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Problem 1
A cantilever beam has been constructed from a steel having Young’s modulus E =
208 GPa, Poisson ratio ν = 0.29, and tensile yield strength σy = 410 MPa. The
length of the beam, from its base to its tip, is L = 1 m, and the uniform cross-
section is rectangular, h = 5 mm thick and b = 30 mm wide.
Two axiallymounted strain gages have been mounted on the top surface (y = h/2).
The precise absolute axial position “x” is not known for either gage; however, it is
known that the two gages are spaced an axial distance of l = 200 mm apart along
the length of the beam.
A tip load of magnitude ‘P’ is applied to the cantilever, acting in a direction
parallel to the yaxis, causing the two gages (#1, which is nearer the base of the
cantilever, and #2, which is nearer its tip) to register the following strain values:
Problem

Figure 1: Schematic of tiploaded cantilever showing positions of strain gages
numbers 1 and 2.
Gage no. Axial coordinate Strain
1 x1 =? 1200 × 10−6
2 x2 = x1 + 200 mm 900 × 10−6
• Evaluate the load P.
• Using the (nowknown) value of P, find the axial locations (x1 and x2) of both
gages.
Problem 2.
There is current interest in the use of microfabricated cantilever beams in
detecting the presence of bacteria in a liquid solution. Following fabrication of
the cantilever, its surface is coated with an antibody that is specific to the
presence of the bacterium of interest, and the free vibration characteristics of
the cantilever (its firstmo de natural frequency, ω0) is recorded experimentally.

Then the coated cantilever is exposed to a liquid medium. If the soughtfor
bacteria are present in solution, they will preferentially attach themselves to the
antibody coating on the surface of the cantilever, in the process increasing the
vibrating mass of the cantilever by an amount Δm = nb mb, where nb is the
number of bacterium cells that attach, and mb is the mass of the bacterium cell.
Image removed due to copyright considerations
Figure 2: Scanning electron microscope images of E. coli bacteria attached to
various microfabricated resonating cantilever beams. (from: Ilic, et al., Applied
Physics Letters, 77, #3, 2000, 450452.
Subsequent testing of the addedmass cantilever should reveal a progressively
decreasing natural frequency as more bacterium cells are added.
The cantilever of interest has been fabricated of silicon nitride, having a mass
density of ρ = 3.1 × 103 kg/m3 and Young’s modulus of E = 100 GPa. It has a
uniform rectangular cross section, of thickness h = 320 nm, width b = 15 µm,
and total length L = 100 µm.

• Assuming that the antibody coating itself does not appreciably affect the
natural frequency of the cantilever, estimate the firstmode natural frequency,
ω0, of the cantileve in the absence of adhered bacteria. (Note: please refer to the
Lab 1 Handout Notes on vibration of cantilever beams for relevant analysis.)
• Assume that the added mass Δm = nbmb is uniformly distributed along the
length of the beam, and further, assume that the presence of the adhered
bacteria does not affect the stiffness of the beam. Let the total mass of the coated
beam be m = ρbhL; assuming that Δm << m, show that the change in frequency
resulting from the added mass, Δω, can be expressed by
– Hint # 1: You might wish to show that the natural frequency of a uniform
beam of total mass ‘m’ can be written as ω ∝ (1/L)2 �EIL/m (neglecting
dimensionless factors).
– Hint # 2: If the change in total mass (Δm) is very small in comparison to the
initial beam mass (m), then the resulting frequency of the perturbedmass system
can be evaluated by taking a Taylor series expansion of the frequency expression
about the reference mass.

• Assume that the cantilever has been covered by nb = 100 bacterium cells. From
the figures, it appears that the diameter, Db, of each bacterium cell is near Db =
1µm; assuming each cell to be spherical and to have a mass density, ρb, equal to
that of water, . justify an estimate of each cell’s mass as mb = 5.24 × 10−16 kg.
Using this estimate, and assuming that the cell “added mass” is uniformly
distributed over the surface of the beam, evaluate the change in natural
frequency that you can expect to see for the bacteriumcoated cantilever.
Problem 3.
Atomic force microscopy resolves the structure of surfaces to near atomiclevel
resolution. A key structural element of an atomic force microscope (AFM) is a
small cantilevered beam fabricated from a material such as silicon using
lithographic technology (e.g., precision etching). The beam has a sharptipped
diamond stylus at its tip, the stylus is tracked across the specimen surface, and
very local surface interactions between the tip and the specimen surface cause
the tip of the cantilever to deflect. The magnitude of the tip deflection is
measured by reflecting a laser beam off its back side and imaging the location of
the reflected light. The tip/surface forces are computed from the deflections
using beam theory, and they are used to map the structure of the surface.

Figure 3 shows a small rectangular crosssection cantilever from an atomic force
microscope (AFM) that has been machined from a single block of silicon using
lithographic technology (e.g., precision etching). The nominal dimensions of this
cantilever are length � = 460 µm, width w = 50 µm, and thickness t = 2.0 µm.
Elastic constants of silicon are Young’s modulus E = 107 GPa, and Poisson ratio ν
= 0.2.
The stiffness k = P/δ of the AFM cantilever is an important design parameter.
• Compute the stiffness of this AFM cantilever.
• Unfortunately, dimensional tolerances in the etching technique inevitably lead
to a range of dimensions over a population of the AFM cantilever products. The
manufacturer quotes the following tolerances for this cantilever: l = 460 µm ±
5µm; w = 50 µm ± 3µm; and tmin = 1.5 µm ≤ t ≤ tmax = 2.5 µm. Accordingly, the
stiffness of any given cantilever can vary from its nominal value, depending on its
actual dimensions. Provide estimates of the minimum and maximum stiffnesses,
kmin and kmax , that can be expected is this cantilever, corresponding to extreme
ranges of the dimensions.

Problem 4
CDL Problem 8.15
Hint/suggestion: Problem 8.15a is the continuation of the “selfweight” cantilever
problem we looked at in the lecture. You should be well on the way to solving for
its moment distribution, and then you should be able to follow the procedures
adopted for the tiploaded cantilever to calculate the corresponding lateral
displacement distribution, v(x), and anything else needed.
Problem 8.15(b) is statically indeterminant, meaning that there is no direct way
to evaluate its bending moment distribution, M(x), without simultaneously
considering the displacement distribution, v(x). There are two straightforward
ways to approach such statically indeterminant beam problems.
• Combine CDL equations 3.11 and 3.12 to obtain
M" (x) − q(x) = 0 M " ⇒ (x) + w = 0
and then use M(x) = EIv " (x) to obtain

From here on, there are 4 integrations, 4 constants of integration, and 4
boundary conditions (on v(0), φ(0), M(L), and v(L)) to determine the constants.
It will work, but the algebra is deadly dull.
• On the other hand, you can use the principle of superposition. In this case,
observe that the stated problem is equivalent to a cantilever, subject to (a) the
distributed loading q(x) = −w, AND to a concentrated upwardsdirected tip load,
of [unknown] magnitude RL. The total vertical tip displacement of the cantilever
is vtotal (L) = 0 and by superposition,
where it is understood that there is a downward tip displacement, vq(L) < 0
(from problem 8.15a), as well as an oppositedirection tip displacement, vRL > 0,
from the unknown positiveupward tip reaction force, RL. This leads to a simple
linear equation to solve for RL; once RL is obtained, all parts of the solution to
8.15b are known by superposition... Try whichever method suits you.

Solution
Solution 1
Figure 1: schematic drawing of Problem 1
This beam is under bending and shear. Strain gauges are located at the top
surface of the beam. On the top surface, there is zero transverse shear.
Thus the beam is only in a stress state due to bending at the location of
the strain gauges. We must know how to compute the stress due to a
bending moment and relate that stress to the strain in order to solve this
problem.
First, the axial stress due to bending is

where I and y on the top of surface of the beam are the following
M can be found via taking a section cut in the beam and balancing the
moments. Be sure to use proper sign convention
M(x) = −P(L − x)
Re-write (1) using (2), (3), (4)

This is in the Linear-Elastic regime so,
σ (x) = E ϵ (x)
Re-write (5) with (6) to get
We have two values of strains σ1 and σ2 at positions x1 and x2 so we now
have two equations (7), (8), with three unknowns x1 , x2 , and P
The third equation needed comes from the geometry condition given in the
problem statement
x2 − x1 = 200 × 10−3 = d
Now we just have 3 unknowns and 3 equations so it can be solved
anyway you like. One was is shown below

To solve for P , combine equations (8), (9), (10)
Now solve (11) for P , and plug in values. We know
then,
Knowing P we can get the positions of the strain gauges. Solve (8) for x1 .

Now get x2
Solution 2
The natural frequency of a simple harmonic oscillator depends on both the
stiffness of the restoring (elastic) member in the system and the mass which is
being accelerated/decelerated. For a rigid mass m connected to a massless spring
of linear stiffness k (dimensions: force/length), having one end grounded while
the other is attached to the moving mass, the natural frequency is simply
For the first-mode natural frequency of continuous uniform beam, δ0 , we have

where ρ = 0.24. To see where it comes from, please refer to section 8 of
Lab No.1 handout.
1). Estimate the first-mode natural frequency, δ0
We are assuming that the antibody coating itself does not appreciably affect
the natural frequency of the cantilever. So, based on the geometry of the
beam, we have the beam stiffness,
where I is the area moment of inertia of the cross-section, which, for
rectangular crosssections of this orientation, is equal to:
with b the width and h the thickness of the beam. The mass of this
cantilever is equal to its volume times its density, which is,

m = pV = pL-b-h = 3.1×103[kg/m3] 100×10−6[m] 15×10−6[m] 320×10−9[m]
= 1.49×10−12[kg]
Substitute eqn 18 and eqn 19 into eqn 17, we have the first mode natural
frequency of this microfabricated cantilever,
2). Frequency change from the added mass, δ According to eqn 18 and eqn
21, the natural frequency of a cantilever is
The sought-for bacteria will preferentially attach themselves to the antibody
coating on (22) the surface of the cantilever, in the process increasing the
vibrating mass of the cantilever by an amount m = nbmb, where nb is the
number of bacterium cells that attach, and mb is the mass of the bacterium
cell. We are assuming the added mass m is uniformly distributed along the length
of the beam, and further, assume that the presence of the adhered bacteria does
not affect the stiffness of the beam. From Eqn 22, the change in frequency
resulting from the add mass can be expressed by

For the function of (1 + m )− 2 , if the change of in total mass (m) is very small in
m comparison to the initial beam mass (m), then it can be expanded by taking a
Taylor series expansion. The general Taylor series expansion has the form of
In our case,
Then, the change in frequency resulting from the adding mass can be expressed
as the following

3). Evaluate the change in natural frequency for the bacterium-coated cantilever
For each bacterium cell, we assume it is spherical, with the diameter, Db =1µm.
Its density, b, is equal to that of water, i.e. 1.0 × 103kg/m3 . An estimate of each
cell’s mass is
The total mass of all the bacterium cells, which is the ”added mass”, is
Figure 2: Scanning electron microscope images of E. coli bacteria attached
to various microfabricated resonating cantilever beams. (from: Ilic, et al.,
Applied Physics Letters, 77, #3, 2000, 450-452.
Using eqn 26, and assuming the cell ”added mass” is uniformly distributed
over the surface of the beam, the change in natural frequency is
Please be noticed that the frequency, f , in units of [cycles/sec] (or [Hz]) is
related to the angular frequency (or radian frequency), δ, by f [Hz] =
δ[rad/s]/(2[rad/cycle]).

Solution 3
In this problem, the CNT(Carbon nanotube) can be assumed to be a cantilever
beam. For elastic response, the lateral displacement of the cantilever v(x) is
related to the bending moment by
where a is the distance from x = 0 to the loading point. In the following
derivation, x will be only in the range of 0 ≤ x ≤ a.
Remember that k , the stiffness, is a function of structure parameters such as
geometry. That is, for the same material, two beams with different widths for
example will have different stiffnesses. Young’s modulus, on the other hand,
depends ONLY on the material and thus is a material property. Finding a way
to correlate these two is a reasonable approach to solving this problem.

From the definition of stiffness,
Furthermore, as we learned in class and in the lab, the following is true:
From Equation 3, it can be concluded that k is proportional to the width and (
thickness 3 ) of the beam and inversely proportional to the ( length 3 ) .
Therefore, to get the minimum stiffness, we need to use the minimum width and
thickness and the maximum length for our calculations using Equation 3 and
vice versa for the maximum stiffness.

More specifically:
Let’s derive the necessary results instead of using Table 8.1 in CDL. For beam
(a), using the sign convention found in CDL, we have:

Let’s use the principle of superposition for part (b)This problem can be broken
into two parts: a beam with a uniformly distributed load and an identical beam
with an unknown tip load RL . We know the deflection 1 and bending moment M
for the first part from part (a) (Equations 4 & 8) and for the second part ( 21 and
2M accordingly) from the lab handout:
Make sure you use consistent sign convention. The assumption here is that this is
an upwards-directed tip load.
By superposition:

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