ME 3671 Design Project Report

The Ohio State University

Department of Mechanical Engineering

Team 10

Mohammad Fahad, Marquita Kenney, Sobin Kwon,

Subramanyam Mamillapalle, Kevin Pripusich,

Robert Sorensen, Alec Temes

2

Table of ContentsTable of Figures...............................................................................................................................3

Table of Tables................................................................................................................................4

Introduction......................................................................................................................................5

Design Procedure.............................................................................................................................6

Beams...........................................................................................................................................6

Dimensioning...........................................................................................................................6

Fatigue: Goodman Analysis...................................................................................................13

Mounting................................................................................................................................15

Shaft...........................................................................................................................................16

Force Analysis........................................................................................................................18

Key.........................................................................................................................................21

Bearings..................................................................................................................................21

Wire Ropes.................................................................................................................................24

Structure.....................................................................................................................................28

Bolts........................................................................................................................................28

Studs.......................................................................................................................................31

User Feedback............................................................................................................................33

Conclusion.....................................................................................................................................34

Appendices....................................................................................................................................36

Appendix A: Initial Design Sketches.........................................................................................36

Appendix B: Final Design Sketches and Diagrams...................................................................39

Appendix C: Bill of Materials and Product Spec Sheets...........................................................47

Appendix D: MATLAB Codes..................................................................................................49

Beam dimensioning, deflection, fatigue analysis, and stress analysis...................................49

Bolt Analysis..........................................................................................................................53

Stud Analysis..........................................................................................................................54

Appendix E: References.............................................................................................................55

Acknowledgments.........................................................................................................................55

3

Table of Figures

Figure 1: Overall Design of Exhibit................................................................................................5

Figure 2: Deflection of the different cross sectioned beams...........................................................8

Figure 3: Horizontal Rectangular Beam..........................................................................................9

Figure 4: Vertical Rectangular Beam..............................................................................................9

Figure 5: Vertical I Beam..............................................................................................................10

Figure 6: Horizontal I Beam..........................................................................................................10

Figure 7: Solid Square Beam.........................................................................................................11

Figure 8: Hollow Square Beam.....................................................................................................11

Figure 9: Solid Circular Beam.......................................................................................................12

Figure 10: Hollow Circular Beam.................................................................................................13

Figure 11: Horizontal rectangular beam with the mount in the back............................................16

Figure 12: Shaft, bearing, and wheel.............................................................................................19

Figure 13: Shaft shear force diagram XY-Plane............................................................................19

Figure 14: Shaft moment diagram XY-Plane................................................................................20

Figure 15: Diameter of the wire rope............................................................................................27

Figure 16: Bolts in the brackets attached to stud and ground........................................................30

Figure 17: Bending stress experienced by peg measured from centroid.......................................32

Figure 18: Pegs of the studs...........................................................................................................33

Figure 19: User feedback textboxes..............................................................................................34

Figure 20: Initial layout of front side.............................................................................................36

Figure 21: Initial layout of back side.............................................................................................36

Figure 22: Initial gear train for lever and shaft system..................................................................37

Figure 23: Initial mounting sketch.................................................................................................37

Figure 24: Initial hook to attach the rope to the beam...................................................................37

Figure 25: Initial overall design of beam in stud with shaft..........................................................38

Figure 26: Horizontal rectangular beam drawing..........................................................................39

Figure 27: Vertical rectangular beam drawing..............................................................................40

Figure 28: Vertical I beam drawing...............................................................................................40

Figure 29: Horizontal I beam drawing...........................................................................................41

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Figure 30: Solid square beam drawing..........................................................................................41

Figure 31: Hollow square beam drawing.......................................................................................42

Figure 32: Solid circular beam drawing........................................................................................42

Figure 33: Hollow circular beam drawing.....................................................................................43

Figure 34: Door wall drawing........................................................................................................43

Figure 35: Pulley of the shaft drawing..........................................................................................44

Figure 36: Input shaft drawing......................................................................................................44

Figure 37: Support wall (side panels) drawing..............................................................................45

Figure 38: Support wall (front and back panels) drawing.............................................................45

Figure 39: Wheel drawing.............................................................................................................46

Figure 41: Bearing and spec sheet.................................................................................................47

Figure 42: Price analysis of materials............................................................................................47

Table of Tables

Table 1: Moment of inertias for different cross sectional areas......................................................7

Table 2: Beams and their corresponding max stress and minimum yield strength required to prevent yielding.............................................................................................................................15

Table 3: Calculations for Factor of safety.....................................................................................20

Table 4: Radial and axial forces at bearings..................................................................................22

Table 5: Dynamic and Static Equivalent Radial Load...................................................................24

Table 6: Summary of Chosen Bearing for location A and C.........................................................24

Table 7: Rating factors for safety for the wire rope.......................................................................25

Table 8: Different radii for bolts and their corresponding stresses experienced...........................31

Table 9: Parameters of the bolts used throughout the exhibit.......................................................31

Table 10: Cost break down............................................................................................................47

5

Introduction

The design of beams has important consequences in a wide range of engineering fields.

Beams are used in a wide variety of applications, from nanostructures all the way to skyscrapers;

naturally, mostly everyone is familiar with beams. While most people are aware how the

material, length, and mass used in creating a beam impacts its function, many do not consider the

cross section and its often misunderstood role.

This exhibit allows users to compare how the cross section of a beam affects their ability

to deflect each with a given force. The beams have been specifically chosen to showcase how the

shape, amount of material, and orientation of the cross section impact its strength, thus fostering

a fundamental understanding of the concept of area moment of inertia. Guests will leave with a

greater insight into why engineers choose to use a certain type of beam in particular applications.

Figure 1: Overall Design of Exhibit

6

Design Procedure

Beams

DimensioningIn order to successfully complete the objective of teaching visitors how the cross section

of a beam affects deflection given a certain force, it was important to choose cross sections that

best displayed this concept. A MATLAB code was written by the team that calculates the

moment of inertia for different cross sections, and given a force and beam length uniform to all

beams, calculates each deflection. Below is a sample calculation showing how the moment of

inertia for the horizontal rectangular beam was determined:

I= 112

b h3= 112

( 4∗13 )=.333 i n4

Below is a sample calculation showing how the deflection of this beam was found, given

the material’s modulus of elasticity, length, and the force applied:

y= F l3

3 EI= (20 lb ) (3∗12inches )3

3(29∗104 lb / in2)(.333 in4)=3.2177 inches

The results were displayed on a bar graph to aid comparison. The main goal of this

process was to isolate other variables so that the team could see the effect on deflection solely

due to the cross section- therefore one consideration was to make each cross section have similar

outside dimensions, just as each beam uses the same material and has the same length. For

example, the circular beam should have a diameter close to the outside width of a square beam.

If these two dimensions were very different, a user might suspect the moment of inertia due to

the cross section doesn’t make a difference, only the amount of material used does. The square

and circle also were recreated with a hollow cross section while maintaining the same outside

diameters. The rectangular beam and I beams each had two variations, which had the exact same

dimensions, only rotated. These help to convey to the user a conceptual understanding of the area

moment of inertia, and the physical meaning behind it.

7

Table 1: Moment of inertias for different cross sectional areas

Figure 2: Deflection of the different cross sectioned beams

The implementation of this MATLAB code served not only to select the dimensions of

each cross section, but also to validate earlier design decisions made, such as the material used in

8

Figure 3: Horizontal Rectangular Beam

9

Figure 4: Vertical Rectangular Beam

Figure 5: Vertical I Beam

10

Figure 6: Horizontal I Beam

11

Figure 7: Solid Square Beam

Figure 8: Hollow Square Beam

12

Figure 9: Solid Circular Beam

13

Figure 10: Hollow Circular Beam

Fatigue: Goodman AnalysisTwo main criteria were the driving factors for the material selection of the beams: easily

deflectable and can withstand a year of fluctuating loads. Since the beams will experience

cyclical loadings, the Goodman Analysis combined with the Fluctuating Loads Analysis was

used to determine the minimum yield strength required by a certain material.

Since the beam needs to deflect easily, metal materials were immediately taken off of the

table and attention was turned towards polycarbonates. To start preliminary calculations,

Acrylonitrile butadiene styrene (ABS) was randomly selected. The following are material

specifications from SOLIDWORKS [2],

Su=Ultimate Strength=10700 psiSy=Yield Strength=7000 psi Roughly 627,833

people visited COSI in 2012 according to The Center of Science and Industry Annual Report [3],

so this number was used as a rough estimate for the number of fluctuations each beam would

experience. This number takes into account people who do not use the exhibit and people who

use it multiple times. The following are further parameters for the Goodman Analysis:

N= fluctuations=627,833 flucuationsSN=Stress at N cycles=1218.32 psi

The SN value was approximated from a graph created by Osswald [5]. The following

calculations are sample calculations for the horizontal cross-section rectangular beam.

Preliminary calculations were done on this beam first it would yield well before any of the other

cross sectional areas.

Dimensional parameters were selected to begin preliminary calculations. The moment of

inertia of the horizontal rectangular beam was calculated to be the following:

b=preliminary base length of cross section=4∈¿

h=prelimnary height of cross section=1∈¿

I= 112

(b∗h3 )= 112

( 4∗13 )=.333i n4 (Table 4.2)

Using this in conjunction with other preliminary parameters, the maximum stress was

determined:

14

F=preliminary applied force=50 lbL=lengthof beam=36∈¿

M=F∗L=50∗36=1800∈∙lbσ max=

M∗yI

=1800∗( 1

2 ).333

=2700 psi (4-4)

50 lbs is used in the calculations because it is an average amount of force a person can

apply [6]. To calculate the minimum stress, the weight of the rope had to be incorporated. This is

due to the fact that when the beam is at rest, the weight of rope will still apply a downwards

force, causing the beam to deflect even though there is no applied force. For further calculations

of the wire rope used, see section Wire Ropes on page 25.W rope=weight of rope=0.1938lb

M=F∗L=0.1938∗36=6.975∈∙lbσ min=

M∗ yI

=6.75∗( 1

2 ).333

=10.4625 psi (4-4)

It can be seen that the yield strength of ABS, 7000 psi, is larger than the max stress the

beam will experience, which is 2700 psi. Though this was a good indication that the beam will

probably not fail at this applied load, further analysis needed to be conducted. To use the

Goodman Analysis, several variables and parameters were calculated.

σ m=mean stress=σmax+σmin

2=2700+10.4625

2=1355.2 psi (Page 242)

σ a=amplitude stress=σ max−σmin

2=2700−10.4625

2=1344.8 psi (Page 242)

mt=Su−SN

Su=10,700−1218.32

10,700=.8861 (5-66)

Rt=σm

σmax=1350

2700=0.5019

(5-67)

Using these variables, the minimum yield strength required was calculated using the

Goodman Analysis.

Sy , min=SN

1−mt∗Rt= 1218.32

1−540∗0.5=2194.3 psi (5-68)

It can be seen that Sy,min, 2194.3 psi, is significantly smaller than the yield strength for

ABS, which is 7000 psi. This means that the horizontal rectangular cross-section beam will not

yield after about one years’ worth of fluctuations. This same analysis was applied to all of the

15

beams, and the values can be seen in the chart below. The code for this can be found in the

Appendix.

Table 2: Beams and their corresponding max stress and minimum yield strength required

to prevent yielding.

Beam Max Stress (psi) Minimum Yield Strength (psi)

Rectangular Vertical 675 2197.3

Rectangular Horizontal 2700 2197.3

I Vertical 851.93 2196.6

I Horizontal 2393.1 2189.8

Solid Square 400 2196.6

Hollow Square 2156.4 2189.6

Solid Circle 679.06 2196.6

Hollow Circle 2915.8 2189.8

Mounting

16

The initial approach to mounting the beams to the studs was to directly attach them by

using brackets, bolts, and washers. However, due to the fact that several of the beams have

circular cross sections, attaching them using a bracket did not seem feasible. Knowing that the

beams would be made out of the polycarbonate ABS, it was then determined to customize a

standard mount that was simply a part of the beam. The mount on the beam was sized to form an

I-shape and will fit into slots in the stud. Stress on the studs can be seen on page 32.

Figure 11: Horizontal rectangular beam with the mount in the back

ShaftIn order to determine the required diameter of the shaft, force and moment diagrams were

drawn which were further used in the shaft equation. Forces at end points of the shaft were

known by multiplying the density of the wheel and the pulley times their respective volumes.

Reaction forces at the location of the two bearings were calculated by using the equilibrium

condition in x-direction, y-direction, z-direction and moment. Force and moment diagrams for

the shaft are shown in the figure (13) and (14) respectively.

The distortional energy required diameter for shafts is given in (8-11) as

17

d=( 16π {2 K fb ( nd )

M a

SN+√ 3

T m

Su })1/3

(8-11)

Where d = required shaft diameter for N cycles

nd = selected design safety factor

Kfb = fatigue stress concentration for bending

Ma = maximum bending moment

Tm = maximum torsional moment

SN = fatigue strength for a design life of N cycles (infinite life)

Equation for Fatigue Analysis

The shaft is designed for infinite life hence, SN is the fatigue strength and is equal to Sf for infinite

life.

SN=S f=k∞ S 'f

k ∞=¿ (5-57)

where S 'f =fatigue strength limit of specimen

k ∞=infinite life adjustment factor

k gr= Grain size and direction = 1.0 (typical value)

k we = Welding = 1 (no welding required)

k f = Geometrical discontinuity = 1.0 (uniform shape)

k sr = Surface condition = 0.75 (From Figure 5.33)

k rs = Residual surface stress = 1.0

k fr = Fretting = 1.0 (no fretting anticipated)

k cr= Corrosion = 1.0 (no corrosion anticipated)

k sz = Size effect = 0.9 (From Table 5.3)

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k sp = operating speed = 1.0 (From Table 5.3)

k r = Strength reliability required = 0.87 (95% reliability)

S ' f=0.5 Sut at N=106 if Sut ≤200 ksi

For Sut=105 ksi, S ' f=0.5 Sut=0.5 (105) = 52.5 ksi

Since the shaft is not experiencing high angular velocity, calculations for critical speed is not

needed.

Force AnalysisEquilibrium condition in x-direction, y-direction, z-direction and moment were used to

calculate forces.

Moment at point C ∑ M C :(50+1.6) (12.75 )+(50−2.9) (2.25 )=12 C y

C y=¿59.67lbs ↑

Forces in y-direction

∑ F y :50−50+59.67−1.6−2.9+A y=0

A y=55.17 ↓

Forces in x-direction ∑ F x : A x+C x=80

Ax=40 lbs

C x=40lbs

∑ F z=0 :

19

Figure 12: Shaft, bearing, and wheel

0 2 4 6 8 10 12 14

-80

-60

-40

-20

0

20

40

60

-55.17

-8.07

51.6

Shaft shear force diagram XY-Plane

Length (in)

Shea

r for

ce (l

bs)

Figure 13: Shaft shear force diagram XY-Plane

20

0 2 4 6 8 10 12 14

-350

-300

-250

-200

-150

-100

-50

0 0

-124.13

-274.13

-330.62

-150

0Shaft moment diagram XY-Plane

Length (in)

Mom

ent (

in-lb

s)

Figure 14: Shaft moment diagram XY-Plane

Table 3: Calculations for Factor of safety

No. Rating factorSelected Rating

Number (RN)

1. Accuracy of loads knowledge 2

2. Accuracy of stress calculations 1

3. Accuracy of strength knowledge 0

4. Need to conserve material -3

5. Seriousness of failure consequences -3

6. Quality of manufacture 0

7. Conditions of operation 0

8. Quality of inspection/maintenance -2

21

t=∑i=1

8

(RN )i (2-85)

t=0+0+(−2 )+(−4 )+(−3 )+0+0+0=−5

nd=1+(10+t )2

100for t ≥−6 (2-86)

nd=1+(10+(−5))2

100=1.25

Now with the factor of safety known, along with our maximum torque and moment, we can

calculate the diameter of the shaft using equation (8-11).

d=( 16π {2(1)(1.25 ) 330.62

26218+√ 3 150

104982 })1/3

d=0.597

Using the next largest nominal size for steel shafts we acquire a shaft diameter of 0.625 inches.

Required shaft (R258-4140 Alloy 4140 Steel round-Annealed) was chosen from

metalsdepot.com which costs $12.50 for 2 feet length.

KeyIn order to secure wheel to the shaft, a square parallel key was used. Diameter of the shaft

was used to determine the required length of the keys using equation (8-30).in.

leq−st=πD

2K tτ (8-30)

Where K tτ=torsional stress concentration factor (Table 8.3¿book )

D = Diameter of the shaft at key

K tτ = 1.7 (for static loading) and the square key size, according to Table 8.1 in the Textbook for a

0.5 in shaft diameter, is 3/16

leq−st=(π ) (0.625 )(2)(1.7)

=0.577∈¿

22

Bearings The radial and axial forces on the shaft were determined for each bearing, as shown in the table

below:

Table 4: Radial and axial forces at bearings.

Bearing Position Radial Force (lbs) Axial Force (lbs)

Bearing 1 (At A) 55.17 40

Bearing 2 (At C) 59.67 40

In order to choose a suitable bearing, the dynamic equivalent radial dynamic and static

loads (Pe)d and (Pe)s, the reliability factor KR, and the impact factor (IF) was found. With a

reliability of 95%, KR was determined using the Table 11.2 which was 0.62. Impact Factor (IF)

for this application was determined using Table 11.3 which was 1.2 assuming light impact for

this application.

Tentatively, deep groove ball bearings were selected which can support moderate radial

loads and thrust loads in both directions. The characteristics of this bearing clearly falls under the

design criteria. Dynamic load factors are Xd1=1, Yd1=0 and Xd2=0.55, Yd2=1.45 and static load

factors are Xs1=1, Ys1=0 and Xs2=0.6, Ys2=0.5. The above values were plugged into the following

equations below.

Dynamic equivalent radial load :(P ¿¿e)d=Xd F r+Y d Fa ¿ (11-3)

Static equivalent radial load: (P¿¿e )s=X s F r+Y s Fa ¿ (11-3)

Sample radial load calculation for bearing at position A:

(P¿¿e )d1=(P ¿¿e)s 1=(1 ) (55.17 )+(0 ) ( 40 )=55.17 lb ¿¿

(P¿¿e )d 2=(0.55 ) (55.17 )+(1.45 ) (40 )=88.34 lb ¿

(P¿¿e )s 2=(6 ) (55.17 )+(0.5 ) ( 40 )=53.07 lb ¿

23

Sample radial load calculation for bearing at position C:

(P¿¿e )d1=(P ¿¿e)s 1=(1 ) (59.67 )+(0 ) ( 40 )=59.67 lb ¿¿

(P¿¿e )d 2=(0.55 ) (59.67 )+(1.45 ) (40 )=90.82 lb¿

(P¿¿e )s 2=(6)(59.67 )+(0.5 ) ( 40 )=55.80 lb¿

From the values calculated above, highest values of (P¿¿e )d ¿ and (P¿¿e )s¿ were selected for

bearings at position A and C respectively.

Next, the design life of the bearing was to be calculated. A three year life is expected (153

weeks) and to be conservative, it was assumed that every single person attending the exhibit will

cause one revolution of the shaft, meaning they will interact with the exhibit twice and cause

maximum beam deflection each time. Using that 658,000 people visited COSI in 2013 we can

calculate the design life using the following equation below.

Ld=153

weeks∗658,000 people51 weeks

∗1rev

person=1.974 x106 revolutions

With the design life and the maximum radial dynamic and static loads calculated, radial dynamic

and static load ratings were calculated using the following equations.

¿ (11-4)

¿ (11-4)

Where ¿=required radial dynamic load rating to give a bearing reliability of R percent

¿=required radial static load rating to give a bearing reliability of R percent

Ld=design life, revolutions

Knowing that a ball bearing will be chosen, therefore a=3. Plugging in all the values, following

results were obtained.

For bearing A;

24

¿

¿

For bearing C;

¿

¿

Table 5: Dynamic and Static Equivalent Radial Load

Bearing Position Cd (lb) Cs (lb)

Bearing 1 (At A) 155.95 97.34

Bearing 2 (At C) 160.47 105.34

In addition to figuring out the key dimensions, the shaft diameter was used to find a suitable

bearing. A suitable deep groove bearing was chosen from Motion Industries (Model SKF RLS 5)

and the cost is only $31.83 [4]. This bearing has the ability to take higher dynamic and static

equivalent radial which makes it suitable for this application.

Table 6: Summary of Chosen Bearing for location A and C

Bearing Internal

Diameter (in)

Outside

Diameter (in)

Width (in) Cd (lbs) Cs (lbs)

SKF RLS 5 5/8 1-9/16 7/16 2150 1068

Wire RopesThe team designed a wire rope that will be attached to the beam’s end on a hook and the

other end of the rope will be wrapped around a pulley with a radius of 3 inches. The pulley is

attached to a 5/8” shaft, where the user will turn a wheel of the same diameter as the pulley to

induce a deflection in the beams.

25

By using this mechanism, the user will effectively apply a direct force of 50 lbs to the

beams without receiving any backlash from directly pulling on the rope [6]. The design of this

rope demanded a rope that is not heavy but still strong enough to accept 100 lbs. of force without

failing. 100 lbs. is twice the magnitude of the estimated force applied by the users on the beam. It

is used to ensure the safety of the rope under extreme circumstances.

The initial guess for the rope is the 6*37 because the rope will be sheaved onto a 6”

drum and that requires the rope to be extremely flexible. A 6*7, or a 6*19 are not as flexible

because of the fewer number of outer strands. 8*19 is thicker than the other three ropes so it will

be harder to wrap around as well. Based on the judgement, it is assumed that the nstatic is 5 and

nfatigue is 2.00. The method for choosing the factor of safety for the wire rope is below

Table 7: Rating factors for safety for the wire rope

No.1 Rating Factor Selected RN

1 Accuracy of loads knowledge -2

2 Accuracy of stress knowledge +1

3 Accuracy of strength knowledge 0

4 Need to Conserve -1

5 Seriousness of failure consequences +3

6 Conditions of operation 0

7 Quality of manufacture +1

8 Quality of inspection and maintenance -2

t=−2+1+0−1+3+0+1−2=0 (2-85)

nd=1+(10−0 )2

100=2.00=n fatigue (2-86)

The next step is to check if the rope selected will work under static conditions. Static

analysis tells that at how much pressure the rope will start stretching.

For static analysis:

26

σ= TA r

=

1000.427

∗1

dr2 =234.19

dr2

(17-28)

σ=Su

nstatic=200000

5=40,000 psi

40000 psi is pressure at which the rope starts to fail and starts stretching. This is the

design constraint that must be taken under consideration. The designed diameter should not let σ

go over 40000 psi.

dr=√ 234.1940000

=0.0765

This diameter can be tested with the equation below:

σ=dw

ds∗Er=

( 0.076518 )∗9.9∗106

18∗0.0765 ¿30562.2 psi (17-31)

Since the 30562.2 psi is within 40000 psi, the diameter of the rope should be fine at

0.0765 inches.

The next test is to see at what diameter the rope doesn’t fail under multiple cycles. It is

estimated by COSI that there are about 625,000 visitors a year on average and assuming that

there will be 6*105 cycles on the rope a year, the following calculations have been done below.

Using Figure 17.17 from the text, the approximate fatigue strength parameter RN ≈0.0018 @

625,000 cycles. RN is an estimate from the table and has some uncertainty associated with it.

¿ (17-33)

¿

¿

From the Table 17.10 for a 6*37 rope on a cast carbon-steel sheave and the allowable

bearing pressure is ¿¿

¿

27

The defining factor is the fatigue diameter and that is 0.1852”. However the smallest

standard size rope diameter is ¼ inch diameter for a 6 *37 IPS fiber core (FC) rope. This is the

rope that will be used in the exhibit.

Figure 15: Diameter of the wire rope

The maximum tension this rope can withstand under axial load is:

T=σ∗Ar=40,000∗0.427∗0.252=1067.6lbs

The rope is intended to withstand at least 100 lbs., but it can take 1067.6 lbs. This means

that the rope is only using 9.4% of its maximum axial load rating.

The next step is to check if the standard size diameter will sustain the static test again:

σ=dw

ds∗Er=

( 0.2518 )∗9.9∗106

18∗0.25 ¿30555.56 psi (17-31)

This value is still less than 40,000 psi, so the new diameter passes the static analysis test.

The maximum tension that the rope can withstand to be under fatigue pressure ¿ can be

calculated as:

T=¿

28

The maximum fatigue tension the rope can withstand is 138 lbs. The rope will only

experience 72.5% of this maximum tension and will not fail under this axial load.

The maximum tension that the rope can withstand under wear would be:

T=¿¿¿

The maximum wear tension the rope can experience is 885 lbs. Under the axial loading

conditions of the exhibit, the rope will only experience 11.3% of this maximum value.

The dominating factor for the wire rope is fatigue condition which allows the rope to take

up to 138 lbs. of force, so with the selected wire rope (6 *37 IPS fiber core (FC) rope) is more

than adequate to support the task.

StructureBoltsSince the bolts will experience fluctuating bending moments, a cyclic stress analysis was applied

to determine the size of the bolts that should be used. Low-carbon alloy steel (AISI 4620) was

chosen to begin preliminary calculations.

SU=Ultimate Strength=87,000 psi First, the stress experienced at N cycles was calculated.

This value represents the maximum stress the bolts can withstand for one year. To do this, the

fatigue strength of a polished specimen was first calculated by using the following:

Sf' =0.4 SU=0.4 (87,000 )=34,800 psi (at N f =106 cycles ) ( Pg 248 of text )

Using this, the fatigue strength was determined using (5-55)

Sf =fatigue strength=k ∞ S f'

Where k∞ is the infinite life adjustment factor. This factor is defined as (5-57).

k ∞=( kgr kwe k f ksr k sz krs k fr kcr k sp k r )∞The following parameters are strength influencing factors. These are determined by

physical parameters of the bolt.

k gr=grain ¿direction=1.0( typical value)k we=welding=1.0 (No welding anticipated)

k f=geometrical discontinuity=1.0 (Uniform shape)k sr=surface condition=0.7(¿ figure 5.33)

k sz=¿¿0.9k rs=fretting=1.0k cr=corrosion=1.0k sp=operating speed=0.9

29

k r=strengthreliability required=0.9 (90 % reliability ) Using this information, the fatigue

strength can be calculated.

k ∞=(1.0 ) (1.0 ) (1.0 ) (0.7 ) (0.9 ) (1.0 ) (1.0 ) (0.9 ) (0.9 )=0.5103Sf =k ∞ S f' =(0.5103 ) (34,800 )=17,758 psi

From this, the stress at N cycles can be calculated by estimating the S-N curve for this

material (Page 245).

N f =cycles at fatigue strength=106 cyclesN=flucuations per year=627,833 cycles

SN=SU−SU−S f

log ( N f )log ( N )=87000−87,000−17,758

log (106 )log (627,833 )=20,091.371 psi This is the

maximum stress that the bolts can experience before failure after one year. Once this value was

determined, the actual stress experienced by the bolts was calculated. First, the bending moment

experienced by the bolts was determined.

F=applied force=50 lbL=length of beam=36∈¿

M=F∗L=50∗36=1800 lb ∙∈¿

Figure 16: Bolts in the brackets attached to stud and ground

The moment of inertia of the front bolts were calculated by doing the following:

I=∑ Ai y i2=Ab (1.5 )2+ Ab (1.5 )2=4.5 Ab(13−38)

30

Where Ab=area of bolt=π rb2

Using this, the bending stress was determined by (4-4).

σ= MyI

=1800∗34.5 Ab

=1200Ab

The radius, rb, was varied from 0.125 inches to 2 inches to determine the stress each bolt

size would experience. The following is a table of the stress experienced in the bolts between

0.125 in to 0.5 in.

Table 8: Different radii for bolts and their corresponding stresses experienced

Radius (in) Stress Experienced (psi)

0.125 3167.013

0.25 770.711

0.375 340.861

0.5 191.406

From this table, it can be seen that the smallest radius experiences a load of 3167.013 psi,

which is significantly smaller than the maximum stress of 20,091.371 psi. Since this is the

maximum stress any of the bolts will experience on the structure, these calculations determined

the parameters of the bolts that will be used throughout the exhibit.

Table 9: Parameters of the bolts used throughout the exhibit

Diameter (in) Material SAE Grade

0.25 Low-carbon steel (AISI) 1

31

A preload force in the bolts was then calculated to help guarantee they do not carry the entire

applied load. The criterion outlined the textbook (Page 498) states that the preload should bring

the fastener to 75% of yield strength:

F=.75∗S y A

From Table 13.3, the yield strength of the chosen material was found to be 36 ksi. With the

diameter of the bolts known, the initial preload was calculated as follows:

F=.75∗S y A=.75∗(36∗103 lbi n2 )¿

StudsA bending stress analysis was done to the pegs of the studs to ensure that the studs do not

snap when the load is applied to the end of the beam. The front face of the pegs have dimensions

of 1.5 inches by 4.5 inches. It can be assumed that the studs each take half of the load that is

being applied. Using this information, and the other parameters of the exhibit, the bending stress

along the pegs can be determined by,

M=(50 lb2 )∗¿I peg=1.5∈¿1.53i n3∗1

12=11.3906 i n4 The stress at each point from the centroid

was then calculated by varying y in the following equation (4-4):

σ= MyI

Where y is the distance from the centroid of the front face of the peg to some point. The

stress at each point can be seen in the following graph.

-2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5-200

-150

-100

-50

0

50

100

150

200Bending Stress vs. Distance from Centroid of Peg

Distance from Centroid (in)

Stre

ss (p

si)

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Figure 17: Bending stress experienced by peg measured from centroid

It can be seen that the top of the peg experiences about 200 psi of tensile pressure and the

bottom of the peg experienced about 200 psi of compressive pressure. These values were

important in allowing the group to determine what material could be used in the pegs and would

not fail.

Figure 18: Pegs of the studs

User FeedbackAs the users turn the wheel to deflect the beam, they will be able to observe their

deflection by comparing it to the tick marks that will be on the Plexiglas. The tick marks on the

Plexiglas will be spaced to show not only how much they are deflecting the beam by, but also

show how much force they are applying. In addition to these markings, there is an optosenor that

is placed near the pulleys of each shaft. On the pulley will be a dark tick mark. When the dark

tick mark passes by the optosenor, it will trigger the optosensor and make the beam’s

corresponding textbox light up. This will teach the users about different applications of the

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beams. Each text box will have some sort of fun fact about the beam of interest. These textboxes

can be seen in the figure below.

34

Figure 19: User feedback textboxes

35

ConclusionThe design process for the beam exhibit began by first deciding the parameters of each

beam’s cross section in order to best convey differences between different shapes. After

establishing the cross sections to be used, the housing for the beams were modeled in

SolidWorks. Each beam was paired with another and together held by a stud which in turn was

bolted to the floor of the exhibit by use of a bracket. The bolts were designed to withstand the

cyclic stress imparted on them, and a .25 in diameter low-carbon steel SAE Grade 1 was chosen

throughout the exhibit. Material for the beams was chosen specifically for its low modulus of

elasticity, making it easy for museum guests to deflect the weaker beams but still relatively

difficult for those such as the solid square beam. Fatigue was analyzed through the Goodman

approach on the beams in order to guarantee they would last a year of typical museum

attendance.

The shaft used to turn the rope was designed with a factor of safety of 1.25 and diameter of the

shaft came out to be 0.597” after fatigue analysis for infinite life. Nominal size shaft of 0.625”

was selected from metalsdepot.com which is made of 4140 alloy steel. Attached to the shaft is a

pulley, held in place on the shaft by a square parallel key. Length of the key was determined to

be 0.577”.

Considering the fact that 658,000 people visited COSI in 2013, design life for the bearing was

calculated to be 1.974 x106revolutions. Bearings were designed to support radial and axial loads

and the most suitable bearings used to hold the shafts was determined to be a deep groove ball

bearings bought from Motion Industries Model SKF RLS 5. The rope connecting the shaft to the

pulley was designed to withstand the force required to bend the beams a maximum distance

corresponding to 50 pounds [6].

The wire rope is another critical component that was designed carefully to withstand large

amount of force, more than twice the force (138 lbs.) than anticipated. The designed wire rope is

a 6*37 IPS (FC) with a standard size diameter of ¼ inches. The rope must be flexible to be

sheaved on a 6” diameter drum and this rope can do it because of its large number of outer

strands.

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Appendices

Appendix A: Initial Design Sketches

Figure 20: Initial layout of front side

Figure 21: Initial layout of back side

37

Figure 22: Initial gear train for lever and shaft system

Figure 23: Initial mounting sketch

38

Figure 24: Initial hook to attach the rope to the beam

Figure 25: Initial overall design of beam in stud with shaft

39

Appendix B: Final Design Sketches and Diagrams

Figure 26: Horizontal rectangular beam drawing

40

Figure 27: Vertical rectangular beam drawing

Figure 28: Vertical I beam drawing

41

Figure 29: Horizontal I beam drawing

Figure 30: Solid square beam drawing

42

Figure 31: Hollow square beam drawing

Figure 32: Solid circular beam drawing

43

Figure 33: Hollow circular beam drawing

Figure 34: Door wall drawing

44

Figure 35: Pulley of the shaft drawing

Figure 36: Input shaft drawing

45

Figure 37: Support wall (side panels) drawing

Figure 38: Support wall (front and back panels) drawing

46

Figure 39: Wheel drawing

Figure 40: Support stud

47

Appendix C: Bill of Materials and Product Spec Sheets

Figure 41: Bearing and spec sheet

Figure 42: Price analysis of materials

58%

11%

16%

9%

2% 3% 1% 0%Price Anlaysis

BeamsGroundPlexiglassBracketsStudKeyBoltsPulley&WheelShaft and bearingsWire Rope

Table 10: Cost break down

Please note that many of the materials such as the beams, Plexiglas, and walls (ground) will be bought in bulk and then machined to size in Team 10’s machine shop.

Appendix D: MATLAB Codes

Beam dimensioning, deflection, fatigue analysis, and stress analysis disp('ME 3671 Design Project')disp('Calculations of beams')disp(' ') quit = 0;defaultans = {'29*10^4','12*3','50','4','1','3','3','.25','3','2.85','3','2.8076','0'}; while quit==0 prompt = {'Elastic Modulus (psi)','Beam Length (inches)','Force (lb)','Rectangular Beam Width','Rectangular Beam Height','I Beam Base Width','I Beam Height','I Beam Thickness','Square Beam Outer Width','Square Beam Inner Width','Circular Beam Outside Diameter','Circular Beam Inside Diameter','Quit?'};dlg_title = 'Input';answer = inputdlg(prompt,dlg_title,1,defaultans);defaultans = answer;E = str2num(answer{1}); % Elastic modulus (psi)l = str2num(answer{2}); % Length of beam (in inches)% Rectangular BeamF = str2num(answer{3}); % Force applied at end of beam (N)br = str2num(answer{4}); % Base length of beam (in inches)hr = str2num(answer{5}); % Height of beam (in inches)% I beambi=str2num(answer{6}); % Base length b of beam. Width (inches)di=str2num(answer{7}); % Total height of the beam (inches)ti=str2num(answer{8}); % Thickness of the middle section of the I. Cannot exceed bi. (inches)% Square Beamas = str2num(answer{9}); % Length or height of the outer square cross section (inches)bs = str2num(answer{10}); % Length or height of the square inner hole (inches)% Circular Beamdoc= str2num(answer{11}); % Outside diameter "do" of circular beam (inches)dic= str2num(answer{12}); % Inside diameter "di" of circular beam (inches)% Quitquit= str2num(answer{13}); % Fatigue Parameters% Goodman analysisN=627833;Su=10700; %[psi] Ultimate tensile strength of ABSSN=1218.32; %[psi] SN if it is a normal beam for 1 year of fluctuationsmt=(Su-SN)/Su;%s_min=0; %minimum bending stress experienced by beam L_rope=2; %[ft] length of ropeW_rope=0.096875*L_rope; %[weight of rope]

50

disp('Goodman Fatigue Analysis') % Rectangular Beam A_rect=br*hr; % cross-sectional area (inches squared) I_rect_vert=(hr*br^3)/12 %inertia for rectangular beam in vertical position (in^4)I_rect_hor=(br*hr^3)/12 %inertia for recangular beam in horizontal position (in^4)y_rect_vert=(F*l^3)/(3*E*I_rect_vert); %vertial displacement of rectangular beam in vert position (inches)y_rect_hor=(F*l^3)/(3*E*I_rect_hor); %vertical displacement of rectangular beam in hor position (inches) s_max_rect_vert=((F*l)*(br/2))/(I_rect_vert) %maximum bending stress experienced by beam in vertical positions_max_rect_hor=((F*l)*(hr/2))/(I_rect_hor) %maximum bending stress experienced by beam in horizontal positions_min_rect_vert=((W_rope*l)*(br/2))/(I_rect_vert) %minimum bending stress experienced by beam in vertical positions_min_rect_hor=((W_rope*l)*(hr/2))/(I_rect_hor) %minimum bending stress experienced by beam in horizontal position s_m_rect_vert=(s_max_rect_vert+s_min_rect_vert)/2 %bending stress magnitude in vert positions_a_rect_vert=(s_max_rect_vert-s_min_rect_vert)/2 %bending stress amplitude in vert positions_m_rect_hor=(s_max_rect_hor+s_min_rect_hor)/2 %bending stress magnitude in hor positions_a_rect_hor=(s_max_rect_hor-s_min_rect_hor)/2 %bending stress amplitude in hor position Rt_rect_vert=s_m_rect_vert/s_max_rect_vertSy_rect_vert=SN/(1-mt*Rt_rect_vert) %minimum yield strength required by beam in vertical position Rt_rect_hor=s_m_rect_hor/s_max_rect_horSy_rect_hor=SN/(1-mt*Rt_rect_hor) %minimum yield strength required by beam in horizontal position % I Beam hi=di-2*ti; % This assumes that the thickness of the middle part is the same as the thickness of the top and bottom platforms A_i=bi*di-hi*(bi-ti); % Cross-sectional area (inches squared) I_i_vert=(bi*di^3-hi^3*(bi-ti))/12; I_i_hor=(2*ti*bi^3+hi*ti^3)/12; % double checkedy_i_vert=(F*l^3)/(3*E*I_i_vert); %vertial displacement of i beam in vert position

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y_i_hor=(F*l^3)/(3*E*I_i_hor); %vertical displacement of i beam in hor position s_max_i_vert=((F*l)*(di/2))/(I_i_vert); %maximum bending stress experienced by beam in vertical positions_max_i_hor=((F*l)*(bi/2))/(I_i_hor); %maximum bending stress experienced by beam in horizontal positions_min_i_vert=((W_rope*l)*(br/2))/(I_i_vert) %minimum bending stress experienced by beam in vertical positions_min_i_hor=((W_rope*l)*(hr/2))/(I_i_hor) %minimum bending stress experienced by beam in horizontal position s_m_i_vert=(s_max_i_vert+s_min_i_vert)/2; %bending stress magnitude in vert positions_a_i_vert=(s_max_i_vert-s_min_i_vert)/2; %bending stress amplitude in vert positions_m_i_hor=(s_max_i_hor+s_min_i_hor)/2; %bending stress magnitude in hor positions_a_i_hor=(s_max_i_hor-s_min_i_hor)/2; %bending stress amplitude in hor position Rt_i_vert=s_m_i_vert/s_max_i_vert;Sy_i_vert=SN/(1-mt*Rt_i_vert); %minimum yield strength required by beam in vertical position Rt_i_hor=s_m_i_hor/s_max_i_hor;Sy_i_hor=SN/(1-mt*Rt_i_hor); %minimum yield strength required by beam in horizontal position % Square Beam A_squ_sol=as^2; % Cross-sectional area of solid rectangle (inches squared)A_squ_hol=as^2-bs^2; % Cross-sectional area of hollow rectangle (inches squared) I_squ_sol=as^4/12; %inertia of solid squareI_squ_hol=(as^4-bs^4)/12; %inertia of hollow squarey_squ_sol=(F*l^3)/(3*E*I_squ_sol); %vertial displacement of solid square beamy_squ_hol=(F*l^3)/(3*E*I_squ_hol); %vertical displacement of hollow square beam s_max_squ_sol=((F*l)*(as/2))/(I_squ_sol); %maximum bending stress experienced by solid beams_max_squ_hol=((F*l)*(as/2))/(I_squ_hol); %maximum bending stress experienced by hollow beams_min_squ_sol=((W_rope*l)*(br/2))/(I_squ_sol); %minimum bending stress experienced by beam in solid beams_min_squ_hol=((W_rope*l)*(hr/2))/(I_squ_hol); %minimum bending stress experienced by beam in hollow beam s_m_squ_sol=(s_max_squ_sol+s_min_squ_sol)/2; %bending stress magnitude in solid beams_a_squ_sol=(s_max_squ_sol-s_min_squ_sol)/2; %bending stress amplitude in solid beam

52

s_m_squ_hol=(s_max_squ_hol+s_min_squ_hol)/2; %bending stress magnitude in hollow beams_a_squ_hol=(s_max_squ_hol-s_min_squ_hol)/2; %bending stress amplitude in hollow beam Rt_squ_sol=s_m_squ_sol/s_max_squ_sol;Sy_squ_sol=SN/(1-mt*Rt_squ_sol); %minimum yield strength required by solid beam Rt_squ_hol=s_m_squ_hol/s_max_squ_hol;Sy_squ_hol=SN/(1-mt*Rt_squ_hol); %minimum yield strength required by hollow beam % Circular Beam A_cir_sol=(pi*doc^2)/4; % Cross-sectional area of solid circle (inches squared)A_cir_hol=(pi*doc^2)/4-(pi*dic^2)/4; % Cross-sectional area of hollow circle (inches squared) I_cir_sol=(pi*doc^4)/64; %inertia of solid circleI_cir_hol=(pi*(doc^4-dic^4))/64; %inertia of hollow circley_cir_sol=(F*l^3)/(3*E*I_cir_sol); %vertial displacement of solid circular beamy_cir_hol=(F*l^3)/(3*E*I_cir_hol); %vertical displacement of hollow circular beam s_max_cir_sol=((F*l)*(doc/2))/(I_cir_sol); %maximum bending stress experienced by solid beams_max_cir_hol=((F*l)*(doc/2))/(I_cir_hol); %maximum bending stress experienced by hollow beams_min_cir_sol=((W_rope*l)*(br/2))/(I_cir_sol); %minimum bending stress experienced by beam in solid beams_min_cir_hol=((W_rope*l)*(hr/2))/(I_cir_hol); %minimum bending stress experienced by beam in hollow beam s_m_cir_sol=(s_max_cir_sol+s_min_cir_sol)/2; %bending stress magnitude in solid beams_a_cir_sol=(s_max_cir_sol-s_min_cir_sol)/2; %bending stress amplitude in solid beams_m_cir_hol=(s_max_cir_hol+s_min_cir_hol)/2; %bending stress magnitude in hollow beams_a_cir_hol=(s_max_cir_hol-s_min_cir_hol)/2; %bending stress amplitude in hollow beam Rt_cir_sol=s_m_cir_sol/s_max_cir_sol;Sy_cir_sol=SN/(1-mt*Rt_cir_sol); %minimum yield strength required by solid beam Rt_cir_hol=s_m_cir_hol/s_max_cir_hol;Sy_cir_hol=SN/(1-mt*Rt_cir_hol); %minimum yield strength required by hollow beam Beams={'Rectanglar Vert.';'Rectangular Hor.';'I Vert.';'I Hor.';'Square Sol.';'Square Hol.';'Circular Sol.';'Circular Hol.'};

53

MaxStress=[s_max_rect_vert;s_max_rect_hor;s_max_i_vert;s_max_i_hor;s_max_squ_sol;s_max_squ_hol;s_max_cir_sol;s_max_cir_hol];MinYield=[Sy_rect_vert;Sy_rect_hor;Sy_i_vert;Sy_i_hor;Sy_squ_sol;Sy_squ_hol;Sy_cir_sol;Sy_cir_hol];Table=table(MaxStress,MinYield,'RowNames',Beams) y = [y_rect_vert,y_rect_hor,y_i_vert,y_i_hor,y_squ_sol,y_squ_hol,y_cir_sol,y_cir_hol];x = ['Vertical Rectangular','Horizonatal Rectangular','Vertical I','Horizontal I','Solid Square','Hollow Square','Solid Circle','Hollow Circle'];closeif quit==0figure(1)bar(y)set(gca,'XTickLabel',{'Vertical Rectangular','Horizonatal Rectangular','Vertical I','Horizontal I','Solid Square','Hollow Square','Solid Circle','Hollow Circle'})ylabel('Deflection (Inches)')end end

Bolt Analysis clear; clc; disp('Bolt Analysis')Su=87000;Sff=Su*0.4ksr=0.7;ksz=0.9;kr=.9;ksp=.9;K=kr*ksp*ksr*kszSf=K*SffNf=10^6;N=627833; SN=Su-((Su-Sf)/log10(Nf))*log10(N);fprintf('Max Stress bolt can experience: %f psi\n',SN) r=[0.25:0.25:4]/2; fprintf(' Radius (in) Stress Experienced (psi) \n')for k=1:1:length(r); sigma_max=1200/(pi*r(k)^2); stress=(sigma_max-(sigma_max/2))/(1-(sigma_max/2)/Su); fprintf(' %f %f \n',r(k),stress)end

54

Stud Analysisclear; clc; disp('Stud Analysis')disp(' ') a_peg=1.5; %[in] front face length of peg on studb_peg=1.5; %[in] side width of peg on studh_peg=4.5; %[in] height of peg on studF=50; %[lb] force being applied to beamL=3*12; %[in] length of beamy=linspace(h_peg,0,200); for k=1:length(y) M=(F/2)*(L+b_peg); I=a_peg*h_peg^3*(1/12); sigma(k)=(M*(y(k)-(h_peg/2)))/I;end plot(y-(h_peg/2),sigma)title('Bending Stress vs. Distance from Centroid of Peg')xlabel('Distance from Centroid (in)')ylabel('Stress (psi)')

55

Appendix E: References

[1] J. A. Collins, et al., Mechanical Design of Machine Elements and Machines: A Failure

Prevention Perspective, 2nd ed. Hoboken, NJ: Wiley, 2010, pp. 123-794.

[2] SOLIDWORKS (2014). Edit Materials: Properties. Waltham, MA.

[3] Center of Science and Industry (2012). Center of Science and Industry Annual Report

[Online]. Available: https://www.cosi.org/downloads/financial/2012-annualReport.pdf

[4] "Motion Industries", Motionindustries.com, 2016. [Online]. Available:

https://www.motionindustries.com/productCatalogSearch.jsp.

[5] T. A. Osswald, G. Menges, Materials Science of polymers for Engineers, 2nd ed. Cincinnati.

OH: Hanse Publishers, 2003, pp. 490-494.

[6] ASH Answers Fact Sheet (2016). Canadian Center for Occupational Health and Safety

[Online]. Available: https://www.ccohs.ca/oshanswers/ergonomics/push1.html

Acknowledgments

Team 10 would like to thank Professor Robert Siston, Professor Blaine Lilly, and Professor Sandra Metzler for their insightful knowledge, input, and help.