# (p.389) Appendix E

# (p.389) Appendix E

**Elementary Elasticity Theory**

(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:*

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

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:

where the quantity *E* is the *Young’s modulus*.

Under the same stress *σ* _{x}, the *y* and *z* dimensions of the bar contract:

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 *Ƭ _{xy}*, and is called a component of the

*shear stress*:

The condition of equilibrium, that is, the absence of a net torque on a small volume element, requires that *Ƭ _{xy}* =

*Ƭ*. A shear stress causes a shear strain, defined as (see Fig. E.2)

_{yx}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]):

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*):

*d*V(*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

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

Consider a small section Δ*z* 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

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

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*,

For a rod of diameter *D*, it is

For a tube with outer radius *D* and inner radius *d*,

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

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

Integrating it again and using the condition that, at *z* = *L*, the torque is zero, *d* ^{2} *u*(*z*)/*dz* ^{2} = 0, we get

Integrating twice and using the condition that, at *z* = 0, *u* = 0 and *du/dz* = 0, we obtain

The deflection at *z* = *L* is

The angle at *z* = *L* is

# 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

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

Denoting

the general solution of Eq. E.26 is

*u*(*z*) = *A* cos *kz* + *B* sin *kz* + *c* cosh*kz* + *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

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

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

The resonance frequencies of the cantilever is determined by

The lowest resonance frequency is

# 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

The shear stress is, according to Eq. E.8,

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

By integrating over the entire area, the total torque is

In terms of diameter *d* = 2*r*, the torque is

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,

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,

The velocity of the elastic wave is

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.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*/2*nD*. For every one half of a coil, the length of wire is

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

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

Substituting these values into Eq. E.40, we obtain

The quantity (*Gd* ^{4}/8*nD* ^{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. 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

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

The diameter of contact is

The total yield, or total relative motion, is

The average normal stress in the contact area is

The maximum normal stress at the point of contact, according to Hertz, is 1.5 times the average normal stress:

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)