Appendix E

APPENDIX E

CALCULATIONS

Introduction:

The following are all calculations, mechanical, computer and electrical for the MARTHA device. The calculations are organized by what system they correspond to. All common assumptions are to be made, unless otherwise stated (i.e. g=9.81 m/s2).

E1

PVC Analysis

PVC Material Specifications:Sut=10500 psiSy=8500 psiE = 490 ksi

Chain Material Specifications: 4130 Normalized alloy steelSut= 97200 psiSy=63100 psiE = 29700 ksi

Screw Specifications: #6-32 ¾” Machine Screwd = 0.138”dr = 0.0974”N = 32At = 0.0119 ¿2

Sut = 25000 psiSp = 22479 psiSy = 22000 psiE = 1490 ksiAB = 0.138(0.5) = 0.069¿2

l = 0.5”P = 15 lbF i = .9Sp A t = (.9)(0.0119)(22479) = 241 lb

Kb=EAt

l+E

Ab

l=

0.0119 (490000)0.5

+0.069 (490000)

0.5=135000

Km=Am

Em

l=

0.069 (490000)0.5

=67600

C=K b

KmK b

= 135000(135000)(67600)

=0.67

Pb=CP=( .67 ) (15 )=10.05 lbPm= (1−C )P= (1−.67 ) 15=4.95 lbFb=F i+Pb=241+10.05=251.05lbFm=F i−Pm=241−4.95=236.05 lb

σ b=Fb

At

=251.050.0119

=21100 psi

σ m=Fm

Ab

=236.050.069

=3421 psi

N y=S y

σb

=2200021100

=1.04

E2

P0=Fi

(1−C)= 241

(1−.67)=730.3 lb

N separation=P0

P=730.3

15=49

Front wheel: (6/15) 13 = 5.2 lbBack wheel: (11/15) 13 = 7.8 lbFront:5.2lb/2 wheels = 2.6 lbσ = F/A = 2.6/0.036 = 72.2 psiBack:7.8lb/2 wheels = 3.9 lbσ = F/A = 3.9/0.036 = 108.3 psi

E3

Chassis Analysis

Chassis material: 6061-AluminumΡ = 0.0975 lb/¿2

Sut=45000 psiSy=40000 psiE = 10000 ksi

Screw: #8-32 1” machine screwd = 0.164”N = 32dr=0.1234

At=0.0140¿2

Sut=25000 psiSy=22000 psiE = 1490 ksiSp=22500 psil = 0.8”Ab=(0.164 ) (0.8 )=0.1312¿2

Chassis-Screw Connection:P = 12 lbF i=.9S p A t=283.5 lb

Kb=EAt

l+E

Ab

l=88935

lb¿

Km=AmEm

l=

(0.044 ) (100000000 )0.8

=440000lb¿

Am=( π4 ) (D−d )2=( π4 ) (0.4−0.164 )2=0.044 i n2

Pb=CP=(12 ) ( 0.17 )=2.04 lbPm= (1−C )P=9.96 lb

P0=F i

(1−C )= 283.5

(1−0.17 )=341.57 lb

N separation=P0

P=28.5

Fb=F i+Pb=285.55 lbFm=F i−Pm=273.54 lb

C=Kb

Km+Kb

=0.17

σ b=Fb

At

=285.550.014

=20396 psi x 28 screws=728.43 psi per screw

E4

σ m=Fm

Ab

=2084.9 psi

N y=S y

σb

= 22000728.43

=30.2

Shear on side screws:

F=mg= (12 lb )( 9.81m

s2 )=117.72N

A=0.014 i n2

σ shear=117.720.014

=8408.6 psi x 8 screws=1051.075 psi per screw

E5

Motor Mount Analysis (Screw Connections)

Material Specifications: 6061-Aluminum

ρ=0.0975lb

i n3

Sut=45000 psiSy=40000 psiE=10000ksi

Bolt Material: ¼” x 1” 307 Grade A Steel – SAE Grade 5, medium carbon, cold drawnHead width: 3/8”lt=0.75Sy=92ksiSut=120ksiSp=85ksiE=30000ksiAt=0.032 in2

Ab=π (0.25 )2=0.2

Connections Relating to Drawing:1. L-Bar to Chassis2. L-Bar to Motor Face Plate3. L-Bar to Bearing4. Top L-bar to Plate with ABS Plastic5. Motor Face Plate to Motor

L-Bar to Motor Al face connection:Same Preload Specs as #8-32 x 1” screw listed aboveC = 0.17P = 1.8 lbF i=283.5 lb

At=0.014 i n2

Ab=0.1312 in2

Pb=CP=0.306 lbPm= (1−C )P=1.494 lbFb=F i+Pb=283.806 lbFm=F i−Pm=282.006 lb

σ b=Fb

At


σ m=Fm

Ab

=2149.44 psi

E6

N y=S y

σb

=1.08

P0=F i

(1−C )=341.57

N separation=P0

P=190

Shear on screw head:F = 1.8 lb x 9.8m/s2=17.66 NA = 0.014 i n2

σ shear=17.600.014

=1261.4 psi x2 screws=630.7 psi per screw

L-Bar to bearing connection:Material: SAE 1840 BronzeBearing housing: 6061-AluminumSame Preload Specs as #8-32 x 1” screw listed aboveP = 1 lbC = 0.95Pb=CP=0.95 lbPm= (1−C )P=0.05 lbFb=F i+Pb=2448.95 lbFm=F i−Pm=2447.95 lb

σ b=Fb

At

=76530 psi x 2 screws=38265 psi per screw

σ m=Fm

Ab

=12239.75 psi

N y=S y

σb

=1.2

P0=F i

(1−C )=37210 lb

N separation=P0

P=48960

L-Bar to Chassis bolt connection:Specifications for bolt preload listed aboveF i=0.9S p A t=2448lblthread=2d+0.25=0.75ls=lbolt−lthread=1−0.75=0.25lt=l−ls=0.75

Kb=EAt

l+E

Ab

l=2.53 x107 lb

¿

Km=AmEm

l=1.25x 106 lb

¿

E7

Am=18x 0.25=0.09375 i n2

C=Kb

Km+kb = 0.95

P = 2 lbPb=CP=1.9lbPm= (1−C )P=0.1lbFb=F i+Pb=2449.5 lbFm=F i−Pm=2447.5 lb

σ b=Fb

At

=76547 psi x2 screws=38273.5 psi per screw

σ m=Fm

Ab

=26107 psi

N y=S y

σb

=1.2

P0=F i

(1−C )=48960 lb

N separation=P0

P=24480

L-Bar to Aluminum Motor face connection with ABS Plastic Spacer:ABS Specifications:Sut=5500 psiSy=3000 psiE = 270 ksiσ

|¿|=Fm

A=6216.8 psi :Compressive ¿

#8-32 ScrewsP = 1 lbF i=.9S p A t=283.5 lb

Kb=EAt

l+E

Ab

l=88935

lb¿

Km=AmEm

l=

(0.044 ) (100000000 )0.8

=23760lb¿

Am=( π4 ) (D−d )2=0.044 i n2

Pb=CP=0.79 lbPm= (1−C )P=0.21lb

P0=F i

(1−C )= 283.5

(1−0.17 )=1350 lb


E8

σ b=Fb

At


N y=S y

σb

=4.65

N separation=P0

P=1350

Aluminum face to Motor connection:#10-24 1.5” screws:d = 0.19”dr=0.1359N = 24At=0.0175 i n2

Sy=22000 psiSut=25000 psiE = 490 ksiAb=Π (d2 )=0.028 i n2

Am=π Deff

2

4=0.042 i n2

Deff=0.23F i=.9S p A t=354.375 lb

Kb=EAt

l+E

Ab

l=17150

lb¿

Km=AmEm

l=

(0.042 ) (100000000 )1.3

=319567lb¿

C=Kb

Km+kb = 0.05

Pb=CP=0.09 lbPm= (1−C )P=1.71lb

P0=F i

(1−C )=373 lb


σ b=Fb

At

=20250 psi x 2 screws=10125 psi per screw

N y=S y

σb

=1.1

N separation=P0

P=207.2

E9

Motor Mount Analysis (Bearing & Motor)

Motor Specs: 48:1 ratioLoad speed = 33 rpmLoad torque = 280 oz-in = 17.5 lb-in.0.5” diameter, 1.75” long motor shaft, Do=0.5 , {D} rsub {i} =0.25

τ=TρJ

=17.5(0.25)3.8 x10−4 =11513.2 lbf=τ max

J=π (Do−Di )4

32=3.8 x 10−4

Set Screws: 10-32 set screws ½” d set screw=0.165

At=0.0175∈.2

d p=d−( 0.645192 )=0.163

Sy=22000 psiAlloy steel: F = 311.376 N

σ t=FAt

=1.78x 104 psi

N y=S y

σ t

=1.24

J=π (0.165 )4

32=7.3 x 10−5

Set Screw¿1 :τ=17.5 ( 0.165

2 )7.3 x 10−5

=19777.4 psi−¿

Set Screw¿2 :τ=17.5 (1 )

7.3 x 10−5=2.4 x 105 psi−¿

Hydrodynamic Bearing Analysis:ld=0.75

Lube thickness: h = 0.0017”Abs. viscosity = η = Uρ = 3.5E-6n = 33 rpm = 1980 rpsA = π (0.25)(0.25) = 0.19i n2

U = πdn = 1555 in/secCd=dn=0.000425

C r=Cd

2=0.0004=e

On=20

E10

ε= eC r

=0.747

ε x=0.21394+0.385170¿

K ε=O n

4 π=1.592

τ x=ηUh

=3.2 psi F=A τ x=6.08 lb

Average oil pressure: Pavg=Pld

=17 psi

Stationary :τ s=ηd3 ln π2

Cd (1−ε2 )0.5=0.45lb−¿

θmax=cos−1 1−√(1¿+24 ε2)4 ε

=159degrees¿

P= ηUr C r

2

l2

43 εsinθ

(1+3cosθ )3=1670.3 psi

ϕ=tan−1( π √1−ε2

4 ε )=35degrees

Rotationg :τ r=τ s+Pe sinϕ=0.8332lb−¿Power lost∈bearing :Φ=2π τ r (925 )=0.006 HP

Coefficient of Friction : μ=2 τ rPd

=0.004

N= 0.80.25

=3.2

E11

Gear Assembly

Bending Stress on Drive Gear:

σ b=W t Pd

FJ

K aKm

K v

x K s Kb K l

For series 116-4 DC Motor: w = 33 rpm, T = 280 in-oz = 17.5 lb-inF = 0.25”N p=8a = 0.3”

d=N p

Pd

Do=N+2P

=Pd+2a

3.5=Pd=2 ( .3 ) Pd=2.9

d= 82.4

=2.76

V t=dw2

=(2.76 ) (33 rpm )

22 π12

=23.85ftmin

Qv=6¿8=7K v=0.96J=0.2Km=1.6K a=1.25 (assume moderate shock terrain )K s=K l=1 (non−idler )

W t=2T p

d p

=2 (17.5 )

2.76=12.68 lbf

σ b=(12.68 ) (2.9 )

( .25 ) ( .2 )(1.25 ) (1.6 )

(0.96 )=1532.16 psi

Al-6061: S f b'=14 x103 psi

S fb=K L

KT K R

S f b'

K r=1.25 KL=1.6831 (N )−0.0323=1.073 K t=1(roomtemperature)

N=30rpm( 60minhr )(3 hrs

day )( 4 daysweek )( 52weeks

year ) (1 year )=1.123 x106 cycles

S fb=1.073

(1 ) (1.25 )(14 x 103 )=12017.6 psi

N fb=S fb

σb

=7.84

E12

Gear Shaft AnalysisSteel 1050: Sut=90ksi S y=50ksiT m=17.5 lb−¿T a=0M a=Mmax

F=Tr= 17.5

2.762

=12.68 lbf

ΣF y :R1−F−F=0 , R1=2 F=25.36 lbf

V=−R1< x−0¿0+F<x−1¿0+F<x−1.5¿0

M= ∫ V=−R1<x−0¿1+F<x−1¿1+F< x−1.5¿1=31.7 lbfSe '=.5Sut=45ksi

C load=1 (bending )C¿ ¿0.869 (d )−0.097 =0.994Csurf=A (S ut )

b=0.569 ¿

C temp=1C reliab=0.814 (assume99 % reliabilitiy )Se=( 45 ) (0.994 ) (0.569 ) (0.814 ) (1 ) (1 )=20.7188ksi

d=((32 N sf

π )(√(K f M a )2+ 34

(K fsT a )2

20.7188k+ √(K fmM a )2+ 3

4(K fsmT a )2

Sut)

13 )

d=0.25 {S} rsub {ut} =90k {N} rsub {sf} =9.19

σ max=MmaxC

I

I= 112

bh2=3.255x 10−4 i n4

σ max=12.2ksiwhich is less than Sut so i t' s safe

τ max=TpJ

J=bh1

12(b2+h2 )=6.51x 10−4

τ max=3.5ksi

E13

Surface Fatigue Internal Square

σ c=Cp√ wb

FIdCaCm

C v

C sC f

F=0.25 {W} rsub {t} =12.68 lb

I=( sinϕcosϕ2 )(N ¿¿ p) 1N gN p

=0.08¿

C p=1950 psi ( steelon aluminum )Cm=K m=1.6Ca=1C v=¿A=50+56 (1−B )B=.25 (12−Q v)

.667=0.7314A=65.04….C v=1.27C s=K s=C f=1 ( conventionalmethods )σ c=33727 psi

N fs=( S fc

σ c)

2

S fc=CLCH

CTCR

S f c '

CL=2.466 (N )−.056=1.13C R=K R=1.25 (99 % reliability )CH=CT=1 (RoomTemperature )S f c '=60¿70ksi=65ksiS fc=58760 psiN fs=3.035Backlash : gap=0.001%T increse onshaft :0.0010.25

=.005=4%

τ=TpJ

pnew=.251

2=.1255J new=( .2512

12 ) (.2512+ .2512)=0.003969

τ=33200 psi , τ old=3360 psi33603320

=101.2 %

E14

Spring Analysis

Spring Constant:19.11 lb = limitRobot max weight = 20 lb

F=−kx x=3.5mmF=50 N k= 50 N3.5mm

=14.28Nmm

Type :extension springsShear on Axle:MaxT∈spring=19.11 lbf

σ=MyI

ΣM=19.1 (7 )+19.1 (11 )−F2 (18 ) F2=19.47 lbfΣF y=−F1+19.1+19.1−19.47 F1=18.73 lbfMmax=131.11

0 2 4 6 8 10 12 14 16 18 20

-30

-20

-10

0

10

20

30

Shear Diagram

Length (inches)

V (

shea

r)

Figure 1

0 2 4 6 8 10 12 14 16 18 20

-140-120-100

-80-60-40-20

0

Moment Diagram

Length (inches)

Mom

ent

Figure 2

E15

σ=(131.11 )( .25

2 )I

I=1.917 x10−4 in4

σ max=85.47ksi

E16

Wheels & Axles Analysis

|Plastic|: Sut=4000 psi S y=1400 psi20lbf

4=5 lbf per wheel

σ= FAπ (5 )2

4=A=19.63 in2

σ=2546 psi per wheel

E17

Wire Cutting Device Analysis

Motor Specs:Produces2300 rpm /voltVoltage given :12VAmpsDrawn :5 A2300 rpm/V x12V=27,600 rpmPower=VxI=12x 5=60watts=0.0805 HPRadius of Cutting Blade=1.5∈¿

Torque=rxF=33000 (HP )

2π (rpm )=0.1535 ft−lb=1.838 lb−¿ .

F=Tr=1.838

1.5=1.225 lbf

Wire Specs:

Awire=0.0075 in2Strength=18 lb

in2

Force ¿cutwire=Strength x Awire=0.135lbf

E18

Soil/Air Collector Analysis

Servo Specs:Torque = 38 oz.-inForce required closing lid of air collector:Lid = 2.1 oz.Alid=π (r )2=π ( .5 )2=0.785 i n2

Servo Arm = 4” long

Torqueused ¿close lid=τ=38oz .in2

4∈¿=9.5oz .−¿ .¿Required τ ¿close lid=1.05oz .−¿ .

E19

MATLAB Code Calculations%%%Tyler Maydew - LAW OF COSINES clcclose all %%%This section calculates the angle. This does not change regardless of%position. a=sqrt((n1-n2)^2+(w1-w2)^2);b=sqrt((n2-n3)^2+(w2-w3)^2);c=sqrt((n3-n1)^2+(w3-w1)^2);A=acos((b^2+c^2-a^2)/(2*b*c)); theta=180-(A*180/pi);%%%This set of code is to determine whether we need to spin clockwise or%counterclockwise.dn32=n3-n2;dn12=n1-n2;dw32=w3-w2;dw12=w1-w2;alpha=atan(dn32/dw32)*180/pi;beta=atan(dn12/dw12)*180/pi;%This will find theta for line 3 off horizontal of 2if dn32>=0 if dw32<=0 theta2=abs(alpha); else theta2=180-alpha; endelse if dw32>=0 theta2=180+abs(alpha); else theta2=360-alpha; endend%this will find theta1 for line 1 off horizontal of 2if dn12>=0 if dw12<=0 theta1=abs(beta); else theta1=180-beta; endelse if dw12>=0 theta1=180+abs(beta); else theta1=360-beta; endend if theta1>=theta2

E20

direc=true;else direc=false;end %%%Distance Calc n=n1-n3;nd=n*0.000277777778;nm=111028.2811574018*nd;w=w1-w3;wd=w*0.000277777778;wm=wd*85803.10402713598;d=sqrt(nm.^2+w.^2);

E21

Documents

Appendix E