445.204
Introduction to Mechanics of Materials (재료역학개론)
Chapter 7: Generalized Hooke’s law (Ch. 6 in Shames)
Myoung-Gyu Lee, 이명규
Tel. 880-1711; Email: [email protected]
TA: Chanmi Moon, 문찬미
Lab: Materials Mechanics lab.(Office: 30-521) Email: [email protected]
Contents
- Hooke’s law in 3D - E vs. G vs. v
- Strain energy/ strain energy density - Energy method
- Plane stress, plane strain
2
Hooke’s law in 3D
The stress σ
xin the x-direction produces 3 strains
1) Longitudinal strain (extension) along the x-axis,
2) Transverse strains (contraction) along the
y and z -axes, due to the Poisson’s effect,
E
x x
ε = σ
E
x x z
y
νε νσ ε
ε = = − = −
Hooke’s law in 3D
The total strain produced along a particular direction can be determined by the principle of superposition
That is, the resultant strain along the x-axis, comes from the strain contribution due to the application of σx, σy and σz
σx
causes: in the x-direction
σycauses: in the x-direction
σzcauses: in the x-direction
Therefore, applying the principle of superposition in the x- axis results in
E
σ
xE
νσ
y−
E
νσ
z−
[ ( ) ]
1
z y
x
x
E σ ν σ σ
ε = − +
Hooke’s law in 3D
In general, by superposition of the components of strain in the x, y, and z directions, the strains can be written as
( )
[
x y z]
x
v
E σ σ σ
ε = 1 − +
( )
[
y z x]
y
v
E σ σ σ
ε = 1 − +
( )
[
z x y]
z
v
E σ σ σ
ε = 1 − +
Hooke’s law in 3D
Shearing stresses acting on the unit cube produce shearing Strains; i.e., no Poisson’s effect
xy
xy
G γ
τ =
yz yz
G γ τ =
xz xz
G γ τ =
ij
ij 2 ε
γ =
Why?
Hooke’s law in 3D
ij kk ij
ij
σ δ
E σ ν
E ν
ε 1 + −
=
−
−
−
−
−
−
=
zx yz xy zz yy xx
zx yz xy zz yy xx
σ σ σσ σ σ
G 0 1
0 0
0 0
G 0 0 1
0 0
0
0 G 0
0 1 0
0
0 0
E 0 1 E
ν E
ν
0 0
E 0 ν E
1 E
ν
0 0
E 0 ν E
ν E
1
γ γ γε ε ε
In Matrix Form
0 0 0
T T T α α α
∆
∆
∆
If non-isothermal
E vs. G vs. v
0 0
xx xy
xy yy
S
S τ τ
τ τ
−
=
• Stress tensor (state) in x-y coordinate system
' ' ' '
' ' ' '
0 0
x x x y
x y y y
S S
τ τ
τ τ
=
• Stress tensor (state) in x’-y’ coordinate system
' ' ' '
(2 )( ) 2 2 (2 )( ) 1 0
y x a t S a t 2 y x S
τ − = →τ =
Force equilibrium in the x’ direction
8
E vs. G vs. v
' ' x y
cb
ε = Ob 2
cos 45o
cb= gb = gb
cos 45o 2
Ob = a = a
1 1
(2 ) (2 ) (1 )
2 xx 2 yy
S S Sa
gb a v a a v a v
E E E
ε ε
= + = + = +
' ' (1 )
x y
S v
ε = E +
' ' (2 ' ') (2)( (1 ))
x y x y
G S G S v
τ = ε ↔ = E +
2(1 ) G E
= v
9
+
Strain energy
• Strain energy increment in the uniaxial tension for infinitesimal volume dv
xxd xxdv τ ε
• More generally for normal stresses
(
τ ε τ εxxd xx + yyd yy +τ εzzd zz)
dv• More generally for normal and shear stresses
(
τ ε τ εxxd xx + yyd yy +τ ε τ γzzd zz + xyd xy +τ γyzd yz +τ γzxd zx)
dv• Energy per unit volume
xx xx yy yy zz zz xy xy yz yz zx zx ij ij
du =τ ε τ εd + d +τ ε τ γd + d +τ γd +τ γd =τ εd
(
2)
xyd xy xy d xy xyd xy yxd yx
τ γ =τ ε =τ ε τ ε+
• Then,
0
ij
ij ij
u = ∫ε τ εd 0ij ij ij
V
U = ∫ ∫ε τ εd dv
Strain energy for linear isotropic elastic solids
( ) yy ( ) ( )
xx zz
xx yy zz yy zz xx zz xx yy
xy yz zx
xy yz xz
du d v d v d v
E E E
d d d
G G G
τ τ τ τ τ τ τ τ τ τ τ τ
τ τ τ τ τ τ
= − + + − + + − +
+ + +
( )
2 2 2 2 2 2
1 1
( ) ( )
2 xx yy zz xx yy yy zz zz xx 2 xy yz zx
u v
E τ τ τ E τ τ τ τ τ τ G τ τ τ
= + + − + + + + +
(
2 2 2)
2 2 2( ) ( )
2(1 )(1 2 ) xx yy zz xx yy zz 2 xy yz zx
Ev G
u G
v v ε ε ε ε ε ε γ γ γ
= + + + + + + + +
+ −
• As functions strain
Strain energy density: physical interpretation
εxx
σxx
xx xx
u = ∫ τ ε d
*
xx xxu = ∫ ε τ d
AreaUnder SS curve
Complementary strain-energy density function
Strain-energy density function
Energy method – Castigliano theorem II
* Derivation in the textbook
k k
U P
∂ = ∆
∂
U
M θ
∂ =
∂
The rate of change of the strain energy for a body with respect to
any statically independent force P
kgives the deflection component
of the point of application of this force in the direction of force
The rate of change of the strain energy for a body with respect to
any statically independent couple M gives the amount of rotation
at the point of application of the point couple about an axis collinear with the couple
moment 13
Volume strain
( )( )( )
( )( )( )
z y
x
x x
x
x x
x
ε ε
ε Δ
strain small
for
ε 1 ε 1
ε 1 1
dz dy dx
dz dy dx dz
dy ε dx
ε 1 ε 1
Δ 1
+ +
=
− +
+ +
=
− +
+
= +
For a rectangular parallelepiped with edges dx, dy and dz The volume in the strained condition is:
(1 + εx)(1 + εy)(1 + εz) dx dy dz
The dilation (or volume strain) ∆ is given as
Plane stress
Plane Stress (σ3 = 0)
This stress state is typically observed in
- a thin sheet loaded in the plane of the sheet, or
- a thin wall tube loaded by internal pressure where there is no stress normal to a free surface.
Set σz = σ3 = 0.
Therefore,
[ ]
[ ]
[ 1 2]
3
3 2
2
2 1
1
1 1 1
σ σ
ν ε
νσ σ
ε
νσ σ
ε
+
−
=
−
=
−
=
E E
E [ ( )]
( )
[ ]
( E)E E
E E
2 2
1
1 2
1 1
1 1 1
1
νε ν
σ
νσ ε
ν σ
ε
−
−
=
+
−
=
1 2 2
1
1 ν σ
νε
ε E
= − +
[ ]
[ 2 1]
2 2
2 2 1
1
1 ,
1
νε ν ε
σ
νε ν ε
σ
− +
=
− +
=
E Similarly
E
Plane strain
Plane Strain (ε3 = 0): This occurs typically when
One dimension is much greater than the other two
Examples are a long rod or a cylinder with restrained ends.
This shows that a stress exists along direction-3 (z-axis) even though the strain is zero.
( )
[ ]
[
1 2]
3
2 1
3
3 1 0
σ σ
ν σ
σ σ
ν σ
ε
+
=
= +
−
= but
E
( )
( )[ ]
( )
( )[ ]
0
1 1 1
1 1 1
3
1 2
2 2
2 1
2 1
=
+
−
−
=
+
−
−
=
ε
σ ν ν
σ ν ε
σ ν ν
σ ν ε
E E
