8.4 †Compact Forms 10 Solving a boundary value problem with 15 unknowns through 15 equations is a formidable task.. 8.4.4 Ellipticity of Elasticity Problems 8.5 †Strain Energy and Extena
Trang 1Draft8.3 Boundary Value Problem Formulation 3
x
C D
E
σ
Note: Unknown tractions=Reactions
t n y u u Γ x u
CD BC
DE EA
t s
Γt 0
? σ
?
?
?
? 0
0
0
Figure 8.2: Boundary Conditions in Elasticity Problems
9 Hence, the boundary value formulation is suumarized by
∂T ij
∂X j + ρb i = ρ
∂2u i
∂t2 in Ω (8.4)
E∗ = 1
2(u∇x+∇x u) (8.5)
T = λI E + 2µE in Ω (8.6)
u = u in Γu (8.7)
t = t in Γt (8.8)
and is illustrated by Fig 8.3 This is now a well posed problem.
8.4 †Compact Forms
10 Solving a boundary value problem with 15 unknowns through 15 equations is a formidable task Hence, there are numerous methods to reformulate the problem in terms of fewer unknows
8.4.1 Navier-Cauchy Equations
11 One such approach is to substitute the displacement-strain relation into Hooke’s law (resulting in stresses in terms of the gradient of the displacement), and the resulting equation into the equation of motion to obtain three second-order partial differential equations for the three displacement components
known as Navier’s Equation
(λ + µ) ∂
2u k
∂X i ∂X k + µ
∂2u i
∂X k ∂X k + ρb i = ρ
∂2u i
∂t2 (8.9) or
(λ + µ) ∇(∇·u) + µ∇2u + ρb = ρ ∂
2u
∂t2 (8.10) (8.11)
Trang 2Draft4 BOUNDARY VALUE PROBLEMS in ELASTICITY
Natural B.C.
t i: Γt
Stresses
T ij
Equilibrium
∂T ij
∂x j + ρb i = ρ dv i
dt
Body Forces
b i
Constitutive Rel.
T = λI E + 2µE
Strain
E ij
Kinematics
E∗= 1
2(u∇x+∇x u)
Displacements
u i
Essential B.C.
u i: Γu
✻
❄
❄
❄
❄
❄
Figure 8.3: Fundamental Equations in Solid Mechanics
8.4.2 Beltrami-Mitchell Equations
12 Whereas Navier-Cauchy equation was expressed in terms of the gradient of the displacement, we can follow a similar approach and write a single equation in term of the gradient of the tractions
∇2T ij+ 1
1 + ν T pp,ij = − ν
1− ν δ ij ∇·(ρb) − ρ(b i,j + b j,i) (8.12) or
T ij,pp+ 1
1 + ν T pp,ij = − ν
1− ν δ ij ρb p,p − ρ(b i,j + b j,i) (8.13)
8.4.3 Airy Stress Function
Airy stress function, for plane strain problems will be separately covered in Sect 9.2
8.4.4 Ellipticity of Elasticity Problems
8.5 †Strain Energy and Extenal Work
13 For the isotropic Hooke’s law, we saw that there always exist a strain energy function W which is
positive-definite, homogeneous quadratic function of the strains such that, Eq 18.29
T ij= ∂W
hence it follows that
W =1
Trang 3Draft8.6 †Uniqueness of the Elastostatic Stress and Strain Field 5
14 The external work done by a body in equilibrium under body forces b i and surface traction t i is equal to
Ω
ρb i u i dΩ +
Γ
t i u i dΓ Substituting t i = T ij n j and applying Gauss theorem, the second term
Γ
T ij n j u i dΓ =
Ω
(T ij u i),j dΩ =
Ω
(T ij,j u i + T ij u i,j )dΩ (8.16)
but T ij u i,j = T ij (E ij+ Ωij ) = T ij E ij and from equilibrium T ij,j =−ρb i, thus
Ω
ρb i u i dΩ +
Γ
t i u i dΓ =
Ω
ρb i u i dΩ +
Ω
(T ij E ij − ρb i u i )dΩ (8.17) or
Ω
ρb i u i dΩ +
Γ
t i u i dΓ
External Work
= 2
Ω
T ij E ij
2 dΩ
Internal Strain Energy
(8.18)
that is For an elastic system, the total strain energy is one half the work done by the external forces
acting through their displacements u i
8.6 †Uniqueness of the Elastostatic Stress and Strain Field
15 Because the equations of linear elasticity are linear equations, the principles of superposition may be
used to obtain additional solutions from those established Hence, given two sets of solution T ij(1), u(1)i ,
and T ij(2), u(2)i , then T ij = T ij(2)− T(1)
ij , and u i = u(2)i − u(1)
i with b i = b(2)i − b(1)
i = 0 must also be a solution
16 Hence for this “difference” solution, Eq 8.18 would yield
Γ
t i u i dΓ = 2
Ω
u ∗ dΩ but the left hand
side is zero because t i = t(2)i − t(1)
i = 0 on Γu , and u i = u(2)i − u(1)
i = 0 on Γt, thus
Ω
u ∗ dΩ = 0.
17 But u ∗ is positive-definite and continuous, thus the integral can vanish if and only if u ∗= 0 everywhere,
and this is only possible if E ij = 0 everywhere so that
E ij(2)= E ij(1)⇒ T(2)
ij = T ij(1) (8.19)
hence, there can not be two different stress and strain fields corresponding to the same externally imposed body forces and boundary conditions1and satisfying the linearized elastostatic Eqs 8.1, 8.14 and 8.3
18 This famous principle of Saint Venant was enunciated in 1855 and is of great importance in applied
elasticity where it is often invoked to justify certain “simplified” solutions to complex problem
In elastostatics, if the boundary tractions on a part Γ1 of the boundary Γ are replaced by a statically equivalent traction distribution, the effects on the stress distribution in the body are negligible at points whose distance from Γ1 is large compared to the maximum distance between points of Γ1
19 For instance the analysis of the problem in Fig 8.4 can be greatly simplified if the tractions on Γ1 are replaced by a concentrated statically equivalent force
1This theorem is attributed to Kirchoff (1858).
Trang 4Draft6 BOUNDARY VALUE PROBLEMS in ELASTICITY
F=tdx
t dx
Figure 8.4: St-Venant’s Principle
20 So far all equations have been written in either vector, indicial, or engineering notation The last two were so far restricted to an othonormal cartesian coordinate system
21 We now rewrite some of the fundamental relations in cylindrical coordinate system, Fig 8.5, as
this would enable us to analytically solve some simple problems of great practical usefulness (torsion, pressurized cylinders, ) This is most often achieved by reducing the dimensionality of the problem from 3 to 2 or even to 1
z
θ r
Figure 8.5: Cylindrical Coordinates
8.8.1 Strains
22 With reference to Fig 8.6, we consider the displacement of point P to P ∗ the displacements can be
expressed in cartesian coordinates as u x , u y , or in polar coordinates as u r , u θ Hence,
u x = u r cos θ − u θ sin θ (8.20-a)
u y = u r sin θ + u θ cos θ (8.20-b)
Trang 5Draft8.8 Cylindrical Coordinates 7
u
u
u
*
x
y
r
θ
u
θ
r
x
y
P
P
θ θ
Figure 8.6: Polar Strains
substituting into the strain definition for ε xx(for small displacements) we obtain
ε xx = ∂u x
∂x =
∂u x
∂θ
∂θ
∂x+
∂u x
∂r
∂r
∂u x
∂θ =
∂u r
∂θ cos θ − u r sin θ − ∂u θ
∂θ sin θ − u θ cos θ (8.21-b)
∂u x
∂r =
∂u r
∂r cos θ − ∂u θ
∂θ
∂x = − sin θ
∂r
ε xx =
− ∂u r
∂θ cos θ + u r sin θ +
∂u θ
∂θ sin θ + u θ cos θ
sin θ
r
+
∂u r
∂r cos θ − ∂u θ
∂r sin θ
Noting that as θ → 0, ε xx → ε rr , sin θ → 0, and cos θ → 1, we obtain
ε rr = ε xx | θ→0=∂u r
23 Similarly, if θ → π/2, ε xx → ε θθ , sin θ → 1, and cos θ → 0 Hence,
ε θθ = ε xx | θ→π/2= 1
r
∂u θ
∂θ +
u r
finally, we may express ε xy as a function of u r , u θ and θ and noting that ε xy → ε rθ as θ → 0, we obtain
ε rθ= 1 2
ε xy | θ→0= ∂u θ
∂r − u θ
r +
1
r
∂u r
∂θ
(8.24)
24 In summary, and with the addition of the z components (not explicitely derived), we obtain
Trang 6Draft8 BOUNDARY VALUE PROBLEMS in ELASTICITY
ε rr = ∂u r
ε θθ = 1
r
∂u θ
∂θ +
u r
ε zz = ∂u z
ε rθ = 1
2
1
r
∂u r
∂θ +
∂u θ
∂r − u θ
r
(8.28)
ε θz = 1
2
∂u θ
∂z +
1
r
∂u z
∂θ
(8.29)
ε rz = 1
2
∂u z
∂r +
∂u r
∂z
(8.30)
8.8.2 Equilibrium
25 Whereas the equilibrium equation as given In Eq 6.17 was obtained from the linear momentum principle (without any reference to the notion of equilibrium of forces), its derivation (as mentioned) could have been obtained by equilibrium of forces considerations This is the approach which we will follow for the polar coordinate system with respect to Fig 8.7
δ θ d θ
θ
d
d r
d r
θθ
r r
δ
δ θ
r +
r
r+dr
θ
θ
d
r
f f
r
θ
δ
δ θθ
r +
+δ δ
r r
r
θ
r
T
rr
rθ
θ
r
T T
T T T
T T
T
+δ
T T
Figure 8.7: Stresses in Polar Coordinates
26 Summation of forces parallel to the radial direction through the center of the element with unit
thickness in the z direction yields:
%
T rr+∂T rr
∂r dr&
(r + dr)dθ − T rr (rdθ)
−%T θθ+∂T θθ
∂θ + T θθ&
dr sin dθ2
+%
T θr+∂T θr
∂θ dθ − T θr
&
dr cos dθ2 + f r rdrdθ = 0
(8.31)
we approximate sin(dθ/2) by dθ/2 and cos(dθ/2) by unity, divide through by rdrdθ,
1
r T rr+
∂T rr
∂r
1 + dr
r
− T θθ
r − ∂T θθ
∂θ
dθ
dr+
1
r
∂T θr
∂θ + f r= 0 (8.32)
27 Similarly we can take the summation of forces in the θ direction In both cases if we were to drop the
dr/r and dθ/r in the limit, we obtain
Trang 7Draft8.8 Cylindrical Coordinates 9
∂T rr
∂r +
1
r
∂T θr
∂θ +
1
r (T rr − T θθ ) + f r = 0 (8.33)
∂T rθ
∂r +
1
r
∂T θθ
∂θ +
1
r (T rθ − T θr ) + f θ = 0 (8.34)
28 It is often necessary to express cartesian stresses in terms of polar stresses and vice versa This can
be done through the following relationships
T xx T xy
T xy T yy
=
cos θ − sin θ
sin θ cos θ
T rr T rθ
T rθ T θθ
cos θ − sin θ
sin θ cos θ
T
(8.35) yielding
T xx = T rrcos2θ + T θθsin2θ − T rθ sin 2θ (8.36-a)
T yy = T rrsin2θ + T θθcos2θ + T rθ sin 2θ (8.36-b)
T xy = (T rr − T θθ ) sin θ cos θ + T rθ(cos2θ − sin2θ) (8.36-c) (recalling that sin2θ = 1/2 sin 2θ, and cos2θ = 1/2(1 + cos 2θ)).
8.8.3 Stress-Strain Relations
29 In orthogonal curvilinear coordinates, the physical components of a tensor at a point are merely the Cartesian components in a local coordinate system at the point with its axes tangent to the coordinate curves Hence,
T rr = λe + 2µε rr (8.37)
T θθ = λe + 2µε θθ (8.38)
T rθ = 2µε rθ (8.39)
T zz = ν(T rr + T θθ) (8.40)
with e = ε rr + ε θθ alternatively,
E rr = 1
E
(1− ν2)T rr − ν(1 + ν)T θθ (8.41)
E θθ = 1
E
(1− ν2)T θθ − ν(1 + ν)T rr (8.42)
E rθ = 1 + ν
E rz = E θz = E zz= 0 (8.44)
8.8.3.1 Plane Strain
30 For Plane strain problems, from Eq 7.52:
σ rr
σ θθ
σ zz
τ rθ
=
E
(1 + ν)(1 − 2ν)
(1− ν) ν 0
ν (1− ν) 0
ν ν 0
0 0 1−2ν2
ε rr
ε θθ
γ rθ
(8.45)
and ε zz = γ rz = γ θz = τ rz = τ θz= 0
31 Inverting,
ε rr
ε θθ
γ rθ
=
1
E
1− ν2 −ν(1 + ν) 0
−ν(1 + ν) 1− ν2 0
ν ν 0
0 0 2(1 + ν
σ rr
σ θθ
σ zz
τ rθ
Trang 8Draft10 BOUNDARY VALUE PROBLEMS in ELASTICITY 8.8.3.2 Plane Stress
32 For plane stress problems, from Eq 7.55-a
σ rr
σ θθ
τ rθ
=
E
1− ν2
1ν ν1 00
0 0 1−ν
2
ε rr
ε θθ
γ rθ
ε zz = − 1
1− ν ν(ε rr + ε θθ) (8.47-b)
and τ rz = τ θz = σ zz = γ rz = γ θz= 0
33 Inverting
ε rr
ε θθ
γ rθ
=
1
E
−ν1 −ν1 00
0 0 2(1 + ν)
σ rr
σ θθ
τ rθ
Trang 9Chapter 9
SOME ELASTICITY PROBLEMS
1 Practical solutions of two-dimensional boundary-value problem in simply connected regions can be accomplished by numerous techniques Those include: a) Finite-difference approximation of the dif-ferential equation, b) Complex function method of Muskhelisvili (most useful in problems with stress concentration), c) Variational methods, d) Semi-inverse methods, and e) Airy stress functions
2 Only the last two methods will be discussed in this chapter
3 Often a solution to an elasticity problem may be obtained without seeking simulateneous solutions to the equations of motion, Hooke’s Law and boundary conditions One may attempt to seek solutions
by making certain assumptions or guesses about the components of strain stress or displacement while leaving enough freedom in these assumptions so that the equations of elasticity be satisfied
4 If the assumptions allow us to satisfy the elasticity equations, then by the uniqueness theorem, we have succeeded in obtaining the solution to the problem
5 This method was employed by Saint-Venant in his treatment of the torsion problem, hence it is often
referred to as the Saint-Venant semi-inverse method.
9.1.1 Example: Torsion of a Circular Cylinder
6 Let us consider the elastic deformation of a cylindrical bar with circular cross section of radius a and length L twisted by equal and opposite end moments M1, Fig 9.1
7 From symmetry, it is reasonable to assume that the motion of each cross-sectional plane is a rigid body
rotation about the x1 axis Hence, for a small rotation angle θ, the displacement field will be given by:
u = (θe1)×r = (θe1)×(x1e1+ x2e2+ x3e3) = θ(x2e3− x3e2) (9.1) or
u1= 0; u2=−θx3; u3= θx2 (9.2)
where θ = θ(x1)
8 The corresponding strains are given by
E11 = E22= E33= 0 (9.3-a)
E12 = −1
2x3
∂θ
Trang 10Draft2 SOME ELASTICITY PROBLEMS
X X
M
2
3
T
T
a
θ
n n
Figure 9.1: Torsion of a Circular Bar
E13 = 1
2x2
∂θ
9 The non zero stress components are obtained from Hooke’s law
T12 = −µx3
∂θ
T13 = µx2 ∂θ
10 We need to check that this state of stress satisfies equilibrium ∂T ij /∂x j = 0 The first one j = 1 is
identically satisfied, whereas the other two yield
−µx3d2θ
dx2 1
µx2d
2θ
thus,
dθ
dx1 ≡ θ = constant (9.6)
Physically, this means that equilibrium is only satisfied if the increment in angular rotation (twist per unit length) is a constant
11 We next determine the corresponding surface tractions On the lateral surface we have a unit normal
vector n = 1a (x2e2+ x3e3), therefore the surface traction on the lateral surface is given by
{t} = [T]{n} = 1
a
T021 T012 T013
T31 0 0
0
x2
x3
=
1
a
x2T12
0 0
12 Substituting,
t = µ
a(−x2x3θ + x2x3θ )e1= 0 (9.8) which is in agreement with the fact that the bar is twisted by end moments only, the lateral surface is traction free
Trang 11Draft9.2 Airy Stress Functions; Plane Strain 3
13 On the face x1= L, we have a unit normal n = e1and a surface traction
t = Te1= T21e2+ T31e3 (9.9) this distribution of surface traction on the end face gives rise to the following resultants
R1 =
T11dA = 0 (9.10-a)
R2 =
T21dA = µθ
x3dA = 0 (9.10-b)
R3 =
T31dA = µθ
x2dA = 0 (9.10-c)
M1 =
(x2T31− x3T21)dA = µθ
(x22+ x23)dA = µθ J (9.10-d)
We note that *
(x2+ x3)2dA is the polar moment of inertia of the cross section and is equal to
J = πa4/2, and we also note that*
x2dA =*
x3dA = 0 because the area is symmetric with respect to
the axes
14 From the last equation we note that
θ = M
which implies that the shear modulus µ can be determined froma simple torsion experiment.
15 Finally, in terms of the twisting couple M , the stress tensor becomes
[T] =
0 −
M x3
J M x J2
− M x3
M x2
16 If the deformation of a cylindrical body is such that there is no axial components of the displacement and that the other components do not depend on the axial coordinate, then the body is said to be in a
state of plane strain If e3is the direction corresponding to the cylindrical axis, then we have
u1= u1(x1, x2), u2= u2(x1, x2), u3= 0 (9.13) and the strain components corresponding to those displacements are
E11 = ∂u1
E22 = ∂u2
E12 = 1
2
∂u1
∂x2 +
∂u2
∂x1
(9.14-c)
E13 = E23= E33= 0 (9.14-d)
and the non-zero stress components are T11, T12, T22, T33 where
T33= ν(T11+ T22) (9.15)
Trang 12Draft4 SOME ELASTICITY PROBLEMS
17 Considering a static stress field with no body forces, the equilibrium equations reduce to:
∂T11
∂x1 +
∂T12
∂T12
∂x1 +
∂T22
∂T33
we note that since T33= T33(x1, x2), the last equation is always satisfied
18 Hence, it can be easily verified that for any arbitrary scalar variable Φ, if we compute the stress components from
T11 = ∂
2Φ
∂x2 (9.17)
T22 = ∂
2Φ
∂x2 (9.18)
T12 = − ∂2Φ
∂x1∂x2 (9.19)
then the first two equations of equilibrium are automatically satisfied This function Φ is called Airy
stress function.
19 However, if stress components determined this way are statically admissible (i.e they satisfy equilibrium), they are not necessarily kinematically admissible (i.e satisfy compatibility equations).
20 To ensure compatibility of the strain components, we express the strains components in terms of Φ from Hooke’s law, Eq 7.36 and Eq 9.15
E11 = 1
E
(1− ν2)T11− ν(1 + ν)T22 = 1
E
(1− ν2)∂
2Φ
∂x2 − ν(1 + ν) ∂2Φ
∂x2
(9.20-a)
E22 = 1
E
(1− ν2)T22− ν(1 + ν)T11 = 1
E
(1− ν2)∂
2Φ
∂x2 − ν(1 + ν) ∂2Φ
∂x2
(9.20-b)
E12 = 1
E (1 + ν)T12=−1
E (1 + ν)
∂2Φ
For plane strain problems, the only compatibility equation, 4.140, that is not automatically satisfied is
∂2E11
∂x2 +∂
2E22
∂x2 = 2 ∂
2E12
substituting,
(1− ν)
∂4Φ
∂x4 + 2 ∂
4Φ
∂x2∂x2 +∂
4Φ
∂x4
or
∂4Φ
∂x4 + 2 ∂
4Φ
∂x2∂x2+∂
4Φ
∂x4 = 0 or ∇4Φ = 0 (9.23)
Hence, any function which satisfies the preceding equation will satisfy both equilibrium, kinematic,
stress-strain (albeit plane strain) and is thus an acceptable elasticity solution
21 †We can also obtain from the Hooke’s law, the compatibility equation 9.21, and the equilibrium
equations the following
∂2
∂x2+ ∂
2
∂x2
(T11+ T22) = 0 or ∇2(T11+ T22) = 0 (9.24)