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Page 1: Chapter 2 Basics of Materials and Mechanicssuemasu/files/chapter2.pdf · Mechanics of Materials 2 2.1 Stress 2.1.1 Definition of stress -PS ü S ü S P-P-PP P ' P 2 ' P ' P 1 Æ V

Mechanics of Materials 1

Chapter 2

Basics of Materials

and Mechanics

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Mechanics of Materials 2

2.1 Stress

2.1.1 Definition of stress

- P

S

ΔSΔS

φ

P- P

- P

P

P

P2

PP1

内力

V

V1

V2

hypothetical internal

surface S Body is in

equilibrium

Sum of the internal

forces on S

=Applied force P

Stress Vector

Quantity to express the

intensity of internal

force

SS

Pp

0lim (1)

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Mechanics of Materials 3

Unit of Stress ・・・・Pa(=N/m2), kgf/m2, psi (=lb/inch2),etc

Stress depends not only the internal force (vector) but

also the direction of the surface considered.

dS

dPp

dS

dPp

2

1

sin

cos

Normal component:Normal Stress

Inplane component:Shear Stress

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Mechanics of Materials 4

Positive stresses are defined as shown above. In the other sides

of the cube where the outward normals are opposite to the

coordinate axes, the directions of positive stresses are opposite.

Definition of Stresses in the Cartician

Coordinate System

x

y

z

P

B

A

O

C dx

dy

dz x xyxz y

yx

yz

z

zx

zy

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Mechanics of Materials 5

Stress in an arbitrary surface

Equilibrium of the force for infinitesimal tetrahedron

Area of ABC=dS

Outward normal of ABC : =(l,m,n)

l, m, n are directional cosines

OA=dx, OB=dy, OC=dz

OBC=dydz/2=dS・l,

OCA=dzdx/2=dS・m,

OAB=dxdy/2=dS・n

n

x

y

z

x

y

z

O A

B

C pdS

-pxldS

-pymdS

-pzndS

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Mechanics of Materials 6

zzyzxz

xzyyxy

xzxyxx

ZYX

,,

,,

,,

,,

p

p

p

p

Stress vectors

Equilibrium of the force when no body force exists.

0 ndSmdSldSdS zyx pppp

ndSmdSldSdS

Z

Y

X

z

zy

zx

yz

y

yx

xz

xy

x

x

y

z

OA

B

C pdS

-pxldS

-pymdS

-pzndS

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Page 7: Chapter 2 Basics of Materials and Mechanicssuemasu/files/chapter2.pdf · Mechanics of Materials 2 2.1 Stress 2.1.1 Definition of stress -PS ü S ü S P-P-PP P ' P 2 ' P ' P 1 Æ V

Mechanics of Materials 7

n

m

l

Z

Y

X

zyzxz

zyyxy

zxyxx

Then

Stress vector of an arbitrary surface

→Nine components are necessary to express the stress state.

Note) From the equilibrium of an infinitesimal cube in moment,

zy=yz , xz=zx, yx=xy

Six components are practically independent.

yxxyxzzxzyyzzyx ,,,,,,,,pp

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Mechanics of Materials 8

2.1.2 Transformation of Coordinate system

Let (x, y, z) and (x', y', z') be two Cartesian coordinate systems

with a common origin. The cosines of the angles between the six

coordinate axes are represented in the following tabular form:

x y z

x’ a11 a12 a13

y’ a21 a22 a23

z’ a31 a32 a33

x

y

z

O

p

i’

j

i

x’ y’

z’ k

k’

j’

Normal to the

x’ axis

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Mechanics of Materials 9

333231

232221

131211

332313

322212

312111

'''''

'''''

'''''

aaa

aaa

aaa

aaa

aaa

aaa

zyzxz

zyyxy

zxyxx

zzyzx

yzyyx

xzxyx

The stresses referring to (x', y', z') coordinate

system can be derived from the stresses

referring to (x, y, z) coordinate system .

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Mechanics of Materials 10

2.1.3 Principal stress

The values of stresses at a point depend on the plane where the

stresses are considered. The direction of the plane where the normal

stress is extremum is called principal axis and the normal stress is

called principal stress. The principal directions and principal stresses

are obtained through the following eigenvalue analysis.

0

zyzxz

zyyxy

zxyxx

222

3

2222

1

xyzzxyyzxzyx

xyzxyzyxxzzy

zyx

J

J

J

The coefficients of the equation are called stress invarient.

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Mechanics of Materials 11

2.1.4 Maximum Shear Stress

Shear Stress becomes maximum at the surface whose unit

normal is 0,2/1,2/1,, nml

212

1

212

1

The normal stress at the surface of maximum shear stress is

where the coordinate system accords to the principal directions. 1 is

maximum principal stress and 2 is minimum principal stress

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Mechanics of Materials 12

2.1.5 Equilibrium Equation

0

0

0

Zzyx

Yzyx

Xzyx

zyzxz

zyyxy

zxyxx

x y

z

P

B

A

O

C dx

dy

dz

dxx

xx

dxx

xyxy

dxxxz

xz

dyy

yy

dyy

yxyx

dyy

yzyz

dzz

zz

dz

zzx

zx

dz

z

zyzy

From the equilibrium of the force acting on the

small rectangular parallelepiped

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Mechanics of Materials 13

A’

B’ C

D’

C’

A B

dz

D

dy

dx

x

y

z

2.2 Strain 2.2.1 Definition of strain

Intensity of Deformation

Normal strain

Shear strain

x

xxx

i

ir

AB

ABBA

''

xy

2

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Mechanics of Materials 14

Normal strain represents change of length.

When the length is increased, it is called

tensile strain. When the length is decreased,

it is called compressive strain. Normal strain

is usually positive if tensile. Strain being

defined as length/length is non-dimensional

value.

Shear strain represents change of shape. It is

defined as reduction of angle between two

coordinate axes which were normal each

other.

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Mechanics of Materials 15

Relations between the strain components and

displacements (u,v,w)

x

w

z

w

x

v

z

v

x

u

z

u

x

w

z

u

z

w

y

w

z

v

y

v

z

u

y

u

z

v

y

w

y

w

x

w

y

v

x

v

y

u

x

u

y

u

x

v

z

w

z

v

z

u

z

w

y

w

y

v

y

u

y

v

x

w

x

v

x

u

x

u

zx

yz

xy

z

y

x

222

222

222

2

1

2

1

2

1

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Mechanics of Materials 16

When u and v is infinitesimal,

x

w

z

u

z

v

y

w

y

u

x

v

z

w

y

v

x

u

zxyzxy

zyx

,,

,,

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Mechanics of Materials 17

2.2.2 Motion of elastic body

Displacement of P

dz

dy

dx

dz

dy

dx

w

v

u

zzyzzx

yzyxy

zxxyx

zxy

xz

yz

drp

2/2/

2/2/

2/2/

0

0

0

0

0

0

0 uuuu

y

u

x

v

x

w

z

u

z

v

y

wzyx

2

1,

2

1,

2

1

where

, : deformation

: rotation

O

/2

P

O'

P'/2

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Mechanics of Materials 18

2.2.3 Transformation of strain component

under change of coordinate system

Two rectangular Cartesian

coordinate systems x, y, z

and x’, y’, z’

333231

232221

131211

21

21

21

21

21

21

332313

322212

312111

'''21

''21

''21

'''21

''21

''21

'

aaa

aaa

aaa

aaa

aaa

aaa

zyzxz

zyyxy

zxyxx

zzyzx

yzyyx

xzxyx

Table of directional cosines

x y z

x’ a11 a12 a13

y’ a21 a22 a23

z’ a31 a32 a33

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Mechanics of Materials 19

zyxzyx

zyxyxz

zyxxzy

zxzx

yzyz

xyxy

xyzxyzz

xyzxyzy

xyzxyzx

zxxz

yzzy

xyyx

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2.2.4 Compatibility Equation

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Mechanics of Materials 20

2.3 Generalized Hooke’s Law and elastic constants

Hooke’s Law refers that normal strain is directly

proportional to the relating normal stress. However, in

general stress state, the whole strain components are

dependent on whole stress state as

xy

zx

yz

z

y

x

xy

zx

yz

z

y

x

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

CCCCCC

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

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Mechanics of Materials 21

For three dimensional isotropic elastic body

Generalized

Hooke’s law

yxzz

xzyy

zyxx

E

E

E

1

1

1

G

G

G

xyxy

zxzx

yzyz

zyxzz

zyxyy

zyxxx

E

E

E

211

211

211

xyxy

zxzx

yzyz

G

G

G

Stresses are expressed by Strains

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Mechanics of Materials 22

One dimensional tension

Young’s modulus

strain Tensile

stress TensileE

L

T

strainTensile

direction rsein transvestraineCompressiv

Poisson’s ratio

Shear modulus

zyx

zyxK

strainVolumetric

3pressureAverage

Bulk modulus

strainShear

stressShearG

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Mechanics of Materials 23

An isotropic elastic material has only two independent

elastic constants.

The four elastic constants mentioned in the preceding

slide are not independent each other. Each elastic constant

can be expressed by the other two constants as follows.

213

12

EK

EG

When we measure the load and longitudinal and transverse

strains in a tensile test of a round bar, we can obtain

Young’s modulus and Poisson ratio simultaneously.

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Mechanics of Materials 24

2.4 Strain Energy

Let’s consider a bar of Young’s modulus E, length l and uniform

cross-sectional area A. The energy done by a force P0 under which the

bar is elongated by l .

lPPEA

ll

l

EA

xdxl

EAPdxU

ll

02

02

00

2

1

2

1

2

1

Stress and strain are used instead of

load and elongation

VE

VEV

lAl

l

A

PU

22

2

1

2

1

2

1

2

1

where V=Al is volume of the bar. Strain energy density

x

P

Strain Energy

Elongation

Exte

rna

l fo

rce

P0

x0

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Mechanics of Materials 25

Strain Energy Density Function A

222

2222

2

2112

2

1

xyyzzx

zyxzyx

xyxyyzyzzxzxzzyyxx

G

E

A

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Mechanics of Materials 26

2.5 Fundamental Principle of Elasticity

2.5.1 Principle of Virtual Work

Assume that the mechanical system is in equilibrium under

applied forces and prescribed geometrical constraints. Then, the

sum of all the virtual work, denoted by ’W, done by the external

and internal forces existing in the system in any arbitrary

infinitesimal virtual displacements satisfying the prescribed

geometrical constraints is zero.

0

S

iiV

iiV

ijij dSuT- dVuX dVeW

0

S

iiV

iiV

dSuT- dVuXdVAW

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Mechanics of Materials 27

2.5.2 Principle of minimum potential energy

The structures will be in

equilibrium when the potential

energy is minimum.

Potential energy

=U+W

S

iV

i dSugdVuGW V

i dVuAU

V

=0

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Mechanics of Materials 28

2.5.3 Saint Venant’s Principle Statically equivalent force systems which act over a

given small portion S on the surface of a body

produce approximately the same stress and

displacement at a point in the body sufficiently far

removed from the region S over which the force

system act.

P

P

P

P

P

P

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Mechanics of Materials 29

A convenient theorem to obtain concentrated forces where the

displacement is prescribed or a displacement at the point where a

concentrated force is applied.

(1) Castigliano’s first theorem

When the total energy of the system A(ui) is expressed as a

function of the prescribed displacement ui, the reaction force Pi in the

direction of ui at the point is

(2) Castigliano’s second theorem

When the total complementary energy B(Pi) of the system is

expressed as a function of the applied load Pi, the displacement ui in

the direction of Pi at the point is

2.5.4 Castigliano’s theorem

ii

u

AP

ii

P

Bu

From Arches Higdon, Edward H. Ohlsen, william B.Stiles,

William F. Riley, Mechanics of Materials, John Wiley & Sons

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Mechanics of Materials 30

E,A,l,

T

Temperature rise → Thermal expansion of the material

(Thermal strain)

Constraint of the deformation → Thermal stress

Appendix Thermal stress

Example A bar of a length l is constrained

at its both ends. Obtain the stress when the

temperature of the bar is elevated by T.

(Solution)

When the bar is not constrained, the thermal strain of the bar is

= T

When the bar does not expand owing to the constraints, the strain

should be zero. Then, stress to cause the mechanical strain of

must occur, that is,

=ET

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