Machine Design 1 Homework 5

8/10/2019 Machine Design 1 Homework 5

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PAH

FLORIDA ATLANTIC

UNIVERSITY

E LM

4500

—

Machine

Design 1

Homework 5

D r.

Guoqiang

C ai

Ionathan

Padilla

0 4 / 2 2 / 2 0 1 4

’ | a E | 9 § ' I Y l J )

L I -1'1 I ‘ - \

I3>Sm;n\:j j



,,/

EML 4500

-

M A C H I N E DE S IG N I

Spring 2014

HOMEWORK 5

(Assigned:

04/10, Due:

04/22)

A

trailer hitch

of

a tractor

h a s a multiview sketch in Fig. 1 . T he

Dimensions

of its

ball

bracket

a r e

given in Fig.

2 . The

weight

of the trailer

is 2 000 kg .

T wo

bolts

a re u s e d

to attach the bracket

to

the tractor. T he

tractor

is capable to

accelerate

to 20 m/s in 1 0 seconds with a kinetic fi' iction

coefficient 0.1. Design the two

bolts using

Class

8 . 8

steel bolt (Table

15.7,

p . 914) fo r all safety

factors

ofat least 1 . 5 , and specify the

tightening

torque

for installation. Assume

1 .

the

thickness

of the tractor channel

is

30

mm,

. - - ~ ¢ - 'r F i . 1

2. the

fatigue

life

is infinite with

50 % reliability. g

Print

the

following

data:

_

\. _

1 .

Bolt diameter, class number, I b o “ ,

1 ,,

l t h d .

2 .

Joint stiffness constant. 1 _ . _ . . . . . . . .. ‘ ~

“ “

'°'° ”'

, _ _ ; E

1

P f ~ * + J

c a

fly )

3 .

The

percentage

of the initial s t r e s s

a s the proof

strength. 5

_

_ £

_ _ , , , , , , _ m _ . , , _ _ - J - 8

4 . Initial, mean and alternating forces a t

e a c h

bolt. ' L s

5 .

S t r e s s

concentration factors

kf

and k f . , , _ 5

,1 ’

A

A g

6 .

Initial,

mean and alternating normal s t r e s s e s . fifill-jg £2 3

7 . All

correction

factors,

uncorrected

and

corrected -

. , ,. , . .\ . -» . » _ ' , . -, » , - ~ = r . .| - L _ _ _ _ _ _ _ “ , _ _ _ _ , _

fatigue strengths.

‘C

X

8 . Three factors of

safety.

9. Tightening

torque.

no ‘

wf___ 351:5

. : ; . . . . . ; . .

I

_ , p W ‘

-I--

5 4 4-

3 ’

2 2

-I--»

I I

1

III

_

/’

\'

i

ht’

2

I I F» .

* ~ . , _

L § . _ i _ _ _ , . »

2- - I l g

‘R E T i t ? - Q ‘ amEmit

' “ ~ ~ - W . 4 5 4 ‘ . - ' . l . . L E { 1 ‘ > F A I M S H

H

1 , , 1 ,, 5

um

, _ _ _



T o begin,

the bolt

diameter, Dbm, the

class

number,

the length

of

the

bolt, Ibo“,

length

of the

shaft

ls, and

the

length ofthe

thread,

l thd. These

design

components can

be determined

b y u sin g th e

following

tables.

a b l e

154 P l °lPa| Dimensions

of

I S O Metric

Standard

Screw Threads

Data C a l c u l a t e d

from

E q u a tio n s 1 5

1-See

R e f e re n c e 4

for

More Information

_ . , , _ A , C ° a ' § § ? T h r e a d s

F f fl e

T519345

_ , ,

Mal° Pitch Minor” Tensile

I

Pitch

Minor

T 8 1 5 1 5 9

Diameter P Diameter S t r e s s Area p

Diameter

Stress Area

dlmm) mm rl ( mm) A,

(MIT12)

mm d, (mm) /1 ; (m m 2)

I7W_ I

I

I T— 77 7 _;-

_____7 ._¢ —— ND

3 0 ‘

3.5

4.0

5.0

6.0

7.0

8.0

10.0

12.0

14.0

16.0

18 .0 -

20.0

.50

0.60

0.70

0.80

1 . 00

1 . 00

1 . 2 5

1 .5 0

1 . 7 5

2.00

2.00

2.50

2.50

2 . 3 9

2 . 7 6

3 . 14

4.02

4 . 77

5 . 77

6.47

8 . 1 6

9 . 8 5

11 . 55

13 . 55

14 . 9 3

16.93

5.03

6.78

8 . 7 8

14 .18

20 . 12

2 8 . 8 6

36.61

5 7 . 9 9

8 4 . 2 7

1 1

5 44

156.67

1 9 2 . 4 7

2 4 4 . 7 9

1.00

1 .25

1 .2 5

1 .50

1.50

1.50

1.50

6.77

8.47

10.47

1 2 . 1 6

14 . 16

16.16.

1 8 . 1 6

3 9 . 1 7

5 1 . 2 0

9 2 . 0 7

1 2 4 . 5 5

1 5 2 . 2 5

2 1 5 6 . 2 3 :

2 7 1 . 5 0

Table 15-7

Metric Specifications an d Strengths for Steel Bolts

Size

Range

M in imum M inimum M inimum

Outside

Proof Yield T ensile

Class

Diameter Strength

Strength

Strerzgh

N u m b e r

( m m )

( M P a )

_

( r a r e ;

_

{ M F ‘ a ) W

‘M a t e r ia l

4.6 M5-M36

4.8 M 1 .6— M1

5. 8

M5— N\24

8 .8

i f

M 3 ? - M 3 6

6

9 .8

M 1

.6— M16

Now that the

initial dimensional values

are

determined,

steps

towards

finding

the joint stiffness

constant

can

now

be taken.

2 2 5

31 0

3 8 0

60 0

65 0

240

340

42 0

660

7 2 0

T he

first

step

is to

solve

fo r

the joint

aspect

ratio, j.

Where

ldamp [m] is the length of

the

clamp.

_ Dbolt

[clamp

400

42 0

52 0

8 3 0

9 00

low or medium

carbon

low or med ium

carbon

low or

medium

carbon

medium carbon.

Q &T

medium carbon, Q &T



Using

th e following tab le to

solve

fo r

the P values.

T a b l fl 15-3 Parameters for Equation

15. 19l15l

0. 1 0

0.20

0.30

0.40

0.50

0.60

0.4389

-0.61 1 8

0.6932

0 .7 3 5 1

0.7580

0.7709

- 0 . 9 1 9 7

- 1 . 1 2 1 - 5

-1 .2426

-1.2612

-1 .2632

-1 .2600

0 .89 0 1

1 . 0 8 7 5

1 . 1 1 7 7

1 . 1 1 1 1

1 . 0 9 7 9

1.0851.

T he

P

values,

p3 ,

p 2 , p l, an d p o ,

can

be found using interpolation.

_ _ _ ]_.ltb1

P

—

Prb, + (Pro,

_

Prbl) jtb

_

1 - t b

2

1

- 0 .31 87

-13.3806

-0 .3845

-0.3l77’9

- 0 .3708

-0.3647

Now

that

the

P

values have been calculated,

joint

stiffness,

C ,

can

now

be

calculated

Where pfl,

is

the

P

value from Table

15-8,

and jtb

is

the joint

aspect

ratio

from

T a ble 1 5 -8

T he

percentage

of the initial stress a s the proof

strength

was found to be 90%.

In

order

to solve fo r the initial force, F i,

can

be

calculated

using.

W h ere A ,

is

the tensile

stress

area from

Table

1 5 -2 , a nd 5, ,

is the

proof strength

from T a ble 1 5 -7

C=P3+P2' l 'P1+Po

1

F,=-§><A,><S, ,

T he normal force, F , , , , , . , , , ,

applied

to

the

trailer

is equivalent

to

the

weight of the trailer.

T he

friction force,

F , - , . , , , , can

be

fou nd b y the

following

equation

Where

,u

is the friction

coefficient.

T he acceleration force can

be

found

by,

Where m T , . a i , , , , . [N] is the mass of the trailer, and a T , . , , C t , , , . is

the accelerat ion

of the tractor

Ffric :1 + Fnorm

Faccel =

mTrailer

X aTractor

T he max force on the

tractor, Fmaxr

[N] can

be

found using the

following

equation.

Fm;- \xT = Faccel + Ffric



T he

max

force

on

the

bolts can b e

calculated

by, ;

F

Hh X _i1?.7i

bolt

F m a x B

=

Nbolts

Where

Hh

is

the

height

of

the

hitch,

and

Nbolts

is

the

number

of

bolts.

f

Now that

the max force on the bolts is found,

alternating

force

can b e

calculated

1

Fa=ExCxFmaxB

The mean force.

F m , can

be

calculated

using the

following

formula

F m = F , + F a

T h e a x ia l fo rc e can b e calculated b y th e following equation.

F 1 ,

= F , +C > < F m a x g

T h e a x ia l

stress can b e found by ,

.

Ub a

T he alternating stress can b e found by ,

I

F0.

act =

It

[Pa]

The mean stress can

be

found by,

I

Fm

am =It

Pa]

The stress

concentration

factors can now

be

found

Kf

=

X X Dbolt

K’ (Utnax

)n0m

<

S)‘

S

I

-K n n

Kfm ”

J f(Ua)

O ’ > Kf(a1nax

)nom > Sv

(am)nom

I

0 , K f ro- , , , , , , ) , , , , , , ( 1 — R ) >

2 5 , .

T he

initial

normal stress

can

be found

by,

F

i

0'; =Kfm X

It

T he alternating normal stress can

be found

by,



0,, = 0 , ; > < K ; [Pa]

T he mean normal stress can be found by,

a m = o ' , ' , , > < Kfm [Pa]

T he

size

correction

factor

is

found to

be

C s i z e =

1

The surface

correction factor is fou nd to b y

c , , , , . , = 4 . 5 1

> <

s , ; , ° - 2 6 5

T he temperature

correction

factor,

C t e m p ,

is

1 because it is

at

room temperature.\

T he reliability correction factor is found to be 1 .

T he

load

correction

factor,

Cload,

is

found to

b e 0.7

Now

that

all

the

correction factors have been found, the correction fatigue strength

can

now be

determined.

Se

:

load

X Cs ize

X

Ctemp X

Creliab

X Csurf X

S e

Next,

the

three

factors

of

safeties can

be

calculated.

T he

fatigue factor

of safety

ca n be

found by ,

N

Sut “Ui

I

S, ,><(a, ,,— a,~)+S, , ,><a, ,

T he safety factor

against yielding

can b e

found by ,

S

N ,

=

- 8

T he

bolt separation

safety

factor

is

found by ,

Sa p _ F m a x B X(1_ C)

Lastly, the

tightening

torque can be

found using the

following

equation.

Tr

X Dbojt X Fi



Wnd

indow

1

Of

1

Homework

5

1 ) Bolt Diameter:

0.018 [m ]

Bolt Class:

8.8

Bolt Length: 0.065 [m]

Shaf t

Length:

0.023

[ m ]

(:\\

Thread Length: 0.042 [m ]

2)

Joint

stiffness

constant:

0.199

3 )

Intia Stress

Percentage as the Proof Strengt :

9 0

Per

ent \\

( £7

4)

Inta force:

57.741

[kN]

(/7

M e a n

force: 58.782

[kN]

Alternating

force: 1.041

[kN} I

5) Stress Concentration Factors

kf = 6.183

kfm

= 2.052

6)

Intia

Normal Stress:

615.476

[MPa]

M e a n

Normal

Stress: 626.569 [MPa]

Alternating

Normal

Stress:

33.431

[MPa]

7 ) Csize: 1.0

Csurf:

0.76

Ctemp:

1.00

Cload:

0.70

Creliab: 1.00

Uncorrected Fatigue

Strength:

415.00 [MPa]

Corrected Fatigue

Strength: 220.6820 [MPaT

i

8) Bolt Fatigue Safety Factor: 1.568

Bolt

Safety

Factor due to Yielding:

2

Bolt Separation

Safety Factor: 6.913

9) Tightening Torque: 218.261 [N*m]

\_J,

l— ' I \ . ) (J O

x M _ ‘ _ _ _ _

>>



___1g_g_,+

_,g__“#_g

20

l\M1

H W 5 . m M

1

of

m g

clear;

clc;

Jonathan

Padilla

8 EML 4500 - Machine Design 1

Homework 5

fPrintf('Homework 5\n\n\n‘);

8

Mass Of traler

MassTrail=2000;

[kg]

Number

of

bolts

NBOlts=2;

Tractor acceleration

aTrac=2;

[m/s“2]

Friction Coeffecient

mu=0.l;

Gravity

g=9.8l;

[m/s“2j

Weight of

the

traler

WTrail

=

MassTrail*g;

[N]

Channel thickness

Tch=0.03;

[m]

8

Height of the

hitch

Hh=0.07;

[m]

Hitch

thickness

Th=0.0l9;

[m]

Hitch base to

center

bolt

length

Lbase=0.031; [m ]

Hitch

bottom

to

center

of bolt length

Lbott=0.02; [m]

From Table

15-2,

major diamter

Db=0.018;

[m]

From

Table 15-7,

proof strength

Spb=600e6; [Pa]

From Table 15-7, yield

strength

Syb=660e6; [Pa]

From Table 15-7

Sutb=830e6;

[Pa

From Table 15-2,

tensile strength

area

Atb=0.000l9247;

[m“2]

Cross

Sectional

area

Ab=(pi/4)*Db“2;

[m“21

Length

of

clamp

Lcb=Th+Tch; [m l

Bolt

length

Lb=0.065;

[ml

Modulus of Easticty

Eb=206.8e9;

[Pal

From

Table 15-8,

Parameters for

stiffness

jtb=[0.l00.20 0.30 0.40 0.50 0.60];

tensile strength

l _ _ _ I §

pOtb=[O.4389

0.6118 0.6932 0.7351 0.7580 .7709];

p1tb=[—0.9l97 —.17l5 —1.2426 —.26l2 —l.2632 —.2600];



p2tb=[0.890l

1.0875

1.1177

1.1111

1.0979 1.0851];

P3tb=l-0.3187 -0.39 06 -0.38 45 ~o.37v9

-0.3708 -0.36471;

Uncorrected

fatigue

strength

Seprime=0.5*Sutb:

Normal

Force

applied to trailer

Fnorm=WTrail; [N3

Friction

force

Ffric=mu*Fnorm: [N]

Acceleration

force

Facc=MassTrail*aTrac; [N]

a

M ax force

on

tractor

FmaxTrac=Facc+Ffric;

[N ]

Max force

on

bolts

FmaxB0lt=(Hh*FmaxTrac/Lbott)/NBolts; [N]

Pre-loa intia

force

FinitBolt b.5* pb*Atb; { N }

Joint

as ratio

j=Db/Lcb;

Determining

P

values by interpolating

p3=interpl(jtb,p3tb,j);

p2=interpl(jtb,p2tb,j);

p1=interp1(jtb,p1tb,j);

pO=interp1(jtb,pOtb,j);

Joint stiffness

C=p3+p2+p1+pO;

Axial

force

Fb=FinitBolt+C*FmaxBolt; [N]

Alternating

Fa=(l/2)*C*(FmaxBolt); [N ]

Mean force

Fm=FinitBolt+Fa;

[N]

Axial

stress

sigmab=Fb/Atb;

Alternating

stress,

prime

sgmaa=Fa/Atb;

Mean

stress, prime

sigmam=Fm/Atb;

Max nominal stess

sigmamax=sigmaa+sigmam;

Stress concentration factor

kf=S.7+0.02682*Db*1e3;

if

kf*sigmamax<Syb

kfm=kf;

else

kfm=(Syb—(kf*sgmaa))/sgmam;

end

Corrected alternating stress

sigmaac=sigmaa*kf;

Corrected mean

stress

sigmamc=sigmam*kfm;

Corrected

inta stress



O

4 / 2 2 / 1 4 3:58

P M

Z:\M D l\M D1 H W 5 . m 3 of 3

sigmaic=kfm*(FinitBolt/Atb);

CSize=1;

Csurf=4.5l*((Sutb*1e—6)“~0.265);

Ctemp=1;

Creliab=1;

Cload=0.7;

Corrected

fatigue

strength

Se=Cload*Csize*Csurf*Ctemp*Creliab*Seprime;

Thread length

lthd=2*Db+0.006;

Bolt shaft length

ls=Lb—thd;

Material thread length

lt=Lcb—ls;

Fatigue

safety

factor

Nf=(Se*(Sutb—sigmaic))/(Se*(sigmamc~sigmaic)+(Sutb*sigmaac));

Safet

factor

against yielding

Ny=Syb/sigmab;

Safety

factor

separation

Nsep=FinitBolt/(FmaxBolt*(l—C));

Tightening torque

Tt=0.21*Db*FinitBolt;

fprintfn

fprintf|

fprintfl

fprintf<

fprintfl

fprintf<

fprintf<

fprintf<

fprintfn

fprintfl

fprintfi

fprintf|

fprintfl

fprintf|

fprintf<

fprintf|

fprntf1

fprintfl

fprntffl

fprntfi

fprntffl

fprntffl

fprntffi

fprntffl

F

i

T

|

F

\

F

i

w— *‘

i,

rfi i

7*

7*‘

‘

fprntffi

fprintf(

11)

I

I

I

J 1 ‘ : L » J f \ J

U 7

G \

\l

- - . - r

I

I

:8}

fprintf(‘9)

Bolt Dameter:

.3f

[m]\n',Db);

Bolt Class:

8.8\n‘);

Bolt

Length:

.3f [m]\n‘,Lb);

Shaft Length:

.3f

[m]\n',ls);

Thread

Length:

.3f [m]\n',lthd);

Joint stiffness

constant:

%.3f\n',C);

Intia Stress

Percentage as

the Proof

Strength:

9 0 Percent

\n')

Inta

force: 4.3f [kN]

\n,

FinitBolt*.OO1);

Mean

force:

4.3f

[kN]

\n, Fm*.0O1);

Alternating force: 4.3f [kN] \n, Fa*.0O1);

Stress

Concentration

Factors\n‘);

kf = 3f\n,kf);

kfm

=

3f\n,kfm;

Intia

Normal

Stress: .3f [MPa]\n',sigmaic/1000000);

Mean

Normal

Stress:

.3f

[MPa]\n',sgmamc/1000000);

Alternating

Normal

Stress: .3f [MPa]\n',sigmaac/1000000);

: %1.lf \n,

Csize);

: %3.2f

\n,

Csurf);

:

3.2f

\n,

Ctemp);

Cload:

3.2f

\n, Cload);

Creliab: 3.2f

\n, Creliab);

Uncorrected Fatigue

Strength: 5.2f [MPa]\n',Seprime/1000000);

Corrected Fatigue

Strength: 5.4f

[MPa]\n',Se/1000000);

Bolt Fatigue Safety Factor:

%.3f\n‘,Nf);

Bolt

Safety Factor

u

to

Yielding: %3f\n‘,Ny);

Bolt Separation Safety Factor: %3f\n',Nsep);

Tightening Torque: .3f

[N*m]\n\n',Tt);

Csize

Csurf

Ctemp

Documents

Machine Design 1 Homework 5