## C. Julian Chen

Print publication date: 2007

Print ISBN-13: 9780199211500

Published to Oxford Scholarship Online: September 2007

DOI: 10.1093/acprof:oso/9780199211500.001.0001

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# (p.389) Appendix E

## Elementary Elasticity Theory

Source:
Introduction to Scanning Tunneling Microscopy
Publisher:
Oxford University Press

(p.389) Appendix E

Elementary Elasticity Theory

Excellent treatises and textbooks on elasticity are abundant, for example, Landau and Lifshitz [353], Timoshenko and Goodier [354], and Timoshenko and Young [355]. It takes a lot of time to read and find useful information. This appendix contains an elementary treatment of elasticity relating STM and AFM.

# E.1 Stress and strain

Imagine that a small bar along the z direction is isolated from a solid body, as shown in Fig. E.1. The force in the x direction per unit area on the x facet is denoted as σ x, a normal stress:

(E.1)
$Display mathematics$

A body under stress deforms. In other words, a strain is generated. With a normal stress σ x, the bar elongates in the x direction. The standard notation to describe strain is by introducing displacements, u, v, and w, in the x, y, and z directions, respectively, as shown in Fig. E.1. The dimensionless quantity

(E.2)
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Fig. E.1. Normal stress and normal strain (a) The normal force per unit area in the x direction is the x component of the stress tensor. (b) The normal stress causes an elongation in the x direction and a contraction in the y and z directions.

(p.390)

Fig. E.2. Shear stress and shear strain (a) The shear force per unit area is a component of the stress tensor. (b) The shear stress causes a shear strain.

is called the unit elongation, which is a component of the strain tensor. Hooke’s law says that the unit elongation is proportional to the normal stress in the same direction:

(E.3)
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where the quantity E is the Young’s modulus.

Under the same stress σ x, the y and z dimensions of the bar contract:

(E.4)
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(E.5)
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This effect was discovered by Poisson, and the dimensionless constant ν is called Poisson’s ratio. For most materials, ν ≈ 0.25.

The x component of the force per unit area on a facet in the y direction is denoted as Ƭxy, and is called a component of the shear stress:

(E.6)
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The condition of equilibrium, that is, the absence of a net torque on a small volume element, requires that Ƭxy = Ƭyx. A shear stress causes a shear strain, defined as (see Fig. E.2)

(E.7)
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The relation between shear stress and shear strain can be established based on the relation between normal stress and normal strain, in Eqs E.3 and E.4. Actually, by rotating the coordinate system 45°, it becomes a problem of normal stress and normal strain. Using geometrical arguments, it can be shown that (see, for example, [354]):

(p.391)

(E.8)
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The quantity G ≡ E/2(1+ν) is called the shear modulus.

# E.2 Small deflection of beams

The deflection of a beam under a vertical distribution of force, as shown in Fig. E.3, is a very basic problem in the theory of elasticity. Under the condition that the deflection is small and there is no force parallel to the main axis, the deflection u as a function of distance z satisfies a simple differential equation, as we will show in this section.

To start with, we show two simple relations in statics; see Fig. E.3. First, consider the equilibrium of the net force in the vertical direction. At each cross section of the beam, there is a vertical force V(z), which should compensate the external force F(z):

dV(z) ≡ V(z + dz) − V(z) = −F(z)dz     (E.9)

Second, at each cross section of the beam, the normal stress forms a torque M(z). The equilibrium condition of a small section of the beam dz with respect to rotation requires (see Fig. E.3)

dM(z) ≡ M(z + dz) − M(z) = V(z)dz     (E.10)

Combining Eqs E.9 and E.10, we have

(E.11)
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Under the influence of a torque, the beam deforms. In other words, the slope tanθ changes with z, as shown in Eq. E.4. For small deflections, tanθ ≈ θ. In other words,

(E.12)
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Consider a small section Δz of the beam. The total force horizontal must be zero. For symmetrical cross sections, such as rectangles, circular bars, and tubes, symmetry conditions require that the neutral line of force must be in the median plane, denoted as x = 0 (see Fig. E.4). The distribution of normal strain is then

(E.13)
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(p.392)

Fig. E.3. Bending of a beam Under the action of a vertical force F distributed along the axis of the beam, the beam bends. The deflection u as a function of position z is determined by the force F(z). (a). the equilibrium of a small section dz of the beam respect to the force in the vertical direction. (b). the equilibrium of a small section dz of the beam with respect to the torque.

Using Eq. E.3 and neglecting the contraction of the width A(x), the total torque acting on a cross section is

(E.14)
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The quantity I is the moment of inertia of the cross section. For simple shapes, the integral can be evaluated easily, which is left as an exercise. For a rectangular bar of width b and height h,

(E.15)
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For a rod of diameter D, it is

(E.16)
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(E.17)
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Combining Eqs E.10 and E.14, we obtain a differential equation for the deflection u(z),

(E.18)
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(p.393)

Fig. E.4. Deformation of a segment of a beam Under the influence of a torque acting on a cross section, a beam bends. For small deformations, the slope is θ = du(z)/dz. The change of slope with distance /dz is connected with a strain distribution in the beam, Δwz. The strain is connected with a distribution of normal stress σ z in the beam. The total torque is obtained by integration over the cross section of the beam.

We will give an example of a concentrated force W acting at a point on the beam, z = L, which is clamped at the origin, z = 0. Thus, F(z) = W δ(z − L). Integrating Eq. E.18 once, we find

(E.19)
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Integrating it again and using the condition that, at z = L, the torque is zero, d 2 u(z)/dz 2 = 0, we get

(E.20)
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Integrating twice and using the condition that, at z = 0, u = 0 and du/dz = 0, we obtain

(E.21)
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(E.22)
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The deflection at z = L is

(E.23)
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The angle at z = L is

(E.24)
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(p.394)

Fig. E.5. Vibration of a beam. One end of the beam is clamped; the displacement and the slope are zero. Another end of the beam is free; the force and the torque are zero. The lowest resonance frequency, which corresponds to a vibration without a node, is determined by Eq. E.32.

# E.3 Vibration of beams

If a deflection exists in a beam in the absence of external force, the elastic force will cause an acceleration in the vertical direction. From Eq. E.18, using Newton’s second law, the equation of motion of the beam is

(E.25)
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where ρ is the density of the material, and A is the cross-sectional area of the beam. For a sinusoidal vibration, u(z, t) = u(z) cos(ωt+α), Eq. E.25 becomes

(E.26)
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Denoting

(E.27)
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the general solution of Eq. E.26 is

u(z) = A cos kz + B sin kz + c coshkz + D sinh kz,     (E.28)

which can be verified by direct substitution. The most important case in STM and AFM is the vibration of a beam with one end clamped and another end free; see Fig. E.5. At the clamped end, z = 0, the boundary conditions are

(E.29)
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At the free end, z = L, the vertical force V and the torque M vanish. The boundary conditions are

(p.395)

(E.30)
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By going through simple algebra (we leave it as an exercise for the reader), the vibration frequency is found to be determined by the equation

cos kL cosh kL + 1 = 0.     (E.31)

Denoting the solutions of Eq. E.31 as kL = r n, where n = 0, 1, 2, 3, …, the first three are

r 0 = 1.875, r 1 = 4.694, r 2 = 7.854     (E.32)

The solutions with n > 2 are almost exactly

(E.33)
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The resonance frequencies of the cantilever is determined by

(E.34)
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The lowest resonance frequency is

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# E.4 Torsion

An important issue in the theory of elasticity related to STM and AFM is the shearing stress and the angles of twist of circular bars and tubes subjected to twisting torque T, as shown in Fig. E.6. Consider a small section of a circular bar with length L and radius r. If the twist angle θ per unit length is not too large, every cross section perpendicular to the axis remains unchanged. A small, initially rectangular volume element at radius ρ will have a shear strain

(E.36)
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The shear stress is, according to Eq. E.8,

(E.37)
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(p.396)

Fig. E.6. Torsion of a circular bar. (a) Under the action of a torque T, a bar of length L is twisted through an angle θ. (b) The twist results in a distribution of shear strain γ in a cross section, which generates a distribution of shear stress Ƭ.

The torque acting on a ring of radius ρ and width is

(E.38)
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By integrating over the entire area, the total torque is

(E.39)
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In terms of diameter d = 2r, the torque is

(E.40)
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Another important case is the torsion of a long thin plate, such as a rectangular cantilever – see Fig. E.7. If the thickness h is much smaller than the width b, the torsional rigidity is,

(E.41)
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Similar to the discussion in the previous section, the frequency of torsional vibration of a rectangular cantilever can be determined as follows. The moment of inertia around the z axis is, see Fig. E.7,

(E.42)
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The velocity of the elastic wave is

(E.43)
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(p.397)

Fig. E.7. Torsion and torsional vibration of a rectangular cantilever (a) By applying a torque on both ends of a rectangular cantilever, a twist of angle θ is generated. (b) Torsional vibration of a rectangular cantilever, and the evaluation of its moment of inertia.

If one end of the cantilever is clamped and the other end is free, the wavelength of the standing wave is n + ¼ times the length of the cantilever, L. The resonance frequencies are

(E.44)
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# E.5 Helical springs

Helical springs are common elements of many mechanical devices. The important parameters of a helical spring are: wire diameter r, coil diameter (from an axis of the wire to another axis) D, number of coils n, and the modulus of elasticity of shear of the material G. To provide an understanding of the helical springs, we give a simple derivation of the total stretch f and the maximum shear stress τ max as a function of the axial load P acting on the spring. The stretch f of a spring is the increase of the pitch h times the number of coils n. To simplify the derivation, we imagine that every coil in the spring is a flat ring at rest; that is, the pitch at rest is zero. Because all the quantities are linear in the pitch, an incremental value of pitch results in an incremental value of axial load and maximum shear stress. When the pitch is increased by h = f/n, the cross section of the wire is twisted by an angle ø = f/2nD. For every one half of a coil, the length of wire is

(E.45)
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which has a total twist angle (see Fig. E.8)

(E.46)
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(p.398)

Fig. E.8. Stiffness of a helical spring (a) A fictitious position with a zero pitch. (b) When the pitch h of a spring increases, the twisting angle of the wire increases. The torque also increases. Using the formula for torsion, the stiffness of a helical spring is obtained.

On the other hand, to generate a torque T in a wire, the axial load at the center of the spring should be

(E.47)
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Substituting these values into Eq. E.40, we obtain

(E.48)
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The quantity (Gd 4/8nD 3) is often called the stiffness of a spring.

In designing a spring, the maximum stress must not exceed the allowable stress or working stress of the material. From Eq. E.37, we find the maximum shear stress in the wire is

(E.49)
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# E. 6 Contact stress: The Hertz formulas

By pressing a sphere upon a planar surface, a deformation occurs near the original point of contact – see Fig. E.9. This problem was first solved by Hertz in 1881 ([353, 354]). The derivations are complicated. We state the results without proof.

Consider a sphere of diameter D of material 1 in contact with a planar surface of material 2. The reaction of the load P is determined by an effective Young’s modulus E*, which is defined as

(E.50)
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(p.399)

Fig. E.9. Contact stress and contact deformation When a sphere (a) is pressed upon a flat surface with a vertical load P, deformation occurs (b). The yield y, the maximum stress σ max, and the diameter of the contact area d were analyzed by Hertz in 1881.

in terms of the Young’s modulus and Poisson’s ratios of material 1 and material 2.

The diameter of contact is

(E.51)
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The total yield, or total relative motion, is

(E.52)
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The average normal stress in the contact area is

(E.53)
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The maximum normal stress at the point of contact, according to Hertz, is 1.5 times the average normal stress:

(E.54)
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These formulas are important in the discussions of forces in AFM, tip and sample deformation, as well as some mechanical design problems of STM and AFM. (p.400)