homework1 solution v5

7/23/2019 Homework1 Solution v5

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2.007 Design and Manufacturing 1

Homework #1

NAME:___School Solution__________________________________________

Date Issued:Tuesday 5 FEB, 11AMDate Due:Thursday 14 FEB, 11AM

Please answer the following 6 questions showing your work to the extent possible withinthe allotted time. Point allocations are listed for each question. The points sum to 100.This should not take more than 6 hours. Each homework counts as 5% of your totalgrade. You will submit your work in hardcopy at the beginning of lecture.

1. (30 points)

a. (5 points) Determine the degrees of freedom of the arm mechanism on the robot depicted below.

Revolutejoints

Upper link

Cableactuayou caconsidthis toseparfrom tmechfor pa

Revolutejoints

Rigidattachment

Lower link

Endeffector

linkBase

Rigidattachment

The mechanism has onedegree of freedom. There arethree components that canmove, the upper arm, thelower arm and the endeffector. I am assuming herethat the base is fixed in place.Three DOF per body in theplane minus two degrees offreedom for each of four

revolute joints 3*3-4*2=1.

If a student assumed the baseis free to move, you could addone more body and one moreconnection (rolling contact withthe floor), so an acceptableanswer would be:4*3-5*2=2 Thats one DOF forthe arm and one DOF formotion back and forth of thebase..


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b. (5 points) Note the configuration of the arm mechanism below. What is this typemechanism called? What might have motivated the choice of this mechanism?

This mechanism is called a parallelogram linkage as described on page 9 of the reading Design ofMechanisms. One key advantage is that the end effector will maintain its orientation even as itranslates during the linkages motions. For example it could keep a fork lift device horizontal.

3 Bodies

4 Joints

Assumethese arefixed


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c. (10 points) Estimate the torque, t, the planetary gear motor must apply to raise the arm at a constant rate if a 1kg mass is held on theend of the arm. In part (b), you may neglect friction (this assumption is to be reconsidered in parts c and d). The links of the armare made of 1/16 inch thick Aluminum sheets bent to 90 degrees and their mass is 0.1kg each.

t

Ruler marked

in inchesshould allow

you to pull o

dimensionsneeded from

the photo

Cable foractuationwhich wrapsaround thespool

t

Planetary gear

motor drives thisspool which has

a 1 inch diameter

Rigid

attachment

~ 25

inches

~ 5

inches


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The load is 1kg applying about 10N downward on the end effector. The end effector about 5times farther out than the attachment point of the cable. By virtual work, we can infer that thevertical component of the cable tension must be 50N. The cable is very nearly vertical. I think itis safe enough to neglect the cosine error.

The volume of the arm to the left of the pivot that is not counterbalanced by similar armgeometry on the right of the pivot is 20in*1/16in*8*in=10*in^3. The density of aluminum is 2.7gr/cm^3. The weight of the unbalanced arm is about 4N. This is about half as far out as the1kg load, so it adds about 20% of additional tension. This is almost negligible, but not quite.

Also, there is the weight of the plate at the end and the fasteners. I will bump up the estimate to65N to 70N of cable tension.

There is a quick and easy check student could do in the lab. The robot was down there allweek. If they put the end of the arm on a scale and set the cable so it was slack, they wouldsee a reading of about 400gr. That would be 4N. Given the distance from the pivot, I wouldestimate almost 20N of cable tension to hold up the arm structure using the cable instead of thescale.

The spool has a 1 inch = 2.5 cm dia or 1.25 cm radius. The torque is therefore60N*1.25cm=0.75Nm of torque.

It is worth noting that the BP-06 planetary gear motor in the jit provides a rated 11 kgf*cm(weird units givin by the manufacturer) which is about 1N*m. This rating is for a 3V supply andwe generally would be able to source 5V from the radio or 7.4V directly from the LiPo batteries.So this arm probably can lift 1kg extrenally applied load if the friction is negligable.


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d. (15 points) Estimate the torque, t, the planetary gear motor must apply to raise the arm at a

constant rate. In this part, you should account for friction. As part of your work,characterize the loading of both the lower and the upper link. The links of the arm rotatedirectly on inch aluminum rods and the coefficient of friction of Aluminum on

Aluminum is about 1.0 if they are poorly lubricated.

The revolute joint in the main pivot of the lower link bears 74 N of upward thrust. If the coefficient offriction is about 1, then there is also 74 N of load perpendicular to that. But this acts at half the holediameter or at 1/8 inch whereas the cable applies its load at 5in or 40 times the distance. So the resultingadditional cable tension needed to overcome friction on the main joint lower arm is about 2N.

The revolute joint in the pivot near the end effector at the lower link bears 10 N of upward thrust. Since allthe revolute joints experience the same angular velocity on this paraalelogram linkage, they eachcontribute to additional cable tension needed proportional to the joints loading So the resulting additionalcable tension needed to overcome friction on the distal joint of the lower arm is about 0.3N.

The revolute joints in the upper link also bear some loading, but it is considerably less. The upper link isa two force member and it will be in tension. The amount of tension depends on the details of where the1kg external load is placed and I will estimate the loads are 10N each joint. So the resulting additionalcable tension needed to overcome friction on the distal joint of the lower arm is about 0.6N.Summing all the additions to cable tension needed, I estimate the total is 63N up from 60N when weassumed no friction. Just a 5% increase in cable loading needed to overcome friction. The motor torquewill go up in the same proportion or about 0.79Nm of torque.

10N

4N

60N

74N


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2. (20 points) Create a 3D solid model of the component depicted below. This is part of a servo

actuated valve you could use to score on the angioplasty task. There will be examples of this

part at the CAD session on Thursday 7 FEB. There will also be people to work with you if

you get stuck.

There are two options for completing this problem (check only one block below):

I have completed this in lecture and my instructor has recorded my result

I posted my models(not screen shots) on Stellar under HOMEWORK/HW1_2/

Advanced exercise (optional): The back side

of the part has a recess to mate with the fittingon the VS-2 servo. Add this feature by

copying geometry from the VS-2 CAD model.

The most common error was to put the main,

larger hole in the wrong orientation like this.

Another common error was to make theseholes 2X too big, probably by interpreting thediameter given on the drawing as a radius.


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3. (10 points) Use Solidworks to make a 2D model (within a sketch) of the window glassmechanism below. Make the sketch move on the screen when you rotate the handle.Convince yourself that it has the degrees of freedom that it should have.

There are two options for completing this problem (check only one block below):

I have completed this in lecture and my instructor has recorded my result

I posted my models(not screen shots) on Stellar under HOMEWORK/HW1_3/

This is all one link

connecting P1A1and B1 Figure from C. H. Suh,

Computer Aided Design

of Mechanisms

Students generally did very well on this. Someof the mechanisms were under-constrained. A

few were over-constrained. A few were

1DOF, but kinematically different than the one

in the figure.


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4. (15 points) Which one of these pulley arrangements can hold the weight (W) with thelowest applied force (T)? Circle one and also explain the reason for your choice in terms anMIT freshman (who has taken 8.01 and 8.02) would understand. Use five sentences at themost for each subsection (a through c). Note: These machines use pulleys, not capstans.

a. (5 points)

This one is just a pulley. Nomechanical advantage,

although the tension is

reoriented by the pulley.

Many pulleys in a simplesequence. No mechanical

advantage, although the

tension is reoriented multipletimes by the pulleys. In fact,

there will be some losses at

each pulley.

Basic block and tackle. 2X

mechanical advantage becausethe tension is applied twice to

the weight.

Splitting the load into two

cables helps by dividing the

load among them. But youstill need to pull both cables.

When you join them again,youre back where you started.


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This one is just a pulley. No

mechanical advantage,

although the tension isreoriented by the pulley.

A pulley attached to a lever.

Nice idea, but the lever hasthe load farther out than the

cable. About 1/2 X

mechanical advantage.

Another good idea that is,

however, implemented

backwards here. With theangle at about 10 deg, T=6W.

Put the weight where the

tension is and vice versa,youve really got something!

A pulley attached to a lever.

This time its right. The lever

has the cable farther out than

the load. About 2 X

mechanical advantage due to

the lever. In addition, theblock and tackle gives

another 2X.


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This one is just a pulley. No

mechanical advantage,

although the tension isreoriented by the pulley.

An inclined plane a good

machine. With the angle at

about 30 deg (maybe a bitmore), TW/2. But the block

is sliding on the plane, so

friction might hurt if the goal

is to raise the load up theslope. Or maybe it helps hold

the load. Since the problem

statement says the goal is tohold the load, I want to say the

- in the is the operative

one.

An inclined plane a good

machine. With the angle atabout 30 deg (maybe a bit

more), TW/2. Ill assumethe rolling elements keep the

losses very low. This will

only work for a shortdistance though, then the

cylinder that comes out that

back needs to be reinserted at

the top.

Basic block and tackle, butgets it backwards. 1/2Xmechanical advantage because

the cable tension is applied

once to the weight, so thecable tension equals the

weight. But then that cable

tension is applied twice to the

pulley so we need T=2W

T=W

T=2W T=W/2 0.87W T=W/2


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5. (10 points) A sketching exercise

A) (5 points) Make a sketch from memory of a bicycle. One view is enough. Artistic rendering isnot needed, but proportion and functional detail is important. Focus on the features that make the

bicycle work properly. For example, be sure to show how the front wheel and fork are attached

to the rest of the frame and how is the chain is attached to the front and rear spockets.

B) (5 points) Go look closely at a bicycle. Critique your own sketch. Annotate with any differencesthat you regard as functionally important. Is there anything about a bicycle you noticed now that

you hadnt appreciated before?

I dont have a solution, but below is one nice student response.


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6. (15 points) Sketch a four bar mechanism that could guide the combination wrench

through the three positions indicated. Please use the attachment points indicated by the

small circle and triangle. Drawing by hand is fine if you do it carefully or use tools suchas a straight edge and compass or use a computer if you wish.

After finding the center points of the two pivots, we wanted you to draw a sensible

looking mechanism. If you just make the links straight, the links might collide with

the edge of the box, so we generally were hoping to see some countermeasure to thator maybe a note indicating a potential risk.

The method I used for the geometricconstruction was intercepting two

perpendicular bisectors.

homework1 solution v5

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