Mechatronics 1 Problem Set 1 (Solutions) · the Mechatronics book on page 190 as Equation 9.44. V =...

1 ELECTRONICS

Mechatronics 1

Problem Set 1 (Solutions)

Joe Foley

12. september 2011

Please write your name on the top of each page. The �rst two digits of your kennitala are X and Y in theproblems. E.g. 123456-7890 would have X = 1 and Y = 2.

This problem set is to be done alone. Try all of the problems. You are allowed to use books and internet webpages, but you must cite your sources properly. Citation can be as simple as a URL or a book title, author, andpage.

Each problem will be evenly evaluated for method and answer. Answer each section to the best of your ability.For full credit, you must answer 3 problems correctly in each section. Extra credit will be given for additional correctproblems. Each question is worth 2 points.

1 Electronics

1.1 Basic

1. Basic components.

(a) What do these symbols mean?

Resistor(left), Capacitor (center), Inductor (right)

(b) What is the equation for current and voltage in each.

V = IR (1)

I = CdV

dt(2)

V = LdI

dt(3)

2. For this circuit, what is Vout? What is the power dissipated by this circuit?

Rev: 8, 12. september 2011 1 ps1-solutions.tex

1.2 Advanced 1 ELECTRONICS

First we use Kircho�'s current law to trace current through the resistors.

I =Vin

1000 + 1000X + 100Y(4)

Now using Eq 4, we can �nd the voltage drop across the XY00 resistor using ohm's law.

Vout = VXY00 = IR (5)

= IRXY00 (6)

= Vin1000X + 100Y

1000 + 1000X + 100Y(7)

1.2 Advanced

1. Varying inputs.

(a) Graph the voltage at Vout for these Vin?

First, we notice that this circuit is the same as the one in the previous section, but with di�erent resistancevalues. Adjusting Eq 7 for these new values we get:

Vout = Vin100X + 10Y

100 + 100X + 10Y(8)

The value to the right of Vin is our scaling constant. Multiply this by the input voltage to get our �nalgraph.


1.2 Advanced 1 ELECTRONICS

(b) Graph the current passing through R3 and R4 for each input.Thanks to KCL, we know that the current is the same through both of them. Ohm's law tells us V = IR,so

I =V

R(9)

=V

R3 +R4(10)

As before, our scaling factor is simply 1R3+R4

, which we multiply by the Vin to �gure out the current (I)

2. Your little brother Guðni knows you are in Mechatronics and wants you to make a device that will explodewires to make a big �ash! He has a 3mm piece of copper wire which is 0.5mm in diameter. Copper has aconductance of 5.96E7 S/m. His wire is the resistor in the following circuit you have designed:

(a) What is the resistance of Guðni's wire? The conductance (inverse of resistance is de�ned):

S =I

V=

1

R(11)

The conductance of a particular geometry is related to the area A and length l:

S1 = SA

l(12)

The wire is a simple cylinder, so the area A using the diameter d is:

A = πr2 =π

4d2 (13)

We can now combine all of these equations:

R1 =1

S1=

L

SA(14)

=4L

Sπd2(15)

(b) Graph the voltage and current for the �rst 100 ms when Guðni �ips the switch from A to B. To e�ectivelygraph this we need to know the equation for an RC circuit. Conveniently, this equation can be found inthe Mechatronics book on page 190 as Equation 9.44.

V = V0e− 1

RC t (16)

1RC is often known as the �time-constant� τ . In the �rst time constant, the circuit is at ≈ 65%, andis at 99.9% after 5τ . The τ here is very very small, so we actually don't see much after the �rst fewnano-seconds.

(c) How much power is dissipated in the wire in the �rst millisecond? Note: This is an error, the questionshould say �energy�. The power dissipated in a resistor is simply:

P = V I =V 2

R(17)


1.3 Challenge 1 ELECTRONICS

So the power dissipated is:

P =V 20

Re−

2RC t (18)

And the energy dissipated will be:

E =

1e−3∫t=0

P dt (19)

=

∫V 20

Re−

2RC t dt (20)

= −V20 RC

2Re−

1RC t (21)

= −V20 C

2e−

1RC t

∣∣∣∣1e−3

0

(22)

(23)

1.3 Challenge

1. Certain nano-crystalline structures might be modeled as an in�nite grid of resistors with value R in triangles.Calculate the e�ective resistance in terms of R from point A to B.

This solution is based upon an excellent writeup at http://stevensholland.com/challenge-problem-solution-from-jan-19th-2007/Consider the hypothetical situation. Attach a 1 amp current input to A, but leave B to do whatever it wishes.By KCL1, the equal resistances will cause 1

6 amp to �ow out equally.

Consider the symmetrical situation: We attach a 1 amp current sink to B, but leave A �oating. KCL tells usthat also 1

6 amp will be �owing into B.

By superposition, when we hook up a combined 1 amp current source and sink across A and B, we will see

Itotal = I1 + I2 (24)

=1

6+

1

6(25)

=1

3amp (26)

Now for the entertaining bit. The de�nition of a Thevenin equivalent current source says:

Rth =V

Ith(27)

We know voltage from ohm's law:V = ItotalR (28)

1Kirchko�'s Current Law


2 SOFTWARE

Since we chose 1 amp as our supply Ith, the math is quite nice:

Rth =ItotalR

Ith(29)

=13amp

1ampR (30)

=1

3R (31)

A much more general solution to this problem can be found at http://www.mathpages.com/home/kmath668/kmath668.htm. An interesting secondary result for a square resistor array is that the corners have a resistanceof 2

π .

2 Software

2.1 Basic

1. Fill in the truth tables for these logic gates. A truth table shows what the output would be for a given set ofinputs.

The �rst gate is an Or gate. It will output a 1 if either of its inputs is 1. If both inputs are 0, then it willoutput 0. The second gate is an And gate. It will output a 1 if both inputs are 1. If either input is 0, thenit will output 0.

2. Convert XY into binary, octal, and hexadecimal. Explain each step of the conversion explicitly. Feeding itinto a webform on the internet is not an acceptable explanation! Each of these conversions divides a numberinto powers of 2, 8, or 16 respectively, called its �base�. The easiest way to do the conversion is to divide bythat power and check for a remainder. This remainder becomes the next number to divide. When you reachless than the base , you read the remainders in reverse order. For the number 99, we would get in binary:Value Divide Remainder Bit76 38 0 038 19 0 119 9 1 29 4 1 34 2 0 42 1 0 51 0 1 6


2.2 Advanced 2 SOFTWARE

So the binary for 76 is 1001100. Because the largest bit is �rst, this is called �big endian�. An alternativewould have the lowest bit �rst: 0011001, and is called �little endian�. For simplicity, I will only show the �bigendian� solutions.

In octal (base 8):Value Divide Remainder Digit76 9 4 09 1 1 11 0 1 2

Resulting in: 411. A standard convention of octal is to start with a leading 0, so the value would be 0411.

In hexadecimal (base 16):Value Divide Remainder Digit76 4 12 04 0 4 1

For values greater than 9, we have A=10, B=11, C=12, D=13, E=14, F=15. 76 is 4C in hexadecimal. Theconvention is to put a �0x� in front of hexadecimal numbers, so it would be 0x4C.

2.2 Advanced

1. Write a logic expression with 4 inputs that will be true if the inputs are X or Y encoded into binary. Theexpression will return false with all other inputs. You are allowed to use AND, OR, NOT, and XOR. Theinputs will be I0,I1,I2,I3. We use the binary conversion scheme mentioned previously. Assume X = 7 and Y= 6 for our example. X in 4-bit binary is 0111 2:

Value Divide Remainder Bit7 3 1 03 1 1 11 0 1 2

Y is 0110 in 4-bit binary.

Value Divide Remainder Bit6 3 0 03 1 1 11 0 1 2

Now we examine the di�erences between these two sets. It appears that:

I0 I1 I2 I31 or 0 1 1 0

There are many ways to encode this. Important things to note:

• We don't care what the I0 is, so we can ignore it.

• The other inputs are a �xed set of values to cause true.

• We test for 0's by using a NOT and an AND.

So we can write this simply as: I1 And I2 And Not I3

2. Write a function in any programming language (C, Python, PSEUDOCODE, etc) that will return 1 if theinputs are X or Y encoded into binary. The function should return 0 if any other input is given. This functionwill take the arguments I0,I1,I2,I3.Again, this can be solved many ways. The simplest is to use the logical operators in that language. In C/C++:

int test_binary(int I0, int I1, int I2, int I3) {

return(I1 || I2 || !I3);

}

2We put 0's in front of the number as �padding� to make it enough digits/bits.


2.3 Challenge 3 MECHANICAL

Material Density Young's Modulus Yield Stress Poisson RatioGlass 3100 kg/m3 70 GPa 33 MPa 0.3

Aluminum 2.70 g/cm3 70 GPa 414 MPa 0.35Rubber 1200 kg/m3 0.1 GPa 15 MPa 0.3

Ta�a 1: Material properties

2.3 Challenge

1. Marel wants you to design a simple conveyor control system that alternates which side a �sh is dispensedon. The detector Fish will be true as the �sh falls through it. There are two detectors at the end of theconveyor Left and Right. Marel also wants a safety switch Panic that will stop the conveyor at any timeand Restart to resume operation. Write a short program in C, C++, Python, Perl, Assembly, Java, orLISP that is able to control this system.

FISH

RIGHTLEFT

Sadly, this problem originally had a component of controlling a stepper motor which had the inputs Step

and Direction, but that got lost in an edit. This problem had many ways to solve it depending upon theprogramming language chosen. Items of consideration for its design were:

(a) Finite state machine with state change detection

(b) Fast control loop that does not block to deal with the Panic and Restart signals

(c) State variable to remember which direction the �sh was last dispensed.

(d) What happens if a �sh falls o� the belt before getting to the sensor?

(e) What happens if another �sh comes through the chute while one is on the belt?

3 Mechanical

A friend who is a furniture designer would like you to check his/her design. The table has a glass top and hollowaluminum legs (Mynd 1). The sides of the glass are simply supported on two more hollow tubes of the same type asthe legs(Mynd 2). On the bottom of each leg is an very thin aluminum cap which sits on top of a rubber foot(Mynd 3).(You can ignore the mass of the cap, it just spreads the load evenly over the foot.) Material parameters are belowin Ta�a 1. The lines joining the legs are stabilizers and can be ignored for this analysis. Explain any assumptionsyou make.

3.1 Basic

1. How much does the glass and aluminum weigh? (In N)Conveniently, the glass and aluminum are in standardized sizes so we �rst calculate the mass. Watch theunits! I converted to MKS for simplicity in the �nal evaluation Doing a symbolic solution is always a goodidea to show your thinking and avoid unit problems.

length of table L = 2000mm = 2m

width of table W = 100X + 10Ymm = 0.1X + 0.01Ym

thickness of table top Tt = 10mm = 0.01m


3.1 Basic 3 MECHANICAL

Mynd 1: Table assembly.

Mynd 2: Aluminum tube extrusion pro�le.

Mynd 3: Rubber foot geometry, unloaded.


3.1 Basic 3 MECHANICAL

height of the legs and mount hl = 500mm = 0.5m

cylinder outer dLiameter and foot diameter Cd = 20mm = 0.02m

cylinder thickness Tc = 0.5mm = 0.0005m

foot thickness hf = 10mm = 0.010m

glass density ρg = 3000kg/m3

aluminum density ρal = 2.70g/cm3 = 2700kg/m3

rubber density ρr = 1200kg/m3

rubber Young's modulus Er = 0.1GN/m2

The glass is a rectangle:mglass = ρgLWTt (32)

The aluminum is composed of two sets of tubes. The horizontal tubes are just W long. The vertical tubesare hl − Cd long 3

mal = ρal (2W + 4(hl − Cd)) (33)

Newton provides us with how to covert this to weight (which is a force from gravity's acceleration −g. )

F = mA (34)

= (mal +mg)(−g) (35)

= − (ρal (2W + 4(hl − Cd)) + ρgLWTt) g (36)

2. What is the stress on the rubber foot? Note that there are 4 feet. Stress is de�ned as force over area (likepressure):

σ =F

A(37)

Our area is the 4 circles of the foot

Af = 4πr2f (38)

= 4πD2

4(39)

= πD2 (40)

So we can now evaluate Equation 37. We note that compressive stress is negative and tensile stress is positivein standard conventions:

σf =F

Af(41)

= − (ρal (2W + 4(hl − Cd)) + ρgLWTt) g

πD2(42)

3. What is the strain on the rubber foot?The Young's modulus (E) gives us the ratio of stress σ to strain ε.

E =σ

ε(43)

Solving for strain

ε =σ

E(44)

We can now evaluate Equation 44.

εf = −σfEr

(45)

=(ρal (2W + 4(hl − Cd)) + ρgLWTt) g

πD2Er(46)

3If you ignored the width of the top cylinder or assumed only the bottom was part of the height, that was acceptable as the diagramwas not clear


3.2 Advanced 3 MECHANICAL

3.2 Advanced

1. What are the dimensions of the foot when it is under load?We use the de�nition of strain to calculate the change in thickness (in line with our stressor).

ε =δL

L=l − LL

(47)

In our case, the l is what we are looking for, so solving Equation 47.

l = (εL) + L (48)

= L(ε+ 1) (49)

Evaluating for our values:

hf,2 = L(ε+ 1) (50)

= hf (εf + 1) (51)

= hf

(1− g ρal (2W + 4(hl − Cd)) + ρgLWTt

πD2Er

)(52)

This takes care of the �nal length, but the rubber will also bulge out. The poisson ratio tells us by how much.Since the ratio for rubber is 0.54, it can be approximated as an isotropic incompressible solid. This meansthat conservation of mass can be approximated as conservation of volume5 So we must equilibriate the volumebefore and after compression.

V0 = V1 (53)π

4D2

0h0 =π

4D2

1h1 (54)

D20h0 = D2

1h1 (55)

D21 = D2

0

h0h1

(56)

D1 =

√D2

0

h0h1

(57)

= D0

√h0h1

(58)

= Cd

√hfhf,2

(59)

= Cd

√√√√ hf

hf

(1− g ρal(2W+4(hl−Cd))+ρgLWTt

πD2Er

) (60)

= Cd

√1

1− g ρal(2W+4(hl−Cd))+ρgLWTt

πD2Er

(61)

2. The legs

(a) What is the stress in the aluminum legs?To simplify this analysis, we will assume that the legs are of negligible weight compared to the heavyglass top. We can use Equation 32 for our loading force. We have a simple hollow cylinder for our area.

Al = 4(Aouter −Ainner) (62)

= 4(π

4C2d −

π

4(Cd − Tc)2) (63)

4This was a mistake in the problem set, it was set at 0.3. If you did the appropriate steel-like poisson calculations, that was alsoaccepted.

5This is also not technically true for materials like rubber, but analysis of hyperelastic materials in compression and tension is outsidethe scope of this course � graduate studies in material mechanics spend a great deal of time on this subject.



= π(C2d − (Cd − Tc)2) (64)

= π(C2d − C2

d − 2CdTc + T 2c ) (65)

= π(T 2c − 2CdTc) (66)

σl =FlAl

(67)

=ρgLWTt

π(T 2c − 2CdTc)

(68)

(b) What is the strain in the aluminum legs?As before, we use the de�nition of Young's modulus (Equation 44)

εl = − σlEal

(69)

= −ρgLWTt

π(T 2c −2CdTc)

Eal(70)

= − ρgLWTtπEal(T 2

c − 2CdTc)(71)

(72)

(c) Where is the angle of principal stress in the leg?This was a question focusing on what the de�nition of principal stress was. Principal stress is thegeometry you would draw in the element being examined where there is no shear forces. At a simplisticexamination, there aren't shear forces at all, so the angle of principal stress is 0 (pure axial compressionand tension).

3. How would you improve this design with minimum modi�cations? (Your friend is very insistent on his/herartistic vision.) Show quantitatively what performance improvements are observed by your change.For this question, many answers would be considered as long as there was supporting stress-strain analysisshowing that the de�ections were less or it addressed a particular shortcoming. Of note, if something heavyis placed on the center of the table, the two bracket cylinders are likely to displace enough that the glass willfall even with the stabilizers.

3.3 Challenge

1. Using a thin beam approximation, how much does the glass bend? Is this likely to cause it to fall out? Explainwhy or why not.Since this is a thin beam, we can use the moment of intertia (I) for �at rectangles with width b and thicknessh.

It =bh3

12(73)

=WT 3

t

12(74)

We also assume that the slope of de�ection is small so we can use the Uniformly Loaded Simply-SupportedBeam equation fromMIT Engineering Mechanics of Solids page 8.7 http://web.mit.edu/emech/dontindex-build/where L is our span, x is the position being considered, and w0 is the force per length.

v(x) =w0L

4

24EI

x

L(1− 2

x

L

2+x

L

3) (75)

Solving this for x = L2 where the stress and bending are at maximum

v|max =5w0L

4

384EI(76)



The only element missing is force over length, which we can calculate from the stress.

w0g =F

l(77)

=−gmglass

L(78)

=−gρgLWTt

L(79)

Combining these all together

v|gmax =5w0gL

4

384EIt(80)

=5−gρgLWTt

L L4

384EgWT 3

t

12

(81)

= − 60gρgL4

384EgT 2t

(82)

We now compare this displacement against our length to see if this is a signi�cant amount of bending.

v|gmaxL

=?? (83)

2. Are the aluminum legs in danger of yielding? Will they buckle? Explain why or why not.If the σl gets close to the yield stress of aluminum, we will have yielding. To consider buckling, we can usethe standard simpli�ed Euler buckling formula for the �rst eigenvalue en.wikpedia.org/wiki/Buckling

Fbuckle =π2EI

k2L2(84)

where k is the column e�ective length constant. The stabilizer isn't part of the analysis, so we can say thatthe top mount is just a pin joint, and the bottom is as well. This means k = 1. We can now use Equation 68

and the moment of a thin walled cylinder with thickness h: If = πh4

4

Fbuckle,l =π2EalIl1hl

(85)

=π2Eal

πT 4c

4

hl(86)

=π3EalT

4c

4hl(87)

Now we can compare against the force on the beam, which is a quarter of the weight from the glass inEquation 32

Fl =−gρgLWTt

4(88)

We can see what the safety factor S is:

Sbuckle,l =

∣∣∣∣Fbuckle,lFl

∣∣∣∣ (89)

=

∣∣∣∣∣−gρgLWTt

4π3EalT 4

c

4hl

∣∣∣∣∣ (90)

=

∣∣∣∣−gρgLWTt4Hl

π3EalT 4c

∣∣∣∣ (91)


Mechatronics 1 Problem Set 1 (Solutions) · the Mechatronics book on page 190 as Equation 9.44. V =...

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Transcript of Mechatronics 1 Problem Set 1 (Solutions) · the Mechatronics book on page 190 as Equation 9.44. V =...