Since all our analysis is only valid in the linear approximation in small sjj’, Eq. (12) is reduced to 3
dV
dV
dV s dV Tr
(7.13)
n
deformatio
after
n
deformatio
before
jj
s,
j 1
where Tr ( trace)3 of any matrix (in particular, any tensor) is the sum of its diagonal elements; in our current case
3
Tr s
( ) s .
(7.14)
jj
j1
The tensor theory shows that the trace does not depend on the particular choice of the coordinate axes; so, the diagonal elements of the strain tensor characterize the medium’s compression/extension.
Next, what is the meaning of its off-diagonal elements? It may be illustrated by the simplest example of a purely shear deformation shown in Fig. 2. (The geometry means to be uniform along the z-
axis normal to the plane of the drawing.) In this case, all displacements (assumed small) have just one Cartesian component, in Fig. 2 along the x-axis: q = n x y (with << 1), so that the only nonzero element of the initial strain tensor qj/ rj’ is qx/ y = , and the symmetrized tensor (9b) is 3 The traditional European notation for Tr is Sp (from the German Spur meaning “trace” or “track”).
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0
/ 2 0
s / 2
0
0 .
(7.15)
0
0
0
Evidently, the change of volume, given by Eq. (13), vanishes in this case. Thus, off-diagonal elements of the tensor s characterize shear deformations.
y
A
F
h
F
Fig. 7.2. An example of pure shear.
x
0
To conclude this section, let me note that Eq. (9) is only valid in Cartesian coordinates. For the solution of some important problems with the axial or spherical symmetry, it is frequently convenient to express six different elements of the symmetric strain tensor in either cylindrical or spherical coordinates via three components of the displacement vector q in the same coordinates. A straightforward differentiation of the definitions of these curvilinear coordinates, similar to that used to derive the well-known expressions for spatial derivatives of arbitrary functions,4 yields, in particular, the following formulas for the diagonal elements of the tensor:
(i) in the cylindrical coordinates:
q
1
q
q
s
,
s
q
,
s
z
.
(7.16)
zz
z
(ii) in the spherical coordinates:
q
1
1
cos
1
r
q
q
s
,
s
,
.
(7.17)
rr
q
r
s
q q
r
r
r
r
sin
sin
These expressions, which will be used below for the solution of some problems for symmetrical geometries, may be a bit counter-intuitive. Indeed, Eq. (16) shows that even for a purely radial, axially-symmetric deformation, q = q()n, the angular element of the strain tensor does not vanish: s = q/.
(According to Eq. (17), in the spherical coordinates, both angular elements of the tensor exhibit the same property.) Note, however, that this relation describes a simple geometric fact: the change of the lateral distance d << between two close points at the same distance from the symmetry axis, at a small change of that keeps the angle d between the directions towards these two points intact.
7.2. Stress
Now let us discuss the forces that cause the strain – or, from a legitimate alternative point of view, are caused by the strain. Internal forces acting inside (i.e. between arbitrarily defined parts of) a 4 See, e.g., MA Eqs. (10.1)-(10.12).
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continuum may be also characterized by a tensor. This stress tensor, 5 with elements jj’, relates the Cartesian components of the vector dF of the force acting on an elementary area dA of an (in most cases, just imagined) interface between two parts of a continuum, to the components of the elementary vector dA = n dA normal to the area – see Fig. 3:
3
Stress
tensor
dF
dA .
(7.18)
j
jj' j'
j' 1
The usual sign convention here is to take the outer normal dn, i.e. to direct dA out of “our” part of the continuum, i.e. the part on which the calculated force dF is exerted – by the complementary part.
dA n dA
n
dF
interface
dA
“our” part
Fig. 7.3. The definition of vectors dA and dF.
of the continuum
In some cases, the stress tensor’s structure is very simple. For example, as will be discussed in detail in the next chapter, static and ideal fluids (i.e. liquids and gases) may only provide forces normal to any interface, and usually directed toward “our” part of the body, so that
Pressure
dF P
dA,
i.e.
P
,
(7.19)
jj'
jj'
where the scalar P (in most cases positive) is called pressure, and generally may depend on both the spatial position and time. This type of stress, with P > 0, is frequently called hydrostatic compression –
even if it takes place in solids, as it may.
However, in the general case, the stress tensor also has off-diagonal terms, which characterize the shear stress. For example, if the shear strain in Fig. 2 is caused by the shown pair of forces F, they create internal forces Fxn x, with Fx > 0 if we speak about the force acting upon a part of the sample below the imaginary horizontal interface we are discussing. To avoid a horizontal acceleration of each horizontal slice of the sample, the forces should not depend on y, i.e. Fx = const = F. Superficially, it may look that in this case, the only nonzero element of the stress tensor is dFx/ dAy = F/ A = const, so that tensor is asymmetric, in contrast to the strain tensor (15) of the same system. Note, however, that the displayed pair of forces F creates not only the shear stress but also a nonzero rotating torque = – Fhn z
= –( dFx/ dAy) Ahn z = –( dFx/ dAy) Vn z, where V = Ah is the sample’s volume. So, if we want to perform a static stress experiment, i.e. avoid the sample’s rotation, we need to apply some other forces, e.g., a pair of vertical forces creating an equal and opposite torque ’ = ( dFy/ dAx) Vn z, implying that dFy/ dAx =
dFx/ dAy = F/ A. As a result, the stress tensor becomes symmetric, and similar in structure to the symmetrized strain tensor (15):
5 It is frequently called the Cauchy stress tensor, partly to honor Augustin-Louis Cauchy who introduced this notion (and is responsible for the development, mostly in the 1820s, much of the theory described in this chapter), and partly to distinguish it from other possible definitions of the stress tensor, including the 1st and 2nd Piola-Kirchhoff tensors. For the small deformations discussed in this course, all these notions coincide.
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0
F / A 0
σ F / A
0
0 .
(7.20)
0
0
0
In many situations, the body may be stressed not only by forces applied to their surfaces but also by some volume-distributed ( bulk) forces dF = f dV, whose certain effective bulk density f. (The most evident example of such forces is gravity. If its field is uniform as described by Eq. (1.16), then f = g, where is the mass density.) Let us derive the key formula describing the summation of the interface and bulk forces. For that, consider again an elementary cuboid with sides drj parallel to the corresponding coordinate axes n j (Fig. 4) – now not necessarily the principal axes of the stress tensor.
( j' )
dF
dAj'
n j
n
( j' )
j"
dF
d( ( j' )
dF )
Fig. 7.4. Deriving Eq. (23).
dr
n
j'
j'
If elements jj’ of the tensor do not depend on position, the force dF( j’) acting on the j’-th face of the cuboid is exactly balanced by the equal and opposite force acting on its opposite face, because the vectors dA( j’) at these faces are equal and opposite. However, if jj’ is a function of r, then the net force d( dF( j’)) does not vanish. (In this expression, the first differential sign refers to the elementary shift drj’, while the second one, to the elementary area dAj’.) Using the expression jj’dAj’ for the j’th contribution to the sum (18), in the first order in dr the j th components of the vector d( dF( j’)) is
d(
( j' )
dF
) d
(7.21)
j
dA
jj'
j'
jj' dr dA
jj' dV ,
r
j'
j'
r
j'
j'
where the cuboid’s volume dV = drj’dAj’ evidently does not depend on the index j’. The addition of these force components for all three pairs of cuboid faces, i.e. the summation of Eqs. (21) over all three values of the upper index j’, yields the following relation for the j th Cartesian component of the net force exerted on the cuboid:
3
3
d( dF ) d( dF ( j' )
jj'
)
dV .
(7.22)
j
j
j' 1
1
j'
rj'
Since any volume may be broken into such infinitesimal cuboids, Eq. (22) shows that the space-varying stress is equivalent to a volume-distributed force dFef = fef dV, whose effective (not real!) bulk density fef has the following Cartesian components
3
jj'
f
,
(7.23) Euler-Cauchy
ef
j
principle
j' 1 rj'
so that in the presence of genuinely bulk forces dF = f dV, the densities fef and f just add up. This is the so-called Euler-Cauchy stress principle.
Let us use this addition rule to spell out the 2nd Newton law for a unit volume of a continuum: Chapter 7
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2q
f f .
(7.24)
2
ef
t
Using Eq. (23), the j th Cartesian component of Eq. (24) may be represented as
Continuum
2 q
3
dynamics:
j
jj'
.
(7.25)
2
f j
equation
t
'1
j
rj'
This is the key equation of the continuum’s dynamics (and statics), which will be repeatedly used below.
For the solution of some problems, it is also convenient to have a general expression for the work W of the stress forces at a virtual deformation q – understood in the same variational sense as the virtual displacements r in Sec. 2.1. Using the Euler-Cauchy principle (23), for any volume V of a medium not affected by volume-distributed forces, we may write6
3
3
3
W f q d r ( f )
3
jj'
3
q
d r
q
d r
(7.26)
j
j
.
ef
ef
j
j 1
r
V
j j
, ' 1
V
V
j'
Let us work out this integral by parts for a volume so large that the deformations qj on its surface are negligible. Then, swapping the operations of the variation and the spatial differentiation (just like it was done with the time differentiation in Sec. 2.1), we get
3
q
j
3
W
d r.
(7.27)
jj'
j, j' 1
r
V
j'
Assuming that the tensor jj’ is symmetric, we may rewrite this expression as
1 3
q
q
j
j
3
W
d r.
(7.28)
2
jj'
j'j
j j
r
, ' 1
r
V
j'
j'
Now, swapping indices j and j’ in the second expression, we finally get 3
1
q
q
3
W j j' 3
d r
3
σ s d r ,
(7.29)
jj'
jj'
2
jj'
jj'
j, j' 1
r
r
V
j'
j
j, j'
1 V
where sjj’ are the elements of the strain tensor (9b). It is natural to rewrite this important formula as 3
Work of
stress
W wr 3
d r,
where wr
s ,
(7.30)
jj'
j'j
forces
V
j, j' 1
and interpret the locally-defined scalar function w(r) as the work of the stress forces per unit volume, at a small variation of the deformation.
As a sanity check, for the pure pressure (19), Eq. (30) is reduced to the obviously correct result
W = – P V, where V is the volume of “our” part of the continuum.
6 Here the sign corresponds to the work of the “external” stress force dF exerted on “our” part of the continuum by its counterpart – see Fig. 3. Note that some texts make the opposite definition of W, leading to its opposite sign.
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7.3. Hooke’s law
In order to form a complete system of equations describing the continuum’s dynamics, one needs to complement Eq. (25) with an appropriate constitutive equation describing the relation between the forces described by the stress tensor jj’, and the deformations q described (in the small deformation limit) by the strain tensor sjj’. This relation depends on the medium, and generally may be rather complicated. Even leaving alone various anisotropic solids (e.g., crystals) and macroscopically-inhomogeneous materials (like ceramics or sand), strain typically depends not only on the current value of stress (possibly in a nonlinear way) but also on the previous history of stress application. Indeed, if strain exceeds a certain plasticity threshold, atoms (or nanocrystals) may slip to their new positions and never come back even if the strain is reduced. As a result, deformations become irreversible – see Fig. 5.
elastic
stress deformation
(reversible)
fracture
point
plastic
deformation
(irreversible)
Fig. 7.5. A typical relation between the
stress and strain in solids (schematically).
0
strain s
Only below the thresholds of nonlinearity and plasticity (which are typically close to each other), the strain is nearly proportional to stress, i.e. obeys the famous Hooke’s law.7 However, even in this elastic range, the law is not quite simple, and even for an isotropic medium is described not by one but by two constants, called the elastic moduli. The reason for that is that most elastic materials resist the strain accompanied by a volume change (say, the hydrostatic compression) differently from how they resist a shear deformation.
To describe this difference, let us first represent the symmetrized strain tensor (9b) in the following mathematically equivalent form:
1
1
s s
(7.31)
jj
jj
jj
jj
'
'
'
Tr s
'
Tr s .
3
3
According to Eq. (13), the traceless tensor in the first parentheses does not give any contribution to the volume change, e.g., may be used to characterize a purely shear deformation, while the second term describes the hydrostatic compression alone. Hence we may expect that the stress tensor may be represented (again, within the elastic deformation range only!) as
1
1
Hooke’s
2 s
Tr s
3 K
Tr s
,
(7.32) law via
jj '
jj'
jj'
jj'
3
3
and K
where K and are constants. (The inclusion of coefficients 2 and 3 into Eq. (32) is justified by the simplicity of some of its corollaries – see, e.g., Eqs. (36) and (41) below.) Indeed, experiments show that 7 Named after Robert Hooke (1635-1703), the polymath who was the first to describe the law in its simplest, 1D
version.
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Hooke’s law in this form is followed, at small strain, by all isotropic materials. In accordance with the above discussion, the constant (in some texts, denoted as G) is called the shear modulus, while the constant K (sometimes denoted B), the bulk modulus. The two left columns of Table 1 show the approximate values of these moduli for typical representatives of several major classes of materials.8
Table 7.1. Elastic moduli, density, and sound velocities of a few representative materials (approximate values) Material
K (GPa)
μ (GPa)
E (GPa)
(kg/m3) v
l (m/s)
v t (m/s)
Diamond(a)
600
450
1,100
0.20
3,500
1,830
1,200
Hardened steel
170
75
200
0.30
7,800
5,870
3,180
Water(b)
2.1
0
0
0.5
1,000
1,480
0
Air(b)
0.00010
0
0
0.5
1.2
332
0
(a) Averages over crystallographic directions (~10% anisotropy).
(b) At the so-called ambient conditions ( T = 20C, P = 1 bar 105 Pa).
To better appreciate these values, let us first discuss the quantitative meaning of K and , using two simple examples of elastic deformation. However, in preparation for that, let us first solve the set of nine (or rather six different) linear equations (32) for sjj’. This is easy to do, due to the simple structure of these equations: they relate the elements jj’ and sjj’ with the same indices, but the tensor’s trace effect. This slight complication may be readily overcome by noticing that according to Eq. (32), 3
1
Tr σ 3 K
.
(7.33)
jj
Tr s,
Tr
that
so
s
Tr σ
j 1
3 K
Plugging this result into Eq. (32) and solving it for sjj’, we readily get the reciprocal relation, which may be represented in a similar form:
1
1
1 1
s
Tr σ
Tr σ
.
(7.34)
jj'
jj '
jj'
jj'
2
3
3 K 3
Now let us apply Hooke’s law, in the form of Eqs. (32) or (34), to two simple situations in which the strain and stress tensors may be found without using the full differential equation of the elasticity theory and boundary conditions for them. (That will be the subject of the next section.) The first situation is the hydrostatic compression when the stress tensor is diagonal, and all its diagonal elements are equal – see Eq. (19).9 For this case, Eq. (34) yields
s
P
,
(7.35)
jj'
3
jj '
K
8 Since the strain tensor elements, defined by Eq. (9), are dimensionless, while the strain, defined by Eq. (18), has the dimensionality similar to pressure (of force per unit area), so do the elastic moduli K and .
9 It may be proved that such a situation may be implemented not only in a fluid with pressure P but also in a solid sample of an arbitrary shape, for example by placing it into a compressed fluid.
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i.e. regardless of the shear modulus, the strain tensor is also diagonal, with all diagonal elements equal.
According to Eqs. (11) and (13), this means that all linear dimensions of the body are reduced by a similar factor, so that its shape is preserved, while the volume is reduced by
Δ
3
V
s
P
.
(7.36)
V
jj
j 1
K
This formula clearly shows the physical sense of the bulk modulus K as the reciprocal compressibility. As Table 1 shows, the values of K may be dramatically different for various materials, and even for such “soft stuff” as water, this modulus is actually rather high. For example, even at the bottom of the deepest, 10-km ocean well ( P 103 bar 0.1 GPa), the water’s density increases by just about 5%. As a result, in most human-scale experiments, water may be treated as an incompressible fluid – the approximation that will be widely used in the next chapter. Many solids are even much less compressible – see, for example, the first two rows of Table 1.
Quite naturally, the most compressible media are gases. For a portion of gas, a certain background pressure P is necessary just for containing it within its volume V, so that Eq. (36) is only valid for small increments of pressure, P : V
P
.
(7.37)
V
K
Moreover, the compression of gases also depends on thermodynamic conditions. (In contrast, for most condensed media, the temperature effects are very small.) For example, at ambient conditions, most gases are reasonably well described by the equation of state called the ideal classical gas: Nk T
V
P Nk T
B
,
i.e. P
.
(7.38)
B
V
where N is the number of molecules in volume V, and k B 1.3810-23 J/K is the Boltzmann constant.10
For a small volume change V at a constant temperature T, this equation gives Nk T
B
P
V
P
V
V
P
,
i.e.
.
(7.39)
T const
V 2
V
V Tconst
P
Comparing this expression with Eq. (36), we get a remarkably simple result for the isothermal compression of gases,
K
,
(7.40)
T const
P
which means in particular that the bulk modulus listed in Table 1 is actually valid, at the ambient conditions, for almost any gas. Note, however, that the change of thermodynamic conditions (say, from isothermal to adiabatic11) may affect the compressibility of the gas..
Now let us consider the second, rather different, fundamental experiment: a purely shear deformation shown in Fig. 2. Since the traces of the matrices (15) and (20), which describe this situation, are equal to 0, for their off-diagonal elements, Eq. (32) gives merely jj’ = 2 sjj’, so that the deformation angle (see Fig. 2) is just
10 For the derivation and a detailed discussion of Eq. (37), see, e.g., SM Sec. 3.1.
11 See, e.g., SM Sec. 1.3.
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F
1
.
(7.41)
A
Note that the angle does not depend on the thickness h of the sample , though of course the maximal linear deformation qx = h is proportional to the thickness. Naturally, as Table 1 shows, μ = 0 for all fluids because they do not resist static shear stress.
However, not all situations, even apparently simple ones, involve just either K or . Let us consider stretching a long and thin elastic rod of a uniform cross-section of area A – the so-called tensile stress experiment shown in Fig. 6.12
l
F
F
A
Fig. 7.6. The tensile stress experiment.
z
Though the deformation of the rod near its clamped ends depends on the exact way forces F are applied (we will discuss this issue later on), we may expect that over most of its length the tension forces are directed virtually along the rod, dF = Fzn z, and hence, with the coordinate choice shown in Fig. 6, xj = yj = 0 for all j, including the diagonal elements xx and yy. Moreover, due to the open lateral surfaces, on which, evidently, dFx = dFy = 0, there cannot be an internal stress force of any direction, acting on any elementary internal boundary parallel to these surfaces. This means that zx =
zy = 0. So, of all elements of the stress tensor only one, zz, is not equal to zero, and for a uniform sample, zz = const = F/ A. For this case, Eq. (34) shows that the strain tensor is also diagonal, but with different diagonal elements:
1
1
s
,
(7.42)
zz
9 K 3
zz
1
1
s s
.
(7.43)
xx
yy
9 K 6
zz
Since the tensile stress is most common in engineering practice (including physical experiment design), both combinations of the elastic moduli participating in these two relations have earned their own names. In particular, the constant in Eq. (42) is usually denoted as 1/ E (but in many texts, as 1/ Y), where E is called Young’s modulus:13
1
1
1
9 K
Y
,
i.e. E
.
(7.44)
m
E
9 K
3
3 K
12 Though the analysis of compression in this situation gives similar results, in practical experiments a strong compression of a long sample may lead to the loss of the horizontal stability – the so-called buckling – of the rod.
13 Named after another polymath, Thomas Young (1773-1829) – somewhat unfairly, because his work on elasticity was predated by a theoretical analysis by L. Euler in 1727 and detailed experiments by Giordano Riccati in 1782.
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As Fig. 6 shows, in the tensile stress geometry szz qz/ z = l/ l, so that Young’s modulus scales the linear relation between the relative extension of the rod and the force applied per unit area:14
l
1 F
.
(7.45)
l
E A
The third column of Table 1 above shows the values of this modulus for two well-known solids: diamond (with the highest known value of E of all bulk materials15) and the steels (solid solutions of
~10% of carbon in iron) used in construction. Again, for all fluids, Young’s modulus equals zero – as it follows from Eq. (44) for = 0.
I am confident that most readers of these notes have been familiar with Eq. (42), in the form of Eq. (45), from their undergraduate studies. However, this can hardly be said about its counterpart, Eq.
(43), which shows that at the tensile stress, the rod’s cross-section dimensions also change. This effect is usually characterized by the following dimensionless Poisson’s ratio:16
s
s
xx
yy
1
1 1
1
1 3 K 2
.
Poisson
(7.46)
s
s
ratio
zz
zz
9 K 6
9 K 3 2 3 K
According to this formula, for realistic materials with K > 0, 0, may vary from (-1) to (+½), but for the vast majority of materials,17 its values are between 0 and ½ – see the corresponding column of Table 1. The lower limit of this range is reached in porous materials like cork, whose lateral dimensions almost do not change at the tensile stress. Some soft materials such as natural and synthetic rubbers present the opposite case: ½.18 Since according to Eqs. (13) and (42), the volume change is V
1 F
l
s s s
(7.47)
xx
yy
zz
1 2 1 2 ,
V
E A
l
such materials virtually do not change their volume at the tensile stress. The ultimate limit of this trend,
V/ V = 0, is provided by fluids and gases, because, as it follows from Eq. (46) with = 0, their Poisson’s ratio is exactly ½. However, for most practicable construction materials such as various steels (see Table 1) the relative volume change (47) is as high as ~40% of that of the length.
Due to the tensile stress dominance in practice, the coefficients E and are frequently used as a pair of independent elastic moduli, instead of K and . Solving Eqs. (44) and (46) for them, we get E
K
E
,
.
(7.48)
1
(
3 2 )
1
(
2 )
14 According to Eq. (47), E may be thought of as the force (per unit area) that would double the initial sample’s length, if only Hooke’s law was valid for deformations that large – as it typically isn’t.
15 E is probably somewhat higher (up to 2,000 GPa) in such nanostructures as carbon nanotubes and monatomic sheets ( graphene), though there is still substantial uncertainty in experimentally measured elastic moduli of these structures – for a review see, e.g., G. Dimitrios et al., Prog. Mater. Sci. 90, 75 (2017).
16 In some older texts, the Poisson’s ratio is denoted , but its notation as dominates modern literature.
17 The only known exceptions are certain exotic solids with very specific internal microstructure – see, e.g., R.
Lakes, Science 235, 1038 (1987) and references therein.
18 For example, silicone rubbers (synthetic polymers broadly used in engineering and physics experiment design) have, depending on their particular composition, synthesis, and thermal curing, = 0.470.49, and as a result combine respectable bulk moduli K = (1.52) GPa with very low Young’s moduli: E = (0.00010.05) GPa.
Chapter 7
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Using these formulas, the two (equivalent) formulations of Hooke’s law, expressed by Eqs. (32) and (34), may be rewritten as
E
s
(7.49a)
jj'
jj'
Tr s jj' ,
Hooke’s
1
1 2
law via
E and
1
s
Tr σ
.
(7.49b)
jj'
jj'
jj'
E
1
The linear relation between the strain and stress tensor in elastic continua enables one more step in our calculation of the potential energy U due to deformation, which was started at the end of Sec. 2.
Indeed, to each infinitesimal part of this strain increase, we may apply Eq. (30), with the elementary work W of the surface forces increasing the potential energy of “our” part of the body by the equal amount U. Let us slowly increase the deformation from a completely unstrained state (in which we may take U = 0) to a certain strained state, in the absence of bulk forces f, keeping the deformation type, i.e. the relation between the elements of the stress tensor, intact. In this case, all elements of the tensor
jj’ are proportional to the same single parameter characterizing the stress (say, the total applied force), and according to Hooke’s law, all elements of the tensor sj’j are proportional to that parameter as well. In this case, integration of Eq. (30) through the variation yields the following final value:19
Elastic
3
1
deformation
U ur 3
d r,
ur
energy
s .
(7.50)
2
jj'
j'j
V
j, j' 1
Evidently, this u(r) may be interpreted as the volumic density of the potential energy of the elastic deformation.
7.4. Equilibrium
Now we are fully equipped to discuss the elastic deformation dynamics, but let us start with statics. The static (equilibrium) state may be described by requiring the right-hand side of Eq. (25) to vanish. To find the elastic deformation, we need to plug jj’ from Hooke’s law (49a), and then express the elements sjj’ via the displacement distribution – see Eq. (9). For a uniform material, the result is20
2
3
2
3
E
q
E
q
j
j'
f 0.
(7.51)
2 1
( )
2
j
j'
r
r r
j
j'
1
2 1
(
1
)(
2 )
'
1
j
j'
Taking into account that the first sum is just the j th component of 2q, while the second sum is the j th component of (q), we see that all three equations (51) for three Cartesian components ( j = 1, 2, and 3) of the deformation vector q, may be conveniently merged into one vector equation Elastic
E
E
continuum:
2
q
( q) f .
0
(7.52)
equilibrium
1
(
2 )
1
(
2 1
)( 2 )
19 To give additional clarity to the arising factor ½, let me spell out this integration for the simple case of a 1D
spring. In this case, Eq. (30) is reduced to U = W = F x, and if the spring’s force is elastic, F = x, the integration over x from 0 to its final value yields U = x 2/2 Fx/2.
20 As it follows from Eqs. (48), the coefficient before the first sum in Eq. (51) is just the shear modulus , while that before the second sum is equal to ( K + /3).
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For some applications, it is more convenient to recast this equation into a different form, using the well-known vector identity21 2q = (q) – (q). The result is
E 1
( )
E
( q)
( q) f .
0
(7.53)
1
( 1
)( 2 )
1
(
2 )
It is interesting that in problems without volume-distributed forces (f = 0), Young’s modulus E
cancels out. Even more fascinating, in this case, the equation may be re-written in a form not involving the Poisson’s ratio either. Indeed, calculating the divergence of the remaining terms of Eq. (53), taking into account MA Eqs. (9.2) and (11.2), we get a surprisingly simple equation
2
( q) .
0
(7.54)
A natural question here is how the elastic moduli affect the deformation distribution if they do not participate in the differential equation describing it. The answer is different in the following two cases. If what is fixed at the body’s boundary are deformations, then the moduli are irrelevant, because the deformation distribution through the body does not depend on them. On the other hand, if the boundary conditions describe fixed stress (or a combination of stress and strain), then the elastic constants creep into the solution via the recalculation of these conditions into the strain. As a simple but representative example, let us calculate the deformation distribution in a (generally, thick) spherical shell under the effect of pressures inside and outside it – see Fig. 7a.
(a)
(b)
R 2
P
P
2
2
R
t
1
P
1
P
R 1
Fig. 7.7. The spherical shell
problem: (a) the general case, and
(b) the thin shell limit.
Due to the spherical symmetry of the problem, the deformation is obviously spherically-symmetric and radial, q(r) = q( r)n r, i.e. is completely described by one scalar function q( r). Since the curl of such a radial vector field is zero,22 Eq. (53) is reduced to
( q) ,
0
(7.55)
This means that the divergence of the function q( r) is constant within the shell. In the spherical coordinates:23
1 d
2 rq
(7.56)
2
const
.
r dr
Naming this constant 3 a (with the numerical factor chosen just for the later notation’s convenience), and integrating Eq. (56) over r, we get its solution,
21 See, e.g., MA Eq. (11.3).
22 If this is not immediately evident, please have a look at MA Eq. (10.11) with f = fr( r)n r.
23 See, e.g., MA Eq. (10.10) with f = q( r)n r
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b
q( r) ar
,
(7.57)
2
r
which also includes another integration constant, b. The constants a and b may be determined from the boundary conditions. Indeed, according to Eq. (19),
P , at r R ,
(7.58)
rr
1
1
P , at r R .
2
2
In order to relate this stress to strain, let us use Hooke’s law, but for that, we first need to calculate the strain tensor components for the deformation distribution (57). Using Eqs. (17), we get
q
b
q
b
s
a 2
,
s s
a
,
(7.59)
rr
3
3
r
r
r
r
so that Tr (s) = 3 a. Plugging these relations into Eq. (49a) for rr, we obtain E
b
2
3 .
(7.60)
rr
a
a
3
1
r 1 2
Now plugging this relation into Eqs. (58), we get a system of two linear equations for the coefficients a and b. An easy solution to this system yields
1 2
3
3
P R P R
1 ( P P ) 3 3
R R
1 1
2
2
a
,
1
2
1
2
b
.
(7.61)
3
3
E
R R
2
3
3
E
R R
2
1
2
1
Formulas (57) and (61) give a complete solution to our problem. (Note that the elastic moduli are back, as was promised.) This solution is rich in physical content and deserves at least some analysis.
First of all, note that according to Eq. (48), the coefficient (1 – 2)/ E in the expression for a is just 1/3 K, so that the first term in Eq. (57) for the net deformation describes the hydrostatic compression. Now note that the second of Eqs. (61) yields b = 0 if R 1 = 0. Thus for a solid sphere, we have only the hydrostatic compression that was discussed in the previous section. Perhaps less intuitively, making two pressures equal also gives b = 0, i.e. the purely hydrostatic compression, for arbitrary R 2 > R 1.
However, in the general case, b 0, so that the second term in the deformation distribution (57), which describes the shear deformation,24 is also substantial. In particular, let us consider the important thin-shell limit, when R 2 – R 1 t << R 1,2 R – see Fig. 7b. In this case, q( R 1) q( R 2) is just the change of the shell radius R, for which Eqs. (57) and (61) (with R 3
3
2 – R 1 3 R 2 t) give
b
P P
1
2 R 2
1 2
1
R 2 1
R
Δ q( R) aR
P P
.
(7.62)
2
1
2
R
t
3
E
2 E
t 2 E
Naively, one could think that at least in this limit the problem could be analyzed by elementary means. For example, the total force exerted by the pressure difference ( P 1 – P 2) on the diametrical cross-section of the shell (see, e.g., the dashed line in Fig. 7b) is F = πR 2( P 1 – P 2), giving the stress, 24 Indeed, according to Eq. (48), the material-dependent factor in the second of Eqs. (61) is just 1/4.
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F
R 2 P
1
P 2
R
P
,
(7.63)
1
P 2
A
2 Rt
t
2
directed along the shell’s walls. One can check that this simple formula may be indeed obtained, in this limit, from the strict expressions for and , following from the general treatment carried out above. However, if we now tried to continue this approach by using the simple relation (45) to find the small change Rsrr of the sphere’s radius, we would arrive at a result with the general structure of Eq.
(62), but without the factor (1 – ) < 1 in the numerator. The reason for this error (which may be as significant as ~30% for typical construction materials – see Table 1) is that Eq. (45), while being valid for thin rods of arbitrary cross-section, is invalid for thin but broad sheets, and in particular the thin shell in our problem. Indeed, while at the tensile stress both lateral dimensions of a thin rod may contract freely, in our last problem all dimensions of the shell are under stress – actually, under much more tangential stress than the radial one.25
7.5. Rod bending
The general approach to the static deformation analysis, outlined at the beginning of the previous section, may be simplified not only for symmetric geometries but also for the uniform thin structures such as thin plates (also called “membranes” or “thin sheets”) and thin rods. Due to the shortage of time, in this course, I will demonstrate typical approaches to such systems only on the example of thin rods.
(The theory of thin plates and shells is conceptually similar but mathematically more involved.26) Besides the tensile stress analyzed in Sec. 3, the two other major types of rod deformation are bending and torsion. Let us start from a “local” analysis of bending caused by a pair of equal and opposite external torques = n y y perpendicular to the rod axis z (Fig. 8), assuming that the rod is
“quasi-uniform”, i.e. that on the scale of this analysis (comparable with the linear scale a of the cross-section) its material parameters and the cross-section A do not change substantially.
(a)
d
(b)
y
x
x
dz dqz
y
y
0
z
0
dz
Fig. 7.8. Rod bending, in a local reference frame (specific for each cross-section). The bold arrows show the simplest way to create the two opposite torques
y: a couple of opposite forces for each torque.
Just as in the tensile stress experiment (Fig. 6), the components of the stress forces dF, normal to the rod’s length, have to equal zero on the surface of the rod. Repeating the arguments made for the tensile stress discussion, we have to conclude that only one diagonal element of the tensor (in Fig. 8, zz) may differ from zero:
25 Strictly speaking, this is only true if the pressure difference is not too small, namely, if P 1 – P 2 >> P 1,2 t/ R.
26 For its review see, e.g., Secs. 11-15 in L. Landau and E. Lifshitz, Theory of Elasticity, 3rd ed., Butterworth-Heinemann, 1986.
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.
(7.64)
jj '
jz
zz
However, in contrast to the tensile stress, at pure static bending, the net force directed along the rod has to vanish:
2
F d r
,
0
(7.65)
z
zz
S
where S is the rod’s cross-section, so that zz has to change its sign at some point of the x-axis, selected to lie in the plane of the bent rod. Thus, the bending deformation may be viewed as a combination of a stretch of some layers of the rod (bottom layers in Fig. 8) with compression of other (top) layers.
Since it is hard to make more conclusions about the stress distribution immediately, let us turn over to strain, assuming that the rod’s cross-section is virtually constant over the length of our local analysis. From the above representation of bending as a combination of stretching and compression, it is evident that the longitudinal deformation qz has to vanish along some neutral line on the rod’s cross-section – in Fig. 8, represented by the dashed line.27 Selecting the origin of the x-coordinate on this line, and expanding the relative deformation in the Taylor series in x, due to the cross-section smallness we may keep just the first, linear term of the expansion:
dq
x
s
z
.
(7.66)
zz
dz
R
The constant R has the sense of the curvature radius of the bent rod. Indeed, on a small segment dz, the cross-section turns by a small angle d y = – dqz/ x (Fig. 8b). Using Eq. (66), we get d y = dz/ R, which is the usual definition of the curvature radius R in the differential geometry, for our special choice of the coordinate axes.28
Expressions for other elements of the strain tensor are harder to guess (like at the tensile stress, not all of them are equal to zero!), but what we already know about zz and szz is already sufficient to start formal calculations. Indeed, plugging Eq. (64) into Hooke’s law in the form (49b), and comparing the result for szz with Eq. (66), we find
x
E .
(7.67)
zz
R
From the same Eq. (49b), we could also find the transverse elements of the strain tensor, and conclude that they are related to szz exactly as at the tensile stress:
s s
s ,
(7.68)
xx
yy
zz
and then, integrating these relations along the cross-section of the rod, find the deformation of the cross-section’s shape. More important for us, however, is to calculate the relation between the rod’s curvature and the net torque acting on a given cross-section S (taking dAz > 0): E
EI
(r dF)
2
2
2
x
d r
x d r
y
,
(7.69)
y
y
zz
R
R
S
S
S
27 Strictly speaking, that dashed line is the intersection of the neutral surface (the continuous set of such neutral lines for all cross-sections of the rod) with the plane of the drawing.
28 Indeed, for ( dx/ dz)2 << 1, the general formula MA Eq. (4.3) for the curvature (with the appropriate replacements f x and x z) is reduced to 1/ R = d 2 x/ dz 2 = d( dx/ dz)/ dz = d(tan y)/ dz d y/ dz.
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where Iy is a geometric constant defined as
I x 2 dxdy .
(7.70)
y
S
Note that this factor, defining the bending rigidity of the rod, grows as fast as a 4 with the linear scale a of the cross-section.29
In these expressions, x has to be referred to the neutral line. Let us see where exactly this line passes through the rod’s cross-section. Plugging the result (67) into Eq. (65), we get the condition defining the neutral line:
xdxdy
.
0
(7.71)
S
This condition allows for a simple interpretation. Imagine a thin sheet of some material, with a constant mass density per unit area, cut in the form of the rod’s cross-section. If we place a reference frame into its center of mass, then, by its definition,
dxdy
r
.
0
(7.72)
S
Comparing this condition with Eq. (71), we see that one of the neutral lines has to pass through the center of mass of the sheet, which may be called the “center of mass of the cross-section”. Using the same analogy, we see that the integral Iy given by Eq. (72) may be interpreted as the moment of inertia of the same imaginary sheet of material, with formally equal to 1, for its rotation about the neutral line
– cf. Eq. (4.24). This analogy is so convenient that the integral is usually called the moment of inertia of the cross-section and denoted similarly – just as has been done above. So, our basic result (69) may be rewritten as
Rod
1
y
.
(7.73) bending:
R
EI
curvature
y
vs. torque
This relation is only valid if the deformation is small in the sense R >> a. Still, since the deviations of the rod from its unstrained shape may accumulate along its length, Eq. (73) may be used for calculations of large “global” deviations of the rod from equilibrium, on a length scale much larger than a. To describe such deformations, Eq. (73) has to be complemented by conditions of the balance of the bending forces and torques. Unfortunately, a general analysis of such deformations requires a bit more differential geometry than I have time for, so I will only discuss this procedure for the simplest case of relatively small transverse deviations q qx of an initially horizontal rod from its straight shape that will be used for the z-axis (Fig. 9a), by some forces, possibly including bulk-distributed forces f =
n xfx( z). (Again, the simplest example is a uniform gravity field, for which fx = – g = const.) Note that in the forthcoming discussion the reference frame will be global, i.e. common for the whole rod, rather than local (pertaining to each cross-section) as it was in the previous analysis – cf. Fig. 8.
First of all, we may write a static relation for the total vertical force F = n xFx( z) exerted on the part of the rod to the left of the considered cross-section – located at point z. The differential form of this relation expresses the balance of vertical forces exerted on a small fragment dz of the rod (Fig. 9a), necessary for the absence of its linear acceleration: Fx( z + dz) – Fx( z) + fx( z) Adz = 0, giving 29 In particular, this is the reason why the usual electric wires are made not of a solid copper core, but rather a twisted set of thinner sub-wires, which may slip relative to each other, increasing the wire flexibility.
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Essential Graduate Physics
CM: Classical Mechanics
(a)
(b)
F( z dz)
x
F
f ( z) Adz
q 0
0
0
τ 0
( z)
y
( z)
y
- F( z)
F 0
q( z)
q( z dz)
F n F
x
0
q 0
τ 0
τ
0
z
z dz
Fig. 7.9. A global picture of rod bending: (a) the forces actin
g on a sm
all fragm
ent of a rod, and (b) two
bending problem examples, each with two typical but different boundary conditions.
dF
x f A ,
(7.74)
dz
x
where A is the cross-section’s area. Note that this vertical component of the internal forces has been neglected in our derivation of Eq. (73), and hence our final results will be valid only if the ratio Fx/ A is much smaller than the magnitude of zz described by Eq. (67). However, in reality, these are exactly the forces that create the very torque = n y y that in turn causes the bending, and thus have to be taken into account in the analysis of the global picture.
Such an account may be made by writing the balance of the components of the elementary torque exerted on the same rod fragment of length dz, necessary for the absence of its angular acceleration: d y + Fxdz = 0, so that
d
y F .
(7.75)
x
dz
These two equations should be complemented by two geometric relations. The first of them is d y/ dz = 1/ R, which has already been discussed above. We may immediately combine it with the basic result (73) of our local analysis, getting:
d
y
y
.
(7.76)
dz
EI y
The final equation is the geometric relation evident from Fig. 9a:
dq
x ,
(7.77)
y
dz
which is (as all expressions of our simple analysis) only valid for small bending angles, y << 1.
The four differential equations (74)-(77) are sufficient for the full solution of the weak-bending problem, if complemented by appropriate boundary conditions. Figure 9b shows the conditions most frequently met in practice. Let us solve, for example, the problem shown on the top panel of Fig. 9b: bending of a rod, “clamped” at one end (say, immersed into a rigid wall), under its own weight. As should be clear from their derivation, Eqs. (74)-(77) are valid for any distribution of parameters A, E, Iy, and over the rod’s length, provided that the rod is quasi-uniform, i.e. its parameters’ changes are so slow that the local relation (76) is still valid at any point. However, just for simplicity, let us consider a Chapter 7
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uniform rod. The simple structure of Eqs. (74)-(77) allows for their integration one by one, each time using the appropriate boundary conditions. To start, Eq. (74) with fx = – g = const yields F gAz const
(7.78)
x
gA z l,
where the integration constant has been selected to satisfy the right-end boundary condition: Fx = 0 at z
= l. As a sanity check, at the left wall ( z = 0), Fx = – gAl = – mg, meaning that the whole weight of the rod is exerted on the supporting wall – fine.
Next, by plugging Eq. (78) into Eq. (75) and integrating, we get
gA
gA
gA
,
(7.79)
y
2 z 2 lz const
2
2
z
2 lz l
z l2
2
2
2
where the integration constant’s choice ensures the second right-boundary condition: y = 0 at z = l – see Fig. 9b again. Now proceeding in the same fashion to Eq. (76), we get
gA ( z l)3
gA
const
(7.80)
y
( z l)3 3 l,
2 EI
3
6 EI
y
y
where the integration constant is selected to satisfy the clamping condition at the left end of the rod: y
= 0 at z = 0. (Note that this is different from the support condition illustrated on the lower panel of Fig.
9b, which allows the angle y to be different from zero at z = 0, but requires the torque to vanish at that point.) Finally, integrating Eq. (77) with y given by Eq. (80), we get the rod’s global deformation law, gA
( z l)4
gA
z l
l
3
(
)4
4
q ( z)
l z const
3
l z
,
(7.81)
x
6 EI
4
6 EI
4
4
y
y
where the integration constant is selected to satisfy the second left-boundary condition: q = 0 at z = 0.
So, the bending law is sort of complicated even in this very simple problem. It is also remarkable how fast does the end’s displacement grow with the increase of the rod’s length:
4
gAl
q ( l)
.
(7.82)
x
8 EI y
To conclude this solution, let us discuss the validity of this result. First, the geometric relation (77) is only valid if y ( l) << 1, and hence if qx( l) << l. Next, the local formula Eq. (76) is valid if 1/ R
= ( l)/ EIy << 1/ a ~ A-1/2. Using the results (79) and (82), we see that the latter condition is equivalent to
qx( l) << l 2/ a, i.e. is weaker than the former one, because all our analysis has been based on the assumption l >> a. Another point of concern may be that the off-diagonal stress element xz ~ Fx/ A, which is created by the vertical gravity forces, has been ignored in our local analysis. For that approximation to be valid, this element must be much smaller than the diagonal element zz ~ aE/ R =
a/ Iy taken into account in that analysis. Using Eqs. (78) and (80), we are getting the following estimates: xz ~ gl, zz ~ a gAl 2/ Iy ~ a 3 gl 2/ Iy. According to its definition (70), Iy may be crudely estimated as a 4, so that we finally get the simple condition a << l, which has been assumed from the very beginning of our solution.
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7.6. Rod torsion
One more class of analytically solvable elasticity problems is the torsion of quasi-uniform, straight rods by a couple of axially-oriented torques = n z z – see Fig. 10.
F
F
τ
τ
z
Fig. 7.10. Rod torsion. Just as
0
z
in Fig. 8, the couples of forces
F are just vivid representations
of the opposite torques .
F
F
This problem is simpler than the bending in the sense that due to its longitudinal uniformity, d z/ dz = const, it is sufficient to relate the torque z to the so-called torsion parameter d
z
.
(7.83)
dz
If the deformation is elastic and small (in the sense a << 1, where a is again the characteristic size of the rod’s cross-section), is proportional to z. Hence our task is to calculate their ratio, Torsional
rigidity:
C
z
z
,
(7.84)
definition
d / dz
z
called the torsional rigidity of the rod.
As the first guess (as we will see below, of a limited validity), one may assume that the torsion does not change either the shape or size of the rod’s cross-sections, but leads just to their mutual rotation about a certain central line. Using a reference frame with the origin on that line, this assumption immediately enables the calculation of Cartesian components of the displacement vector dq, by using Eq. (6) with d = n zd z:
dq yd ydz
,
dq xd xdz
,
dq .
0
(7.85)
x
z
y
z
z
From here, we can calculate all Cartesian elements (9) of the symmetrized strain tensor:
s s s ,
0
s s ,
0
s s
y, s s
x .
(7.86)
xx
yy
zz
xy
yx
xz
zx
yz
zy
2
2
The first of these equalities means that the elementary volume does not change, i.e. we are dealing with purely shear deformation. As a result, all nonzero elements of the stress tensor, calculated from Eqs.
(32), are proportional to the shear modulus alone:30
,
0
,
0
y
,
.
x
(7.87)
xx
yy
zz
xy
yx
xz
zx
yz
zy
30 Note that for this problem, with a purely shear deformation, using the alternative elastic moduli E and would be rather unnatural. If needed, we may always use the second of Eqs. (48): = E/2(1 + ).
Chapter 7
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Now it is straightforward to use this result to calculate the full torque as an integral over the cross-section’s area A:
r F
(7.88)
z
d
( xdF
ydF )
( x
y
)
.
dxdy
z
y
x
yz
xz
A
A
A
Using Eq. (87), we get z = Iz, i.e.
C I ,
where I ( x 2
y 2 ) dxdy .
(7.89) C at axial
z
z
symmetry
A
Again, just as in the case of thin rod bending, we have got an integral, in this case Iz, similar to a moment of inertia, this time for the rotation about the z-axis passing through a certain point of the cross-section. For any axially-symmetric cross-section, this has to be its central point. Then, for example, for the practically important case of a uniform round pipe with internal radius R 1 and external radius R 2, Eq. (89) yields
2
R
3
C 2 d
4
4
R R .
(7.90)
2
1
2
1
R
In particular, for the solid rod of radius R (which may be treated as a pipe with R 1 = 0 and R 2 = R), this result gives the following torsional rigidity
4
C
R
,
(7.91a)
2
while for a hollow pipe of small thickness t << R, Eq. (90) is reduced to C
R 3
2 t .
(7.91b)
Note that per unit cross-section area A (and hence per unit mass of the rod) the thin pipe’s rigidity is twice higher than that of a solid rod:
C
C
1
2
2
R
R
.
(7.92)
pipe
round
thin
rod
round
solid
A
A
2
This fact is one reason for the broad use of thin pipes in engineering and physical experiment design.
However, for rods with axially-asymmetric cross-sections, Eq. (89) gives wrong results. For example, for a narrow rectangle of area A = w t with t << w, it yields the expression C = tw 3/12
[WRONG!], which is even functionally different from the correct result – cf. Eq. (104) below. The reason for this error is that the above analysis does not describe possible bending qz( x, y) of the rod’s cross-section in the direction along the rod. (For axially-symmetric rods, such bending is evidently forbidden by the symmetry, so that Eq. (89) is valid, and the results (90)-(92) are absolutely correct.) Let us describe31 this counter-intuitive effect by taking
q ( x, y),
(7.93)
z
31 I would not be terribly shocked if the reader skipped the balance of this section at the first reading. Though the calculation described in it is very elegant, instructive, and typical for the theory of elasticity (and for good physics as a whole!), its results will not be used in other chapters of this course or other parts of this series.
Chapter 7
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CM: Classical Mechanics
(where is some function to be determined), but still keeping Eq. (87) for two other components of the displacement vector. The addition of does not perturb the equality to zero of the diagonal elements of the strain tensor, as well as of sxy and syx, but contributes to other off-diagonal elements:
s s
(7.94)
xz
zx
y
,
s s
x
,
2
x
yz
zy
2
y
and hence to the corresponding elements of the stress tensor:
(7.95)
xz
zx
y
,
x
.
yz
zy
x
y
Now let us find the requirement imposed on the function ( x,y) by the fact that the stress force component parallel to the rod’s axis,
dA
x
dAy
dF dA dA
dA
,
(7.96)
z
zx
x
zy
y
y
x
x dA
y dA
has to vanish at the rod’s surface(s), i.e. at a cross-section’s border. The coordinates { x, y} of any point at the border may be considered as unique functions, x( l) and y( l), of the arc l of that line – see Fig. 11.
dAy
dA
y
dy
dA
x
l
0
dx
