kl < mg; unstable

kl = mg; critical, new equilibrium possible

Many structures can exhibit an elastic loss of stability - shells, beams, etc. Elastic instability can becatastrophic, but also useful. Simple devices such as hair clips or light switches rely on an elastic loss ofstability, or, more accurately, on the existence of multiple distinct stable elastic configurations separated byunstable regions. Thesimplest practical case is stability of columns under axial compression. Instability insuch cases is calledbuckling.

4.5.1. Buckling of columns

Consider a column with some unspecified BC, under compressive loadP, in acritical state:

M

2

13P

M

P

Recall that our bending analysis in Sec. 4.4, p. 12 did not account for the possibility of an axial load.Wenow hav eto revisit that analysis with the added axial loadP. Note that we assume thatP||x1 at all times.Also, we assume that there are no transverse loads and zero shear force in any cross section.

dw

dx1

2

1

PM M+dMP

σ11

3

The moment equlibrium gives:

dM = Pdw (51)

We hav eto add contraction due to axial compression tou1 in Eqn. (19):

u1 = x2dw

dx1+ γ x1 + C (52)

whereC is a constant determined from the BC andγ = const isa factor determined from the problem ofuniaxial compression.

If the cross section area isA and the Young’s modulus of the material of the column isE, then stressdue to axial compression istcomp = P/A. Strain due to axial compression is

-21-

ε comp11 =

tcomp

E=

P

EA

Displacementu1 is found by integration:

ucomp1 = ∫ ε comp

11 dx1 = ε comp11 x1 =

P

EAx1

Soγ = P/EA.

Note that we use elastic superposition to solve this problem.We split the total stress/strain states intothat due to bending and that due to uniaxial compression.We discuss this in more detail in Sec. 5.7.

From Eqn. (52):

ε11 = x2d2w

dx21

+ γ (53)

The bending moment is:

M = ∫AtdAx2 = E ∫A

ε11x2dA = E ∫A(x2

d2w

dx21

+ γ )x2dA = Ed2w

dx21

∫Ax2

2dA+ Eγ ∫Ax2dA

The first integral on the right hand side isI , the second moment of area. The second integral isi , the firstmoment of area. Given that the analysis is considered wrt the neutral line,i = 0. Sothe differential equa-tion of bending is identical to Eqn. (29) from Sec. 4.4:

M = EId2w

dx21

(54)

By differentiating Eqns. (51) and (54) over x1:

dM

dx1=

d

dx1

EI

d2w

dx21

=d

dx1(Pdw)

Finally, assumingP = const ,E = const andI = const ,one obtains a 3rd order linear ODE:

EId3w

dx31

= Pdw

dx1

or

d3w

dx31

−P

EI

dw

dx1= 0 (55)

where by introducing:

z2 = −P

EI

( P is compressive, hence negative ):

d3w

dx31

+ z2 dw

dx1= 0 (56)

the solution to which is:

w = C1 coszx1 + C2 sinzx1 + C3 (57)

Let’s find the integration constants for a simply supported column from both ends:

P

132

P

The length of the column isl . There are plenty of BC we can use:w(x1 = 0) = 0, w(x1 = l ) = 0,

-22-

w′(x1 = l /2) = 0, w′′(x1 = 0) = 0, w′′(x1 = l ) = 0.

w′′(x1 = 0) = 0 ⇒ C1 = 0

w(x1 = 0) = 0 ⇒ C1 + C3 = 0

w′′(x1 = l ) = 0 ⇒ C2 sinzl = 0

C2 = 0 is a trivial solution for a straight, unbuckled, column. This solution is of no interest to us. Hence itmust be that:

sinzl = 0

or

zl = nπ

Finally, the magnitude of the critical load is

Pcrit = EI

nπl

2

The lowest critical load corresponds ton = 1, i.e. when the deformed shape of the column is half of the sinewave:

Plowestcrit = π 2 EI

l2(58)

Note that we cannot fitC2 from the BC. This means that anyC2 fits the BC, provided it’s not toohigh to violate the small deflection assumption. The deflection is

w = C2 sinnπ x1

l(59)

(Ex. prob. 20, 21, 22, 23).

5. Three-dimensional stress/strain analysis

5.1. Motivation

We hav ealready defined one dimensional stress vector ast = f/A, Eqn. (5) in Sec. 4.1.1 on p. 3, andrefined it ast = lim∆A→0 f/∆A, Eqn. (22) in Sec. 4.4 on p. 12. Why do we need anything else?

Because the above vector definitions lead to simple scalar axial stress expressions,t = f /A ort = lim∆A→0 f /∆A, only for very specific CS.We will now illustrate what happens when different CS ischosen, e.g. for a problem of a cable under tension.

t0

θ

t cosθt sinθ

21

t0

ft0 =

fS0 f

f

θ

t0

f

1

2t =

fS

=f cosθ

S0= t0 cosθ nt

The cable is loaded by forcef along the axis. The cross section of the cable, cut normal to the axis, isS0.

-23-

The stress vector acting on an element of surface normal to the axis ist0 = f/S0. This is the left diagramabove.

Consider now what happens if I choose a different CS, rotated in the plane of the drawing by angleθ .This is the middle diagram above. The cross section area has changed. It is nowS = S0/ cosθ . HenceS ≥ S0. Howev er, from equilibrium, the force acting on this element of surface is stillf. Hence the stressvector is

t =fS

=f cosθ

S0= t0 cosθ

Clearlyt||t0 andt ≤ t0.

Moreover, the stress vector in the new CS, t, is no longer normal to the cross section, given by thenormal vectorn. Hence we can split this vector into anormaland ashearcomponents This is the right dia-gram above. The normal stress is

tn = t cosθ = t0 cos2 θ

The shear stress is

ts = t sinθ = t0 cosθ sinθ =1

2t0 sin 2θ

Note the use of the double angle in the shear stress expression. Thiswill be significant in Sec. 5.6.

Conclusion: the values of the normal and the shear stresses acting on an element of surface dependon its orientation. However, the stressstateis a physical property which cannot depend on the orientationof CS.

For illustration, let’s calculate the values of the normal and the shear stresses for certain values ofθ .

θ cosθ sinθ cos2 θ cosθ sinθ0 1 0 1 0

π /6 √3/2 1/2 3/4 √3/4π /4 √2/2 √2/2 1/2 1/2π /3 1/2 √3/2 1/4 √3/4π /2 0 1 0 0

normal =t0

θ = 0

t0

t0

shear = 0 shear =t0√3/4

θ =π6

t0

t

normal =t03/4

shear =t0/2

t0

t

θ =π4

normal =t0/2

θ =π3

t0

t

normal =t0/4

shear =t0√3/4

n

t0

t0

normal = 0

shear = 0

θ =π2

It turns out that to fully describe the stress state at a point one has to know stress vectors acting onthree elements of surface passing through that point, each with different normal.Together these 3 vectors,e.g. t1, t2, t3, form stresstensor, which is the subject of the next chapter.

-24-

5.2. Stress

Consider a body:

fn

1

n

∆SP2

3

xP

The following analysis is originally due to Cauchy (1822). We will use a Cartesian CS,xi , to giv ecoordinates to points in this body. Point P is defined by vectorxP = xP

i , i = 1, 2, 3.

Consider an imaginary planar cut across the body, with normaln = ni , through point P. Imagine asmall area of surface,∆S, on this plane, that includes point P. Finally, imagine thatfn = f n

i is the force act-ing on∆S. This is aninternal force, as opposed to any surface loadings. Superscriptn denotes that this isthe force acting on the element with normaln.

The stress vector definition, introduced by Eqn. (22) in Sec. 4.4 on p. 12, is still adequate. The onlyminor change is the superscriptn, to emphasise where the force and the stress are acting.

tn = lim∆S→0fn

∆S(60)

In generaltn will depend on the position of point P, and also on the orientation of∆S, i.e. onn. Inother wordstn = tn(xP, n). How does this function look like? To answer this question, we will use thecon-servation of linear momentumlaw (Newton’s second law) applied to a small body of mass∆m. Deforma-tion changes the density and volume of the body, but the mass is maintained constant.

f = ∆mdvdt

= ρ∆Vx (61)

wheref is the force acting on a body moving with velocityv, t is time,ρ is density,V is volume, dotdenotes time derivative and x is acceleration.

Let’s apply this law to a pyramid (tetrahedron) built from point P with edges along the coordinateaxes. h is the height of the pyramid from P to the base, alongn. tn is the stress acting on the base.

-25-

t32 A

1

2

3

B

C

P

h

t1

t11

t21

tn

n

t13

t33t3

t31

t23

t2

t22

t12

Let ∆S be the area of ABC. The volume of pyramid PABC is∆V =1

3h∆S.

Let’s now write down (and draw) all forces acting on the pyramid. We hav eto assume that there areforces, or stress vectors, also on the other three faces - ABP, BCP, ACP. We denote the forces asf i , i = 1, 2, 3,so that e.g.f1 is the force acting on the element of surface with normal along axis 1, i.e. BCP.The stresses on the faces areti , so that e.g.t2 is the stress acting on the element of surface with normalalong axis 2, i.e. ACP.

The force and the stress vectors can be split into components along the coordinate axes. Thusf1 = ( f 1

1 , f 12 , f 1

3 ), where f 11 is the force along direction 1 acting on the element of surface with normal

along 1, f 12 is the force along direction 2 acting on the element of surface with normal along 1 andf 1

3 is theforce along direction 3 acting on the element of surface with normal along 1.t2 = (t2

1, t22, t2

3).

So the superscript denotes where the stress or force is acting, and subscript denotes in which direc-tion it acts.

We adopt the following sign convention. Onplanes with normals pointing in the positive coordinatedirection the stresses are positive if they also point in the positive coordinate direction. On planes with nor-mals pointing in the negative coordinate direction stresses are positive if they also point in the negativecoordinate direction.For normal stresses this simply means that if they point in the same direction as thenormal, then they are positive; if they point in the opposite direction to the normals, then they are negative.Let’s assume positive forces for simplicity.

The force along 1 is

− f 11 − f 2

1 − f 31 + f n

1 = −t11∆S1 − t2

1∆S2 − t31∆S3 + tn

1∆S

where∆S1, ∆S2 and∆S3 are the respective surface areas. or, since∆S1 is the projection of∆S on the ele-ment of surface with normal along 1, then

∆S1 = ∆S|n ⋅ n1| = ∆Sn1 (62)

-26-

(Ex. prob. 24)

wheren1 is the normal to∆S1. With that the force along 1 is:

−t11∆Sn1 − t2

1∆Sn2 − t31∆Sn3 + tn

1∆S = (−t11n1 − t2

1n2 − t31n3 + tn

1)∆S

Also, we have to allow for the possibility of abody force, b, which acts on a volume, as opposed tof,which acts on a surface. Theunit of the body force is force per unit volume. With that the total force along1 is:

(−t11n1 − t2

1n2 − t31n3 + tn

1)∆S+ b1∆V

Finally the conservation of linear momentum along 1 looks like:

(−t11n1 − t2

1n2 − t31n3 + tn

1)∆S+ b1∆V = ρ∆Vx

A critical step - the linear momentum must be conserved for any arbitrary volume, including an infinitesi-mal volume, i.e. whenh → 0. ∆V = O(h∆S) hence∆V = o(∆S), meaning that as∆S → 0, ∆V/∆S → 0. Insimple words∆V tends to zero faster than∆S, so the∆V terms can be dropped, so finally the conservationof linear momentum along 1 looks like:

t11n1 + t2

1n2 + t31n3 = tn

1

and similar equations for directions 2 and 3:

t12n1 + t2

2n2 + t32n3 = tn

2

t13n1 + t2

3n2 + t33n3 = tn

3

The last 3 equations can be written as one vector equation:

Σ3i=1 t i

j ni = t ij ni = tn

j

where we used the Einstein’ssummation conventionwhich says that if an index repeats exactly twice in aterm of an expression, then summation is assumed over this index. Thissummation index is calleddummy.

By convention the 9 scalarst ij are written asσ ij . The classical form of the above equation is:

σ ij ni = tnj (63)

or in tensor notation:

σσ ⋅ n = tn (64)

The significance of this equation is that thestress stateat any point in a body is described by 9 scalars.Wetherefore see that stress analysis requires mathematical tools beyond vectors. Infact the 9 scalarsσ ij form arank 2 tensor(R2T) described in sec. 5.3 on p. 29.

Stress tensor can be shown graphically as a 3× 3 matrix containing the normal stresses on main diag-onal.

σσ = σ ij =

σ11

σ21

σ31

σ12

σ22

σ32

σ13

σ23

σ33

The symmetry of R2T means thatσ ij = σ ji , or, in tensor notationσσ = σσ T . So this means thatσ12 = σ21,σ23 = σ32, σ31 = σ13, and the matrix can be written as:

σσ =

σ11

σ12

σ13

σ12

σ22

σ23

σ13

σ23

σ33

which is symmetric.

Symmetric matrices are sometimes shown with only the upper triangle. All elements below the maindiagonal are replaced by the wordsymor simply by blank space:

-27-

σσ =

σ11

sym

σ12

σ22

σ13

σ23

σ33

=

σ11 σ12

σ22

σ13

σ23

σ33

5.2.1. Elementary cube of material

We hav esaid that stress state is a continuous field, and hence stress is defined at a point. It is usefulto imagine an infinitesimal volume of material centred at a point.We call this anelementary cube of mate-rial.

By definition,σ11 , σ22 , σ33 are callednormal stresses, and all otherσ ij , i ≠ j are calledshearstresses. Byconvention, positive normal stresses aretensileand negative normal stresses arecompressive.The normal stresses are shown on the elementary cube of material with arrows normal to the cube faces.The shear stresses are shown with arrows parallel to the cube faces:

13

2

3

11

1

32

33

31

2321

22

12

In this exampleσ11 andσ31 = σ13 are negative. All other stress components are positive.

We will show later that the stress tensor is symmetric, i.e.σ ij = σ ji .

( Ex. prob. 25.)

5.2.2. Conservation of linear momentum

Let’s see what the conservation of linear momentum means for an arbitrary body of volumeV andsurfaceS:

∫Vbi dV + ∫S

tni dS= ∫V

ρ xi dV (65)

(63) → (65):

∫Vbi dV + ∫S

σ ij n j dS= ∫Vρ xi dV

or using the Green’s theorem,

∫Vbi dV + ∫V

σ ij , j dV = ∫Vρ xi dV

where subscripts after a comma denote spatial differentiation:σ ij , j = ∂σ ij /∂x j .

Further:

∫V(bi + σ ij , j − ρ xi )dV = 0

which must hold∀V, so the integrand is zero:

σ ij , j = −bi + ρ xi (66)

or in tensor notation:

∇ ⋅ σσ = −b + ρ x (67)

-28-

If the acceleration and body forces are small and can be neglected, then the equation is particularlysimple.

σ ij , j = 0 (68)

or in tensor notation:

∇ ⋅ σσ = 0 (69)

Theseequilibrium equationsare one of the few fundamental building blocks of the continuum solidmechanics.

(Ex. probs. 26, 27)

5.2.3. Conservation of angular momentum

First read sec. 5.3 until you understand the alternating tensor.

The conservation of linear momentum law effectively says: force changes velocity. Similarly, theconservation of angular momentum law says: moment of force changes angular momentum.

For a body of massm with velocityv, the angular momentum is

L = mv × x (70)

wherex is the vector from the axis of rotation to the centre of mass of the body and× is thecross productof velocity and the position vectors. NotethatL is directed along the axis of rotation:L⊥v, L⊥x. If theforcef is applied to a deformable body of unchanging mass, then the conservation of the angular momen-tum law is:

md(v × x) = f × xdt

or

md

dt(v × x) = f × x (71)

For a deformable body we have to apply this law to every element of massdm = ρdV.

We now switch to index notation, in whichv × x is expressed aseijk v j xk, whereeijk is the alternatingtensor explained in Sec. 5.3.With that Eqn. (70) can be rewritten as

dLi = ρdVeijk v j xk

or, for the whole body:

Li = ∫Vρeijk v j xkdV

We need to take into account the moments of the body and the surface forces. So the conservation law is:

∫Vρ

d

dt(eijk v j xk)dV = ∫V

eijk b j xkdV + ∫Seijk tn

j xkdS (72)

Using Eqn. (63) and the Green’s theorem the last integral in Eqn. (72) can be converted to a volume inte-gral:

∫Seijk tn

j xkdS= ∫Seijkσ jpnpxkdS= ∫V

eijk (σ jp xk),pdV = ∫Veijk (σ jp,pxk + σ jp xk,p)dV

it is easy to show thatσ jp xk,p = σ jk . Using the equilibrium equations, Eqn. (66):

= ∫Veijk (−b j xk + ρ x j xk + σ jk)dV

Expanding the first integral in Eqn. (72):

∫Vρ

d

dt(eijk v j xk)dV = ∫V

eijk (ρ v j xk + ρv j xk)dV

So that Eqn. (72) can be rewritten as:

-29-

∫Veijk (ρ v j xk + ρv j xk − b j xk + b j xk − ρ x j xk − σ jk)dV = 0

This equation must be valid for any arbitrary volume, hence the integrand is zero. After cancelling termswe have:

eijk (ρv j vk − σ jk) = 0

It is easy to show thateijk v j vk = 0, hence finally we have

eijkσ jk = 0

and this means thatσ is symmetric:

σ jk = σ kj (73)

or, in tensor notation:

σσ = σσ T (74)

(Ex. probs. 37, 38, 39, 40, 41, 42, 43).

5.3. Tensors

Imagine two orthonormal CS:xi andxi ′, with the common origin. In other words, the primed,′, CSis rotatedwith respect to the original, unprimed, CS.We call any such rotation acoordinate transformationor CT for short. So any vector →a will have componentsai = (a1, a2, a3) in the original CS andai ′ = (a1′, a2′, a3′) in the new, rotated, CS. The vector itself does not change with any CT. What is chang-ing are its components.We say that a vector is an object that isinvariant to CT.

By definition, scalarsAij , i , j = 1. . . N, whereN is the dimensionality of space, form arank 2 tensoror R2T, if f or any 2 arbitrary vectors,bi andci , the product:Aij bi c j is invariant to CT:

if ∀bi , ci : Aij bi c j = Aij ′bi ′c j ′ ∴ A is R2T (75)

(Ex. prob. 28)

In 2D space R2T has 22 = 4 components. In3D space R2T has 32 = 9 components. Ingeneral R2Ttensor hasN2 components inN-dimensional space.

R2T transforms a vector into another vector:

ci = Aij b j or c = A ⋅ b (76)

(Ex. prob. 29)

Rotation tensor,Rij = R, is the key. R rotates vectora, leaving its magnitude intact. The rotated vec-tor is denoteda′. We use′, to denote a vector in a new, rotated, CS. So the basis vectors are rotated fromei

into ei ′: It is easy to show that for a CT eachij component ofR is cos∠(ei ′, e j ). Finally R is orthogonal:R−1 = RT , where superscript−1 means theinverseand superscriptT denotes atransposeof a R2T, andRT ⋅ R = I = δ ij , whereI = δ ij is the rank 2identitytensor, also called Kronecker delta tensor. Its matrixanalogue is the unit matrix.

If R rotates vectora into a′, thenRT rotatesa′ back intoa. Remember that the vector→a does notchange! Whatis changing are it’s components in two different CS.

(Ex. probs. 30, 31, 32)

It is also easy to show that components of R2T change like this:

Tij ′ = Rip RjqTpq or T′ = R ⋅ T ⋅ RT (77)

which can be taken as an alternative definition of R2T. To enable manipulations with tensors using matri-ces, one must have identical last subscript of the first tensor and the first subscript of the second tensor, i.e.A ⋅ B = ApqBqr , B ⋅ A = BpqAqr , and

(B ⋅ A)T = AT ⋅ BT = ATpqBT

qr = (BpqAqr )T

-30-

(Ex. prob. 33)

Similar to Eqns. (75) and (77) we can define tensors of arbitrary rank.For example:

if ∀bi : ai bi = ai ′bi ′ ∴ ai is rank1 tensor ,i. e. vector (78)

if ∀bi , ci , di : Mijk bi c j dk = Mijk ′bi ′c j ′dk′ ∴ Mijk is rank3 tensor (79)

if ∀bi , ci , di , ei : Qpqrsbpcqdr es = Qpqrs′bp′cq′dr ′es′ ∴ Qpqrs is rank4 tensor (80)

from where we can write down how components of tensors of various ranks change with CT. For ranks 1, 3and 4 these transformations respectively are:

ai ′ = Rij a j (81)

Mijk ′ = Rip Rjq Rkr M pqr (82)

Qijkl ′ = Rip Rjq Rkr RlsQpqrs (83)

Tw o tensors of the same rank areequalif f all corresponding components are equal.For example,rank 4 tensorsB andD, are equal iffB1111 = D1111, B2111 = D2111, B3111 = D3111, . . ., B3333 = D3333.

A R2T A is symmetriciff A = AT , or, in index notation, iff Aij = A ji .

A R2T isskew-symmetricor anti-symmetriciff A = −AT , or, in index notation, iff Aij = −A ji .

Any two tensors of the same rank can be added or subtracted. The result is a tensor of the same rank,e.g. Aij + Bij = Cij , M pqr − Npqr = Zpqr.

Any R2T can be represented as sum of a symmetric and a anti-symmetric tensors:Aij = sym(Aij ) + asym(Aij ).

(Ex. prob. 34, 35, 36).

A helpful artificial rank 3 tensor is thealternating tensor, eijk , also called permutation or Levi-Civitatensor in some books. By definition:

eijk =

+1,

−1,

0,

ijk = 123, 231, 312

ijk = 132, 321, 213

otherwise

(84)

eijk is useful for writing down cross products and determinants.

(Ex. probs. 44, 45, 46, 47, 48.)

5.3.1. Eigenvalue / vector

Let T be a symmetric R2T. The eigenvalue/eigenvector problem is to find 3 scalar valuesλα and 3vectorsxα , α = 1⋅⋅⋅3, so that:

T ⋅ xα = λα xα (85)

Think about the meaning of Eqn. (85).We said before that a R2T transforms a vector into another vector.Sinceλα x||x, we see thatxα are chosen so that the action ofT on them does not rotate but scales them.λαare called theprincipal valuesor eigenvaluesand the vectorsxα are called theprincipal vectorsor eigen-vectors.

Eqn. (85) has non-trivial solutions iff

det(T − λI) = 0 (86)

which is called thecharacteristic equation.In 3D space this is a cubic equation forλ :

−λ3 + I Tλ2 + 12 (II T − (I T)2)λ + III T = 0 (87)

whereI T, II T, III T are called thefirst, secondandthird invariantsrespectively. By definition

-31-

I T = trT = Tii (88)

II T = T: T = Tij Tij (89)

III T = detT (90)

where tr is called thetraceof R2T. This operator is defined only for R2T. It is the sum of all diagonalcomponents.

Some texts define the second invariant as12 (T: T − (trT)2). Ex.prob. 51 shows why.

If T is symmetric then it is possible to prove that all 3 eigenvalues are real. After Eqn. (86) is solved,eachλα is substituted back into Eqn. (85) and eachxα is found. By convention:

λ1 ≥ λ2 ≥ λ3 (91)

If only one principal value is non-zero, and the other two values are zeros, the tensor is calleduniax-ial.

If two principal values are non-zero, and one is zero, the tensor is calledbiaxial. If the two non-zerovalues are equal, the tensor is calledequi-biaxial.

If all three principal values are non-zero, the tensor is calledtriaxial. If, in addition, all three valuesare identical, the tensor is calledhydrostaticor equi-triaxial.

(Ex. probs. 49, 50, 51, 52, 53, 54, 55, 56, 57).

5.4. Strain

When stresses are applied to a body it deforms. Solid mechanics is the mechanics ofdeformablebodies. Ifthe bodies do not deform, then the subject is called mechanics of rigid bodies, which is a differ-ent subject completely. Analysis of deformation andmotionof solids, sometimes also calledkinematicsisthe subject of this section.

In the analysis of motion one must distinguish theoriginal (also calledundeformedor reference) con-figuration fromcurrent(also calleddeformed) configuration. We use capital letter to refer to the originalconfiguration, and small to refer to the deformed configuration, wherever possible.

u + du

1

xX

P

p

qQ

2

3

u

dX dx

Consider pointP with coordinatesX in the original configuration, Imagine that after loading pointP dis-placedby distanceu and now has coordinatesx, and we call itp. Consider another pointQ some short dis-tancedX from P in the original configuration. Its location after deformation is labelledq, a distancedxfrom p. Displacement vector connectingQ andq is u + du.

One of the major assumptions of the continuum mechanics is that in general the motion of a bodyunder load can be described by continuous, smooth, and therefore differentiable functions:

-32-

x = x(X, t) (92)

wheret is time or some other parameter. The motion that is very slow is said to bequasi-static.

These functions arenon-linearin general, however their differentials are related linearly:

dx =∂x∂X

dX (93)

or in index notation:

dxi =∂xi

∂X jdX j (94)

By definition

F = Fij =∂xi

∂X j=

∂x∂X

(95)

is called thedeformation gradienttensor. With that Eqn. (93) can be rewritten:

dx = F ⋅ dX (96)

Deformation must bere versible,meaning theinverseof the deformation gradient must always exist:

F−1 = F−1ij =

∂Xi

∂x j=

∂X∂x

(97)

The deformation gradient is a special type of R2T, calledtwo-pointtensor. By the definition of the inverseof R2T

F−1 ⋅ F = F−1ij F jk = F ⋅ F−1 = Fij F

−1jk = I = δ ik (98)

The square of the length ofPQ is (dL)2 = dX ⋅ dX, and of pq is (dl)2 = dx ⋅ dx. The change in the squareof the length, due to motion,∆L2, is

∆L2 = dx ⋅ dx − dX ⋅ dX = dxi dxi − dXi dXi (99)

or using Eqn. (96):

∆L2 = Fij dX j Fik dXk − dXi dXi = Fij dX j Fik dXk − dX j dXkδ jk = dX j dXk(Fij Fik − δ jk) (100)

or finally

∆L2 = dX j dXk(Fij Fik − δ jk) = (dX⊗dX): (FT ⋅ F − I) (101)

A change in length is a measure ofstretch.

By definition

C = C jk = Fij Fik = FT ⋅ F (102)

is called theleft Cauchy-Green tensor.

Stretch is measured in units of length. What we want is arelative dimensionlessmeasure of defor-mation, which we callstrain. There are many different definitions of strain.We can adopt one suggestedby Eqn. (101):

E = ½(FT ⋅ F − I) = E jk = ½(Fij Fik − δ jk) (103)

The factor ½ will be explained later.

Strain can be expressed via displacements as:

E jk = ½(ui , j + u j ,i + uk,i uk, j ) = E = ½(∇u + (∇u)T + (∇u)T ⋅ (∇u)) (104)

If displacement gradients∇u are small such that their squares can be neglected:|(∇u)T ⋅ (∇u)| = o(|∇u|), thensmall strain formulationwill result:

-33-

ε ij = ½(ui , j + u j ,i ) = εε = ½(∇u + (∇u)T ) (105)

Strain is a symmetric R2T, hence all previous analysis regarding the properties of R2T applies tostrain. Inparticularε11, ε22 andε33 are callednormal strains. Positive values mean elongation and nega-tive mean contraction.ε12, ε23 andε31 are calledshear strains. Shear strains describe change of shape, i.e.change in angle between two arbitrary straight lines before and after the motion.

The principal strains are:

ε1 ≥ ε2 ≥ ε3 (106)

The first invariant of strain tensor,I ε = ε ii , quantifiesvolumetric strain. This is a measure of changein volume for a unit cube of materials - ifI ε > 0, then the volume is increasing; ifI ε < 0 then the volume isdecreasing. IfI ε = 0, then the strain is calledincompressible. If the mass conservation is in force, thensimilar conclusions can be made about density, e.g. I ε > 0 would mean decreasing density.

(Ex. probs. 58, 59, 60, 61, 62, 63, 64).

5.5. Maximum shear value

Now we understand that stress and strain states are expressed via symmetric R2T. Therefore exactlythe same tensor manipulations can be done with both stress and strain. In particular it’s easy to show thatthe maximum shear stress,τmax, and the maximum shear strain,γmax have identical expressions:

τmax =σ1 − σ3

2; γmax =

ε1 − ε3

2(107)

whereσ1, σ3, ε1 andε3 are the maximum and the minimum principal values of stress and strain respec-tively.

The maximum shear exists on planes with the normal at 45to the principal directions 1 and 3.

The above expressions show that in hydrostatic cases, i.e. when the maximum and the minimum prin-cipal values are equal, there is zero shear everywhere.

(Ex. probs. 65, 66).

5.6. Mohr’s diagram

The Mohr’s diagram, also called the Mohr’s circle, is a 2D graphical representation of a symmetricR2T,Tij . Consider that such tensor has principal valuesT1, T2 andT3. Then the Mohr’s diagram is:

normal

shear

T1T2T3

(T2 + T3)/2

(T1 + T2)/2

(T1 + T3)/2

(T1 − T2)/2

(T1 − T3)/2

(T2 − T3)/2

-34-

The axes are the normal and the shear values. Thediagram consists of three circles centred on the normalaxis. Theradii of the circles are equal to the maximum shear values in each principal plane. The circlesintersect the normal axis at principal values. Theadmissiblevalues are within the large circle, but outsideof the two smaller circles. Note that the diagram is symmetric wrt the normal axis, so often only a half isdrawn, with positive shear values.

The Mohr’s diagram is only useful for CS in which at least one coordinate axis is the principal direc-tion. Thismeans that only stress states where at least one normal stress is the principal stress can be repre-sented on the diagram. In other words rotations about only a single coordinate axis, from the principal CS,can be visualised with the Mohr’s diagram. Hereis an example.

We start from the principal orientations (left). Imagine that we start rotating the elementary cube ofmaterial by angleα about axis 3 (middle). Shear componentT12 appears and grows to the maximum valueatα = π /4 (right). Thecorresponding matrix representation is shown below each diagram.

T1T3 T2

T12

T3 T22 T11

2α

Tmax12

T3 T11 = T22

T1

0

0

0

T2

0

0

0

T3

T11

T21

0

T12

T22

0

0

0

T3

Tmin11

Tmax21

0

Tmax12

Tmax22

0

0

0

T3

Note that the Mohr’s diagram can represent at most 4 components of R2T. As long as the other com-ponents are zero, the representation in meaningful. This clearly shows the limitations of the Mohr’s visualaid.

If we continue rotating our cube further about axis 3, then the shear valueT12 starts decreasing (leftand middle) until it vanishes again atα = π /2, and we are back to the principal CS (right).

T12

T3

2α

T11T22

T12

T3

2α

T22 T11 T2T3

2α

T1

T11

T21

0

T12

T22

0

0

0

T3

T11

T21

0

T12

T22

0

0

0

T3

T1

0

0

0

T2

0

0

0

T3

Conclusion: Mohr’s diagram usefulness is limited to only the most simple stress states.We do not talkabout it further.

(Ex. probs. 67, 68, 69).

5.7. Elasticity

So far we have defined two symmetric R2T - stress,σσ , and strain,εε , which describe the states ofloading and of deformation respectively, at every point in the material. This section explains how the twotensors are related to each other.

Up to now the theory we have been developing was almost exclusively mathematical. The only bitsof physics we have used were the fundamental conservation laws. However, we said absolutely nothingaboutmaterial behaviour. It turns out that material behaviour is indeed the link between the strain and thestress tensors.

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Mathematically material behaviour, sometimes also calledconstitutive behaviouror constitutivemodelis just some tensor-valued function linking stress to strain:

σσ = f (εε ) ; εε = g(σσ ) (108)

There are probably infinitely many of such functions. But do any of them describe a real material?

The simplest solid material behaviour that we know of is calledelasticity. There are many ways elas-ticity can be defined.

σσ = C: εε (109)

whereC = Cmnop = const isa rank 4 tensor, is one such definition.C is called theelasticor thestiffnesstensor.

Another definition is to say that if strains areproportionalto stresses, then such material is elastic.This definition is, of course, closest to that discovered by Robert Hooke in 1660. Theword "discovered" isimportant. It’s one thing to have a mathematical description, but quite another to confirm that some realmaterials, under some conditions, obey this mathematical description.

The two above definitions are strictly speaking restricted to alinear elastic behaviour.

Yet another, a more general, definition of elasticity, includingnon-linearelasticity, is to postulate theexistence of anelastic potential, W, such that

σσ =∂W

∂εε=

∂W

∂ε ij= σ ij (110)

or equally

εε =∂W

∂σσ=

∂W

∂σ ij= ε ij (111)

Other definitions of elasticity can be given based on the key properties of this behaviour.

Elastic deformation isrecoverableor re versible, meaning that if loads are applied to a body and thanremoved, then the body returns to the initial configuration.

Elastic deformation ispath independent, meaning that stress at the end of the deformation dependsonly on strain at the end of the deformation. The path by which this state was reached is immaterial.

Elastic behaviour isconservative, meaning that the energy is conserved. If some amount of externalwork was needed to deform a body, then exactly the same amount of energy with be recovered when exter-nal loading is removed.

Elastic deformation islossless, meaning there are no energy losses during a cycle of elastic load-ing/unloading.

All above characteristics of real elastic behaviour put certain constraints on the type of a mathemati-cal functions describing elasticity in Eqn. (108). These must besingle-valued, meaning that only a singlestress tensor corresponds to each strain tensor, and vice versa. Thefunctions must beinversible, meaningthat if it possible to calculate stress from strain, then it must also be possible to calculate strain from stress.All these requirements are satisfied by our first definition, Eqn. (109). In particular inverting Eqn. (109)one can write:

εε = S: σσ (112)

whereS = Smnop = constis a rank 4 tensor called thecompliancetensor. ClearlyS is the tensor inverse ofC. We will show later what it means.

The following analysis can be made with either definition of Eqn. (109) or that of Eqn. (112). Let’sstick with Eqn. (109).

C has 34 = 81 components (number of spatial dimensions to the power of rank). However, only 21 ofthose are independent in the most general case, due to symmetries of the strain and the stress tensors anddue to the existence of the elastic potential:

-36-

Cijpq = C jipq = Cijqp = Cpqij (113)

In the most general case Eqn. (109) can be written explicitly in matrix form as:

σ11

σ22

σ33

σ12

σ23

σ31

=

C1111

sym

C1211

C2311

C3111

C1122

C2222

C1222

C2322

C3122

C1133

C2233

C3333

C1233

C2333

C3133

2C1112

2C2212

2C3312

2C1212

sym

2C1123

2C2223

2C3323

2C1223

2C2323

2C1131

2C2231

2C3331

2C1231

2C2331

2C3131

ε11

ε22

ε33

ε12

ε23

ε31

(114)

or

σ11

σ22

σ33

σ12

σ23

σ31

=

C1111

sym

C1122

C2222

C1133

C2233

C3333

C1112

C2212

C3312

C1212

C1123

C2223

C3323

C1223

C2323

C1131

C2231

C3331

C1231

C2331

C3131

ε11

ε22

ε33

2ε12

2ε23

2ε31

(115)

Both forms can be used for calculation, but neither form can be rotated. This is because the tensor nature ofthe Eqn. (109) is lost after the introduction of factors 2.

It is possible to show that the required number of independent components inC can be reduced fur-ther if material symmetryis exploited. Inthe simplest case there are only 2 independent components inC.The main characteristic of this special case is that material properties are the same in all directions. Materi-als possessing this property are calledisotropic. Isotropic materials are a tiny minority of materials foundin nature, but due to simplicity of their analysis, they form the majority of man-made materials. All othermaterials are calledanisotropic. The vast majority of pure solids (single crystals), minerals, and compositematerials are anisotropic, as the table below shows.

Symmetry type Name Con-stants

Examples

no symm. plane triclinic 21 anorthosite, turquoise

1 plane monoclinic 13 orthoclase (KAlSi3O8), igneous rocks

3 planes orthotropic, orthorombic 9 Ga, U, rolled plate, wood, some composites

stretched cubic tetragonal, trigonal orrhombohedral

6 or 7 7: dolomite (CaMg(CO3)2); 6: In, Sn, As,Bi, Hg, Se, Te, quartz (SiO2)

1 plane, 1 axis hexagonal 5 Be, Cd, Zn, Zr, carbon fibre composites

3 axes cubic or isometric 3 Fe, Po, Al, Cr, Cu

point of symm. isotropic 2 normalised steels, Al alloys

The topic of anisotropy is beyond the scope of this course. From now on we restrict the course to lin-ear isotropic elasticity only. In this caseC is vastly simplified:

σ11

σ22

σ33

σ12

σ23

σ31

=

2µ+ λ

sym

λ2µ+ λ

λλ

2µ+ λ

0

0

0

2µ

0

0

0

0

2µ

0

0

0

0

0

2µ

ε11

ε22

ε33

ε12

ε23

ε31

(116)

whereλ and µ are called theLamé elastic constants.The tensor form of linear isotropic elasticity is:

-37-

σσ = 2µεε + λ trεε I ; σ ij = 2µε ij + λε kkδ ij (117)

So that it’s easy to see that

Cijkl = 2µδ ikδ jl + λδ ij δ kl (118)

The inverse of a rank 4 tensor is defined as follows.

C−1 is an inverse ofC ↔ C: C−1 = I ; Cijkl C−1klmn = Iijmn (119)

where by definition

I = Iijkl = δ ikδ jl (120)

is calledrank 4 identity tensor.It’s meaning is clear from Eqn. (116) - its main diagonal components are 1,and all other components are 0.

So the fact thatS is an inverse ofC means that

C: S = S: C = I ; Cijkl Sklmn = Sijkl Cklmn = Iijmn (121)

Elastic law must be valid for any CS:

σσ ′ = C′: εε ′ (122)

whereC′ is the elastic tensor in a rotated CS. It rotates as any other rank 4 tensor:

Cijkl ′ = RimRjn RkoRlpCmnop (123)

However, it is possible to prove that isotropic elastic tensor is the same in any CS:

Cijkl ′ = Cijkl (124)

Indeed, that is was the term "isotropic" mean.

Different applications of elasticity suggest different optimal pairs of isotropic elastic constants. Thusthe Lamé constants can be recast asE, theYoung’s modulus,andν , the Poisson’s ratio, or asK , thebulk orcompressionmodulus, andG, theshearmodulus. Theseare inter-related as follows:

E =µ(3λ + 2µ)

λ + µ; ν =

λ2(λ + µ)

(125)

λ =ν E

(1 + ν )(1 − 2ν ); µ =

E

2(1+ ν )(126)

K = λ +2

3µ =

E

3(1− 2ν ); G = µ (127)

One remarkable property of any elastic behaviour can be drawn out if one considers Eqn. (109) fortwo different strain states,εε A andεε B. Each strain tensor has a corresponding stress state:σσ A = C: εε A andσσ B = C: εε B. Imagine a composite strain state:εε C = εε A + εε B. It immediately follows that its correspondingstress state is

σσ C = C: εε C = C: (εε A + εε B) = C: εε A + C: εε B = σσ A + σσ B

So that the stress state resulting from a composite deformationA followed byB is a sum orsuperpositionof stress states resulting from deformationsA an B. This fact is of great importance for the solution of elas-tic problems. It means that complex loadings can be represented as a superposition of multiple simpleloadings. Whenthe elastic problem for each simple loading has been solved, then the complete solution isobtained by the superposition of individual simple solutions.

(Ex. probs. 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81.)

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5.8. Solving solid mechanics problems

A constitutive model, of which elasticity is the simplest example, completes a set of fundamentalconcepts and equations required to pose and solve a solid mechanics problem.

The fundamental quantities are

• x = xi - the material point position vector

• tn = tni - the stress vector on an element of surface with normaln = ni

• b = bi - the body force (force per unit volume) field

• ρ - the scalar density field

With these concepts we have derived the Cauchy stress tensor, Eqn. (63), the small strain tensor, Eqn.(105), and the equations of equilibrium, Eqn. (67).

BC can be of two types. Thefirst type is prescribed traction (loading). LetΓ denote the whole of theboundary of a body. Then atraction BCis a constraint where on part of the boundaryΓt tractiontn is pre-scribed astn = tn. A displacement BCis a constraint where onΓu displacementu is given as u = u.

If all BC are of a displacement type and an elastic behaviour is assumed, then nothing else is neededto pose a solid mechanics problem.

∇ ⋅ σσ = −b + ρ xσσ = C: εεu = u on Γu

(128)

Using the elastic relationsσσ is expressed as functions ofεε , and hence ofu: σσ = σσ (u). Whenthese areinserted back into the equilibrium equations, the result is 3 PDE for 3 unknown functionsu. These equa-tions are then solved subject to the BC. Onceu is found,εε is immediately available by the differentiationof displacements, andσσ is found from the elasticity relation.

If all BC are of a traction type, then the equilibrium equations can be solved for stress directly. Oncestress fields are found, strain is obtained from the elastic relationship, and displacements are calculated bythe integration of strain. However, the strain-displacement relationship can be thought of as 6 PDEs for 3unknownsu. Such system might not have a solution, unless the strain functions arecompatible. The fol-lowing compatibilityequations are easily obtained:

ε ij ,kl + ε kl,ij = ε ik, jl + ε jl ,ik (129)

and the solid mechanics problem can be formulated like this:

∇ ⋅ σσ = −b + ρ xtn = tn on Γt

εε = S: σσε ij ,kl + ε kl,ij = ε ik, jl + ε jl ,ik

(130)

Finally, if both traction and displacement BC are applied, the solid mechanics problem is ofmixedtype, the solution method for which is a combination of solution methods for pure traction and pure dis-placement BC.

There is no solution to a general 3D solid mechanics problem, even for the simplest case of linearisotropic elasticity with no body forces and under quasi-static deformation (negligible rates and accelera-tions). Solutionsare available only for specific classes of problems, typically limited by geometry or by atype of loading. Solid mechanics is still as much art as it is science.

(Ex. prob. 82, 83, 84, 85).

6. Special cases

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6.1. Two-dimensional stress/strain problems

6.1.1. Plane stress

Stress state where one principal stress is zero is calledplane stress.

The stress tensor, therefore, looks like this:

σσ =

σ1

sym

0

σ2

0

0

0

Importantly, there is no rotation about axis 3 that can give rise toσ3 ≠ 0. Henceone can remove from theanalysisσ31, σ32 andσ33 completely. What remains is the two-dimensional stress tensor

σσ =

σ11

sym

σ12

σ22

Accordingly we can use an elementary square, not cube, to visualise a two-dimensional stress state,because nothing happens along the third direction. Here is an example of a stress state in principal (left)and some other CS (right).

σ22 = σ1

1

2

σ11 = σ2

σ222 σ11

1

σ12

σ21

Note that the corresponding strain tensor is three-dimensional:

εε =

ε11

sym

ε12

ε22

0

0

ε33

whereε33 is calculated from Eqn. (151) in ex. prob. 79:

ε33 = −νE

(σ11 + σ22)

after the two-dimensional stress-strain problem has been solved.

A typical example of plane stress state is a very thin flat film:

L

t

1

Tn2Tn

Tn1

3

2

where the film thickness,t << L, whereL is some characteristic dimension in the plane of the film.

Let 1 and 2 be axes in the plane of the film, and axis 3 normal to it. Then if the side surfaces are trac-tion free, and the traction is applied only on the film edges, such thatTn

3 = 0, thenσ13 = σ23 = σ33 = 0 onthe free surfaces. Becausethe film is very thin, it is then assumed that even if σ33 ≠ 0 somewhere in the

-40-

interior of the film it is very small and can be neglected. Thusa plane stress state is assumed in such cases.

As a side note, any traction free surface is in plane stress state.A typical example is a roller, a longcylinder under compressive load normal to its axis.

planeσ1

2

3

planeε

On both free ends of the roller plane stress state is established.

To wards the symmetry plane of the roller another two-dimensional state can be assumed. Due tofriction forces material there is constrained and cannot freely move in the axial direction. Henceε33 ≈ 0,andplane strainstate is established.

(Ex. prob. 86, 87).

6.1.2. Plane strain

Strain state where one principal strain is zero is calledplane strain.The strain tensor therefore lookslike:

εε =

ε1

sym

0

ε2

0

0

0

Similar to plane stress state, direction 3 becomes of no interest, and a two-dimensional strain tensor can beused:

εε =

ε11

sym

ε12

ε22

Importantly the stress tensor is three-dimensional:

σσ =

σ11

sym

σ12

σ22

0

0

σ33

However, σ33 is calculated from Eqn. (117):

σ33 = λ(ε11 + ε22)

after the two-dimensional stress-strain problem has been solved.

Plane strain state is typically associated with thick components, specifically if there are factors whichconstrain deformation in thickness direction. One of such factors could be surface friction. Collectively,the extent of such factors is calledconstraint. Four point bending of slender beams is an example oflowconstraintgeometry, where deformation in thickness direction, between the central rollers, is not con-strained.

-41-

In contrast, the hot rolling press roller, discussed above, is one example ofhigh constraintgeometry.Another, similar, example is bending of non-slender beams, where height and thickness are comparable tolength,L ≈ H ≈ t.

plane

H

Lt

symm

Here the friction between the rollers and the block will constrain deformation in thickness,t, direction.Hence, on the through thickness symmetry plane, plane strain state can be assumed.

Note that in this example the stress state on the front and the rear surfaces is plane stress, becausethose are traction free. Hence problems like these are sometimes simplified to the analysis of two extremecases - plane stress, representative of the surface, and plane strain, representative of the symmetry plane.The stress/strain state anywhere else in this body can be assumed to lie between these two extremes.

(Ex. prob. 88).

6.1.3. Axisymmetric

If the geometry and the loading share a symmetry axis, then such problems are calledaxisymmetric.In this case every cross section passing through the symmetry axis is identical, and a three-dimensionalproblem is reduced to a two-dimensional problem of the cross section.

sym axis

12

3

Consider a CS where axes 1 and 2 are in plane of a cross section. Due to symmetry, there can be no shearon planes normal to 3:

σ31 = σ32 = ε31 = ε32 = 0

Moreover, because all cross sections are identical,σ33 cannot depend onx3. Hence

σ33,3 = 0

Likewise, there cannot be a body force, or an acceleration, or a velocity along 3. Hence the whole righthand side of Eqn. (66) is zero along 3.

With these constraints Eqn. (66) along 3 becomes identically zero, and only two equilibrium equa-tions remain. The problem therefore becomes two-dimensional.

However, both the stress and the strain tensors are still three-dimensional:

-42-

σσ =

σ11

sym

σ12

σ22

0

0

σ33

; εε =

ε11

sym

ε12

ε22

0

0

ε33

2

13

ε33 is found from another constraint. Consider a view along the symmetry axis.

x32

1

1

x3 + u3

x1

B′B

A u1A′

Let AB be a small element of lengthx3. After the deformation this element becomesA′B′. It also movesalong 1 byu1. From similar triangles one obtains:

x3

x1=

x3 + u3

x1 + u1

So that

u3 =x3(x1 + u1)

x1− x3 =

x3

x1u1

Thus the axisymmetry constraint means thatu3 is not independent, but rather is a function ofu1.

By definition

ε33 =∂u3

∂x3=

∂

x3

x1u1

∂x3=

u1

x1(131)

After the two-dimensional stress-strain problem has been solved, andε11 has been found,u1 is calculatedas:

u1 = ∫x1

0ε11dx1

with the axisymmetric BC thatu1(x1 = 0) = 0. Thenε33 is found from Eqn. (131). Thenσ33 is found fromEqn. (117):

σ33 = 2µε33 + λ trεε

6.1.4. Torsion

Torsion is probably the easiest two-dimensional problem, for which a full analytical solution is read-ily available. Considera straight rod of circular cross section with radiusR, loaded by torqueT and oneend, and fixed at the other.

-43-

3

T

1

2 α

1

3

2

The geometry of the problem is axisymmetric, but the loading is skew symmetric:

3

F

F F

F

F

FF

F

R 12

Hence the torsion problem is not axisymmetric.

The assumptions that make solution to this problem particularly simple are:

• Flat cross sections initially normal to axis 2 remain flat and normal to axis 2 throughout the deforma-tion. Thisassumption is strictly true only for axisymmetric (circular) cross sections.

• The length of the rod does not change.

• The twist angle,α , changes linearly withx2, i.e.

α ′ =dαdx2

= const (132)

The first two assumptions mean that displacements along the axis are not possible:

u2 = 0

Therefore for any cross section containing the symmetry axis, the only non-zero displacements are normalto this cross section.

Let’s choose the cross section in 23 plane. Then

u3 = 0

To find u1 let’s consider the motion of a "slice" of heightdx2, cut normal to the axis:

dx2

2

3

1

B B′

A CA′O

Point B, located distancex3 from the axis of symmetry (0≤ x3 ≤ R ), moves to B′. The total displacementof point B is u = (u1, 0, 0). PointA moves to A′. The total displacement of pointA is u = (u1 + du1, 0, 0),i.e. both points move only along 1. Angle∠AOC = α , and ∠COA′ = dα , so that

-44-

du1 = dα x3

Eqn. (132) can be rewritten as

dα = α ′dx2

so that

du1 = α ′x3dx2

Then from Eqn. (105)

ε12 =1

2

∂u1

∂x2=

α ′x3

2(133)

ε13 =1

2

∂u1

∂x3= α ′dx2 → 0

All other components of the strain tensor are zero. The strain tensor is:

ε =

0

sym

ε12

0

0

0

0

The strain state is pure shear, so thatε1 = ε12, ε2 = 0 and ε3 = −ε12.

The maximum strain magnitude is achieved on the outer surface of the rod, wherex3 = R.

The equations of linear elasticity reduce to

σ12 = 2Gε12 (134)

and the stress tensor is:

σ =

0

sym

σ12

0

0

0

0

Both the strain and the stress tensors can be visualised as:

31

2

The conservation of angular momentum leads to

T = ∫Aσ12dAx3

whereσ12dA is the elementary force andx3 is the shoulder.

σ12 =x3

Rσ max

12

where from Eqns. (133) and (134):

σ max12 = 2Gε max

12 = Gα ′R (135)

so that the torque is:

T = Gα ′ ∫Ax2

3dA (136)

wherex3 is understood as the radius, because the cross section in 23 plane is chosen arbitrarily. Thereforethe integral in Eqn. (136) is the second polar moment of area, defined by Eqn. (42). Finally one obtains this

-45-

expression for the torque:

T = Gα ′J (137)

from where the twist angle is found by integration:

α = ∫x2

T

GJdx2 (138)

Note that Eqn. (138) is valid for variableT, G, J.

(Ex. prob. 89).

6.2. Application of tensor theory to properties of areas

The second moments of area form a symmetric R2T in 2D space. This means, apart from other prop-erties, that there is always a CS such thatI12 = 0. Otheruseful properties of R2T can be used to simplifythe analysis of cross sections.

(Ex. prob. 90, 91, 92, 93, 94.)

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7. Example problems

P1.

l

A

BC

h

fθ

Show that the moment off about B can be expressed asm = fl sinθ .

P2. A cylindrical column has 3 segments of lengthl , each with different radius.

r1

l

l

l

r3

r2

Draw the axial stress profile.

P3. A conical column of lengthl and base radiusr0, made of a very light material, is compessedwith an axial force at the end.

l

F

r0

Draw the axial stress and displacement profiles.

P4. A 10 m long sandstone column of 1 m diameter is lying flat on the ground. What will be it’sheight when it’s put vertical? Whatwill be it’s diameter at the bottom? Assume that sandstone density is2,000 kg / m3, the Young’s modulus is 50 GPa, the Poisson’s ratio is 0.2.

P5. A l ift of massm = 1 ton is travelling down with velocityv = 1 m / s. When the cable length isl0 = 5 m, there is an emergency stop of the pulley wheel.

m

l0 v

Caclulate the maximum increase in the stress in the cable and the maximum extension of the cable due toemergency stop. Assumethe cable is made of steel with the Young’s modulus ofE = 200 GPa and thePoisson’s ratio ofν = 0. 33and the initial diameter of the cable isd0 = 5 mm.

P6. Calculate the vertical displacement of the force application point, assuming both ropes are madeof the same material, and are of the same initial length and cross section:

-47-

α

F

P7. ForceF is applied to a rigid beam hanging on three identical wires (identical material, length,cross section). Find strains in each wire.

F

d d

d/2

rigid

P8. Calculate the maximum deflection in three-point bending.

P9. Given I11 and I22, find I r .

P10. Find howI11, I22, I12, change with the shift of origin:x j ′ = x j − Sj .

P11. Find I11, I22, I12 for a rectangle with sidesW andH .

P12. Find I11, I22, I12 of a right angled triangle with with sidesW andH .

P13. Compare a rectangular cross section with that of a triangle obtained from this rectangle by cut-ting along a diagonal.

P14. Calculate the second moments of area for a circle.

P15. Calculate the second moments of area for a circular ring.

P16. Quantify the benefits of a circular ring over solid circle cross section.

P17. Find an optimum cross section with fixed areaA, that must fit into a square boxW × W, whereW2 = 4A.

P18. Calculate the second moments for this cross section:

T

W

W

T

P19. A pipe of 20mm outer diameter and 2mm wall thickness, made from a material withσY = 500MPa, is loaded in pure bending. What is the maximum bending moment it can support beforeyielding?

P20. Prove that Eqn. (57) is indeed a solution to Eqn. (56).

P21. Calculate the critical buckling load for a column with clamped/clamped BC.

P22. Draw buckled shapes for columns with pin/pin and clamped/clamped BC forn = 2, 3, 4,. . ..

P23. Draw a stress profile across the cross section of a buckled column.

-48-

P24. For a planar element of sizeS with normaln find its projection on planeP with normalnp.

P25. Illustrate this stress tensor on the elementary cube of material:

σσ =

−100 200

300

−400

500

−600

P26. Show that∇ ⋅ σσ = σ ij , j .

P27. For a tensor of arbitrary rank,Z, explain the difference between∇Z and∇ ⋅ Z.

P28. Prove that stress in R2T.

P29. In 2D space you have vector ai = (1, 1). Show how it is transformed by tensors

Bij =

1

0

0

1; Cij =

0

−1

1

0; Dij =

1

1

1

1

P30. Write down components ofR for a CT consisting of rotating about 3 byβ .

P31. Prove that the rotation tensorR is orthogonal.

P32. Show thata = RT ⋅ a′.P33. Prove that R2TT changes with CT as:

Tij ′ = Rip RjqTpq or T′ = R ⋅ T ⋅ RT

P34. How many components do tensors of ranks 1 to 4 have in 2D and in 3D spaces?

P35. Show that any R2T can be represented as a sum of a symmetric and a skew-symmetric R2T.

P36. Find sym(A) and asym(A) for this R2T:

A =

1

4

7

2

5

8

3

6

9

P37. Prove thatσ jp xk,p = σ jk .

P38. Prove thateijk v j vk = 0.

P39. Prove thateijkσ jk = 0 means that the stress tensor is symmetric.

P40. For a body in free fall use the equilibrium equations to find the stress state.

P41. Calculate stress state in a freely standing column of constant cross section.

P42. Calculate stress state in a cylindrical rocket accelerated by a force at one end.

P43. A block of lengthL, height H , thicknessT and densityρ, lying on a smooth (no friction) flatsurface, is pulled at one end by the weight of massm via a pulley. Find the stress state in the body.

m

L

H

P44. Prove that detR = 1.

P45. Given vectorsa, b in 3D space, write their cross product using vector components notation andindex notation witheijk .

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P46. Prove thatepqr = −eprq.

P47. Prove thateijk is a rank 3 tensor.

P48. Prove thateijk is isotropic.

P49. Calculate the determinant of a symmetric R2T,T.

P50. Show that for a symmetric R2T,T, detT = eijk Ti1T j2Tk3.

P51. Using the definitions of R2T invariants, Eqns. (88), (89), (90), expand the characteristic equa-tion, Eqn. (86), into a cubic equation, Eqn. (87).

P52. Prove that I T is invariant to CT.

P53. Prove that II T is invariant to CT.

P54. Prove that III T is invariant to CT.

P55. Express R2T invariants via the principal values.

P56. For a uniaxial stress state, find the principal stresses and directions.

P57. Draw this stress state in the original and the principal CS:

σσ =

−100

sym

200

300

−200

400

100

P58. Prove that deformation gradient is R2T.

P59. Express the strain tensorE as a function of displacementsu.

P60. Prove that strain is R2T.

P61. Analyse this motion:x = X, exceptx1 = X1 + tX2, t = const .

P62. Draw the principal strains and their directions for the strain tensor from ex. prob. 61.

P63. Apply this rotation tensor to the previous example and validate thatF′ = R ⋅ F ⋅ RT :

R =

0

−1

0

1

0

0

0

0

1

P64. Convert the strain gauge rosette measurements into a strain tensor.

P65. Calculate the maximum shear values and orientations.

P66. Draw planes with maximum and minimum shear values for these stress and strain tensors:

σσ =

300

sym

0

50

0

0

−500

; εε =

10−3

sym

0

−2 × 10−3

0

0

10−3

P67. Draw the Mohr’s diagram for stress stateσσ = 0, exceptσ12 = 200Mpa.

P68. Draw the Mohr’s diagrams for tensors in ex. prob. 66.

P69. Use the Mohr’s diagram to find the principal values of this strain tensor:

εε =

10−2

sym

2 × 10−2

−3 × 10−2

0

0

ε33

P70. Show how the symmetry ofσσ andεε leads to symmetries inC andS.

P71. Show how the existence of elastic potential leads to major symmetries inC andS.

P72. Explain factors of 2 when elastic law is written in matrix form, Eqns. (114) and (115).

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P73. Explain why uniaxial tension/compression and pure shear require only a single elastic constant.

P74. Explain what’s so special about Poisson’s ratio of 0.5.

P75. Show that in anisotropic materials shear strains cause normal stresses and vice versa.

P76. Show that in anisotropic materials principal directions of stress and strain do not coincide.

P77. Show that in isotropic materials shear stresses cause only shear strains, and normal stressescause only normal strains.

P78. Show that in isotropic materials principal directions of stress and strain coincide.

P79. CalculateS starting from Eqn. (117).

P80. Give an example of loading leading to a uniaxial strain state.

P81. Explain quantitatively the difference between the uniaxial stress and uniaxial strain states.

P82. Assuming linear isotropic elasticity express the equilibrium equations viau.

P83. Derive the strain compatibility equation, Eqn. (129).

P84. Prove that stress states are additive.

P85. Prove that strain states are additive.

P86. Prove that for plane stress condition, no rotation in the plane of two non-zero principal stressescan give rise to stress components acting in the plane of zero principal stress.

P87. Calculate reduction in wall thickness in a spherical air balloon under pressure.

P88. Under what conditions plane stress and plane strain states are identical?

P89. Do a full stress/strain/displacement analysis of this angle bracket:

2

P

l

31l

P90. Show that the coordinates of the centroid form a vector.

P91. Show that the first moments of area vanish for any Cartesian CS with origin at centroid.

P92. Show that the second moments of area form symmetric R2T in 2D space.

P93. For a right angled triangular cross section find the principal values and directions of the secondmoment of area tensor.

P94. Show that for a square cross section the second moment about any axis passing through thecentroid is an invariant.

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8. Solutions to example problems

S1. Refer to the drawing:

l

A

B

C

h

f sinθ

θ

ff cosθ

From triangle ABC:h/l = sinθ , or h = l sinθ . Then starting fromm = fh one obtainsm = fl sinθ .

The importance of this result is thatf sinθ is the component of the force acting normal to AB, Hencethe moment can be defined also as the product of the length between the point where the force is applied,A, and the point about which the moment is sought, B, by the component of the force normal to that line:m = f sinθ ⋅ l .

Clearly component of the force acting along AB does not contribute to the moment.

S2. We need a free body diagram.

x1

R

mg

I choose the origin at the base of the column and direct axisx1 up. I do an imaginary cut at some heightx1

and focus on the upper part of the column.I replace the action of the bottom part by a reaction forceR.The only other force acting on the column is the force of gravity,mg. From equilibrium

R = mg = ρVg

whereρ is the density of material andV is the volume of the part of the column above the cut. Now I justneed to think how to expressV as a function ofx1. Depending on where I do the cut there will be threerelationships.

V =

(3l − x1)π r 21

lπ r 21 + (2l − x1)π r 2

2

lπ r 21 + lπ r 2

2 + (l − x1)π r 23

x1 ≥ 2l

l ≤ x1 ≤ 2l

x1 ≤ l

So the force profile has three linear segments, with gradients of each line increasing from top to bottomeach lower segment of the column is heavier than the previous.

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ρglr 2

1 + r 22

r 23

l

l

l

r3

r2

r1

R t

ρglr 2

1

r 22

ρgl

ρgl(1 +r 2

1

r 22

)

ρgl(1 +r 2

1 + r 22

r 23

)

The stresst is the reaction force divided over the cross section area.

t =R

A=

ρg(3l − x1)

ρg(lr 2

1

r 22

+ (2l − x1))

ρg(lr 2

1

r 23

+ lr 2

2

r 33

+ (l − x1))

x1 ≥ 2l

l ≤ x1 ≤ 2l

x1 ≤ l

or, moving l outside of the brackets:

t = ρgl ×

(3 −x1

l)

(r 2

1

r 22

+ (2 −x1

l))

(r 2

1 + r 22

r 33

+ (1 −x1

l))

x1 ≥ 2l

l ≤ x1 ≤ 2l

x1 ≤ l

Note that the stress profile isdiscontinuous, where the cross section changes abruptly. Howev er, the gradi-ent oft is the same in all three segments. Canyou see why?

S3. The free body diagram is trivial.

u1 → −∞

F

x1

FF

π r 20

t

t → −∞

u1

There is a constant forceF in every cross section. The axial stress is negative (compressive).

t = −F

A= −

F

π r 2

whereA is the cross section area andr is the radius of the cone at any cross section. So the stress at the tipof the cone tends to infinity.

The straine11 has the same profile as the stress.

e11 = −F

Eπ r 2

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The displacementu1 is obtained by integration ofe11 over x1. So first we need to expressr as afunction ofx1. From similar triangles we obtain:

r0

l=

r

l − x1→ r =

r0

l(l − x1)

Then

e11 = −Fl2

Eπ r 20(l − x1)2

u1 = ∫x1

0e11dx1 = −

Fl2

Eπ r 20∫

x1

0

1

(l − x1)2dx1 = −

Fl2

Eπ r 20

×1

l − x1+ C

C is found from the boundary conditions - whenx1 = 0 thenu1 = 0.

0 = −Fl2

Eπ r 20

×1

l+ C → C =

Fl2

Eπ lr 20

Finally

u1 =Fl2

Eπ r 20

(1

l−

1

l − x1)

Sou1 → −∞ at the tip. What is the physical interpretation of infinite stress and displacement at the tip?

S4. First we need a stress profile, for which we need a free body diagram.

t

R

mmg

h0

A0

−ρgh0

1

0

x

h0 − x

I set the origin at the bottom of the column.I direct axis 1 (x) upward. I imaginary cut the column at somearbitrary value ofx and consider the part of the column above the cut. I substitute the lower part of the col-umn by thereaction force. The only other force acting on the upper part of the column is the force of grav-ity. This is the free body diagram.

The bending moment in the cross section is zero. Why?

If the mass of the upper part of the column ism, then

R = mg = ρVg = ρ(h0 − x)A0g

whereV is the volume of the upper part of column,ρ is the density of sandstone,A0 is the original, unde-formed, cross sectional area andh0 is the original, undeformed, length (height) of the column.

The axial stress is compressive, hence negative:

t = −R

A0= −

ρ(h0 − x)A0g

A0= ρg(x − h0)

The stress is a linear function ofx. t = 0 at the top, wherex = h0. t = −ρgh0 at the bottom, wherex = 0.

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Note that this is exactly the same equation as for a hydrostatic pressure in a liquid.

Putting the numbers in (I convert all lengths in m):

t = −2000× 10× 10 = −2 × 105

Since the lengths were in m, the stress is in Pa. InMPa it will be 10−6 times less, i.e. 0.2 MPa. Sothe max-imum absolute value of axial stress, 0.2 MPa, is at the bottom of the column.

The axial strain,e11, is

e11 =t

E=

ρg(x − h0)

E

whereE is the Young’s modulus. Soaxial strain also has a linear profile along 1 (x).

Displacementu1 is the integral of strain:

u1 = ∫x

0e11dx = ∫

x

0

ρg(x − h0)

Edx =

ρg

E ∫x

0(x − h0)dx =

ρg

E(x2

2− h0x) + C

whereC is the integration constant found from the boundary conditions. The boundary condition isu1 = 0at the bottom, wherex = 0. SoC = 0. Finally

u1 =ρgx

E(x

2− h0)

u1 at the top is

u1 = −ρg

E

h20

2

or, substituting the numbers (I convert all units of length to m):

u1 = −2000× 10

50× 109×

102

2= −0. 2× 10−4

This is the value in m, or−0. 02mm. Thetop of the column will move down (negative) 0.02 mm, so thecolumn will be 20 µm shorter than when lying flat on the ground.

The transverse strain is

e22 = −ν e11 = νρg(h0 − x)

E

The transverse strain is positive, i.e. tensile. This means the diameter of the column will increase. sincee22

depends onx, the degree of change of the diameter will vary withx. Howev er, for eachx, e22 is constantacross the whole cross section.

At the bottom

e22 = νρgh0

E

Displacement,u2 is found by integratinge22 along the radius

u2 = ∫r

0ν

ρgh0

Edr = ν

ρgh0

Er

The maximum displacement is atr = r0, the radius of the original, undeformed, column.

u2 = νρgh0

Er0

Substituting the numbers in (again keeping all lengths in m):

u2 = 0. 2×2000× 10× 10

50× 1090. 5= 0. 4× 10−6

in m, or 0. 4× 10−3 mm or 0.4 µm. So∆d = 2u2 = 0. 8µm and the diameter of the column at the bottomwill be 1,000.0008 mm.

-55-

S5. The key to answering the question is understanding of the physics. Whenthe lift is travelling ata constant speed, it has a certain kinetic energy,K . When the pulley wheel stopped turning, the lift con-tinue travelling down, decelerating and extending the cable. Thus kinetic energy is decreasing and thestored elastic energy,H , is increasing. Eventually, the lift will slow down to a stop. Thisis point when allkinetic energy will have transferred into stored elastic energy. The stress in the cable will reach maximumat that moment.

So we need to write down expressions forK andH for the whole body.

K =mv2

2

Hbody =1

2e11tV

At the moment when the lift is stationary,H = K , so

mv2 = e11tV (139)

hereV is the cable volume when lift has stopped moving.

Let’s first get an estimate assuming no change in volume of the cable. In this case the solution is par-ticularly simple. Using the Hooke’s law t = Ee11 I get from above

mv2 =t2

EV

from where

tmax = v√ mE

V

Substituting the numbers in (because of the square root, it’s easiest to convert all units of length to m.E = 200 GPa= 2 × 1011 N / m2, radius of the cable is 2. 5× 10−3 m.)

tmax = 1 ×√ 103 × 2 × 1011

5 × π × (2. 5× 10−3)2≈ 1, 427× 106

or tmax = 1. 427GPa. Thisstress is probably too high for most steels, so the cable will break and the liftwill fall.

Now let’s take into account the volume change. Using the Hooke’s law I rewrite (139) as

mv2 = Ee211V (140)

We also need to expressV via e11.

V = lA = lπ d2

4

wherel , A andd are the length, the cross section area and the diameter of the cable when the lift hasstopped.

l = l0 + ∆l = l0 + e11l0 = l0(1 + e11)

d = d0 − ∆d = d0 + e22d0 = d0(1 + e22) = d0(1 − ν e11)

where I used the definition of the Poisson’s ratioν = −e22/e11. Substituting all this back into (140) I obtain:

mv2 = Ee211l0(1 + e11)

π4

(d0(1 − ν e11))2

This is a 4th order equation fore11. I solve it numerically and getemax11 ≈ 7. 128× 10−3. This strain is above

0. 2%,which is the typical elastic proof strain (see the Properties of Materials course), so the cable willeither deform plastically or break.

-56-

tmax = Eemax11 ≈ 2 × 105 × 7. 12810−3 = 1, 426

So the max. stress calcuated with a more accurate approach istmax = 1. 426GPa, which differs by less than1% from that calculated with no regard to volume change.

S6. Make a good free body diagram. Using the symmetry of the problem, we need to draw only asingle rope loaded with half the force,F /2. The axial force is

R =F

2 cosαHence the axial stress in each rope is

t =R

A=

F

2AcosαwhereA is the cross section. The axial strain in each rope is

e =t

E=

R

EA=

F

2EAcosαThe axial displacement is

u = Le =LF

2EAcosαand the new length of each rope is

Lnew = L(1 + e) = L(1 +F

2EAcosα)

Note that whenα → π /2 thenR, t, e, u → +∞.

The vertical component of displacement is

uv = ucosα =FL

2EA

Now we need to take the rigid body rotation into account. Refer to the diagram below.

dF

α

F

L

BLnew

L cosα

L sinα

Both ropes elongate, but the point where the force is applied stays in the middle, due to symmetry. Henceboth ropes rotate about their supports to satisfy displacement compatibility. The total vertical displacement,d, can be calculated from the triangles of the diagram as follows:

d = B − L cosα

where

B2 = (Lnew)2 − (L sinα )2 = L2((1 + e)2 − sin2 α ) = L2(cos2 α + 2e+ e2)

At this point let’s introduce some realistic numbers to this problem. Let’s imagine each rope is 1 mlong and is made of 2 mm diameter steel wire.E = 200 GPa,α = 45 andF = 1 kN. Thenthe strain in

-57-

wires will be:e = 1. 125× 10−3. This strain is below 0.2%, which is usually taken as the yield (proof) strainfor steel. Hence this is a purely elastic problem.

Sincee2 is 3 orders of magnitude smaller thane, we can neglect this term. Displacement,d, in mmis then:

d = L((cos2 α + 2e)½ − cosα ) = 1. 591

S7.

2

F

d/2dd

R2R3 R1

1

The problem is immediately clear from the free body diagram of the beam - there are only two scalar equa-tions of equilibrium which are useful, but three unknowns:

R1 + R2 + R3 = F (141)

R33

2d + R2

d

2= R1

d

2(142)

This problem cannot be solved just from the static equations of equilibrium, other information is required.This is a classicalstatically indeterminatesystem.

In this particular example, the extra information comes from the fact that the beam is declared asrigid. We will use this fact to obtain the required third equation.

Consider the beam after the deformation of the wires has occurred. Remember that the beam is rigid,i.e. it does not deform, just rotates as a rigid body:

d

u3u2

u1

d

From the triangles:u2 − u3

d=

u1 − u2

d

or

2u2 = u1 + u3

That’s all! The rest is just algebra. The state of stress in each wire is uniaxial.

t =R

A

e =t

E

x2 is a coordinate along the axis of each wire.

u = ∫L

0edx2 =

tL

E=

RL

EA

whereA is the cross section,L is the length of each wire, andE is the Young’s modulus of the wire

-58-

material.

With this the displacement compatibility equation can be rewritten as:

2R2 = R1 + R3 (143)

(141)+(143):

3R2 = F ⇒ R2 =F

3

From (141):

R1 = F − R2 − R3

From (142):

R1 = 3R3 + R2

subtracting these two equations:

R3 =F − 2R2

4=

F −2

3F

4=

F

12

Finally

R1 =7

12F ; R2 =

4

12F ; R3 =

1

12F

One can check that the force and moment equilibrium are satisfied.

After all reaction forces are known,t, e andu are immediately available from equations above.

S8. Three-point bending usually means a problem like this:

3

P

1

2

where the beam rests on two rollers and the load is applied via a third roller.

Let’s assume the load is applied at half length of the beam. Then the problem is symmetric aboutx1 = 0 plane. Thefree body diagram and the shear force and the bending moment profiles are:

M

P/2 P/2

P

L

P/2

-P/2

PL/4

F

L is the beam length, i.e. the distance between the bottom rollers. Note that the shear force has a jump (dis-continuity) of magnitudeP, at a point where the load is applied via the top roller.

-59-

Need to integrate Eqn. (29) to calculate deflection:

dw

dx1= ∫ M

IEdx1 + C

where integration constantC is found from BC.

From the moment profile:

M =

Px1

2,

P(L − x1)

2,

0 ≤ x1 ≤ L/2

L/2 ≤ x1 ≤ L

so for 0≤ x1 ≤ L/2:

IEdw

dx1= ∫

x1

0

Px1

2dx1 + C

or

2IE

Pw′ =

x21

2

x1

0

+ C =x2

1

2+ C

where prime,′, denotes function derivative wrt to its only argument, in this casew′ = dw/dx1.

The integration constantC is found from BC. From symmetryw′(x1 = L/2) = 0, so:

0 =L2

8+ C → C = −L2/8

so that

w′(0 ≤ x1 ≤ L/2) =P

2IE

x21

2−

L2

8

For L/2 ≤ x1 ≤ L:

IEdw

dx1= ∫

x1

L/2

P(L − x1)

2dx1 + C

or

2IE

Pw′ = −

(L − x1)2

2

x1

L/2

+ C = −(L − x1)2

2+

(L − L/2)2

2+ C =

L2

8−

(L − x1)2

2+ C

The integration constantC is found from BC. From symmetryw′(x1 = L/2) = 0, soC = 0 and

w′(L/2 ≤ x1 ≤ L) =P

2IE

L2

8−

(L − x1)2

2

Let’s calculate the slope (gradient) at both ends of the beam:

w′(x1 = 0) =P

2IE(02/2 − L2/8) = −

1

16

PL2

IE

w′(x1 = L) =P

2IE((L2/8 − (L − L)2/2) =

1

16

PL2

IE

The values are equal in magnitude and of opposite signs of course, due to symmetry. The slope profilelooks like this:

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PL2/16IEw′

−PL2/16IE

To obtain deflection, need to integrate slope:

w = ∫ w′dx1

For 0 ≤ x1 ≤ L/2:

w =P

2IE ∫x1

0

x21

2−

L2

8dx1 =

P

2IE

x31

6−

L2

8x1

x1

0

+ D =P

2IE

x31

6−

L2

8x1

+ D

The BC for this case isw(x1 = 0) = 0, henceD = 0 and finally:

w(0 ≤ x1 ≤ L/2) =P

2IE

x31

6−

L2

8x1

For L/2 ≤ x1 ≤ L:

w =P

2IE ∫x1

L/2

L2

8−

(L − x1)2

2dx1 =

P

2IE

L2

8x1 +

(L − x1)3

6

x1

L/2

+ D

=P

2IE

L2

8x1 +

(L − x1)3

6−

L3

16−

(L − L/2)3

6

+ D =P

2IE

L2

8x1 +

(L − x1)3

6−

L3

12

+ D

The BC for this case isw(x1 = L) = 0, hence:

0 =P

2IE

L3

8−

L3

12

+ D

so that

D = −PL3

48IE

and finally:

w(L/2 ≤ x1 ≤ L) =P

2IE

L2

8x1 +

(L − x1)3

6−

L3

12−

L3

24

=P

2IE

L2

8x1 +

(L − x1)3

6−

L3

8

One can than deploy the extremum finding tools of differential calculus. However, due to symmetry, it isclear that the maximum deflection is in the middle, atx1 = L/2. We can satisfy ourselves with checkingthat the deflection expressions from the left and from the right side of the beam match. From the left:

wLmax(x1 = L/2) =

P

2IE

L3

48−

L3

16

= −PL3

48IE

And from the right:

wRmax(x1 = L/2) =

P

2IE

L3

16+

L3

48−

L3

8

= −PL3

48IE

The deflection profile looks like this:

wwmax

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S9. r 2 = x21 + x2

2, so I r = I11 + I22.

S10. Simply substitute Eqn. (34) to Eqns. (39)-(42).

I11′ = ∫ (x2′)2dA = ∫ (x2 − S2)2dA = ∫ (x22 − 2x2S2 + S2

2)dA = I11 − 2S2i1 + S22 A

I22′ = ∫ (x1′)2dA = ∫ (x1 − S1)2dA = ∫ (x21 − 2x1S1 + S2

1)dA = I22 − 2S1i2 + S21 A

I12′ = ∫ x1′x2′dA = ∫ (x1 − S1)(x2 − S2)dA = ∫ (x1x2 + S1S2 − x1S2 − x2S1)dA = I12 + S1S2A − S2i2 − S1i1

I r ′ = I11′ + I22′ = I11 + I22 − 2(S2i1 + S1i2) + (S21 + S2

2)A

If the new CS has centroid at the origin, then one can use Eqns. (37) and (38) to obtain:

i1 = S2A

i2 = S1A

so that

I11′ = I11 − S22 A (144)

I22′ = I22 − S21 A (145)

I12′ = I12 − S1S2A (146)

I r ′ = I11 + I22 − (S1 + S2)A (147)

S11. Refer to the diagram below.

C

2

W

H

dA = Hdx1

dA = Wdx2

1

The centroid is at the intersection of the two lines of symmetry. Taking the centroid as the origin:

I11 = ∫ x22dA

It is natural to takedA = Wdx2, so that

I11 = ∫H /2

−H /2x2

2Wdx2 = Wx3

2

3

H /2

−H /2

= WH3

31

8+

1

8

=WH3

12

From symmetry

I22 =HW3

12

It is clear that from symmetryI12 = 0. Butone can check it as

-62-

I12 = ∫ x1x2dA = ∫H /2

−H /2 ∫W/2

−W/2x1dx1x2dx2 = ∫

H /2

−H /2

x21

2

W/2

−W/2

x2dx2 = ∫H /2

−H /2

W2

21

4−

1

4x2dx2 = 0

S12. Refer to the diagram below.

S1H

W1

2

dA = adx2

S2

ax2

C1′

2′

First need to find the centroid.

i1 = ∫ x2dA

It is easiest to representdA as shown in the diagramdA = adx2, wherea is the length of the hatched strip.From triangles

W

H=

a

H − x2

or

a =W

H(H − x2)

So that

i1 = ∫H

0x2

W

H(H − x2)dx2 =

W

H ∫H

0

x2H − x2

2dx2 =

W

H

H2

2H −

H3

3

=WH2

6

From symmetry

i2 =HW2

6

Hence the centroid is at

S1 =i2

A=

HW22

6HW=

W

3

Again from symmetry

S2 =H

3

When calculating the second moments one has a choice: (a) calculating wrt to the centroid directly, or (b)calculating wrt the original CS, and then translating to the centroid using solution to ex. prob. 10.

First, (a) calculating directly. Because the origin has shifted one now has

-63-

W

H=

a

2/3H − x2

or

a =W

H2

3H − x2

So

I11 = ∫2/3H

−H /3x2

2W

H2

3H − x2

dx2 =

W

H ∫2/3H

−H /3

2

3Hx2

2 − x32dx2 =

W

H

2

3H

x32

3−

x42

4

2/3H

−H /3

=W

H

2

9H

8

27H3 +

1

27H3

−

1

416

81H4 −

1

81H4

= WH3

2

27−

15

4 ⋅ 81

=WH3

36

From symmetry

I22 =HW3

36

Now, (b) calculatingI11 in the original CS:

I11 = ∫H

0x2

2W

H(H − x2)dx2 =

W

H ∫H

0

Hx2

2 − x32dx2 =

W

H

H4

3−

H4

4

=WH3

12

and then shifting it to the centroid using expressions from ex. prob. S10:

I11′ = I11 − S22 A =

WH3

12−

H

3

2 WH

2= WH3

1

12−

1

18

=WH3

36

which matches the answer found by method (a). The same can be obtained forI22.

Finally, calculating I12 directly. The most important point is to choose the limits of integration right.Refer to the diagram.

I12 = ∫x2=2/3H

x2=−H /3 ∫x1=a−W/3

x1=−W/3x1x2dx1dx2

Using the expression fora above

a −W

3=

2

3W −

1

3W −

W

Hx2 =

W

3−

W

Hx2

Because the upper integration limit forx1 depends onx2, the integration must be done in the correct order

I12 = ∫x2=2/3H

x2=−H /3

∫

x1=W/3−Wx2/H

x1=−W/3x1dx1

x2dx2

The inner integral is

x21

2

W/3−Wx2/H

−W/3

=W2

21

9−

2

3

x2

H+

x22

H2−

1

9

=W2

2H2x2

2 −2

3Hx2

So that

I12 =W2

2H2 ∫x2=2/3H

x2=−H /3

x2

2 −2

3Hx2

x2dx2 =

W2

2H2

x42

4−

2

3H

x32

3

2/3H

−H /3

=W2H2

2

1

416

81−

1

81

−2

9

8

27+

1

27

=W2H2

2

15

4 ⋅ 81−

2

27

=W2H2

2⋅

15− 2 ⋅ 4 ⋅ 3

4 ⋅ 81= W2H2 ⋅

15− 24

8 ⋅ 81= −

W2H2

72

Note thatI12 is negative!

-64-

And now, calculatingI12 in the original CS and then shifting it to the centroid. Note how the integra-tion limits change.

I12 = ∫H

0 ∫W

H(H−x2)

0x1x2dx1dx2 = ∫

H

0

∫

W

H(H−x2)

0x1dx1

x2dx2

The inner integral is

x21

2

W

H(H−x2)

0

=W2

2H2H2 − 2Hx2 + x2

2

So that the outer integral becomes

I12 =W2

2H2 ∫H

0

H2x2 − 2Hx2

2 + x32dx2 =

W2H2

21

2− 2

1

3+

1

4

=W2H2

2

6 − 8 + 3

12=

W2H2

24

From ex. prob. 10:

I12′ = I12 − S1S2A =W2H2

24−

WH

9⋅

WH

2= W2H2

1

24−

1

18

= W2H2 3 − 4

72= −

W2H2

72

which matches the answer found by the direct method.

S13. Make sure to solve ex. probs. 11 and 12 before attempting this.

The area, and hence the mass per unit length, of the triangular cross section is 1/2 of that of the rec-tangular. Howev er, I11 and I22 for a triangle are 3 times smaller. In addition xmax in a rectangle isH /2, but2/3H in a triangle. So the maximum (or minimum) axial stress in a beam with a triangular cross section is

3 ×2/3

1/2= 4 times higher than in a beam of a rectangular cross section. An engineer will have to make a

judgment on whether the saving in mass is justified.

In addition, note that the triangular cross section has no axis of symmetry, which is reflected by thefact thatI12 ≠ 0.

S14. Due to axisymmetry, the centroid is at the centre of the circle. First calculate the polar secondmoment.

drr

R

I r = ∫ r 2dA

dA = 2π rdr

so that

I r = ∫R

02π r 3dr =

π R4

2

-65-

From ex. prob. 9:

I11 = I22 = ½I r =π R4

4

Due to axisymmetryI12 = 0 in any CS with the origin at centroid.

S15. Use the expressions for the circle, ex. prob. 14.

Ro

Ri

The outer radius of the ring isRo andRi is the inner radius.

Since the area integrals are additive, use superposition:

I r = I circr (Ro) − I circ

r (Ri ) =π2

(R4o − R4

i )

and

I11 = I22 =π4

(R4o − R4

i )

Alternatively can calculate directly as in ex. prob. 14, but with different BC:

I r = ∫Ro

Ri

2π r 3dr = 2πR4

4

Ro

Ri

=π2

(R4o − R4

i )

S16. Use the expressions for the second moments from ex. probs. 14 and 15. Let

Ri = α Ro

where 0≤ α < 1, so that atα = 0 the ring becomes a circle. Let further assume that the outer radii of thering and of the circle are equal,Ro = R. Then

I ring

I circ=

R4 − (α R)4

R4= 1 − α 4

Note that this expression is valid for all second moments, so we don’t need to specify the subscripts.

Ratio of mass per unit length for beams of both cross sections is:

mring

mcirc=

Aring

Acirc=

R2 − (α R)2

R2= 1 − α 2

Because the outer radii are equal in both cases, only change inI needs to be considered when calculatingtmax. From Eqn. (28):

t ringmax

tcircmax

=I circ

I ring=

1

1 − α 4

One can plot these ratios againstα to get a complete picture. Here we examine only two values ofα .

-66-

αt ringmax

tcircmax

mring

mcirc

0.5 1.07 0.750.9 2.91 0.19

So forα = 0. 9,i.e. pipe wall thickness of 0.1 of the outer radius, the beam will be over five timeslighter. The price of this saving in mass is that the maximum stress will be almost three times higher. Thepotential for optimisation is clear.

S17. Refer to the diagram:

WA

W

1. solidcircle. A = π r 2 → r = √ A/π .

I11 =π4

A2

π 2=

1

4πA2

tmax =Mr

I11=

4π √ A

A2√ πM = 4√ π MA−3/2

tmax = 7. 090× MA−3/2

2. solidsquare. width = height =√ A. I11 =A2

12.

tmax =M√ A/2

I11= 6 × MA−3/2

3. ring.Ring area:

A = π (R2o − R2

i )

AssumeRo = W/2 → R2o = A. HenceR2

i = A − A/π .

I11 =π4

A2 − A2(1 − 1/π )2

=

π A2

4

2

π−

1

π 2

=2π − 1

4πA2

tmax =MRo

I11=

4π √ A

(2π − 1)A2M =

4π2π − 1

MA−3/2 = 2. 379× MA−3/2

4 square box. Assume the box shape as this:

Y

W

W

Y

-67-

Abox = Aouter − Ainner = W2 − Y2 = A

SoY2 = W2 − A = 3A.

I11 =1

12(W4 − Y4) =

1

12(16A2 − 9A2) =

7

12A2

tmax =MW/2

I11=

12√ A

7A2M =

12

7MA−3/2 = 1. 714× MA−3/2

5. I-section.

1

W

t

W

The area is approx.A = 3Wt, wheret is thickness of the web. So t ≈ √ A/6.

For this section

I11 = I square11 − 2I rect

11 =W4

12− 2

((W − t)/2)(W − 2t)3

12=

1

12

16A2 − (2√ A −

√ A

6)(2√ A −

2√ A

6)3

=A2

12

16− (2 −

1

6)(2 −

1

3)3

=A2

12

16−

11

6(5

3)3

= 0. 626A2

tmax =MW/2

I11= 1. 597× MA−3/2

So we can achieve over 4 times reduction in stress by choosing the cross section wisely.

Note that, in contrast to the first four sections, I-section works well in only one orientation. If I-sec-tion is fitted by mistake in the opposite orientation, then:

I22 = 2tW3

12+

(W − 2t)t3

12= A2

8

6 ⋅ 6+

2 − 1/3

12 ⋅ 63

= 0. 223A2

tmax =MW/2

I22= 4. 484× MA−3/2

So the maximum stresses in this orientation will be nearly 3 times greater than in the best oreintation. Or,in other words, the I-section oriented badly is still better than the solid square, but worse than ring.

Other popular cross sections used in construction:

S18. First I need to find the centroid of the cross section.I use a CS with the origin at the outer cor-ner of the cross section.I split the section into two parts, A and B, and find centroids and the secondmoments of each part wrt the chosen CS.I use expressions from ex. prob. 10.

-68-

C

T

2

A

B

CA

W

W

SA2

1

T SB2

CB

SC−CA2

SC−CB2

SC−CA1

SC−CB1

In that CS coordinates of the centroids of A and B are easily found, e.g. along axis 2:SA2 =

W

2, SB

2 =T

2.

The total area isA = WT + (W − T)T = T(2W − T).

If W = 40 mm andT = 4 mm, thenA = 304 mm2 or ≈ 3 cm2.

From (46) the coordinate of the centroid of the whole cross section along axis 2 is:

S2 =1

A(SA

2 AA + SB2 AB) =

W

2WT +

T

2(W − T)T

T(2W − T)=

W2 + T(W − T)

2(2W − T)

The second moments of areas A and B wrt their centroids are:

I A11(centroid)=

TW3

12; I B

11(centroid)=(W − T)T3

12

We use (144) to calculate the second moments of these areas wrt the chosen CS. Note that in (144) theprimed CS is the one that passes through the centroid, and the unprimed CS is the starting CS. In our casewe need to do inverse calculation - we know the second moment in the CS that passes through the centroid,but want to find out the second moments in the chosen (global) CS. So we need to original CS is :

I A11 = I A

11(centroid)+ (SA2 )2AA =

TW3

12+

W2

4WT =

TW3

3

I B11 = I B

11(centroid)+ (SB2 )2AB =

(W − T)T3

12+

T2

4(W − T)T =

(W − T)T3

3

The second moment for the whole cross section, in the chosen (global) CS, about axis 1 is

I11 = I A11 + I B

11 =T

3(W3 + (W − T)T2)

Now we can find the second moment of the whole cross section wrt to the centroid of the whole crosssection, about axis 1. From (144):

-69-

I11′ = I11 − S22 A =

T

3(W3 + (W − T)T2) −

W2 + T(W − T)

2(2W − T)

2

T(2W − T)

=T

3

W3 + (W − T)T2 − 3

(W2 + T(W − T))2

4(2W − T)

For our values ofW andT I11 = 4. 6× 104 mm 4 or 4.6 cm4.

From symmetry,I22 = I11.

For I12 it is easier to calculate firstI A12′ and I B

12′ and than calculateI12′ as I12′ = I A12′ + I B

12′. This isbecause for rectanglesI12 = 0 when 1 and 2 are symmetry axes passing through centroid. Note that we donot use the global (unprimed) CS forI12 at all. From (146):

I A12′ = −SC−CA

1 SC−CA2 AA

I B12′ = −SC−CB

1 SC−CB2 AB

where superscriptsC − CA andC − CB and refer to the distances between the centroids of A and B respec-tively, and the centroid of the whole cross section, C.

SC−CA1 =

W2 + T(W − T)

2(2W − T)−

T

2

SC−CA2 =

W2 + T(W − T)

2(2W − T)−

W

2

For the values above, I A12′ = −1. 3× 104mm4.

SC−CB1 =

W2 + T(W − T)

2(2W − T)−

T +

W − T

2

SC−CB2 =

W2 + T(W − T)

2(2W − T)−

T

2

For the values above, I A12′ = −1. 4× 104 mm4.

Finally I12′ ≈ −2. 7× 104 mm4.

The following makes sense only after completing Sec. 6.2.

The principal values and directions from Lapack DSYEV:

Original tensor0.00000000000E+00 0.00000000000E+00 0.00000000000E+000.00000000000E+00 4.60000000000E+00 2.70000000000E+000.00000000000E+00 2.70000000000E+00 4.60000000000E+00

The DSYEV eigenvalues in increasing order0.00000000000E+00 1.90000000000E+00 7.30000000000E+00

The DSYEV orthonormal eigenvectors (columns)1.00000000000E+00 0.00000000000E+00 0.00000000000E+000.00000000000E+00 -7.07106781187E-01 7.07106781187E-010.00000000000E+00 7.07106781187E-01 7.07106781187E-01

The angles (deg)0.00000000000E+00 9.00000000000E+01 9.00000000000E+019.00000000000E+01 1.35000000000E+02 4.50000000000E+019.00000000000E+01 4.50000000000E+01 4.50000000000E+01

So the principal values of the second moments are 7. 3× 104mm4 and 1. 9× 104mm4 and the angle is 45,as expected forI11 = I22.

-70-

S19. Use Eqn. (49) and an expression forI from ex. prob. 15:

Mmax =σY I11

xmax2

=σYπ (R4

o − R4i )

4Ro=

500× 3. 14× (104 − 84)

40= 231, 732N mm ≈ 232 N m

S20. Need to differentiatew three times:

w′ =dw

dx1= −zC1 sinzx1 + zC2 coszx1

w′′ =d2w

dx21

= −z2C1 coszx1 − z2C2 sinzx1

w′′′ =d3w

dx31

= z3C1 sinzx1 − z3C2 coszx1

Putting all this into Eqn. (56):

d3w

dx31

+ z2 dw

dx1= z3C1 sinzx1 − z3C2 coszx1 + z2(−zC1 sinzx1 + zC2 coszx1) = 0

S21. A clamped/clamped buckled column will look like this:

P

13

2

P

M

There are five BCs:

1. w(x1 = 0) = 0

2. w′(x1 = 0) = 0

3. w′(x1 = l /2) = 0

4. w(x1 = l ) = 0

5. w′(x1 = l ) = 0

From Eqn. (57) in Sec. 4.5.1:

w = C1 coszx1 + C2 sinzx1 + C3

Using BC 2:

C2 = 0

Using BC 1:

C1 + C3 = 0

Using BC 4:

C1 coszl + C3 = 0

From the last two equations:

C1 coszl − C1 = 0

C1 = 0 is a trivial solution corresponding to an unbuckled, straight, column. This solution is of no interest.Hence we must have:

coszl = 1

or

-71-

zl = 2nπ

The magnitude of the critical load is:

Pcrit = (2nπ )2 EI

l2

The lowest critical load is whenn = 1:

Plowestcrit = 4π 2 EI

l2

Note that the lowest critical load is 4 times higher for the clamped/clamped BC compared to the pin/pinBC, Eqn. (58) in Sec. 4.5.1.

S22. For a pin/pin column, from Eqn. (59):

w = C sinnπ x1

l

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

n = 1

n = 2

n = 3n = 4

w

C

x1

l

For a clamped/clamped column, from ex. prob. 21:

w = C(cos2nπ x1

l− 1)

00.10.20.30.40.50.60.70.80.9

1

0 0.2 0.4 0.6 0.8 1

n = 1

n = 2

n = 3n = 4

w

2C

x1

l

Then = 1 shapes are drawn to scale, i.e. the maximum values of deflection arew/C = 1. All other shapesare drawn with smaller magnitude to ease understanding.

-72-

S23. From Eqn. (53) in Sec. 4.5.1:

ε11 = x2d2w

dx21

+P

EA

or, sinceP < 0, we can rewrite this as:

ε11 = x2d2w

dx21

−|P|

EA

The stress state is uniaxial, so

t = Eε11 = Ex2d2w

dx21

−|P|

A

or using Eqn. (54):

t =Mx2

I−

|P|

A

or, since M = |P|w:

t = |P|(wx2

I−

1

A)

Buckling with Pcrit producesneutralequilibrium. Inother words anyw value is possible, provided thedeflections remain small. Hence we cannot know the exact answer, and only a qualitative answer is possi-ble.

The stress profile consists of two superimposed fields - bending and compression. The magnitude ofcompression stress is fixed, but the magnitude of the bending stress can vary. Hence we can have a situa-tion where the maximum stress is positive or neg ative.

n.l.=+

bending compression sum

++ -- - + +

=+

compressionbending sum

+ +- --

S24. Refer to the drawing:

-73-

np

n

S

Sp

P

A surface element is defined by its area and its unit normal. One can combine the two into a single vector:

S = Sn

A projection is clearly determined by the relative orientation of the normals of the two planes:

Sp = |S ⋅ np| = S|n ⋅ np|

Let’s see if this is correct on two extreme cases: ifn||np→Sp = S, which is correct. Ifn⊥np→Sp = 0,which is also correct. Finally, what if np matches a basis vector, i.e.np = ei? If np = e1 = (1, 0, 0)T then|n ⋅ np| = n1 andSp = Sn1. Similarly for the two other cases: ifnp = e2 = (0, 1, 0)T thenSp = Sn2. Ifnp = e3 = (0, 0, 1)T thenSp = Sn3.

S25. Remember that the stress tensor is symmetric. Remember to use the sign convention, sec. 5.2.

500 100

600

300

200200

1

2

3

400

400

500

S26. ∇ is a differential operator, a vector calledgradient, defined as:

∇ =∂

∂x1e1 +

∂∂x2

e2 +∂

∂x3e3

whereei are the basis vectors. Consultyour maths lectures and books for more.

So∇ is a vector operator.∇ ⋅ is a dot (or inner) product, meaning that∇ is multipliedby another vec-tor or tensor. In case of a vectora = ai :

∇ ⋅ a =

∂∂x1

,∂

∂x2,

∂∂x3

a1

a2

a3

=∂a1

∂x1+

∂a2

∂x2+

∂a3

∂x3= ai ,i

-74-

So the result is a scalar, or a rank 0 tensor. Hence,∇ ⋅ reduces the rank of the result by one, compared tothe rank of the tensor to which it is applied.

For a R2T, T = Tij :

∇ ⋅ T =

∂∂x1

,∂

∂x2,

∂∂x3

T11

T21

T31

T12

T22

T32

T13

T23

T33

=

∂T11

∂x1+

∂T21

∂x2+

∂T31

∂x3∂T12

∂x1+

∂T22

∂x2+

∂T32

∂x3∂T13

∂x1+

∂T23

∂x2+

∂T33

∂x3

or, for a transposed tensor:

∇ ⋅ TT =

∂∂x1

,∂

∂x2,

∂∂x3

T11

T12

T13

T21

T22

T23

T31

T32

T33

=

∂T11

∂x1+

∂T12

∂x2+

∂T13

∂x3∂T21

∂x1+

∂T22

∂x2+

∂T23

∂x3∂T31

∂x1+

∂T32

∂x2+

∂T33

∂x3

Note that from R2T we got rank 1 vector. So

∇ ⋅ σσ =

∂σ11

∂x1+

∂σ21

∂x2+

∂σ31

∂x3∂σ12

∂x1+

∂σ22

∂x2+

∂σ32

∂x3∂σ13

∂x1+

∂σ23

∂x2+

∂σ33

∂x3

On the other hand, inσ ij , j j is a dummy (summation) index. Onlya single index remains,i . Hence theresult of this spatial derivation is a rank 1 tensor - a vector:

σ ij , j = σ i1,1 + σ i2,2 + σ i3,3 =

σ11,1 + σ12,2 + σ13,3

σ21,1 + σ22,2 + σ23,3

σ31,1 + σ32,2 + σ33,3

Differentiating instead over the first subscript gives:

σ ij ,i = σ1 j ,1 + σ2 j ,2 + σ3 j ,3 =

σ11,1 + σ21,2 + σ31,3

σ12,1 + σ22,2 + σ32,3

σ13,1 + σ23,2 + σ33,3

Sinceσσ = σσ T , both derivatives giv e the same vector.

In vector calculus∇ ⋅ a is calleddivergenceof vector fielda. The index equivalent of∇ ⋅ operator isb...p...,p, whereb...p... is a tensor of arbitrary rank.

S27. The difference is easy to see if one uses the index notation:

∇ ⋅ Z = Z...p...,p

whereas

∇Z = Z...,p

In the first casep is a dummy index, so the resulting tensor is of rankr − 1, wherer is the rank ofZ. In thesecond casep is a live index, so the rank of the resulting tensor isr + 1.

SayZ is a rank 1 tensor, i.e. avector, zi : Then

-75-

∇ ⋅ z = zi ,i =∂z1

∂x1+

∂z2

∂x2+

∂z3

∂x3

which is a scalar, whereas

∇z = zi , j =

∂z1

∂x1∂z2

∂x1∂z3

∂x1

∂z1

∂x1∂z2

∂x1∂z3

∂x1

∂z1

∂x1∂z2

∂x1∂z3

∂x1

which is a R2T.

For a scalar fieldz, ∇z is thegradientof the field, a vector pointing in the direction of the maximumincrease ofz.

S28. Multiply Eqn. (63) by an arbitrary vectora j :

σ ij ni a j = tnj a j

The right hand side is a projection of one vector to another. This is a scalar and an invariant. Hencethe lefthand side must also be constant. Since bothni andai are arbitrary, then by definitionσ ij is a R2T.

S29. Operations on tensor of rank 1 and 2 can be done with conventional column and row vectorsand square matrices. So

bi = Bij a j =

1

0

0

1

1

1

=

1

1

= ai

so Bij is aunit tensorwhich transforms a vector into itself. The standard index notation for a unit tensor isδ ij and the standard tensor notation isI, which clearly matches the notation for a unit matrix.

ci = Cij a j =

0

−1

1

0

1

1

=

1

−1

So |c| = |a|, henceCij is an example of a purerotation tensor, one that changes the orientation of a vector,but leaves its length intact.

di = Dij a j =

1

1

1

1

1

1

=

2

2

So thatd = 2a. ThusDij is an example of astretch tensor, one that changes the length of a vector, but notits orientation.

1a = bc

d2

S30. Let’s assume that we rotate the CS clockwise:

-76-

a1

β

β

1

2 2′

3, 3′

a2

a1′

a

1′

a2′

The cosines of angles between the new and the old basis vectors are then:

cos∠(e′1, e1) = cosβ ; cos∠(e′1, e2) = cos(π /2 + β ) = − sinβ ; cos∠(e′1, e3) = 0

cos∠(e′2, e1) = cos(π /2 − β ) = sinβ ; cos∠(e′2, e2) = cosβ ; cos∠(e′2, e3) = 0

cos∠(e′3, e1) = 0; cos∠(e′3, e2) = 0; cos∠(e′3, e3) = 1

So

R =

cosβsinβ

0

− sinβcosβ

0

0

0

1

We must have a′ = R ⋅ a. Let’s check:

R ⋅ a =

cosβsinβ

0

− sinβcosβ

0

0

0

1

a1

a2

0

=

a1 cosβ − a2 sinβa1 sinβ + a2 cosβ

0

Looking at the diagram above, this is correct.

Remember the form ofR for a simple rotation about only one coordinate axis.

S31. Remember that the basis vectors are orthogonal in orthonormal CS:

ei ⋅ e j = I = δ ij

In a rotated CS:

ei ′ ⋅ e j ′ = Ripep ⋅ Rjqeq = Rip Rjqep ⋅ eq = Rip Rjqδ pq = Rip Rjp = Rip RTpj = δ ij

So

R ⋅ RT = I = R ⋅ R−1↔RT = R−1

S32. By definition:

a′ = R ⋅ a

Multiply both sides byRT from the left:

RT ⋅ a′ = RT ⋅ R ⋅ a

From ex. prob. 31: RT ⋅ R = I, hence:

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RT ⋅ a′ = a

In index notation the same proof looks like this:

ai ′ = Rij a j

Multiply both sides byRTpi from the left:

RTpiai ′ = RT

pi Rij a j

From ex. prob. 31: RTpi Rij = δ pj , hence:

RTpiai ′ = δ pja j = ap

S33. Start from Eqn. (75). IfT is R2T then∀ai , bi :

Tij ai b j ≡ const

That means that this product will be same in any other CS:

Tij ′ai ′b j ′ = Tij ai b j

From ex. prob. 32: ai = RTipap′, b j = RT

jqbq′, so that

Tij ′ai ′b j ′ = Tij RTipap′RT

jqbq′

Both indices on the left are dummy, and hence can be changed at will.I want to changei → p, j → q:

Tpq′ap′bq′ = Tij RTipap′RT

jqbq′

Provided thatai ′ ≠ 0 and bi ′ ≠ 0, we can divide by these vectors:

Tpq′ = Tij RTip RT

jq = Rpi RqjTij

In tensor notation the same proof looks like this:

(T ⋅ a) ⋅ b ≡ const

Explain why the brackets are needed in the above expression.

(T′ ⋅ a′) ⋅ b′ = (T ⋅ a) ⋅ b

From ex. prob. 32: a = RT ⋅ a′, b = RT ⋅ b′, so

(T′ ⋅ a′) ⋅ b′ = (T ⋅ RT ⋅ a′) ⋅ RT ⋅ b′

Provided thatb′ ≠ 0, we can divide by it:

T′ ⋅ a′ = (T ⋅ RT ⋅ a′) ⋅ RT

Then think of tensors as of matrices, i.e. the order matters. Multiply both sides byR from the right:

T′ ⋅ a′ ⋅ R = (T ⋅ RT ⋅ a′) ⋅ RT ⋅ R

Remember thatR is orthogonal, so thatRT ⋅ R on the right disappears:

T′ ⋅ a′ ⋅ R = T ⋅ RT ⋅ a′

Both sides are vectors. Transpose of a vector is the same vector:aT = a. Hence we can transpose just oneside of the equation, not both. Let’s transpose the right hand side:

T′ ⋅ a′ ⋅ R = a′T ⋅ R ⋅ TT

Now multiply both sides byRT from the right:

T′ ⋅ a′ = a′T ⋅ R ⋅ TT ⋅ RT

Finally transpose the right hand side again:

T′ ⋅ a′ = R ⋅ T ⋅ RT ⋅ a′

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divide bya′ and that’s it.

S34. Just think about the meaning of the subscripts, e.g.Mijk is a tensor of rank 3, each componentof which has a unique combination of subscripts. Each subscript can take values between 1 and the dimen-sionality of space, i.e. in 2D space only 1 or 2, and in 3D space 1, 2 or 3. The number of possible combina-tions is number of possible states (dimensionality of space,N ) to the power which is the number of indices(rank,R). Sothe number of components isNR. So in 2D space a vector has 21 = 2 components, a rank 2tensor has 22 = 4 components, a rank 3 tensor has 23 = 8 components and, a rank 4 tensor has 24 = 16 com-ponents. In3D space a vector has 31 = 3 components, a rank 2 tensor has 32 = 9 components, a rank 3 ten-sor has 33 = 27 components and, a rank 4 tensor has 34 = 81 components.

S35. Let’s split R2T A like this:

A =1

2A +

1

2A +

1

2AT −

1

2AT

Now, on the right hand side, put together terms 1 and 3, 2 and 4:

A =1

2A +

1

2AT +

1

2A −

1

2AT

or

A =1

2(A + AT ) +

1

2(A − AT )

The first term is always symmetric, and the second is always anti-symmetric. This might be easier to see inthe index notation:

(Aij + A ji )T = A ji + Aij = Aij + A ji

and

(Aij − A ji )T = A ji − Aij = −(Aij − A ji )

Note that for anti-symmetric tensor all diagonal components are zero:A11 = A22 = A33 = 0.

By definition,

sym(A) =1

2(A + AT )

asym(A) =1

2(A − AT )

S36. Use the expressions from ex. prob. 35.

AT =

1

2

3

4

5

6

7

8

9

sym(A) =1

2(A + AT ) =

1

2

2

6

10

6

10

14

10

14

18

=

1

3

5

3

5

7

5

7

9

asym(A) =1

2(A − AT ) =

1

2

0

2

4

−2

0

2

−4

−2

0

=

0

1

2

−1

0

1

−2

−1

0

Let’s check:

sym(A) + asym(A) =

1

3

5

3

5

7

5

7

9

+

0

1

2

−1

0

1

−2

−1

0

=

1

4

7

2

5

8

3

6

9

= A

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S37. First:

xk,p =∂xk

∂xp

which in 3D space, i.e.k, p = 1, 2, 3,is:

xk,p =∂xk

∂xp=

∂x1

∂x1∂x2

∂x1∂x3

∂x1

∂x1

∂x2∂x2

∂x2∂x3

∂x2

∂x1

∂x3∂x2

∂x3∂x3

∂x3

xi = →x = x is a vector giving the coordinate of a point in 3D space. Critically, all 3 coordinates can be cho-sen independently, i.e. x1, x2, x3 are 3 independent functions. Since they are independent, by definition, aderivative of one function over another is zero, and a derivative of any function over itself is 1. Hence alldiagonal components are 1, and all off-diagonal components are 0:

xk,p =

1

0

0

0

1

0

0

0

1

= I = δ kp

Henceσ jp xk,p = σ jpδ kp = σ jk .

S38. For eachi there are two non-zero terms in this expression:ijk andikj , always with oppositesigns, so they always cancel each other giving 0.For example, fori = 1, we have:

e1 jk v j vk = e123v2v3 + e132v3v2

all other terms are 0, due to repeated indices.

= v2v3 − v3v2 = 0

S39. For eachi this equation means thatσ jk − σ kj = 0, which meansσ jk = σ kj . For example, fori = 3:

e3 jkσ jk = e312σ12 + e321σ21 = σ12 − σ21 = 0 ↔ σ12 = σ21

S40. First make a good drawing:

dmg

1

P

dV

A

n = (1, 0, 0)

We only need one axis in this example, along the direction of the gravitational pull. Let’s call it 1. Next weneed to draw all forces acting on an element of volumedV. Remember that we are in continuum mechan-ics, where stress is afield, and hence we cannot just draw a force due to gravity in the centre of mass of thebody. Instead, consider an arbitrary pointP, and volumedV centred on this point. There is only one forceacting on this element:dmg = ρdVg. This force acts along 1.

The exact placement of axes 2 and 3 is not important. The important fact is that since those are nor-mal to 1, there are no forces acting in 23 plane. Hence immediately we can conclude that

σ22 = σ33 = σ23 = 0

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Next we note that the force field isuniform, meaning exactly the same force is acting on any elementdV. Hence there can be no shear on other planes either:

σ12 = σ13 = 0

Finally, using Eqn (73) we see that only non-zero terms left in the equilibrium equation, Eqn. (66),are:

σ11,1 = −b1 + ρ x1

whereb1 is the body force per unit volume:b1 = ρg. For a free fall x1 = g. Hence these two terms canceleach other:

σ11,1 = 0

or

σ11 = σ11(x2, x3)

And since the body force is uniform, we conclude thatσ11 is also uniform, i.e. does not depend on coordi-nate:

σ11 = const

To find this integration constant we use the boundary conditions, Eqn. (63):

σ ij ni = tnj

We apply this condition to pointA on the boundary, where the normaln = (1, 0, 0). The traction every-where on the boundary is zero:tn

j = 0, hence:

σ11n1 = σ11 = 0

Thereforeσσ = 0 everywhere in the body. Since the stress state is the same in all points, we call it auniformstress field.

This might seem like a trivial exercise. However, this is the first solid mechanics problem that wehave just solved using the full formalism of the theory, including (1) the equilibrium equations, (2) theboundary conditions and (3) the fact that the stress tensor is symmetric.

S41. First step - make afree body diagram, meaning replace all bodies of no interest by their interac-tions on the body of interest. In this example we replace the ground by the distributed reaction forcemg/A,wherem is the mass of the column,A is its cross section, andg is the gravitational acceleration.

Tip: always draw a CS:

1

mg/A

b

Given that there are no forces in 2 and 3 directions, we conclude that only non-zero stress isσ11. Suchstress states are calleduniaxial.

The only remaining equilibrium equation is:

σ11,1 = −b1 + ρ x1

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Given that this problem is static, the acceleration, ¨x = 0, hence:

σ11,1 = −b1

The body force is solely due to gravity. The body force acting on an element of volumedV isdmg= ρdVg. So the body force per unit volume isb1 = ρg.

By integrating one obtains:

σ11 = −ρgx1 + C

The integration constant is found from the traction boundary condition, Eqn. (63):

σ ij ni = tnj

On the top surface of the columnni = (−1, 0, 0)andtnj = 0, i.e. no traction. Hence

σ11n1 + σ21n2 + σ31n3 = −σ11 = 0

So

σ11(x1 = 0) = 0→C = 0

Finally

σσ = 0 except σ11 = −ρgx1

The stress field isuniaxialbecause there is only a single normal stress component in the body.

The stress field isnon-uniformbecause it changes from point to point.

The minimum stress is at the bottom of the column. If the length of the column isL, then this stressis σ min

11 = −ρgL. Clearly this must match the boundary condition at the bottom.m = ρ LA hencemg/A = ρgL, which is correct.

Note that we have used our sign convention here. The stresses at the bottom face point in the oppo-site direction to the normal, and hence are negative.

S42. As in the previous examples, need to make a good free body diagram:

-F/A

1

F

Note that the problem is very similar to the column. The difference is that there is no body force,b = 0.Again the only non-zero stress isσ11:

σ11,1 = ρ x1

Let’s assume that the acceleration is constant, and calculate it from the dynamics of rigid bodies. Note thatforce, and acceleration, point in the negative direction:

x1 =−F

m=

−F

ρV=

−F

ρ LA

whereL is the length of the rocket,A is its cross-section andρ is its density. With that:

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σ11,1 =−F

LA

or by integrating:

σ11 = −F

LAx1 + C

As before we find the integration constant from the traction boundary condition, Eqn. (63):

σ ij ni = tnj

On the top surface of the rocketni = (−1, 0, 0)andtnj = 0, i.e. no traction. Hence

σ11n1 + σ21n2 + σ31n3 = −σ11 = 0

So

σ11(x1 = 0) = 0→C = 0

Finally

σσ = 0 except σ11 = −F

LAx1

The stress field isuniaxialbecause there is only a single normal stress component in the body.

The stress field isnon-uniformbecause it changes from point to point.

At the bottom end of the rocket, the stresses must match the boundary conditions, which is easy tocheck by puttingx1 = L in the above equation.

S43. First step - make a free body diagram. In this example we are only interested in the block onthe table.We remove the flat surface and replace it by a vertical reaction force per unit arear. We removethe rope and replace it by a horizontal force per unit areamg/HT. There is also a body forceb, due togravity. Note that we now hav eonly stresses applied to the body.

We need 2 coordinate axes in this example.

r

mgHT

1

2 b

Note that we have made an assumption that the force from the rope is distributed equally across the wholeof the cross section of the block. This assumption is, of course, incorrect, and in practice there will be acomplex stress distribution around the point where the rope is attached to the block.

As in the previous example, we note that there are no forces acting along 3, through the plane of thedrawing, hence

σ33 = 0

Also, since we assume a 2D problem, the shear stresses must be zero two:

σ31 = σ32 = 0

Let’s write the two remaining equilibrium equations, Eqn. (66), explicitly:

σ11,1 + σ12,2 = −b1 + ρ x1

σ21,1 + σ22,2 = −b2 + ρ x2

Let’s assume that the shape of the block remains rectangular. That means there’s no shear deformation, andhence

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σ12 = σ21 = 0

Also, note that along 1 there is no body force, and along 2 there is no acceleration:

σ11,1 = ρ x1

σ22,2 = −b2

These equations are now independent and can be solved separately.

To solve along 1, we proceed as in the previous example and find the acceleration from the dynamicsof rigid bodies:

x1 =F

mblock=

mg

ρV=

mg

ρ HTL

Every term is constant, so by integrating we get:

σ11 =mg

HTLx1 + C

On the left surface of the blockni = (−1, 0, 0)andtnj = 0, i.e. no traction. Hence

σ11n1 + σ21n2 + σ31n3 = −σ11 = 0

So

σ11(x1 = 0) = 0→C = 0

so that

σ11 =mg

HT

x1

L

It’s easy to see that atx1 = L, σ11 matches the boundary value. Stressalong 1 is tensile everywhere.

Along 2 we proceed exactly as in the column example, to find

σ22 = −ρgx2

The mininumσ22 is at the bottomx2 = H , whereσ22 = −ρgH. This value matches the boundary condition:

r =mblockg

LT=

ρVg

LT= ρgH. Stress along 2 is compressive everywhere.

Finally

σσ = 0 except σ11 =mg

HT

x1

L, σ22 = −ρgx2

or we can write the stress tensor in the matrix form:

σσ =

mg

HT

x1

L0

−ρgx2

0

0

0

Note that the stress state changes from point to point, hence this is an example ofnon-uniformstressfield.

Since there only 2 non-zero normal stresses in the body when all shear stresses are zero, this is anexample of abiaxial stress field.

S44. Use the fact thatR is orthogonal:

R ⋅ RT = I

So that

det(R ⋅ RT ) = 1 = detR detRT = (detR)2↔ detR ± 1

We leave it without a proof that detR = −1 corresponds to a change from the right-handed to the left-

-84-

handed CS, or vice versa, in addition to rotation. If the handedness is to be preserved, then

detR = 1

S45. The cross product of vectorsa = (a1, a2, a3)T andb = (b1, b2, b3)T by definition is

a × b = det

e1

a1

b1

e2

a2

b2

e3

a3

b3

=

e1

a1

b1

e2

a2

b2

e3

a3

b3

whereei are the basis vectors. Expandingthe determinant by the first row:

a × b = (a2b3 − b2a3)e1 − (a1b3 − b1a3)e2 + (a1b2 − b1a2)e3

or as a column vector

a × b =

a2b3 − b2a3

a3b1 − b3a1

a1b2 − b1a2

On the other hand:

eijk a j bk =

e123a2b3 + e132a3b2

e231a3b1 + e213a1b3

e312a1b2 + e321a2b1

=

a2b3 − a3b2

a3b1 − a1b3

a1b2 − a2b1

So

a × b = eijk a j bk

S46. This simply follows from the definition.To get from pqr to prq one has to swap two neigh-bouring indices. After that it is impossible to obtain the original sequencepqr by cyclic permutation:prq → rqp → qpr → prq ≠ pqr. Hence the sign of all components ofepqr will change.

S47. In 3D space 3 non-coplanar vectorsa, b, c define a parallelepiped the volume of which is

V = a ⋅ (b × c) = b ⋅ (c × a) = c ⋅ (a × b)

(BTW, explain why the meaning ofa × (b ⋅ c) = b × (c ⋅ a) = c × (a ⋅ b) is completely different.)

Using the index notation for the cross product (see Ex. prob. 45):

V = eijk ai b j ck

Note that this notation is superior to the× notation because now the equation is completely symmetric wrta, b, c and no brackets are needed.

Volume cannot change with the change in CS, hence

V = eijk ai b j ck = eijk ′ai ′b j ′ck′

or using the vector rotation law:

eijk RTipap′RT

jqbq′RTkr cr ′ = eijk ′ai ′b j ′ck′

or swappingijk and pqr on the left:

epqr RTpiai ′RT

qjb j ′RTrkck′ = eijk ′ai ′b j ′ck′

or sincea, b, c are arbitrary, we can cancel those from both sides:

epqr RTpi R

Tqj R

Trk = eijk ′

finally:

-85-

eijk ′ = Rip Rjq Rkr epqr

which is the definition of a rank 3 tensor. Howev er, a simpler proof is to note that a volume is the same inany CS, hence, by definition,eijk is a rank 3 tensor.

S48. Starting from the rotation law for a rank 3 tensor:

eijk ′ = Rip Rjq Rkr epqr

Let’s see how the non-zero components transform, e.g. 123:

e123′ = R1pR2qR3r epqr = detR = 1

and the same clearly for any cyclic permutation.For the negative components, e.g. 132:

e132′ = R1pR3qR2r epqr = R1pR2r R3qepqr

or usingeprq = −epqr (see ex. prob. 46):

e132′ = R1pR2r R3q(−eprq) = − detR = −1

Now let’s see how the zero components are transformed, e.g. 112:

e112′ = R1pR1qR2r epqr

Note that for anyr there are only two non-zero terms, with opposite signs, which cancel each other:

R1pR1qR2r − R1qR1pR2r = 0

The same is obviously true for any other combination with repeated indices, hence:

eijk ′ = eijk

S49. Writing T as a 3× 3 symmetric matrix

T =

T11

T21

T31

T12

T22

T32

T13

T23

T33

we can use row 1 to calculate the determinant:

detT = T11

T22

T32

T23

T33

− T12

T21

T31

T23

T33

+ T13

T21

T31

T22

T32

= T11T22T33 − T2

23

− T12(T21T33 − T23T31) + T13(T21T32 − T22T31)

= T11T22T33 + 2T12T23T31 − T11T223 − T22T

231 − T33T

212

S50. Remember that the alternating tensoreijk = +1 ↔ ijk = 123, 231, 312,eijk = −1 ↔ijk = 132, 321, 213,and zero otherwise. So

eijk Ti1T j2Tk3 = T11T22T33 + T21T32T13 + T31T12T23 − T11T32T23 − T31T22T13 − T21T12T33

= T11T22T33 + 2T12T23T31 − T11T223 − T22T

231 − T33T

212

Now compare the last expression to that from ex. prob. 49.

S51. Using detT = eijk Ti1T j2Tk3, see ex. prob. 50, we get from (86)

det(T − λI) = (T11 − λ)(T22 − λ)(T33 − λ) + 2T12T23T31 − (T11 − λ)T223 − (T22 − λ)T2

31 − (T33 − λ)T212

= −λ3 + λ2T11 + λ2T22 + λ2T33 − λT11T22 − λT22T33 − λT33T11 + T11T22T33

+2T12T23T31 − T11T223 − T22T

231 − T33T

212 + λT2

12 + λT223 + λT2

31

-86-

= −λ3 + λ2(T11 + T22 + T33) + λ(T212 + T2

23 + T231 − T11T22 − T22T33 − T33T11)

+T11T22T33 + 2T12T23T31 − T11T223 − T22T

231 − T33T

212

or using the definition ofI T, Eqn. (88), and solution to ex. prob. 50:

det(T − λI) = −λ3 + λ2I T + λ(T212 + T2

23 + T231 − T11T22 − T22T33 − T33T11) + III T

To better understand theλ term, let’s expandII T, from Eqn. (89):

II T = T: T = Tij Tij = T211 + T2

22 + T233 + 2T2

12 + 2T223 + 2T2

31

note factors 2 due to symmetry. Now let’s see how (I T)2 look like:

(I T)2 = T211 + T2

22 + T233 + 2T11T22 + 2T22T33 + 2T33T11

We are looking to cancel the first 3 terms in the above two expressions, so that:

II T − (I T)2 = 2T212 + 2T2

23 + 2T231 − 2T11T22 − 2T22T33 − 2T33T11

Clearly theλ term in the cubic equation above is a half of the last expression, so that

det(T − λI) = −λ3 + λ2I T + λ 12 (II T − (I T)2) + III T

S52. Remember that R2TT rotates with CT as

T′ = R ⋅ T ⋅ RT = Tij ′ = RikTkpRTpj = Rik RjpTkp

Also remember thatR is orthogonal, which meansRik Rjk = Rki Rkj = δ ij . So

trT′ = Tii ′ = Rik RipTkp = δ kpTkp = Tpp

S53. As in ex. prob 52

T′: T′ = Tij ′Tij ′ = Rik Rjl Tkl RimRjnTmn = δ kmδ lnTklTmn = TklTkl

S54. As in ex. prob. (88)

III T′ = eijk ′T1i ′T2 j ′T3k′ = Rip Rjq Rkr epqr RisT1sRjt T2t RkuT3u

Using the orthogonality ofR:

III T′ = δ spδ tqδ ruepqrT1sT2tT3u = estuT1sT2tT3u = III T

S55. Just rewrite the tensor using the principal values:

T =

λ1

0

0

0

λ2

0

0

0

λ3

Then

I T = λ1 + λ2 + λ3

II T = λ21 + λ2

2 + λ23

III T = λ1λ2λ3

S56. Consider a rod under tension:

-87-

d

P

P

2 3

1

The rod of∅d is loaded by force P. The average stress is therefore:

σ =P

A=

P

π r 2=

4P

π d2

If we are clever and choose the right handed CS as shown above then the stress tensor will be:

σ ij =

σ0

0

0

0

0

0

0

0

HenceI σσ = σ , II σσ = σ 2, III σσ = 0. Thecharacteristic equation then gives λ1 = σ , λ2 = λ3 = 0. Let’scheck:

λ1 + λ2 + λ3 = σ = I σσ

λ21 + λ2

2 + λ23 = σ 2 = II σσ

λ1λ2λ3 = 0 = III σσ

Let’s find the corresponding principal directions from Eqn. (85):

(1) λ1 = σ gives:

0 ⋅ x1 = 0

−σ x2 = 0

−σ x3 = 0

So x2 = x3 = 0, x1 = ±1. It is up to us to choose one. Let’s take+1. Sothe first principal vector is:

x1 = (1, 0, 0)T

(2) λ2 = 0 giv es:

σ x1 = 0

0 = 0

0 = 0

So thatx1 = 0, andx2, x3 are free. There is an infinite number of solutions.

(3) λ3 = 0 giv es exactly the same:x1 = 0, andx2, x3 are free.

What does this mean?x2, x3⊥(1, 0, 0)T . Can choose any values for free components, but! must makesure the basis vectors are orthogonal and the CS is right handed, e.g.:

x2 = (0, 1, 0)T , x3 = (0, 0, 1)T

or

x2 = (0, 0, 1)T , x3 = (0,−1, 0)T

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The first solution corresponds to the CS drawn, which means it is already a principal CS. The second solu-tion corresponds to the original CS rotated about axis 1 byπ /2 clockwise if looking along it.

Now imagine that we were dumb, and chose this CS:

3′

P

P

2′ 1′

d π /4

What would the stress tensor be in this CS? There are at least two solution methods. In the first methodone can transform the stress tensor from the old CS to the new. In the second method one can define thestress tensor directly in the new CS. Let’s start with the fist method.

The rotation tensor is

R =

cosπ /4

cosπ /4

0

0

0

1

cosπ /4

cos 3π /4

0

=

1/√2

1/√2

0

0

0

1

1/√2

−1/√2

0

The stress tensor in the new CS isσσ ′ = Rσσ RT , so:

Rσσ =

1/√2

1/√2

0

0

0

1

1/√2

−1/√2

0

σ0

0

0

0

0

0

0

0

=

σ /√2

σ /√2

0

0

0

0

0

0

0

Rσ RT =

σ /√2

σ /√2

0

0

0

0

0

0

0

1/√2

0

1/√2

1/√2

0

−1/√2

0

1

0

=

σ /2

σ /2

0

σ /2

σ /2

0

0

0

0

Alternatively the stress tensor can be constructed directly in the new CS following the definitions ofthe components.

σ ′12

σ

σ /√2σ ′11

π /41′

σ /√2

d d

2′

3′

σ

σ ′22 σ ′21

It is clear thatσ ′3i = σ ′i3 = 0. Fromgeometry

σ ′11 = σ ′12 = σ ′21 = σ ′22 =σ

√2cos

π4

= σ /2

which matches the stress tensor components calculated via rotation.

The rotated tensor is symmetric, of course.We now hav enon-zero shear stress, and the stress state isbiaxial. The invariants are:I σ = σ /2 + σ /2 = σ , II σ = 4 ⋅ σ 2/4 = σ 2, III σ = 0. Notethat these, of course,match the values found in the old CS. Clearly this means that the principal values are the same too. Ofcourse; these are invariant to CT too! What about the principal directions? As before, using Eqn. (85):

-89-

(1) λ1 = σ gives:

(σ /2 − σ )x1

σ /2x1

+σ /2x2

+(σ /2 − σ )x2

= 0

= 0

−σ x3 = 0

So x3 = 0. Subtractingthe first 2 equations we getx1 = x2. The length of the basis vector

x ⋅ x = xi xi = 1 = x21 + x2

2 + x23

hencex21 + x2

2 = 1, or x1 = x2 = ±1/√2. We can choose any valid combination, say:

x1 = (1 /√2, 1/√2, 0)T

(2) λ2 = 0 giv es:

σ2

x1

σ2

x1

+σ2

x2

+σ2

x2

= 0

= 0

0 = 0

So thatx1 = −x2, x3 is free. We can arbitrarily setx1 = x2 = 0, x3 = 1, so that:

x2 = (0, 0, 1)T

It’s easy to check thatx1⊥ x2.

(3) λ3 = 0 giv es as before:x1 = −x2, x3 is free. Since the first two new basis vectors have been cho-sen already, we are restricted in our choice of the components of the third, because of the orthonormalityconstraints:

x3 ⋅ x1 = 0 → x31

1

√2+ x3

21

√2+ x3

3 ⋅ 0 = 0 → x31 = −x3

2

x3 ⋅ x2 = 0 → x31 ⋅ 0 + x3

2 ⋅ 0 + x33 ⋅ 1 = 0 → x3

3 = 0

x3 ⋅ x3 = 1 → x31 = −x3

2 = ±1

√2

Finally let’s choose one of the valid combinations:

x3 = (1 /√2,−1/√2, 0)T

Let’s draw the principal CS:

1′′

P

P

d2′

3′

1′

2′′ 3′′

S57. The invariants are:

I σσ = 300

-90-

II σσ = 1002 + 3002 + 1002 + 2 × 2002 + 2 × 2002 + 2 × 4002 = 59× 104

III σσ = −100

300

400

400

100

− 200

200

−200

400

100

− 200

200

−200

300

400

= 13× 106 − 20× 106 − 28× 106 = −35× 106

So that

1

2(II σσ − (I σσ )2) =

1

2(59× 104 − 9 × 104) = 25× 104

From Eqn. (86) the characteristic equation looks like

−λ3 + 300λ2 + 25× 104λ − 35× 106 = 0

I solve this problem numerically using LAPACK l ibrary, routine DSYEV, with Fortran.

Original tensor-1.00000000000E+02 2.00000000000E+02 -2.00000000000E+022.00000000000E+02 3.00000000000E+02 4.00000000000E+02-2.00000000000E+02 4.00000000000E+02 1.00000000000E+02Invariants:3.00000000000E+02 5.90000000000E+05 -3.50000000000E+07

The DSYEV eigenvalues in increasing order are:L3 L2 L1

-4.42895748878E+02 1.28655464368E+02 6.14240284510E+02DSYEV invariants check:3.00000000000E+02 5.90000000000E+05 -3.50000000000E+07

The DSYEV orthonormal eigenvectors (columns) are:x3 x2 x1

-6.34587302398E-01 -7.70838350041E-01 -5.57422078993E-024.91831418220E-01 -3.47155034107E-01 -7.98489347672E-01-5.96155023200E-01 5.34126970299E-01 -5.99422695527E-01The angles (deg) are:1.29389382175E+02 1.40429231753E+02 9.31954495260E+016.05389733835E+01 1.10313402693E+02 1.42986086857E+021.26595014954E+02 5.77152759561E+01 1.26828562438E+02

Soλ1 = 614,λ2 = 129,λ3 = −443,x1 = (−0. 0557,−0. 798,−0. 599),x2 = (−0. 771,−0. 347,−0. 534)andx3 = (−0. 635, 0. 492,−0. 596). The principal CS can be drawn based on the angles above.

The initial stress state looks like this:

300

1

2

3

400

200

100

-100

-200

The principal stress state looks like this:

-91-

1′

3′

614

-443129

2′

2

1

3

S58. Start from the definition ofF:

F =∂x∂X

In another CS:

F′ =∂x′∂X′

By definition:

x′ = R ⋅ x ; X′ = R ⋅ X

Using chain rule:

F′ =∂x′∂X′

=∂x′∂X

⋅∂X∂X′

=∂(R ⋅ x)

∂X⋅

∂(RT ⋅ X′)∂X′

= R ⋅ F ⋅ RT

This is one of the definitions of R2T.

S59. Start from Eqn. (103).We’l l use the index notation, as it is more explicit:

E jk = ½(Fij Fik − δ jk)

or using Eqn. (95):

E jk =1

2

∂xi

∂X j

∂xi

∂Xk− δ jk

or using

xi = Xi + ui

we obtain

E jk =1

2

∂(Xi + ui )

∂X j

∂(Xi + ui )

∂Xk− δ jk

=1

2(δ ij +

∂ui

∂X j)(δ ik +

∂ui

∂Xk) − δ jk

= ½(u j ,k + uk, j + ui , j ui ,k)

S60. Tw o proofs are possible. Proof 1 is based on first proving thatF is R2T (ex. prob. 58). Thenuse Eqn. (103):

E = ½(FT ⋅ F − I)

In another CS:

-92-

E′ = ½(F′T ⋅ F′ − I)

Remember thatI′ = I. Giv en thatF is R2T:

E′ = ½(F′T ⋅ F′ − I) = ½((R ⋅ F ⋅ RT )T ⋅ R ⋅ F ⋅ RT − I) = ½(R ⋅ FT ⋅ RT ⋅ R ⋅ F ⋅ RT − I)

R ⋅ RT = I so

= ½(R ⋅ FT ⋅ F ⋅ RT − I) = R ⋅ ½(FT ⋅ F − I) ⋅ RT = R ⋅ E ⋅ RT

Here we usedI = R ⋅ I ⋅ RT .

The second proof proceeds similarly to that in ex. prob. 58. We start from (104):

E jk ′ = ½(ui , j ′ + u j ,i ′ + uk,i ′uk, j ′)

To prove thatEij is R2T, it is sufficient to prove thatui , j is R2T.

We useui ′ = Rij u j , xi ′ = Rij x j . Using the chain rule:

ui , j ′ =∂ui ′∂x j ′

=∂(Rikuk)

∂xp

∂xp

∂x j ′= Rikuk,p

∂(RTpmxm′)

∂x j ′= Rikuk,pδ mjR

Tpm = Rikuk,pRT

pj

= Rik Rjpuk,p

which is one of the definitions of R2T. In tensor notation

∇u′ = R ⋅ ∇u ⋅ RT

S61. Written explicitly the motion is:

x1 = X1 + tX2

x2 = X2

x3 = X3

The motion can be drawn like this:

α

2

13

t

1

1

We can use just a 2D drawing, because there is no motion along direction 3. What is shown in an unde-formed unit cube of material (solid lines), and its deformed shape (dashed lines). Note that the motion andthe deformation arelinear, meaning that straight lines stay straight throughout the deformation.

Let’s calculateF:

F =∂x∂X

=

1

0

0

t

1

0

0

0

1

Note that the deformation gradient is not symmetric.

The strain tensor is:

E = ½(FT ⋅ F − I)

First

-93-

FT ⋅ F =

1

t

0

0

1

0

0

0

1

1

0

0

t

1

0

0

0

1

=

1

t

0

t

t2 + 1

0

0

0

1

So that

E =

0

t/2

0

t/2

t2/2

0

0

0

0

Note that the strain tensor is symmetric, unlike the deformation gradient.

Alternatively we can calculate strain from displacement field:

u1 = x1 − X1 = tX2

u2 = x2 − X2 = 0

u3 = x3 − X3 = 0

So that

Eij = ½(ui , j + u j ,i + uTi ,kuk, j )

First

ui , j =

0

0

0

t

0

0

0

0

0

So that

uTi ,kuk, j =

0

t

0

0

0

0

0

0

0

0

0

0

t

0

0

0

0

0

=

0

0

0

0

t2

0

0

0

0

Finally

Eij =1

2

0

0

0

t

0

0

0

0

0

+

0

t

0

0

0

0

0

0

0

+

0

0

0

0

t2

0

0

0

0

=

0

t/2

0

t/2

t2/2

0

0

0

0

as before.

This example makes the distinction between thesmalland thefinitestrain formulations very clear.Let’s assume that the strains stay below 1%, which is typical forelasticstrains in engineering structures.That means thatt ≈ 10−2, which means thatt2 ≈ 10−4, i.e. two orders of magnitude smaller. Hence, in thiscase thet2 term can be neglected, and the small strain tensor will be:

εε =

0

t/2

0

t/2

0

0

0

0

0

Note that now there are no normal strains, only shear. This means that the lengths of line elements don’tchange, only the angle between them. If you look at the drawing and imagine thatt is small, you will seethat this is correct. Indeed ift is small, thent = tanα ≈ α . Therefore small scale pure shear deformation issimply change of angles between straight lines. Moreover, if the deformation is redrawn symmetrically wrtthe original shape, the meaning of ½ factors becomes clear:

-94-

t/2

t/2

In contrast, ift is not small compared to 1, as is the case drawn above, then thet2 term is non-negli-gible and cannot be dropped. Indeed, for rubber-like materialst can be many times greater than 1, giving avery larget2 term, meaning thatε22, the elongation along the initial direction 2 will be large. Again, usethe drawing to visualise this.

This deformation type is calledsimple shearor pure shear.

Finally, I ε = 0, hence the deformation isincompressible.

S62. The invariants are:I ε = 0, II ε = t2/2, III ε = 0. Thecharacteristic equation, Eqn. (87), will looklike this:

−ε 3 +t2

4ε = 0

The roots in decreasing order are:ε1 = t/2, ε2 = 0, ε3 = −t/2, and the strain tensor in the principal directionswill look lik e this:

εε =

t/2

0

0

0

0

0

0

0

−t/2

The principal directions are found from Eqn. (85).

(1) λ3 = −t/2:

t

2x1 +

t

2x2 = 0

t

2x1 +

t

2x2 = 0

t

2x3 = 0

The solution isx1 = −x2, x3 = 0. We can choose any solution, as long asxi ⊗x j = I (or in index notationxi

k x jk = δ ij , wherexi

k is kth component ofi th principal (basis) vector). Let’s choosex3 = (√2/2, −√2/2, 0).

(2) λ2 = 0:

t

2x2 = 0

t

2x1 = 0

0x3 = 0

The solution isx1 = x2 = 0, x3 = ±1. Again, we are free to choose either solution. Let’s choosex2 = (0, 0, 1).

(3) λ1 = t/2:

-95-

−t

2x1 +

t

2x2 = 0

t

2x1 −

t

2x2 = 0

t

2x3 = 0

The solution isx1 = x2, x3 = 0. Sincethe other 2 vectors have been set already, we hav eno freedom left.Provided we want to keep the CS right-handed we must choose:x1 = (√2/2, √2/2, 0). Checkthatx1 ⋅ x3 = 0for orthonormality.

We can now draw the deformation in principal coordinates:

1-t/2

13, 2′

2 1′

3′

1

1+t/2

1

Note that in the principal coordinates angles between any pairs of straight lines stay constant. In contrastthe lengths change.

This drawing illustrates well that the deformation is incompressible. The volume of the undeformedunit cube isV = 1. Thevolume of the deformed cube is

v = (1 + t/2)(1 − t/2) = 1 − t2/4

If t is small, i.e. the small strain theory is assumed, thenv ≈ 1 = V, which is the meaning of incompressibledeformation.

S63. The rotation tensor corresponds to rotating about axis 3 byπ /2 counter clockwise. This can bedrawn as:

2, 1′

13

t

1

1

2′

Let’s first calculateF′ directly:

F′ =∂x′∂X′

where

x′ = R ⋅ x

X′ = R ⋅ X

Calculating:

-96-

x′ =

0

−1

0

1

0

0

0

0

1

X1 + tX2

X2

X3

=

X2

−X1 − tX2

X3

Check on the drawing that these expressions are correct.

Similarly

X′ =

0

−1

0

1

0

0

0

0

1

X1

X2

X3

=

X2

−X1

X3

Combining the last 2 lines:

x′ =

X1′X2′ − tX1′

X3′

Again check on the drawing that these expressions are correct.

Finally with simple differentiation:

F′ =

1

−t

0

0

1

0

0

0

1

To verify thatF is R2T, just need to do 2 matrix multiplications:

R ⋅ F =

0

−1

0

1

0

0

0

0

1

1

0

0

t

1

0

0

0

1

=

0

−1

0

1

−t

0

0

0

1

Finally

R ⋅ F ⋅ RT =

0

−1

0

1

−t

0

0

0

1

0

1

0

−1

0

0

0

0

1

=

1

−t

0

0

1

0

0

0

1

which agrees with the answer obtained directly, henceF is indeed R2T.

S64. We assume a popular rosette where the three strain gauges are oriented at 0, 45 and 90:

2′

1

2

0

45

90

1′

We can accept the orientations of 0and 90 gauges as a natural choice of CS.

We denote strains measured along these directions asε0, ε45 andε90. With thatε11 = ε0, ε22 = ε90.All we need to do is to resolveε45 into our CS.

The rotation tensor that transforms the 12 CS into the 1′2′ CS is

R =

√2/2

−√2/2

0

√2/2

√2/2

0

0

0

1

-97-

In 12 CS the strain tensor looks like:

εε =

ε11

ε12

ε13

ε21

ε22

ε23

ε31

ε32

ε33

In vast majority of cases the strain gauge is applied to the free surface, i.e. there are no surface tractions.This meansε13 = ε23 = 0. With this the strain tensor will be

εε =

ε0

ε12

0

ε12

ε90

0

0

0

ε33

The strain gauge rosette measurement says nothing aboutε33. This remains unknown and has to be foundfrom other experiments. However, we know that this strain is unchanged by rotating CS about axis 3. Thisallows us findingε12 without knowingε33.

So

ε45 = ε11′ = R1i ε ij RTj1 =

√2/2

√2/2

0

ε0

ε12

0

ε12

ε90

0

0

0

ε33

√2/2

√2/2

0

=

√2/2

√2/2

0

√2/2(ε0 + ε12)

√2/2(ε12 + ε90)

0

= ½(ε0 + 2ε12 + ε90)

From where

ε12 = ε45 − ½(ε0 + ε90)

S65. Let’s work with a stress tensor. Exactly the same analysis applies to the strain tensor, just swapσ for ε .

Start from the tensor is principal directions:

σ =

σ1

0

0

0

σ2

0

0

0

σ3

Next we need to understand how shear stress is related to other components of stress tensor. Refer to thediagram

sn

p

t

An element of surface with normaln has the stress vectort, which can be decomposed into the normal,p,and the the shear (tangential),s vectors:

s = t − p

where using Eqn. (63) and the fact thatσσ is diagonal:

-98-

t = σσ ⋅ n =

σ1n1

σ2n2

σ3n3

p has the magnitude of the projection oft on n and the direction ofn. Let’s calculate this projection first.We denote itz:

z = t ⋅ n = σ1n21 + σ2n2

2 + σ3n23

Sop is

p = z

n1

n2

n3

and

s =

(σ1 − z)n1

(σ2 − z)n2

(σ3 − z)n3

Note thats is a function ofn only. We are interested in the magnitude of maximum shear stress. Sowe can work with a dot products ⋅ s.

To find the extremum values ofs ⋅ s, we create a scalarobjective function,

f = s ⋅ s + L(n ⋅ n − 1)

whereL is theLagrange multiplier. The expression in brackets expresses the constraint that the normalvectors are of unit length:n ⋅ n = 1. Themethod of Lagrange multipliers belong to the area namedcon-strained optimisation. Using Lagrange multipliers is a trick to bring the equality constraints into the objec-tive function.

From basic differential calculus we know that a necessary condition for an extremum off is df = 0.Using the chain rule:

df =∂ f

∂n1dn1 +

∂ f

∂n2dn2 +

∂ f

∂n3dn3 +

∂ f

LdL

Note that the last term simply recovers the constraint.n1, n2 andn3 are treated as three independent vari-ables (subject only to the above constraint). Thismeans that fordf = 0, all three partial derivatives abovemust vanish:

∂ f

∂n1= 0 ;

∂ f

∂n2= 0 ;

∂ f

∂n3= 0

Let’s find these derivatives. For this we need to calculates ⋅ s first:

s ⋅ s = (σ1 − z)2n21 + (σ2 − z)2n2

2 + (σ3 − z)2n23

= σ 21 n2

1 + σ 22 n2

2 + σ 23 n2

3 + z2n21 + z2n2

2 + z2n23 − 2σ1zn2

1 − 2σ2zn22 − 2σ3zn2

3

= σ 21 n2

1 + σ 22 n2

2 + σ 23 n2

3 + z2(n21 + n2

2 + n23) − 2z(σ1n2

1 + σ2n22 + σ3n2

3)

= σ 21 n2

1 + σ 22 n2

2 + σ 23 n2

3 + z2 − 2z2

= σ 21 n2

1 + σ 22 n2

2 + σ 23 n2

3 − (σ1n21 + σ2n2

2 + σ3n23)2

With that:

-99-

∂ f

∂n1=

∂(s ⋅ s)∂n1

+ 2Ln1 = 0

or

2n1σ 21 − 2z2σ1n1 + 2Ln1 = 0

This equation splits into 2 branches:

n1 = 0 or σ 21 − 2zσ1 + L = 0 (148)

Because all expressions are symmetrical wrtσσ andn, we can straight away write the other 2 equations:

n2 = 0 or σ 22 − 2zσ2 + L = 0 (149)

n3 = 0 or σ 23 − 2zσ3 + L = 0 (150)

Let’s first consider the cases where two projections of the normal are zero.For example, ifn1 = n2 = 0,then from the unity constraintn3 = ±1. However, we immediately see that this direction is one of the prin-cipal directions, wheres = 0. Thiscan also be checked from the expression fors ⋅ s. So these directionsgive the orientations on which shear stress magnitude isminimumi.e. zero.

Next let’s explore cases when only one component ofn is zero, e.g.n1 = 0, n2≠0, n3≠0. In this caseEqns. (149)and (150) will be:

σ 22 − 2zσ2 + L = 0

σ 23 − 2zσ3 + L = 0

By subtracting one from the other we get rid ofL:

σ 22 − σ 2

3 + 2z(σ3 − σ2) = 0

or

(σ2 − σ3)(σ2 + σ3) + 2z(σ3 − σ2) = 0

Again we have 2 branches:

σ2 = σ3 or σ2 + σ3 − 2z = 0

Let’s follow the second branch now. Using the expression forz, with n1 = 0, we obtain:

σ2 + σ3 − 2σ2n22 − 2σ3n2

3 = 0

From the unity constraint:n22 = 1 − n2

3. With that the above equation is rewritten as:

σ2 + σ3 − 2σ2(1 − n23) − 2σ3n2

3 = 0

or

σ3 − σ2 + 2n23(σ2 − σ3) = 0

Since in this branchσ2≠σ3, then

2n23 = 1

or

n3 = ±√2/2

hence from the orthonormality constrain

n2 = ±√2/2

but with the opposite sign ton3. One solution is thusn = (0,±√2/2, +−√2/2). Thefactors are sin and cos of45 . So this vector is in 23 plane and makes 45angles with both 2 and 3 axes.

-100-

What is the magnitude of this extremum shear stress?We put the foundn into the expression fors ⋅ s:

s ⋅ s =σ 2

2

2+

σ 23

2− (

σ2

2+

σ3

2)2 = (

σ2 − σ3

2)2

Hence the magnitude of this shear value is

|s| =σ2 − σ3

2

Following exactly the same logic for the other 2 cases, whenn2 = 0 and n3 = 0, we obtain these directionsand extreme values:

n = (±√2/2, 0, +−√2/2) ; |s| =σ1 − σ3

2

n = (±√2/2, +−√2/2, 0) ; |s| =σ1 − σ2

2

Given our convention thatσ1 ≥ σ2 ≥ σ3, the absolute maximum shear stress is

τmax =σ1 − σ3

2

It remains to check the other solution branches, for examplen1 = 0 andσ2 = σ3. By inserting theseconditions into thes ⋅ s expressions we obtain:

s ⋅ s = σ 22(n2

2 + n23) − (σ2(n2

2 + n23))2 = σ 2

2 − σ 22 = 0

So shear stresses on all planes parallel ton1 are zero. Stress states with any 2 principal stresses equal arecalledequi-biaxial. The distinguishing feature of these stress states is that there is no shear stress in theequi-biaxial plane.

Finally we need to see if there are any solutions withn1≠0, n2≠0 and n3≠0. For this we subtractthree pairs of Eqns.: (148) - (149), (149) - (150), (150) - (148):

σ 21 − σ 2

2 + 2z(σ2 − σ1) = 0

σ 22 − σ 2

3 + 2z(σ3 − σ2) = 0

σ 23 − σ 2

1 + 2z(σ1 − σ3) = 0

Solution is possible only iff

σ1 = σ2 = σ3

which is calledhydrostaticstress state. It is easy to see from the expression fors ⋅ s that in this case shearstress is zero on any plane, which is the distinguishing feature of the hydrostatic stress state.

If the principal stresses are different, then one obtains 3 expressions forz from 3 equations above:

2z =σ 2

2 − σ 21

σ2 − σ1= σ2 + σ1

2z =σ 2

3 − σ 22

σ3 − σ2= σ3 + σ2

2z =σ 2

1 − σ 23

σ1 − σ3= σ1 + σ3

However, since the left hand sides are the same, all stresses must be equal, which is a contradiction.Weconclude that there are no more possible solutions.

S66. Stress state:

-101-

300

2

3

1

50

-500

These are also the only 3 planes with zero shear stress.

Maximum shear stress planes:

400

2

3

1

275

2

3

1

125

2

3

1

Strain state:

0.001

2

3

1

0.001

-0.002

These are also three of the zero shear planes. However, other zero shear planes exist too.

1

2

3

Indeed, sinceε1 = ε3, any plane parallel to direction 2 is zero shear. The strain state is equi-biaxial and 13is the equi-biaxial plane.

Maximum shear strain planes:

0.0015

2

3

1 0.0015

2

3

1

S67. The stress tensor in the matrix form looks like this:

-102-

σσ =

0

200

0

200

0

0

0

0

0

First need to recognise thatσ12 is at maximum when all normal stresses are zero. If it were not, then theMohr’s diagram shows that there would have to be a non-zero normal stress:

200

0

normal

shear

So the diagram must look like this:

100

0-200normal

200

shear 200

The maximum shear is the radius of the big circle. Since it is centred at the origin, the two principalstresses are immediatelyσ3 = −200MPa, andσ1 = 200MPa. Theremaining principal stressσ2 = 0 MPa.That fits the radii and the centres of the two smaller circles.

We can now use the diagram. What will happen if we rotate the cube about axis 3 so thatσ12 dropsto 100MPa?

-173 0normal

shear

100

2α

173

From geometry: cos 2α = 100 / 200= 0. 5→ 2α = 60 → α = 30 . The normal stresses are200 sin 2α = 100√3 ≈ 173MPa. Thestress tensor will look like:

σσ =

±173

100

0

100

+−173

0

0

0

0

Note that extra sign conventions are needed to be sure what signs to use. Clearly the signs will flip if thesign ofα will change.

If α = 45 , then we get to the principal state:

-103-

2000normal

shear

2α

-200

σσ =

±200

0

0

0

+−200

0

0

0

0

From the principal stress state we can now rotate about another axis, say 1. Ifα = 15 , then thestress state will look like this:

2α

13normal

shear

-200 187

50

The shear stressσ23 = 100 sin 2α = 50MPa. Thedistance of normal stresses from the centre of this circle is100 cos 2α ≈ 87MPa. With that the smaller normal stress is 100− 87 = 13MPa and the larger normal stressis 100+ 87 = 187MPa. Writtenas a matrix the stress state is:

σσ =

−200

0

0

0

187

50

0

50

13

where we arbitrarily set the signs to match the Mohr’s diagram.

A warning, one more time: for the Mohr’s diagram to make any sense, one can rotate only about asingle principal axis at a time, relative to the principal orientations. If more complex analysis is required,then the Mohr’s circle must not be used.

S68. The stress tensor:

125

normal

shear

-500 300

275

400

50

The strain tensor:

0.0015

normal

shear

-0.002 0.001

-104-

Note that one circle shrunk to a point. Hence the two remaining circles coincide. This is a direct conse-quence of having two equal principal strains. No shear strains exist in the plane of those two principaldirections, irrespective of any rotation about the other principal direction.

S69. Note that sinceε13 = ε23 = 0 we conclude thatε33 is the principal strain. Hence we are onlyinterested in 12 plane, and we can use a two-dimensional Mohr’s diagram, i.e. just a single circle.

The centre of the circle is at ½(ε11 + ε22) = −10−2. The radius of the circle,γmax, - the maximumshear strain, is calculated from the right angled triangle:

γ 2max =

ε11 − ε22

2

2

+ ε 212 = 8 × 10−4

so thatγmax = 2√2 × 10−2 ≈ 0. 0283.With that the Mohr’s diagram can be drawn:

0.02

normal

-0.01 0.0183-0.03 0.01-0.0383

shear

So the principal strains are−3. 83× 10−2 and 1. 83× 10−2. Because we don’t know the value of the thirdprincipal strain, we cannot give these values definite label.

Finally note that we have obtained a general expression for principal values in two-dimensionalcases:

γ1,γ2 =ε11 + ε22

2±

ε11 − ε22

2

2

+ ε 212

½

which is nothing more than a solution to a quadratic equation, which is what the characteristic equation isreduced to in 2D case.

S70. Symmetry of stress tensor meansσ st = σ ts. Hence one can swap subscriptst ands in Eqn.(109):

σ st = Cstabε ab = Ctsabε ab = σ ts

which means that

Cstab = Ctsab

This is calledminorsymmetry ofC. We will show later that rank 4 tensors can also havemajorsymmetry.

There are thus three redundant combinations ofst: C12ab = C21ab, C23ab = C32ab andC31ab = C13ab.Each of these equations has 32 = 9 unique combinations ofab. Hence the total number of redundant com-ponents ofC is 3× 9 = 27. Thisreduces the number of required constants inC from 34 = 81 to 54.

Exactly the same analysis can be made regarding strain:ε ab = ε ba. Hence

Cstab = Cstba

which is another minor symmetry inC. There are 3 redundant combinations ofab: Cst12 = Cst21,Cst23 = Cst32 andCst31 = Cst13. Each of these combinations has 6 unique combinations ofst. We say 6 andnot 9 because we already have taken the symmetry ofσσ into account. The total number of redundant com-ponents ofC is thus 3× 6 = 18. Thisfurther reduces the number of required independent components inCfrom 54 to 36.

-105-

Both minor symmetries inC can be summarised as:

Cstab = Ctsab = Ctsba = Cstba

The above discussion can be equally applied to the compliance tensorS to yield:

Sstab = Stsab = Stsba = Sstba

S71. From Eqn. (110)elastic energy densitycan be calculated as

W = ∫ σ abdε ab

The units ofW are pressure, e.g. MPa, which is the same as energy per volume.

Using elasticity this can be rewritten as

W = ∫ Cabstε stdε ab = ½Cabstε stε ab

or swappingst andab we get:

W = ½Cstabε abε st

which means

Cstab = Cabst

This property is called themajorsymmetry.

After taking the symmetry ofσσ andεε into account we now hav e6 unique combinations ofst and 6unique combinations ofab. Excluding 6 combinations ofab = st, we hav e6 × 6 − 6 = 30 combinations ofstab, half of which are redundant due to major symmetry. Finally C has 15+ 6 = 21 independent compo-nents.

Exactly the same logic can be applied toS to show that

Sstab = Sabst

S72. Choose an arbitrary stress component, sayσ 23. Expressit as a function of stress with Eqn.(109). Itwill involve a double summation over both strain subscripts - 3× 3 = 9 terms in total:

σ23 = C2311ε11 + C2322ε22 + C2333ε33 + C2312ε12 + C2321ε21 + C2323ε23 + C2332ε32 + C2331ε31 + C2313ε13

due to symmetry in strainε12 = ε21 etc., so:

σ23 = C2311ε11 + C2322ε22 + C2333ε33 + 2C2312ε12 + 2C2323ε23 + 2C2331ε31

When written in the matrix form, these factors of 2 can be made a part of the elastic matrix, as in Eqn.(114), or a part of the strain vector, as in (115).

S73. First we need to explain the meaning of the Poisson’s ratio,ν . Consider a uniaxial tension of arod.

12

3

The dashed lines show the original, undeformed, shape of the rod. Solid lines show the shape of the rodafter some tensile loading was applied at one end. The other end is considered constrained in this illustra-tion.

-106-

Experiments show that as the rod elongates its cross section is reduced, although there are some veryexotic artificial materials which show an increase of the cross section with rod elongation. ThePoisson’sratio quantifies the cross section change relative to change in length.

If axis 1 is along the axis of the rod and axes 2 and 3 are in the cross section, then these clearly arethe principal strain axes. Thestrain tensor in these coordinates will look like:

εε =

ε11

sym

0

ε22

0

0

ε33

By definition, the Poisson’s ratio is

ν = −ε22

ε11= −

ε33

ε11

In isotropic materials there is only a single Poisson’s ratio. Thisdefinition means that a positive Poisson’sratio means that material will shrink in the directions normal to the pulling direction. And inversely, underuniaxial compression, the material will bulge out in the directions normal to the compression direction, ifthe Poisson’s ratio is positive.

Although the Poisson’s ratio is defined on a uniaxial test, it can be used in any arbitrary deformation.

Back to the original question. The only non-zero principal stress isσ11, which can be calculatedfrom Eqn. (117):

σ11 = 2µε11 + λ(ε11 + ε22 + ε33)

or using the Poisson’s ratio:

ε22 = ε33 = −ν ε11

so that

σ11 = 2µε11 + λ(ε11 − ν ε11 − ν ε11) = (2µ+ λ(1 − 2ν ))ε11

From Eqn. (125):

ν =λ

2(λ + µ)

so that

σ11 = (2µ+ λ(1 −2λ

2(λ + µ)))ε11 =

µ(3λ + 2µ)

λ + µε11 = Eε11

So if one is conducting a uniaxial test, and the only quantities of interest are axial strain and stress, thenonly a single material property is required - the Young’s modulus.

This example is, of course, the main justification for using the pair of the Young’s modulus and thePoisson’s ratio as the two linear elastic isotropic material constants.

For the pure shear example refer to ex. prob. 61. Notethat the strain tensor can be expressed as:

εε =

0

sym

ε12

0

0

0

0

from Eqn. (117) one immediately obtains that the only non-zero stress is:

σ12 = 2µε12

or, in shear modulus notation:

σ12 = 2Gε12

from which the motivation for the shear modulus is obvious. Thismaterial property describes resistance toshear.

-107-

S74. From ex. prob. 73, in uniaxial stress state

ε22 = ε33 = −ν ε11

So that

I ε = trεε = ε11(1 − 2ν )

If ν = 0. 5then I ε = 0, and the deformation is incompressible. Because in this case incompressible defor-mation is solely due to a special material property, materials withν = 0. 5are calledincompressible materi-als. Rubbers haveν ≈ 0. 5. The assumption of incompressibility simplifies analysis of such materials.

S75. Can use any of Eqns. (109), (112), (114) or (115). Choose any normal stress, and express it viastrain, e.g. forσ33:

σ33 = . . . + 2C3312ε12 + 2C3323ε23 + 2C3331ε31

which clearly shows that the normal stresses depend on shear strains.

Now choose any shear stress, e.g.σ12:

σ12 = C1211ε11 + C1222ε22 + C1233ε33 + . . .

which clearly shows that the shear stresses depend on normal strains.

If one usedS instead, then normal strains are shown to depend on shear stresses and vice versa.

S76. As in the previous example can use any of Eqns. (109), (112), (114) or (115). Now imagine thestrain tensor in principal coordinates.Try expressing any shear stress component via strain, e.g. forσ13:

σ13 = C1311ε11 + C1322ε22 + C1333ε33

so in general there will be non-zero shear stresses corresponding to principal strains.

If one usesS instead, then non-zero shear strains are shown to exist, corresponding to principal stressstate.

S77. Can use either Eqns. (116) or (117). Choose any normal stress, e.g.σ22:

σ22 = 2µε22 + λ(ε11 + ε22 + ε33)

so normal stresses depend only on normal strains. Choosing any shear stress, e.g.σ12:

σ12 = 2µε12

As before, if one usesS instead, then stresses and strains change places, but the conclusion is thesame.

S78. The answer is immediately clear from Eqns. (116) or (117), or from ex. prob. 77. Assumethestrain tensor in principal coordinates. Since shear strains are zero and shear stresses depend only on shearstrains, then shear stresses are zero too. Therefore we conclude that the stress tensor is also in the principalCS. Hencethe principal directions of strain and stress coincide.

S79. The aim is to express the strain tensor via the stress tensor. First, use (117) to calculatetrσσ = σ kk.

σ kk = σ11 + σ22 + σ33 = (2µε11 + λε kk) + (2µε22 + λε kk) + (2µε22 + λε kk)

= 2µ(ε11 + ε22 + ε33) + 3λε kk = 2µε kk + 3λε kk = (2µ+ 3λ)ε kk

or using the bulk modulus notation,K :

σ kk = 3Kε kk

This expression is notable for it relates the first invariant of strain,I εε = trεε = ε kk, to the first invariant ofstress,I σσ = trσσ = σ kk. We hav esaid earlier that the meaning ofI εε is volumetric strain. Now we can givethe meaning toI σσ . Its negative is calledpressure:

p = −σ kk

-108-

so that positive pressure represents negative stress, which is consistent with our sign convention. Thephys-ical meaning ofp is precisely hydrostatic pressure, exceptp can also be tensile.p can also be described asequi-triaxial stress state. The meaning ofI σσ = 3KI εε is that pressure produces only a change of volume,never a change of shape. This is whyK is called thebulk modulus.

With this Eqn. (117) can be rewritten as

σ ij = 2µε ij +λ

3Kσ kkδ ij

Now we can expressε ij as:

ε ij =1

2µσ ij −

λ3K

σ kkδ ij

or moving to theE andν pair of constants, and using Eqns. (126) and (127), we obtain:

ε ij =1 + ν

Eδ ipδ jq −

ν1 + ν

δ ij δ pqσ pq (151)

So thatS can be written as:

Sijpq =1

E(1 + ν )δ ipδ jq − νδ ij δ pq

S80. Imagine a silicon sealant gun. Assume that the material of the pressure vessel is much stiffer(higher Young’s modulus) than silicon, which is a reasonable assumption. Then the walls of the pressurevessel can be considered rigid. Hence no strain in the direction normal to the axis of the vessel is possible.Consider point A somewhere sufficiently far from the nozzle. The only strain is compressive axial strain:

A

1

2

3

ε3

εε =

0

sym

0

0

0

0

ε3

The stress tensor is found from Eqn. (117):

σσ =

λε3

sym

0

λε3

0

0

(2µ+ λ)ε3

S81. A uniaxial strain state example is shown in ex. prob. 80. Ina uniaxial stress state the stress ten-sor is:

σσ =

σ11

sym

0

0

0

0

0

-109-

if the stress is tensile.

From Eqn. (151) in ex. prob. 79 the corresponding strain tensor will be:

εε =1

E

σ11

sym

0

−νσ11

0

0

−νσ11

Observe that the factors connecting stress and strain components differ between uniaxial stress and uniaxialstrain states. In particular, for directions of non-zero stress and strain we have for uniaxial strain:

σ33 = (2µ+ λ)ε33

and for uniaxial stress:

σ11 = Eε11

However, E ≠ 2µ+ λ , see Eqn. (125).

Similarly, for strains in directions where there are no stresses, and for stresses in directions wherethere are no strains one has for uniaxial strain:

σ11 = λε33

and for uniaxial stress:

σ11 = −E

νε22

Not only λ ≠ E/ν , see Eqn. (126), but here even the signs are opposite.

S82. First expressσσ via u:

σ ij = 2µε ij + λε kkδ ij

where

ε ij = ½(ui , j + u j ,i )

so that

σ ij = µ(ui , j + u j ,i ) + λuk,kδ ij

and

σ ij , j = µ(ui , jj + u j ,ij ) + λuk,kjδ ij

or using the properties of dummy indices and of the Kronecker delta tensor:

σ ij , j = µ(ui , jj + uk,ki) + λuk,ki = µui , jj + (µ + λ)uk,ki

or in tensor notation:

∇ ⋅ σσ = µ∇2u + (µ + λ)∇∇ ⋅ u

Finally the equilibrium equations will be written as:

µui , jj + (µ + λ)uk,ki = −bi + ρ xi

or in tensor notation:

µ∇2u + (µ + λ)∇∇ ⋅ u = −b + ρ x

S83. Remember that the order of differentiation is not important. Consider first components of thestrain tensor in a particular plane, e.g. 12:

ε11 = u1,1

ε12 = ½(u1,2 + u2,1)

-110-

ε22 = u2,2

Differentiateε11 over x2 twice:

ε11,22 = u1,122

Now note that the same term is obtained ifε12 is differentiated over x1 andx2:

ε12,12 = ½(u1,212+ u2,112)

Now note that the second term in the last expression is obtained ifε22 is differentiated twice over x1:

ε22,11 = u2,211

Comparing the last 3 expressions one sees that:

ε11,22+ ε22,11 = 2ε12,12 (152)

A further two expression of this kind are obtained if one considers strains in 23 and 31 planes. Alterna-tively one can just do a cyclic permutation of indices:

ε22,33+ ε33,22 = 2ε23,23 (153)

ε33,11+ ε11,33 = 2ε31,31 (154)

Now consider differentiatingε11 over x2 andx3:

ε11,23 = u1,123 = u1,312

The same term can be obtained ifε13 is differentiated over x1 andx2:

ε13,12 = ½(u1,312+ u3,112) = ½(u1,312+ u3,211)

Again looking at the last term, one sees that it can be obtained ifε32 is differentiated over x1 twice:

ε32,11 = ½(u3,211+ u2,311) = ½(u3,211+ u2,131)

Again looking at the last term we can see that it is obtained ifε21 is differentiated over x3 andx1:

ε21,31 = ½(u2,131+ u1,231)

Note that the very last term is the same as the starting term, inε11,23. Thus the chain is complete and wecan link all four second derivatives of these strain components:

ε13,12− ε32,11 = ½(u1,312− u2,131)

and

ε11,23− ε21,31 = ½(u1,312− u2,131)

so that

ε13,12− ε32,11 = ε11,23− ε21,31

or

ε13,12+ ε12,13 = ε23,11+ ε11,23 (155)

As before, another 2 expressions can be obtained by cyclic permutation of all indices:

ε21,23+ ε23,21 = ε31,22+ ε22,31 (156)

ε32,31+ ε31,32 = ε12,33+ ε33,12 (157)

We leave it without proof that no further independent expressions linking derivatives of the compo-nents of the strain tensor exist.

If Eqns. (152)-(154) are rewritten as:

ε11,22+ ε22,11 = ε12,12+ ε12,12

-111-

ε22,33+ ε33,22 = ε23,23+ ε23,23

ε33,11+ ε11,33 = ε31,31+ ε31,31

then a pattern emerges. The6 equations can be summarised as:

ε ij ,kl + ε kl,ij = ε ik, jl + ε jl ,ik

Note that of these 81 equations 75 are redundant, i.e. repeated or trivial, as fori = j = k = l = 1.

S84. The key to the solution is to remember that the stress tensor is linked to the stress vector, andvectors are additive.

From Eqn. (64):

σσ ⋅ n = tn

Consider an element of the boundary of the body under analysis,Γt, where traction is applied:

t(2)

n tnt(1)

When tractiont(1) is applied, the stress tensor at that point is:

σσ (1) ⋅ n = t(1)

When tractiont(2) is applied, the stress tensor at that point is:

σσ (2) ⋅ n = t(2)

Traction vectors can be added to produce the total traction

tn = t(1) + t(2) = σσ (1) ⋅ n + σσ (2) ⋅ n = (σσ (1) + σσ (2)) ⋅ n

On the other hand the stress tensor corresponding totn is

σσ ⋅ n = tn

From the last two equations one immediately obtains:

σσ = σσ (1) + σσ (2)

which means that the stress states are additive.

An alternative proof is based on the equilibrium equations, Eqn. (67):

∇ ⋅ σσ = −b + ρ x

The key property of this PDE is that it islinear, meaning that ifσσ (1) andσσ (2) are two solutions to thesePDEs, thenσσ = σσ (1) + σσ (2) is a solution too. This is a consequence of the fact that the differential operatoris linear:

∇ ⋅ σσ = ∇ ⋅ σσ (1) + ∇ ⋅ σσ (2)

S85. The proof proceeds similar to ex. prob. 84. Needto remember that displacements are vectorsand therefore additive.

Consider a complex displacement path - displacementu(1), followed by displacementu(2):

-112-

u

u(1)u(2)

u = u(1) + u(2)

Each displacement vector will have a corresponding strain tensor:

εε (1) = ½(∇u(1) + (∇u(1))T )

εε (2) = ½(∇u(2) + (∇u(2))T )

The total displacement will give rise to its own strain tensor:

εε = ½(∇u + (∇u)T )

The gradient (and derivative in general) is a distributive operator, meaning

∇(a + b) = ∇a + ∇b

or in index notation

a,i + b,i =∂a

xi+

∂b

xi=

∂(a + b)

xi= (a + b),i

hence by adding strains from displacementsu(1) andu(2) one obtains:

εε (1) + εε (2) = ½(∇u(1) + ∇u(2) + (∇u(1))T + (∇u(2))T ) = ½(∇(u(1) + u(2)) + (∇u(1))T + (∇u(2))T )

The transposition is a distributive operator too, meaning

(A + B)T = AT + BT

so that

εε (1) + εε (2) = ½(∇(u(1) + u(2)) + (∇(u(1) + u(2)))T ) = ½(∇u + (∇u)T ) = εε

Hence strain states are additive.

Note that the proof in index notation is easier in this case:

ε (1)ij = ½(u(1)

i , j + u(1)j ,i )

ε (2)ij = ½(u(2)

i , j + u(2)j ,i )

So that

ε (1)ij + ε (2)

ij = ½(u(1)i , j + u(2)

i , j + u(1)j ,i + u(2)

j ,i ) = ½(ui , j + u j ,i ) = ε ij

S86. The stress tensor in principal directions is:

σσ =

σ1

sym

0

σ2

0

0

0

The principal direction with zero stress is 3.A rotation tensor with a single rotation about 3 is:

R =

cosθ− sinθ

0

sinθcosθ

0

0

0

1

The rotated stress tensor is:

σσ ′ = Rσσ RT

-113-

In the following we are not interested in the exact values, only whether they are zero or not.We use Greeksymbols for non-zero values.

Rσσ =

αγ0

βδ0

0

0

0

and

Rσσ RT =

ηκ0

ζω0

0

0

0

So we proved that for any rotation in 12 planeσ13 = σ23 = σ33 = 0.

S87. This problem has a point symmetry. Cut a sphere along a diameter into two halves, and useforce equilibrium along the axis normal to the cut, 1:

2

n

n

v

S1 S2 1

3

We assume that the membrane thickness,t << r , wherer is the radius of the sphere, so that the membranethickness can be neglected when the sphere diameter is calculated.S1 is the half sphere surface.S2 is thesurface of the cross section circle.

The horizontal force acting on the membrane due to pressure is

∫S1

pv ⋅ ndS

wheren is the outward normal vector toS1. v is the unit vector pointing in−1 direction. Notethat thisintegral can be interpreted asfluxof vectorpv throughS1.

Flux of this vector throughS2 is

∫S2

pv ⋅ ndS= −pπ r 2

becauseS2 is just an area of a circle, andp = const .

The total flux thoughS1∪S2 is

∫S1∪S2

pv ⋅ ndS= ∫S1

pv ⋅ ndS− pπ r 2

the left hand side can be transformed into a volume integral by Green’s theorem:

∫Vpvi ,i dV = ∫S1

pv ⋅ ndS− pπ r 2 = 0

whereV is the volume bounded byS1∪S2. The volume integral is zero becausev = const . Hence

∫S1

pv ⋅ ndS= pπ r 2

The other horizontal force acting on the membrane is due to stress in the membrane:

-114-

1

σ11

σ11

pr

pπ r 2 = 2π rtσ11

So that

σ11 =pr

2t

From symmetry,σ22 = σ11 and all shear stresses are zero. So the stress state is equi biaxial tension. At anypoint in the membrane any two orthogonal directions in the plane of the membrane are principal. The thirdprincipal direction is through thickness.

The stress tensor is:

σσ =pr

2t

1

sym

0

1

0

0

0

Strain through thickness is calculated from Eqn. (151) in ex. prob. 79:

ε33 = −νE

(σ11 + σ22) = −ν pr

Et

The reduction in thickness is:

tε33 = −ν pr

E

Note that this solution is very crude, because it uses linear elasticity and small strain, linear, formulation.Neither is true in the case of a rubber balloon. Hence this solution might give unrealistic prediction.Forexample, ifE = 10MPa,ν = 0. 5,r = 20mm,t = 0. 2mm,and p = 0. 2MPa, i.e. about twice the atmosphericpressure, thenε33 = −1 and the reduction in thickness is−0. 2mm,meaning the thickness after the deforma-tion is zero.

S88. In general, plane stress state produces non-zero third principal strain. From Eqn. (151) in ex.prob. 79:

ε3 = −νE

(σ1 + σ2)

For this to be zero one must haveσ1 = −σ2.

Plane strain state gives rise to non-zero third principal stress:

σ3 = λ(ε1 + ε2)

This is zero ifε1 = −ε2.

Both conditions can be rephrased as

trσσ = trεε = I σσ = I εε = 0

meaning that the spherical parts of both the strain and the stress tensors are zero.Yet in other words, - theremust be neither volumetric strain, no pressure stress. Such stress/strain state ispure shear. The stress andstrain tensors in principal coordinates will be:

σσ =

σ1

sym

0

−σ1

; εε =

ε1

sym

0

−ε1

or in the coordinates of maximum shear value:

-115-

σσ =

0

sym

σ12

0

; εε =

0

sym

ε12

0

Therefore, pure shear is a special case of both plane strain and plane stress conditions.

S89. The bracket has a right angle and is clamped at the wall. Thelengths of each part of the bracketarel . The view in 13 plane is:

O

l

1

3l

P A

B

Beam AB is along axis 3, and beam BO is along axis 1. It is clear that AB is loaded in bending, while BOis under a combined bending and torsion. Therefore AB seems easier to analyse so we start from it.

The bending moment in AB looks like this:

x3

Pl

B A

M = P(l − x3)

From Eqn.(29) in Sec. 4.4

w′′ =d2w

dx23

=M

EI

or

w′′ =P(l − x3)

EI

By integrating it twice one obtains:

w′ =P

EI ∫ (l − x3)dx3 + C1 =P

EI(lx3 −

x23

2) + C1

w =P

EI ∫ ∫ (l − x3)dx3dx3 + C1x3 + C2 =P

EI(l

x23

2−

x33

6) + C1x3 + C2

The integration constants are found from the BC:

w(x3 = 0) = 0 ⇒ C2 = 0

Note that this BC is correct only when one considers AB in isolation. ClearlywB ≠ 0 for the completeproblem, because it arises from the deflection of BO. However, wB can be added as a rigid body motion toall points in AB, after it is calculated from the analysis of BO.

We now see that there is no easy second BC to fitC1. Indeedw′B is not known. TheBC at point B isa compatibility condition joining the two beams together. This means this problem is of a statically indeter-minate type, meaning that extra information must be used to calculate the BC for AB. Clearly, torsion ofBO determinesw′B. So let’s now move to the analysis of BO.

Let’s start with bending.

-116-

x1

PlM = P(l − x1)

BO

The solution forw is identical to that for AB. One only needs to substitutex1 for x3:

wBO =P

EI(l

x21

2−

x31

6) + C1x1 + C2

However, now we hav etwo easy BC:

wBO(x1 = 0) = 0 ⇒ C2 = 0

w′BO(x1 = 0) = 0 ⇒ C1 = 0

so that

wBO =Px2

1

6EI(3l − x1)

The torsion of BO is due to forceP applied off axis. Thetorque isT = Pl. From Eqn. (138) in Sec. 6.1.4the twist angle is:

θ = ∫ T

GJdx1 = ∫ Pl

GJdx1

so that

θmax =Pl2

GJ

In the following, let’s assume an axisymmetric cross section, for whichI11 = I22, and hence (see ex. prob.9):

J = 2I

Also, from Eqn. (126):

G = µ =E

2(1+ ν )

so thatθmax can be rewritten as:

θmax =Pl2

EI(1 + ν )

The extra BC is:

w′AB(x3 = 0) = θmax

so

CAB1 =

Pl2

EI(1 + ν )

Finally, w on AB, wAB is:

wAB =P

EIl

x23

2−

x33

6+ l2x3(1 + ν )

We can now calculate displacement in the complete system.For this one needs to addwmaxBO to wAB :

wmaxBO =

Pl3

3EI

so thatwAB is:

-117-

wAB =Pl3

EI

−

1

6

x3

l

3

+1

2

x3

l

2

+ (1 + ν )x3

l+

1

3

The maximum displacement in the system is at A:

wmaxAB = wAB(x3 = l ) =

Pl3

EI(5

3+ ν )

The deformed shape of the bracket can now be drawn. Theinitial (undeformed) shape is shown withdashed lines. The new shape is shown with solid lines. The dotted line iswmax

BO , which is added towAB as arigid body motion. The magnitudes are shown in units ofPl3/EI .

5/3+ ν

P

3

1

2

1/3

Let’s now turn to the analysis of stress, first in AB. From Eqn. (28) in Sec. 4.4 the maximum axialstress in the cross section is:

σ33 =Mxmax

2

I

and the maximum moment is atx3 = 0. Let’s also useRo for xmax2 , since we have already assumed an

axisymmetric cross section.

σ max33 =

PlRo

I

We hav estudied pure bending already, it is enough to say that the stress (and the strain) states are uniaxial,with the maximum principal stress

σ1 = σ max33 =

PlRo

I

Stress in BO is a superposition (see ex. probs. 84, 85) of stress tensors due to bending and torsion:

σσBO = σσ bendBO + σσ torsion

BO

It is clear that the sum is maximised atx2 = ±Ro, where the torsion stress acts in 13 plane. In all otherpoints eitherσ11 or the shear stresses, or both, are smaller:

-118-

σ max11 ,σ max

13

1

2

3

σ11 = 0

σ min11 ,σ min

13

hence:

σσBO =

σ11

sym

0

0

0

0

0

+

0

sym

0

0

σ13

0

0

=

σ11

sym

0

0

σ13

0

0

Along x1 the stress due to torsion is constant, and that due to bending is highest atx1 = 0, where it is thesame as for AB:

σ max11 =

PlRo

I

From Eqns. (135) and (137) in Sec. 6.1.4 the maximum torsion shear stress is

σ max13 = Gθ ′Ro =

GPlRo

GJ=

PlRo

2I

Finally

σσBO =PlRo

I

1

sym

0

0

½

0

0

Note that we solved this problem so easily only because it could be reduced to a 2D superposition problem.

The stress tensor can be shown on the elementary cube of material, in the units ofPlRo/I , as:

1/2

1

2

3

1

To find the principal values and directions I use again Lapack DSYEV routine:Original tensor1.00000000000E+00 0.00000000000E+00 5.00000000000E-010.00000000000E+00 0.00000000000E+00 0.00000000000E+005.00000000000E-01 0.00000000000E+00 0.00000000000E+00

The DSYEV eigenvalues in increasing order-2.07106781187E-01 0.00000000000E+00 1.20710678119E+00The DSYEV orthonormal eigenvectors (columns)3.82683432365E-01 0.00000000000E+00 9.23879532511E-01-0.00000000000E+00 -1.00000000000E+00 0.00000000000E+00-9.23879532511E-01 0.00000000000E+00 3.82683432365E-01The angles (deg)

-119-

6.75000000000E+01 9.00000000000E+01 2.25000000000E+019.00000000000E+01 1.80000000000E+02 9.00000000000E+011.57500000000E+02 9.00000000000E+01 6.75000000000E+01

The stress tensor in the original and the principal CS is best shown in 2D representation:

1/2

1.2-0.2

22.5 deg67.5 deg

1′

3′

31/2

1

1

σ1 ≈ 1. 2,andσ3 ≈ −0. 2,both in units ofPlRo/I , and the rotation is 22. 5, as expected. Sothe maximumprincipal stress in the bracket is at point O:

σ O1 ≈ 1. 2

PlRo

I

Alternatively one can use the Mohr’s diagram, as the stress state is two-dimensional:

normal

10 0.5

˜0.7

0.5

-0.2 1.2

shear

Note that some very complex stress/strain state will exist around point B, where all our assumptionsof bending and torsion will be violated.Typically portions of designs which are too hard to analyse aremade deliberately stronger than the rest.

Finally, let’s get a feel for the values of displacement and stress for typical engineering values. Con-sider a ring cross section with the outer radiusRo = 20mm and wall thicknesst = 2mm, a typical steel withE = 200GPa andν = 0. 33,the bracket sizel = 500mm, loaded byP = 1kN. Then

wmax = wA ≈ 29mm

σ max1 = σ O

1 ≈ 278MPa

S90. To show thatS1 andS2 form a vector, one has to show that they change with CT as componentsof a vector.

In matrix form coordinates of the centroid can be written as:

S1

S2

=1

A

i2

i1

S1′ =1

Ai2′ =

1

A ∫ R1 j x j dA =1

AR1 j ∫ x j dA =

1

A(R11i2 + R12i1)

Similarly

-120-

S2′ =1

A(R21i2 + R22i1)

Hence

S1′S2′

=1

A

R11

R21

R12

R22

i2

i1

=

R11

R21

R12

R22

S1

S2

HenceS = Sj = (S1, S2) is a vector.

S91. The easiest logic follows from ex. prob. 90. For any CS with origin at centroidS = 0. Hencei1 = i2 = 0.

Note that the first moments of area strictly donot form a vector. This is because:

i1′ = ∫ R2 j x j dA = R2 j ∫ x j dA ≠ R1 j ∫ x j dA

The transformation is similar to that of vector components, but not the same.

S92. Slight complication arises because

I ij =

I11

I21

I12

I22

is nota tensor. Indeed

I11′ = ∫ ∫ x2′x2′dA = ∫ ∫ R2i xi R2 j x j dA = R2i R2 j ∫ ∫ xi x j dA = R2i R2 j I ij

which is not how R2T components should transform. However:

Yij =

I22

I21

I12

I11

is indeed R2T, which is easily proved:

Ymn′ =

I22′I21′

I12′I11′

For example forY22 one obtains:

Y22′ = I11′ = ∫ ∫ x2′x2′dA = ∫ ∫ R2i R2 j xi x j dA = R2i R2 j ∫ ∫ xi x j dA

= R21R21 ∫ ∫ x1x1dA+ R21R22 ∫ ∫ x1x2dA+ R22R21 ∫ ∫ x2x1dA+ R22R22 ∫ ∫ x2x2dA

= R21R21I22 + R21R22I12 + R22R21I21 + R22R22I22

= R21R21Y11 + R21R22Y12 + R22R21Y21 + R22R22Y11 = R2i R2 jYij

which is how components of R2T should transform. The same can be shown for the other components.

Y = Yij is symmetrical, because∫ ∫ x1x2dA = ∫ ∫ x2x1dA.

We call Y thesecond moment of area tensor.

One can argue that the historical definitions of the second moments of area are wrong, and it would

be more logical to defineI11 as∫ ∫ x1x1dA, etc. In that caseI ij would be R2T, and none of this confusion

would arise.

S93. Construct the second moment of area tensorY using the expressions from ex. prob. 12:

-121-

Y =

HW3

36

−W2H2

72

−W2H2

72WH3

36

=HW

72

2W2

−HW

−HW

2H2

Note thatW is the side length along 1 andH is the side length along 2.

One can construct the quadratic characteristic equation:

(2W2 − λ)(2H2 − λ) − W2H2 = 0

or

λ2 − 2λ(H2 + W2) + 3W2H2 = 0

So the principal values are:

λ1,2 = H2 + W2±(H2 − W2)2 + W2H2)

½

Expressions for the principal directions become quite long, so to make the example more instructive, let’sassume

H = 2W

which means the other two angles in the triangle are 30and 60. ThenY will simplify to:

Y =W4

18

1

−1

−1

4

I solve the eigenvalue/eigenvector problem numerically with LAPACK l ibrary, routine DSYEV:

Original tensor0.00000000000E+00 0.00000000000E+00 0.00000000000E+000.00000000000E+00 1.00000000000E+00 -1.00000000000E+000.00000000000E+00 -1.00000000000E+00 4.00000000000E+00

The DSYEV eigenvalues in increasing order0.00000000000E+00 6.97224362268E-01 4.30277563773E+00

The DSYEV orthonormal eigenvectors (columns)1.00000000000E+00 0.00000000000E+00 0.00000000000E+000.00000000000E+00 -9.57092026489E-01 -2.89784148688E-010.00000000000E+00 -2.89784148688E-01 9.57092026489E-01

The angles (deg)0.00000000000E+00 9.00000000000E+01 9.00000000000E+019.00000000000E+01 1.63154966237E+02 1.06845033763E+029.00000000000E+01 1.06845033763E+02 1.68450337630E+01

The principal directions are shown below.

-122-

3′

2

3

3′ 17 deg

C

2′

107 deg

163 deg

1 = −1′

2

330 deg

1

2

Y11 = I22

Y22 = I11

Y12 = I12

Ymax = Imax

Ymin = Imin

2′

Note that the result is not obvious. Inparticular, the principal axis is not || to the hypotenuse.

The second moment of area tensor in principal directions is:

Y =W4

18

4. 303

0

0

0. 697

The practical importance of this example is that it allows finding the beam orientation for maximumresistance to bending.

S94. Use the fact that the second moments form R2T. If symmetry axes are used as CS thenI11 = I22 and I12 = 0. HenceI1 = I2 = I11, i.e. both principal values are identical. Therefore any axis pass-ing through centroid is a principal axis.

Prepared withgroff, Xfig andgnuplot.