Post on 25-May-2022
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Chapter 5
Stress–Strain Relation
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Chapter 5 Stress–Strain Relation
Contents
5.1 General Stress–Strain System
5.2 Relations Between Stress and Strain for Elastic Solids
5.3 Relations Between Stress and Rate of Strain for Newtonian Fluids
Objectives
- Understand tensor systems of stress and strain
- Study difference between displacement and deformation
- Study solid mechanics to deduce stress-rate of strain relations for fluid
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5.1 General Stress–Strain System
Parallelepiped, cube → infinitesimal C.V.
5.1.1 Surface Stress
Surface stresses: normal stress – σx
shear stress – τxy, τxz
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5.1 General Stress–Strain System
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5.1 General Stress–Strain System
0lim
( )x
xxx x A
x
x
FA
A y z
σ σ→
∆= =
∆∆ = ∆ ∆
0lim
x
yxy A
x
FA
τ∆ →
∆=
∆ 0lim
x
zxz A
x
FA
τ∆ →
∆=
∆
0lim
( )y
xyx A
y
y
FA
A x z
τ∆ →
∆=
∆
∆ = ∆ ∆0
limy
yyy y A
y
FA
σ σ→
∆= =
∆
0lim
y
zyz A
y
FA
τ∆ →
∆=
∆
0lim
( )z
xzx A
z
z
FA
A x y
τ∆ →
∆=
∆∆ = ∆ ∆
0lim
z
yzy A
z
FA
τ∆ →
∆=
∆ 0lim
z
zzz z A
z
FA
σ σ∆ →
∆= =
∆
, ,x y zF F F∆ ∆ ∆ F∆
where = component of force vector
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σx : subscript indicates the direction of stress
τxy : 1st - direction of the normal to the face on which τ acts
2nd - direction in which τ acts
5.1 General Stress–Strain System
• subscripts
• general stress system: stress tensor
~ 9 scalar components
xx xy xz
yx yy yz
zx zy zz
σ τ ττ σ ττ τ σ
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5.1 General Stress–Strain System
[Re] Tensor
~ an ordered array of entities which is invariant under
coordinate transformation; includes scalars & vectors
~ 3n
0th order – 1 component, scalar (mass, length, pressure, energy)
1st order – 3 components, vector (velocity, force, acceleration)
2nd order – 9 components (stress, rate of strain, turbulent diffusion)
At three other surfaces,
' xx x x
xσσ σ ∂
= + ∆∂
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5.1 General Stress–Strain System
' yy y y
yσ
σ σ∂
= + ∆∂
' zz z z
zσσ σ ∂
= + ∆∂
' xyxy xy x
xτ
τ τ∂
= + ∆∂
' yxyx yx y
yτ
τ τ∂
= + ∆∂
' zxzx zx z
zττ τ ∂
= + ∆∂ (5.1)
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5.1 General Stress–Strain System
◈ Shear stress is symmetric.
→ Shear stress pairs with subscripts differing in order are equal.
→ τxy = τyx
[Proof]
In static equilibrium, sum of all moments and sum of all forces equal zero for
the element.
First, apply Newton's 2nd lawduF mdt
=∑
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where I = moment of inertia = r2mr = radius of gyration
= angular acceleration
5.1 General Stress–Strain System
Then, consider torque (angular momentum), T
2( ) ( ) ( )d d d dT rmu r m I Idt dt dt dt
ωω ω= = = =∑
ddtω
Thus,
2 dT mrdtω
=∑ (A)
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5.1 General Stress–Strain System
Now, take a moment about a centroid axis in the z-direction
( ) ( ) ( )2 2 2xy yx xy yxx y x y zy z x zLHS T τ τ τ τ∆ ∆ ∆ ∆ ∆∆ ∆ ∆ ∆= = − = −∑
2 2d dRHS dvolr x y z rdt dtω ωρ ρ= = ∆ ∆ ∆
( ) 22xy yxdx y z x y z rdtωτ τ ρ−∴ ∆ ∆ ∆ = ∆ ∆ ∆
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5.1 General Stress–Strain System
After canceling terms, this gives
22xy yxdrdtωτ τ ρ− =
2
, , 0lim 0
x y zr
∆ ∆ ∆ →→
0xy yxτ τ− =
xy yxτ τ∴ =
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5.1 General Stress–Strain System
[Homework Assignment-Special work]
Due: 1 week from today
1. Fabricate your own “Stress Cube” using paper box.
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5.1 General Stress–Strain System
5.1.2 Strain componentsMotion and deformation of fluid element
translation (displacement)
Motion
rotation
linear deformation
Deformation
angular deformation
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5.1 General Stress–Strain System
(1) Motion
1) Translation: ξ, η
, du dt udtξξ = =
, dv dt vdtηη = =
2) Rotation ← Shear flow
12x
v ux y
ω ∂ ∂
= − ∂ ∂
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5.1 General Stress–Strain System
(2) Deformation
1) Linear deformation – normal strain
x xξε ∂
=∂
y yηε ∂
=∂
i) For compressible fluid, changes in temperature or pressure cause
change in volume.
ii) For incompressible fluid, if length in 2-D increases, then length in
another 1-D decreases in order to make total volume unchanged.
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○ Strain normal strain: ε ← linear deformation
shear strain: γ ← angular deformation
5.1 General Stress–Strain System
2) Angular deformation– shear strain
xy x yη ξγ ∂ ∂
= +∂ ∂
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5.1 General Stress–Strain System
Consider a small element OABC
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i) Displacement (translation): ξ, η, ζ
ii) Deformation: due to system of external forces
(1) Deformation
Normal strain, ε
5.1 General Stress–Strain System
( ) ( )' , ,, ,O O x y zx y z ξ η ζ→ + + +
' ' ' 'OABC O A B C→
change in lengthoriginal length
ε =
( ), ,
' , ,
C x x y y z z
C x x x y x z xx x xξ η ζξ η ζ
→+ ∆ + ∆ + ∆
∂ ∂ ∂ + ∆ + + ∆ + + ∆ + + ∆ ∂ ∂ ∂
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~ ε is positive when element elongates
under deformation
5.1 General Stress–Strain System
( )0 0
O'C'lim limx x x
xxx x xOC x
OC x x
ξ ξξξε
∆ → ∆ →
∂ − − ∆++ ∆ + + ∆ − ∂∂ = = =∆ ∂
( )
0 0
O'A'lim limy y y
y y y yyyOA
OA y y
ηη ηηε
∆ → ∆ →
∂ + ∆ + + ∆ − − ∆+ ∂− ∂ = = =∆ ∂
z zζε ∂
=∂
' 'O C OC
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2) Shear strain, γ
~ change in angle between two originally
perpendicular elements
5.1 General Stress–Strain System
xyFor –plane
( ) ( ), 0 , 0lim lim tan tanxy c A c Ax y x y
γ θ θ θ θ∆ ∆ → ∆ ∆ →
= + ≅ +
, 0limx y
xx
x xx
η
ξ∆ ∆ →
∂ ∆∂=
∂∆ + ∆∂
yy
y yy
ξ
η
∂ ∆∂+
∂∆ + ∆∂
x yη ξ
∂ ∂ = + ∂ ∂
x xxξ∂ ∴ ∆ < ∆
∂
C’D A’E
O’D O’E
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5.1 General Stress–Strain System
(5.2)
(5.4)
(5.3)
xy x yη ξγ ∂ ∂
= +∂ ∂
yz y zζ ηγ ∂ ∂
= +∂ ∂
zx z xξ ζγ ∂ ∂
= +∂ ∂
(2) displacement vector δ
i j kδ ξ η ζ= + +
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5.1 General Stress–Strain System
(3) Volume dilation
change of volume of deformed elementoriginal volume
e =
( )y y x y zx x z z
yd x zVeV x y z
ηξ ζ∂ ∂ ∂ ∆ + ∆ − ∆ ∆ ∆∆ + ∆ ∆ + ∆ ∂∂ ∂ ∆ = =∆ ∆ ∆ ∆
x y zx y zξ η ζ ε ε ε∂ ∂ ∂
≅ + + = + +∂ ∂ ∂
x y ze ε ε ε= + +
ex y zξ η ζ δ∂ ∂ ∂
= + + = ∇ ⋅∂ ∂ ∂
--- divergence
(5.5)
(5.6)
(5.7)
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in which E = Young's modulus of elasticity
= elongation in the x- dir due to normal stress, σx
5.2 Relations between Stress and Strain for Elastic Solids
5.2.1 Normal StressesHooke's law: Stress is linear with strain.
x xEσ ε=
1x xE
ε σ= (5.8)
xε
. : yyy dir
Eσ
ε− =
. : zzz dir
Eσε− =
x xσ ε∝
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where n = Poisson's ratio
5.2 Relations between Stress and Strain for Elastic Solids
Now, we have to consider other elongations because of lateral contraction of
matter under tension.
'xε .x dir− yσ''xε .x dir− zσ
= elongation in the due to
= elongation in the due to
Now, define
(5.9)' yx yn n
Eσ
ε ε= − = −
'' zx zn n
Eσε ε= − = −
(5.10)
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5.2 Relations between Stress and Strain for Elastic Solids
Poisson's ratio: n = 0~0.5
water~0.5; metal~0.3
cork: Poisson's ratio is high.
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Thus, total strain εx is
5.2 Relations between Stress and Strain for Elastic Solids
(5.11)
(5.12)
( ) ( )1' '' xy z y zx x x x x
n nE E E
σ σ σ σ σε ε ε ε σ+ + = + + = − = −
( )1y z xy n
Eσ σ σε − +=
( )1x yz z n
Eσ σε σ + = −
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5.2 Relations between Stress and Strain for Elastic Solids
5.2.2 Shear Stress
~ Hooke’s law τxy = G γxy
xyxy G x y
τ η ξγ ∂ ∂= = +
∂ ∂
yzyz G y z
τ ζ ηγ ∂ ∂= = +
∂ ∂
zxzx G z x
τ ξ ζγ ∂ ∂= = +
∂ ∂ (5.13)
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where G = shear modulus of elasticity
5.2 Relations between Stress and Strain for Elastic Solids
( )2 1EG
n=
+(5.14)
■ Volume dialation
( )
( )
( )
1
1
1
y zx y z x
y z x
x yz
e nE
nE
nE
σ σε ε ε σ
σ σ σ
σ σσ
+ = + + = −
− ++
+ + −
( )( )11 2 x y znE
σ σ σ+ + = − (5.15)
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■ = arithmetic mean of 3 normal stresses
Combine Eqs. (5.12), (5.14) and (5.15)
5.2 Relations between Stress and Strain for Elastic Solids
σ
( )13 x y zσ σ σσ + += (5.16)
21 2x x
neGn
σ ε = + − (5.17)
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5.2 Relations between Stress and Strain for Elastic Solids
Therefore
23x xeGσ σ ε − = −
23y yeGσ σ ε − = −
23z zeGσ σ ε − = −
(5.18)
xy yx Gx yη ξ
τ τ∂ ∂ += = ∂ ∂
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5.2 Relations between Stress and Strain for Elastic Solids
zy yz Gy zζ η
τ τ∂ ∂ += = ∂ ∂
xz zx Gz xξ ζτ τ ∂ ∂ = = +
∂ ∂ (5.19)
[Proof] Derivation of Eqs. (5.17) & (5.18)
(A)(5.15) →
(5.12) →
(5.14) →
( ) ( )11 2 x y znE
e σ σ σ+ +−=
( )1y zx x n
Eσ σε σ + = −
2 (1 )2(1 )
EG E G nn
= → = ++
(C)
(B)
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5.2 Relations between Stress and Strain for Elastic Solids
i) Combine (A) and (B)
( ) ( )( ) ( ) ( )
( )
1 21 2 1 2
1
x y z x y z
y zx x
n n nneE En n
nE
σ σ σ σ σ σ
σ σε σ
− + + + +× = =+ −+
+ = −
( )1
1 2 x xn ne
Enε σ+
+ =−
( )1 1 2xxnE e
n nεσ +∴ = + −
(D)
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5.2 Relations between Stress and Strain for Elastic Solids
Substitute (C) into (D)
( )2
1 2xxn eG
nεσ +∴ = −
→ Eq. (5.17)
ii) Subtract (5.16) from (5.17)
( ) ( )1231 2 x y zxx
n eGn
σ σ σεσ σ + ++− = − − (E)
Substitute (A) into (E); ( )
( ) :1 2x y z
EA en
σ σ σ+ + =−
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5.2 Relations between Stress and Strain for Elastic Solids
( ) ( )1( ) 231 2 1 2
xn EeRHS of E G e
n nε +∴ = − − −
( )( )
( ) ( ) ( )
12 1 2 12 2 331 2 1 2 1 2 1 2x x
nGn G n nG e G en n n n
ε ε + + −= + = − − − − +
( )
( )
11 22 3
1 2x
nG en
ε
− −= + −
123xG eε = −
→ Eq. (5.18)
36/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
Experimental evidence suggests that, in fluid, stress is linear with time rate
of strain.
( )stress straint
∂∝
∂→ Newtonian fluid (Newton's law of viscosity)
[Cf] For solid,
→
stress strain∝
37/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
5.3.1 Normal stressFor solid, Eq. (5.18) can be used as
Hookeian elastic solid:2
23x xeF
Lσ σ ε − = −
GBy analogy,
Newtonian fluid: 2
23x xeFt
tLσ σ ε
∂ − = − ∂
Time rate of strain
2FtL
µ ≡
(5.20)
Now set = dynamic viscosity
38/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
Then,
223
xx
et t
εσ σ µ µ∂ ∂− = −
∂ ∂ (5.21)
By the way,
,x exξε δ∂
= = ∇⋅∂
Therefore,
x ut t x xx t
ε ξ ξ∂ ∂ ∂ ∂∂ ∂ = = = ∂ ∂ ∂ ∂∂ ∂ (5.22)
, , ( , , displacement)u v wt x tξ η ζ ξ η ζ∂ ∂ ∂
= = = =∂ ∂ ∂
39/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
e u v wqt t x y z
δ∂ ∂ ∂ ∂ ∂= ∇ ⋅ = ∇ ⋅ = + +
∂ ∂ ∂ ∂ ∂
(5.23)
i j kδ ξ η ζ= + +
q ui v j wktδ∂
= = + +∂
u v wqx y z
∂ ∂ ∂∇ ⋅ = + +
∂ ∂ ∂
Eq. (5.21) becomes
( )223x
uqx
σ σ µ µ∂= + − ∇ ⋅
∂
40/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
For compressible fluid,
( )223x
uqx
σ σ µ µ∂= + − ∇ ⋅
∂
( )223y
vqy
σ σ µ µ∂= + − ∇ ⋅
∂
( )223z
wqz
σ σ µ µ∂= + − ∇ ⋅
∂
(5.24)
For incompressible fluid,
0de qdt
= ∇ ⋅ =
← time rate of volume expansion=0
41/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
0q→ ∇ ⋅ =
→ Continuity Eq.
Therefore, Eq. (5.24) becomes
2xux
σ σ µ ∂= +
∂
2yvy
σ σ µ ∂= +
∂
2zwz
σ σ µ ∂= +
∂
42/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
5.3.2. Shear stressBy following the same analogy
2xyFtG
x y x ytLη ξ η ξ
τ∂ ∂ ∂ ∂∂ + += = ∂ ∂ ∂ ∂∂
v ux yx t y t
η ξµ µ∂ ∂∂ ∂ ∂ ∂ += + = ∂ ∂∂ ∂ ∂ ∂
µ
vtη∂
=∂ u
tξ∂
=∂
43/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
xy yx
zy yz
xz zx
v ux y
w vy z
u wz x
τ τ µ
τ τ µ
τ τ µ
∂ ∂ += = ∂ ∂ ∂ ∂ += = ∂ ∂ ∂ ∂ = = + ∂ ∂
(5.25)
[Appendix 1]
xy yxv ux y
τ τ µ∂ ∂ += = ∂ ∂
44/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
45/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
, 'xy xyτ τⅰ)
46/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
ⅱ) , 'yx yxτ τ
47/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
ⅲ) composition
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1) Assume viscous effects are completely represented by the viscosity µ for
incompressible fluid
▪ Relation between thermodynamic pressure p and mean normal stress
~ minus sign accounts for pressure (compression)
2) For compressible fluid
in which µ ‘ = 2nd coefficient of viscosity associated solely with dilation
= bulk viscosity
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
σ
( )13 x y zp σ σ σσ + += − = (5.26)
( )'p qσ µ= − + ∇ ⋅
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Since, dilation effect is small for most cases
For zero-dilation viscosity effects ( µ ‘ = 0), (5.24) becomes
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
( )' 0qµ →∇ ⋅
pσ∴ = −
( )223x
up qxσ µ µ∂ = − + − ∇ ⋅ ∂
( )223y
vp qyσ µ µ∂ = − + − ∇ ⋅ ∂
( )223z
wp qzσ µ µ∂ = − + − ∇ ⋅ ∂
Normalstress pressure
Viscouseffects
(5.29)
(5.28)
(5.27)
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For zero viscous effects ( µ = 0) → inviscid fluids in motion and for all fluids
at rest
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
■ Shear stresses in a real fluid
xy yxv ux y
τ τ µ∂ ∂ += = ∂ ∂
yz zyw vy z
τ τ µ∂ ∂ += = ∂ ∂
xz zxu wz x
τ τ µ ∂ ∂ = = + ∂ ∂
(5.30)
x y z pσ σ σ σ= = = = −
0xy yz zxτ τ τ= = =
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[Appendix 2] Bulk viscosity and thermodynamic pressure
→ Boundary-Layer Theory (Schlichting, 1979) pp. 61-63
If fluid is compressed, expanded or made to oscillate at a finite rate, work
done in a thermodynamically reversible process per unit volume is
~ dissipation of energy
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
( )'p qσ µ= − + ∇ ⋅
deW p q pdt
= ∇ ⋅ =
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where µ ‘ = bulk viscosity of fluid that represents that property which is
responsible for energy dissipation in a fluid of uniform temperature during
a change in volume at a finite rate
= second property of a compressible, isotropic, Newtonian fluid
[Cf] µ = shear viscosity = first property
Direct measurement of bulk viscosity is very difficult.
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
' 0 , pµ σ= = −
' 0 , pµ σ≠ ≠ −
53/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
[Appendix 3] Normal stress
Normal stress = pressure + deviation from it
'x xpσ σ= − +
'y ypσ σ= − +
'z zpσ σ= − +
Thus, stress matrix becomes
'0 0'0 0
'0 0
x xy xz
yx y yz
zx zy z
pp
p
σ τ ττ σ ττ τ σ
− +− −
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where λ = compressibility coefficient
5.3 Relations between Stress and Rate of Strain for Newtonian Fluids
Normal stresses are proportional to the volume change (compressibility)
and corresponding components of linear deformation, a, b, c.
Thus,
( )
( )
( )
2
2
2
x
y
z
p aa b c
p ba b c
p ca b c
σ λ µ
σ λ µ
σ λ µ
= − + ++ +
= − + ++ +
= − + ++ +
55/615.3 Relations between Stress and Rate of Strain for Newtonian Fluids
1. (5-3) Consider a fluid element under a general state of stress as illustrated in
Fig. 5-1. Given that the element is in a gravity field, show that the equilibrium
requirement between surface, body and inertial forces leads to the equations
yxx zxx xg a
x y zτσ τ ρ ρ
∂∂ ∂+ + + =
∂ ∂ ∂xy y xy
y yg ax y x
τ σ τρ ρ
∂ ∂ ∂+ + + =
∂ ∂ ∂yzxz z
xg ax y z
ττ σ ρ ρ∂∂ ∂
+ + + =∂ ∂ ∂
Homework Assignment # 5
Due: 1 week from today