EE 147/247A Prof. Pister
Fall 2017
Homework Assignment #4
Due by online submission Friday 9/29/2017 (Saturday 9am)
1. The structure on the left below consists of a rigid body attached to the end of a beam of length L, width a, and thickness b. A force vertical FY acts at a horizontal distance r from the end of the beam, with r positive to the right. The tip deflection y and force are positive in the downward direction.
a. [2] Write an expression for the deflection and rotation of the tip of the beam as a function of the moment arm r.
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Deflection:
yL=L33EIFy+L22EIM0
Where M0=rFy
yL=L33EIFy+L22EIrFy
Rotation of the tip:
y'L=L22EIFy+LEIM0
Where M0=rFy
y'L=L22EIFy+LEIrFy
b. [2] Solve for the value of r that sets the tip deflection to 0.
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Solving for r by setting the deflection equation above equal to zero, we obtain:
Tip deflection:
yL=0
Solve for value of r in terms of L:
yL=0=L33EIFy+L22EIrFy
0=L3+r2
r=-23L
c. [2] Solve for the value of r that sets the tip rotation to 0.
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Solving for r by setting the rotation equation above equal to zero, we obtain:
Tip rotation:
y'L=θL=0
Solve for value of r in terms of L:
y'L=0=L22EIFy+LEIrFy
0=L2+r
r=-12L
d. [2] Consider the vertical deflection of the point at which FY is attached. Compare the stiffness of the mechanism in parts b and c to the simple beam (i.e. Fy applied at r=0).
2 pts. total
1 pt. for comparison with part b
1 pt. for comparison with part c
Part b
The deflection in part b was set equal to zero. Applying a load at r=-23L, we get an infinite stiffness. For a regular cantilever with a point force atr=0we knowk=3EIL3.
So, we see that the mechanism with the lever arm atr=-23Lis infinitely stiffer.
Part c
Assume the lever arm mass is rigid so it doesn't bend. Applying the load atr=-12L we get
yr=-L2=L33EIFy+L22EI-L2Fy
yr=-L2=L33EIFy-L34EIFy
yr=-L2=L312EIFy
We see the stiffnessFyyis
k=12EIL3
For a regular cantilever with a point force atr=0we knowk=3EIL3.
So we see that the mechanism with the lever arm atr=-12Lis 4 times stiffer. By setting y’(L)=0, we enforce the “guided beam” end condition, and expect to see the same S-shaped deflection curve that gives a 4 times stiffer beam.
2. In the structure on the right below, the two beams both have a width a and thickness b.
a. [2] Find L2 such that the spring constants in the x and y directions are equal.
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Deflection in the y direction:
y=L133EI Fy
Deflection in the x direction:
x=L233EI Fx-θL1L2
where
θL1= L1EI (-L2Fx)
Therefore:
x=L233EI Fx+ L1EI L22Fx
Make the spring constants equal to each other:
kx=ky
Fyy=Fxx
Solve for L2 in terms of L1:
0=13L23+L22L1-13L13
Find three solutions, only real one is valid (this answer is rounded):
L2≈0.52L1
b. [2] With that value of L2, how much of the x deflection is due to bending in beam 2, and how much is due to rotation of the tip of beam 1?
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
With that value of L2, the deflection in the x direction is:
x=0.523L133EI Fx+ 0.522L13EI Fx
About 15% of the deflection is due to bending in beam 2, and 85% is due to rotation of the tip of beam 1.
c. [4] What is the y deflection due to a force in x, and the x deflection due to a force in y?
4 pts. total
For the y deflection:
1 pt. for effort
1 pt. for approximately right answer
y=M0L122EI
where
M0=-FxL2
Therefore:
y=-FxL2L122EI
For the x deflection:
1 pt. for effort
1 pt. for approximately right answer
x=-θL1L2
where
θL1= L1EI L2Fy
Therefore:
x=- L22L1EI Fy
3. For a minimum width beam on POLY1 in the polyMUMPS process, assume that E=150GPa and emax=1%
a. [2] calculate the normal stress and strain due to a 1nN axial force
2 pts. total
1 pt. for the correct normal stress
1 pt. for the correct normal strain
σ=FA
σ=250Nm
ε=σE
ε=1.66 nano strain
b. [2] calculate the axial force, and corresponding deflection, at which the beam will fracture
2 pts. total
1 pt. for the correct fracture axial force
1 pt. for the correct corresponding deflection
Fracture force:
σmax=Eεmax
Ffracture=σmaxA
Ffracture=6mN
Corresponding deflection:
ε=ΔLL=xL
x=0.01L m
c. if the beam is 100um long, with a transverse force of 1nN at the tip,
i. [2] calculate the bending moment at the base of the beam, M(0)
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
M0=FL-x=F(L-0)
M0=10-13Nm
ii. [2] calculate the radius of curvature, r(x), as a function of position along the beam
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
1ρx=MxEI=FL-xEI
ρx=Ea3b12FL-x
ρx=200µ100µm-x meters
iii. [2] calculate the strain and stress as a function of position in the cross-section at x=0
2 pts. total
1 pt. for the correct strain
1 pt. for the correct stress
εz,x=zρx
ε=0.5z strain
σ=Eε
σ=75z GPa
iv. [2] calculate the spring constant and deflection of the tip of the beam, y(L)
2 pts. total
1 pt. for the correct spring constant
1 pt. for the correct deflection
k=Ea3b4L3
k=0.6Nm
yL=Fk
yL=1.6nm
v. [2] assuming no geometry-based stress concentration, and ignoring any nonlinearities, calculate the transverse force at which the beam will break
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Ffracture=εmaxEa2b6L
Ffracture=20µN
vi. [2] based on our linear model, what is the expected deflection and angular deflection at which the beam will break due to a transverse force?
2 pts. total
1 pt. for expected deflection
1 pt. for angular deflection
y(L)=Ffracturek
yL=33µm
y'L=FfractureL22EI
y'L=0.5 radians
vii. [1] Why is our linear model not a good model to use when calculating deflection near the fracture limit?
1 pt. total for approximately correct explanation
The deflection the linear model predicts is about 33µm, which is a third of the length of the beam. Important nonlinearities are being ignored because we are assuming a small angle deflection. However, the angle of 0.5 radians corresponds to almost 30 degrees, which is not a small angle. Our linear model would not be as accurate anymore since the small angle assumption can no longer be true.
4. [8] In the SOIMUMPS process,
a. [4] design a comb drive that will give 10uN of force at 15V, and allow at least +/-10 um of travel.
4 pts. total
2 pt. for effort: 1 if you did a calculation to get Ng, 1 if you did a calculation or drew a figure to get comb overlap
1 pt. for getting the number of gaps right (2,000 for 10um film, 800 for 25um)
1 pts. For getting comb finger length right (greater than 20um. I’ll give you full credit for 20um, even though your fingers will contact and explode if you drive them 10um)
F=Ng12ε0V2tg
Choosing 10µm for the SOIMUMPs thickness of the device layer. Also the minimum feature size, or gap width is 2µm.
F=Ng1nN10µm2µm
Choose Ng=2000 to obtain 10uN of force.
Pick length of comb fingers to be greater than 20µm to allow for at least ±10µm of travel: >=10um of initial overlap, > 10um on each side to allow 10um of motion.
b. [2] What is the minimum area of your comb drive, ignoring the anchor?
2 pts. total
1 pt. for effort
1 pt. for approximately right answer
Ignoring the anchor, the minimum width is 2000 gaps, each at 2µm wide, and 2000 fingers (1000 moving, 1000 fixed), also each at 2µm wide, or 8mm. The length of the fingers plus distance of travel total at least 30um
The total area is about:
A=8x10-3*30x10-6[m2]= 0.24µ [m2]
c. [2] If you wanted to fit it into the smallest possible square, how would that change your design (you may make multiple electrical contacts to fixed electrodes, but there should only be one moving electrode)
1 pt. for effort – did you draw anything?
1 pt. for getting at least two columns of comb fingers.
To fit into the smallest possible square, we need to figure out how to get multiple columns of comb fingers in parallel. Something along the lines of the structure below lets you build comb drives in arbitrary aspect ratios. Tough to make that many electrical contacts to the fixed electrodes in a simple SOI process, but easy in two-layer processes.
5. [4] In the polyMUMPS process, design a spring support that will be very stiff in rotation, fit in a 100x100um2 area, and have a stiffness of 10 N/m, while allowing the maximum deflection before fracture.
4 pts. total
1 pt. for effort
1 pt. for choosing correct spring support
1 pt. for doing a spring constant calculation
1 pt. for getting it right
Spring support that is stiff in rotation shown below (mass, comb fingers, and anchor added just as an example application):
Spring constant of spring support:
k=2Ea3bL3=10N/m
Solving for L using minimum width beams, L≈78µm.
6. Calculate the deflection of a cantilevered beam of dimension (a, b, L) and density r due to its own weight
a. If you cut the beam at position x along its length, what is the weight of the portion from x to L? How far is it from x to the center of mass of that portion? Write an expression for M(x), the moment on the cross-section at location x.
4 pts. total
1 pt. for the correct weight
1 pt. for the distance from x to the center of mass of that portion
2 pts. for the M(x) expression
The weight of the portion from x to L:
mx=ρV
mx=ρabL-x
Fweight=ρabL-xg
where g is gravity.
The distance from x to the center of mass of that portion is:
rx=L-x2
The expression for M(x) is therefore:
M(x)=rxFx
M(x)=ρbaL-x22g
b. Write the differential equation with boundary conditions that describe the deflection of the beam, and solve it for y(x).
4 pts. total
2 pts. for the correct differential equation with boundary conditions
2 pts. for integrating correctly with boundary conditions (you can still get 2 pts here if you got the wrong differential equation)
The boundary conditions are set at the anchor of the beam, where x=0.
The y-deflection at x=0 is zero:
y0=0
The angle at x=0 is zero:
y'0=0
The differential equation is:
EId2ydx2=M(x)
EId2ydx2=ρbaL-x22g
Simplifying:
EId2ydx2=gρba2(L2-2Lx+x2)
Integrating once:
EIdydx=gρba2L2x-Lx2+x33+C1
At y’(0)=0, x=0, and therefore C1=0.
Integrating again to obtain the deflection:
EIy(x)=gρba2L2x22-Lx33+x412+C2
At y(0)=0, x=0, and therefore C2=0.
The deflection equation is:
yx=gρbax224EI6L2-4Lx+x2
c. For a POLY1 beam in polyMUMPS, what length is necessary to get a deflection equal to the film thickness? What is the aspect ratio (length over thickness) of that beam?
2 pts. total
1 pt. for the approximately correct length
1 pt. for the aspect ratio
At x=L, the deflection is:
yL=gρbaL224EI6L2-4L2+L2
yL=gρbaL224EI3L2
Solving for L and plugging in the equation for the moment of inertia:
L=2yLEa23gρ1/4
The deflection y(L) is equal to 2µm, a=2µm, g=10m/s2, and ρ=2300kg/m3.
Therefore L would have to be:
L=2.7mm
This corresponds to an aspect ratio of about 1350:1. Picture a 1cm square cross-section metal rod that has the same aspect ratio (13.5 m long) Would it bend more than its own thickness in a 1g field? You bet it would!
7. For a Tang-style comb drive resonator on POLY1 in polyMUMPS with K=1N/m, m= 10-10kg, b=10-7, and 100 electrostatic gaps per side
a. Calculate the resonant frequency and the quality factor
2 pts. total
1 pt. for correct ω0
1 pt. for correct Q
ω0=km=1Nm10-10kg= 10101s2
ω0=105rad/s
Q=kbω0=1Nm10-7Nsm105rads
Q=100
b. Calculate the DC deflection with 1.5, 15, and 150V applied
3 pts. total
1 pt. for each correct force. Not a big deal if you didn’t include the 2x from the fringing fields. Just keep that in mind.
12ϵ0V2={0.01nN, 1nN, 100nN} at {1.5V, 15V, 150V}
Assuming the comb drive is at minimum feature size in polyMUMPS, tg=1.
Also, due to fringing fields we have effectively added a factor of g2 on either side of the capacitor, so we actually get a factor of 2 from the geometry.
With 100 gaps we would get 20012ϵ0V2={2nN, 200nN, 20μN}, and xDC={2nm, 200nm, 20μm} for {1.5V, 15V, 150V}.
c. With a 1.5V sine wave at 1 Hz applied to one comb drive input while the structure is biased at -15V,
i. Calculate the magnitude of the DC, 1 Hz, and 2 Hz forces
3 pts. total
1 pt. for each correct force
From a driving a comb drive with both an AC and DC signal we get
F=20012ϵ0VDC2+VAC22+2VDCVACsinωt-VAC22cos2ωt
For the DC term (DC offset force), the force from the applied VDC term would cancel out, since it is being applied on both the left and right side combs. Therefore, we take only the VAC22 term into account.
FDC=20012ϵ0VAC22≅2000.005nN≅1nN
For the ω term (Force coming out at 1Hz)
Fω=20012ϵ02VDCVAC=20021012ϵ0VAC2
Fω=40000.01nN=40nN
For the 2ω term (Force coming out at 2Hz)
F2ω=20012ϵ0VAC22=2000.005nN=1nN
ii. Calculate the deflection (zero to peak) due to those forces
3 pts. total
1 pt. for each displacement
Since we are well below the resonant frequency (1 Hz vs. 16 kHz) each frequency dependent forcing term is acting in the region dominated by the spring of the resonators frequency response. FDC is always only forcing the spring.
xDC=FDCk=1nN1Nm=1nm
xω ω≪ω0=Fωk=40nN1Nm=40nm
x2ω 2ω≪ω0=F2ω=1nN1Nm=1nm
d. With the 1.5V sine wave input at the resonant frequency of the structure, and the same DC bias as above
i. Calculate the deflection at wn and 2wn .
2 pts. total
1 pt. for each displacement
The ω forcing term will now be at the resonant frequency. The deflection will now be at Q times the low frequency deflection