MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
CAMBRIDGE, MA 02139
2.002 MECHANICS AND MATERIALS II
Spring, 2004
Creep and Creep Fracture: Part II
Stress and Deformation Analysis in Creeping
Structures
c
L. Anand
�
Steady-State Bending of Viscoplastic Beams
1. Kinematics
v(x, t)
trans. displ. of the neutral axis
. ∂ 2v(x,t)
κ(x, t) = ∂x2
�(x, y, t) = −κ(x, t)y
�˙ (x, y, t) − κ̇(x, t)y
. ∂ 2v̇(x,t)
κ̇(x, y, t) = ∂x2
curvature of the neutral axis
longitudinal strain
longitudinal strain rate
curvature rate
|�˙ |sgn(˙�) =
=
Therefore, |�˙ | =
sgn(˙�) =
−|κ˙ |sgn(κ̇)|y |sgn(y)
|κ̇||y | [−sgn(κ̇)sgn(y)]
|κ̇||y | and
−sgn(κ̇)sgn(y)
2. Constitutive Relation
�˙ = �˙ c
= �˙ 0
�
σ = s
�
⇒ σ(x, y, t) = s
�
|σ|
s
|�˙ |
�˙ 0
�n
sgn(σ)
�1/n
sgn(˙�)
|κ̇(x, t)||y|
�˙ 0
�1/n
sgn(˙�)
⇒ σ(x, y, t) = {−sgn(κ̇)sgn(y)s}
�˙ 0
s>0
�
|κ(x,
˙
t)|
�˙ 0
reference strain rate
reference stress
�1/n
|y|1/n
3. Moment-Curvature Rate Relation
M (x, t) = −
�
A
yσ(x, y, t)dA
= {sgn(κ̇)s}
with
In ≡
�
A
�
|κ̇(x, t)|
�˙ 0
�1/n �
A
sgn(y) |y|1+1/n dA
�
|κ̇(x, t)|
M (x, t) = {sgn(κ̇)s}
�˙ 0
�1/n
In
sgn(y)|y|1+1/ndA
4. Equation for Stress
�
|κ̇(x, t)|
σ(x, y, t) = − {sgn(κ̇)s}
�˙ 0
M (x, t) 1/n
sgn(y)
⇒ σ(x, y, t) = −
|y|
In
�1/n
|y |1/n(sgn(y))
5. Differential Equation for Lateral Displacement
�
�1/n
|κ(x,
˙
t)|
|M (x, t)| =
sIn
�˙
�n
� 0
|M (x, t)|
⇒ κ̇(x, t) = �˙ 0
sgn(M (x, t))
sIn
�
�n
2
|M (x, t)|
∂ v̇(x, t)
= �˙ 0
sgn(M (x, t))
∂x2
sIn
Example Problem: Cantilever Beam
M (x, t) − P (t)(L − x) = 0
M (x, t) = P (t)(L − x) 0 ≤ x ≤ L
�
�n
2
∂ v̇
|M |
κ̇(x) =
=
�
˙
sgn(M )
0
2
∂x
sIn
�
�n
|P (t)(L − x)|
= �˙ 0
sIn
�
�n
|P (t)
|
= �˙ 0
(L − x)n
sIn
Example Problem: Cantilever Beam (cont.)
∂ v˙ = 0
Boundary conditions: (1) v̇ = 0 at x = 0 and (2) ∂x
at x = 0 (Assume P (t) > 0)
�
�n �
|P (t)|
∂v̇
1
= �˙ 0
−
(L − x)n+1 + C1
∂x
sIn
n+1
�
�n
|P (t)|
Ln+1
C1 = �˙ 0
(Using BC (2))
sIn
n+1
�
�n
�
∂v̇
|P (t)|
1 �
n+1
n+1
= �˙ 0
−(L − x)
+L
∂x
sIn
n+1
�
�n
�
�
n+2
|P (t)|
1
(L − x)
+ Ln+1x + C2
v̇ = �˙ 0
sIn
n+1
n+2
�
�n
�
�
n+2
|P (t)|
1
L
C2 = −�˙ 0
(Using BC (1))
sIn
n+1 n+2
�
Example Problem: Cantilever Beam (cont.)
�
|P (t)|
v̇ = �˙ 0
sIn
�n
�
(L − x)n+2
Ln+2
�
1
+ Ln+1x −
n+2
n+1
n+2
�
�n
|P (t)|
Ln+2
(Tip deflection rate)
δ̇ = |v̇(x = L)| = �˙ 0
sIn
n+2
|P (t)| =
�
(δ̇/�˙ 0)(n + 2)
Ln+2
�1/n
sIn
Three-Dimensional Generalization of Constitutive
Equations for Elastic-Viscoplastic Materials
1. Strain Rate Decomposition:
�˙ ij = �˙ eij + �˙ cij
�˙ ij
�˙ e
ij
�˙ c
ij
total strain rate
elastic strain rate
creep or viscoplastic strain rate
2. Constitutive Equations for �˙ eij :
⎡
⎛
⎞
⎤
�
1⎣
e
⎝
�˙ ij =
(1 + ν)σ˙ ij − ν
σ̇ kk ⎠ δij ⎦
E
k
E
ν
Young’s modulus
Poisson’s ratio
2. Constitutive Equations for �˙ cij :
� /σ} creep strain rate components
�˙ cij = �˙ c(3/2){σij
�˙ c = �˙ 0 {σ/s}n equivalent tensile creep rate
⎛
⎞
�
�
⎝
σkk ⎠ δij stress deviator components
σij = σij − (1/3)
k
�
�
� σ�
σ =
(3/2)
σij
Mises equivalent tensile stress
ij
i,j
�
�
�
�
2
2
2
= �(1/2) (σ11 − σ22) + (σ22 − σ33) + (σ33 − σ11) +
�
��1/2
�
2
2
2
3 σ12 + σ23 + σ31 �
n
�˙ 0
s
creep exponent
reference strain rate
reference stress
• Note that in uniaxial tension when σ11 = σ, with all
� = (2/3)σ, σ � = σ � =
other σij = 0, we have σ11
22
33
−(1/3)σ, and σ = |σ|. Therefore, the constitutive
equation for �˙ cij yields
�˙ c11 = �˙ 0 {σ/s}n sgn(σ)
�˙ c22 = �˙ c33 = −(1/2)˙�c11
�˙ cij = 0 otherwise,
as it should.
• For the case of rigid-viscoplastic materials, the elastic strains and strain rates are neglected:
. c
�ij = �ij
;
� /σ}
�˙ ij = �˙ cij = �˙ c(3/2){σij
�˙ c = �˙ 0 {σ/s}n
Example Problem: Torsion of Thin-walled Tube
• Consider a thin-walled tube of radius r, wall thickness
t and length L. One end of the tube is fixed, while on
the other a constant twisting moment Mt is applied.
The tube is at high homologous temperatures (creep
conditions prevail). Calculate the twisting rate φ̇ for
the tube.
Example Problem: [Thin-Walled] Torsion
• The angle of twist φ is a function of time, i.e., φ =
φ(t). The angle of twist per unit length is denoted by
α = φ/L = α(t)
• Displacement field:
ur = 0
uθ = αzr
uz = 0
• Strain field:
1 ∂uθ
1 ∂uz
1
�θz =
+
= αr,
2 ∂z
r ∂θ
2
with all other �ij = 0
�
�
• Therefore the strain rate is
1
α̇r
2
1 φ̇
�˙ θz =
r
2L
�˙ θz =
(1)
• Constitutive equation: Since the applied moment M (t)
is constant, the elastic strain rate �˙ eij = 0. (Strictly,
we need to verify that σ̇ ij = 0; however, the thinwalled tube in torsion has one constant non-zero stress
component, σθz , that is directly proportional to twisting moment, Mt (see ‘Equilibrium’, below)). Therefore,
⎧
⎫
� � �
�
⎨
⎬
σ
3
3 σθz
ij
�˙ ij = �˙ c
⇒ �˙ θz = �˙ c
2 ⎩ σ ⎭
2 σ
�˙ c = �˙ 0
� �n
σ
s
⎛
⎞
1 ⎝�
�
σij = σij −
σkk ⎠ δij
3
k
� = σ .
• The only non-zero stress component is σθz
θz
√
Also, σ = 3 |σθz | from the definition of the equivalent
tensile stress. Therefore,
�˙ θz =
√
3
�˙ 0
2
�√
3 |σθz |
s
�n
sgn(σθz )
(2)
• Equilibrium: The applied torque should balance the
internal torque of the only non-zero stress component,
σθz :
dMt = rdF = rσθz dA = rσθz (rdθt) = σθz tr2 dθ
Mt =
� 2π
0
σθz tr2 dθ = 2πtr 2σθz
⇒ σθz =
Mt
2πtr2
(3)
• Finally,
√
�√
�n
3
3 |σθz |
�˙ 0
sgn(σθz )
2
s
√
�√
�n
1 φ̇
3
3 |Mt|
⇒
r =
sgn(Mt)
�˙ 0
2
2L
2
s 2πtr
�√
�n � �
√
3|Mt|
L
sgn(Mt)
⇒ φ̇ =
3˙�0
r
2πtr2 s
�√
�n � �
√
L
3Mt
⇒ φ̇ =
3˙�0
for Mt > 0
2
2πtr s
r
�˙ θz =
Example Problem: Torsion of Thick-walled Tube
• Recall
√
1
3
α̇r = �˙ θz (r) =
�˙ 0
2
2
�√
3 |σθz (r)|
s
(for a ≤ r ≤ b)
�n
sgn(σθz (r))
For simplicity, let σθz (r) > 0. Therefore,
√
�√
�n
�
�1/n
3
3 |σθz |
s
2
√
⇒ σθz (r) = √
�˙ θz (r)
;
�˙ θz =
�˙ 0
2
s
3
3˙�0
�˙ zθ (r) = αr/2
˙
⇒ σzθ (r) ≡ Ar 1/n,
where
s
φ˙
√
A≡√
3
3˙�0L
�
�1/n
With dMt = σθz (r)r 2 dr dθ = Ar 2+1/n dr dθ,
Mt = A
Let
Jn =
� b � 2π
a
0
� b � 2π
a
0
r 2+1/ndθdr =
r2+1/ndθdr
�
�
2π
3+1/n
3+1/n
b
− a
3 + 1/n
Since A = Mt/Jn,
Mt 1/n
r
σθz (r) =
Jn
and
�
φ̇
s
√
√
3
3˙�0L
⇒ φ˙
=
√
3˙�0
�1/n
�√
Mt
=
Jn
3 Mt
s Jn
�n
L