Mechanics of Materials Labcourses.washington.edu/me354/lecture/MOM_Lect_06.pdfMechanical Behavior of...

Jiangyu Li, University of Washington

Lecture 6Strain and Elasticity

Mechanical Behavior of Materials Sec. 6.6, 5.3, 5.4

Jiangyu LiUniversity of Washington

Mechanics of Materials Lab

Strain:Fundamental Definitions

• "Strain" is a measure of the deformation of a solid body

• There are two "types" of strain; normal strain (ε) and shearstrain (γ)

ε = (change in length)(original length)

; units = in / in, m / m, etc

γ = (change in angle ) ; units = radians 28

Prof. M. E. TuttleUniversity of Washington

Strain Within a Plane

+x

+y

a

bc

lx

ly

∠ abc = π/2 radians

Original Shape

+x

+ya

b

c

lx + ∆lx

ly + ∆ly

∠ abc < π/2 radians

Deformed Shapeεxx = ∆lx/lx

εyy = ∆ly/ly

γxy = ∆(∠ abc)


29

Strain Sign Convention• A positive (tensile) normal strain is associated with an

increase in length

• A negative (compressive) normal strain is associated with adecrease in length

• A shear strain is positive if the angle between two positivefaces (or two negative faces) decreases

32

+x

+y

+x

+y

All Strains Positive ε Positiveε and γ Negative

xx

yy xy

3-Dimensional Strain States

• In the most general case, six components of strain exist "ata point": εxx, εyy, εzz, γxy, γxz, γyz

• Strain is a 2nd-order tensor; the numerical values of theindividual strain components which define the "state ofstrain" depend on the coordinate system used

• This presentation will primarily involve strains which existwithin a single plane

29


Strain Within a Plane

• We are often interested in the strains induced within asingle plane; specifically, we are interested in two distinctconditions:

• "Plane stress", in which all non-zero stresses lie withina plane. The plane stress condition induces fourstrain components: εxx, εyy, εzz, and γxy

• "Plane strain", in which all non-zero straincomponents lies within a plane. By definition then,the plane strain condition involves three straincomponents (only): εxx, εyy, and γxy 30


Visualization of StrainF

F

+x

+yεyy > 0εxx < 0γxy = 0

(Loading)

33


Visualization of Strain

+x'+y'εx'x' = εy'y' > 0γx'y' > 0

(Loading)

F

F

45 deg

34


Strain Transformations

• Given strain components in the x-y coordinate system (εxx, εyy,γxy), what are the corresponding strain components in the x'-y'coordinate system?

lx + ∆lx

+x

+y

ly + ∆ly

εxx = ∆lx/lx

εyy = ∆ly/ly

γxy = ∆(∠ abc)

+x'+y'

εx'x' = ?εy'y' = ?γx'y' = ?

θ?

35


Strain TransformationEquations

• Based strictly on geometry, it can be shown:

ε x' x ' =ε xx + ε yy

2+

ε xx − ε yy

2cos2θ +

γ xy

2sin2θ

ε y' y' =ε xx + ε yy

2−

ε xx − ε yy

2cos2θ −

γ xy

2sin2θ

γ x'y'

2= −

ε xx − ε yy

2sin 2θ +

γ xy

2cos2θ

36


"Strain":Summary of Key Points

• ε = (∆ length)/(original length) γ = (∆ angle)

• Six components of strain specify the "state of strain"

• Strain is a 2nd-order tensor; numerical values of individualstrain components depend on the coordinate system used

• Strain is defined strictly on the basis of a change in shape;definition is independent of:

material properties stress

• "Suprisingly," the stress and strain transformation equationsare nearly identical

38


"Constitutive Models"

• The structural engineer is typically interested in the state ofstress induced in a structure during service

• The state of stress cannot be measured directly...

• The state of strain can be measured directly...

• Hence, we must develop a relation between stress andstrain...this relationship is called a "constitutive model," andthe most common is "Hooke's Law"

39


3-D Form of Hooke's Law:Apply Principle of Superposition

• The total strain εxx caused by all stresses appliedsimultaneously is determined by "adding up" the strain εxxcaused by each individual stress component

ε xx = σxxE

+

−νσ yy

E

+

−νσ zzE

+ 0[ ] + 0[ ] + 0[ ]

or

ε xx = 1E

σ xx − νσ yy − νσ zz[ ]51


Hooke's Law:3-Dimensional Stress States

• Following this procedure for all six strain components:

ε xx = 1E

σ xx − νσ yy − νσ zz[ ] γ yz =2(1+ ν)τ yz

E

ε yy = 1E

−νσ xx + σ yy − νσ zz[ ] γ zx = 2(1+ ν)τ zxE

ε zz = 1E

−νσ xx − νσ yy + σzz[ ] γ xy =2(1+ ν)τ xy

E

52



ε xx

ε yy

εzz

γ yz

γ zx

γ xy

= 1E

1 −ν −ν 0 0 0−ν 1 −ν 0 0 0−ν −ν 1 0 0 00 0 0 2(1+ ν) 0 00 0 0 0 2(1+ ν) 00 0 0 0 0 2(1+ ν)

σxx

σ yy

σzz

τ yz

τ zx

τ xy

53



54


σ xx = E(1+ ν)(1− 2ν)

(1− v)εxx + νεyy + νεzz[ ] τyz =Eγ yz

2(1+ ν)

σyy = E(1+ ν)(1− 2ν )

vε xx + (1− v)εyy + νε zz[ ] τ zx = Eγ zx2(1 + ν)

σ zz = E(1+ ν )(1− 2ν)

vεxx + νε yy + (1− v)ε zz[ ] τxy =Eγ xy

2(1 + ν )


σ xx

σ yy

σ zz

τyz

τzx

τxy

=E

(1+ ν)(1− 2ν)

(1− ν) ν ν 0 0 0ν (1− ν) ν 0 0 0ν ν (1− ν) 0 0 00 0 0 (1− 2ν)

20 0

0 0 0 0 (1− 2ν)2

0

0 0 0 0 0 (1− 2ν)2

εxx

εyy

ε zz

γ yz

γ zx

γ xy

55



• "Shorthand" notation:σ{ } = D[ ] ε{ }

where:

D[ ] = E(1 + ν)(1− 2ν)

(1− ν) ν ν 0 0 0ν (1− ν) ν 0 0 0ν ν (1− ν) 0 0 00 0 0 (1− 2ν)

20 0

0 0 0 0 (1− 2ν)2

0

0 0 0 0 0 (1− 2ν)2

56


Hooke's Law:Plane Stress

• For thin components (e.g., web or flange of an I-beam,automobile door panel, airplane "skin", etc) the "in-plane"stresses are much higher than "out-of-plane" stresses:(σxx,σyy,τxy) >> (σzz,τxz,τyz)

• It is convenient to assume the out-of-plane stresses equalzero...this is called a state of "plane stress"

+x

+y

+z

57



σxx

τxy

σyy

εxx = 1E

(σxx − νσyy )

εyy = 1E

(σyy − νσxx )

γ xy = τxy

G=

2(1+ ν)τxy

E

εzz = −νE

(σxx + σyy )

γ xz = γ yz = 0 58



σxx

τxy

σyy

σxx = E(1- ν2 )

(εxx + νεyy )

σyy = E(1- ν2 )

(εyy + νεxx )

τxy = Gγ xy = E2(1+ ν)

γ xy

σzz = τ xz = τ yz = 059



σ xx

σ yy

τ xy

= E(1− ν2 )

1 ν 0ν 1 00 0 1− ν

2

ε xx

ε yy

γ xy

60



• "Shorthand" notation:

σ{ } = D[ ] ε{ }where

D[ ] = E(1− ν2 )

1 ν 0ν 1 00 0 1− ν

2

61


Hooke's Law:Plane Strain

• For thick or very long components (e.g., thick-walled pressurevessels, buried pipe, etc) the "in-plane" strains are muchhigher than "out-of-plane" strains: (εxx,εyy,γxy) >> (εzz,γxz,γyz)

• It is convenient to assume the out-of-plane strains equalzero...this is called a state of "plane strain"

+x

+y

+z

62



σ xx

τxy

σyy

63


σ xx = E(1 + ν)(1− 2ν)

(1− ν )ε xx + νε yy[ ]σyy = E

(1+ ν)(1− 2ν)νε xx + (1− ν)ε yy[ ]

σzz = νE(1+ ν)(1− 2ν)

ε xx + ε yy[ ] = ν σ xx + σyy[ ]

τ xy = Gγ xy =Eγ xy

2(1+ ν )τ xz = τyz = 0


σ xx

σ yy

τ xy

= E(1+ ν)(1− 2ν)

(1− ν) ν 0ν (1− ν) 00 0 (1− 2v)

2

ε xx

ε yy

γ xy

64




σ{ } = D[ ] ε{ }where

D[ ] = E(1+ ν)(1− 2ν)

(1− ν) ν 0ν (1− ν) 0

0 0 (1− 2v)2

65


Hooke's LawPlane Strain

66

σ xx

τxy

σyy

ε xx = 1 − ν2

Eσxx − ν

1− ν

σ yy

ε yy =1− ν 2

Eσ yy −

ν1− ν

σ xx

γ xy = 2(1+ ν)E

τxy

ε zz = γ xz = γ yz = 0

....and: σzz = ν σ xx + σ yy[ ]

Hooke's Law:Uniaxial Stress

• Truss members are designed to carry axial loads only (i.e.,uniaxial stress)

• In this case we are interested in the axial strain only (eventhough transverse strains are also induced)

67


Hooke's Law:Uniaxial Stress

• For: σyy = σzz = τyz = τxz = τzy = 0,

"Hooke's Law" becomes:


σ xx = Eε xx

σ{ } = D[ ] ε{ } where D[ ] = E[ ]68


Hooke's LawSummary of Key Points

• Hooke's Law is valid under linear-elastic conditions only

• The mathematical form of Hooke's Law depends on whetherthe problem involves:

• Structures for which 3-D stresses are induced

• Structures for which (due to loading and/or geometry)plane stress conditions can be assumed

• Structures for which (due to loading and/or geometry)plane strain conditions can be assumed

• Structures for which (due to loading and/or geometry)uniaxial stress conditions can be assumed

69


Jiangyu Li, University of Washington

Assignment

• Mechanical behavior of materialsHW 5.11, 5.12, 5.37

Mechanics of Materials Labcourses.washington.edu/me354/lecture/MOM_Lect_06.pdfMechanical Behavior of...

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