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Page 1: Me307 machine elements formula sheet

1. Stress Analysis

Moment of Inertias

1. Atalet moment of inertia; 2. Polar moment of inertia;

2

xI y dA 2

yI x dA 2 2

( )zJ x y dA

Shape Ix Iy J

Rectangle bh3/12 hb3/12 2 2

12

bhb h

Triangle bh3/36 hb3/36 2 2

18

h bbh

Circle πd4/64 πd4/64 πd4/32

Stresses

Normal Stresses Shear Stresses

Axial

Tensile F

A

Torsional

Tr

J

3

16 /T d for solid circular beam Compression F

A

Bending

b

Mc

I

3

32b

M

d

for solid circular beam

Transverse

(Flexural)

VQ

Ib , Q A y

max 4 / 3V A for solid circular beam

max 2 /V A for hollow circular section

max 3 / 2V A for rectangular beam

Principle stresses

2

2

1,22 2

x y x y

xy

2

tan 2xy

x y

Max. and min shear stresses

2

2

1,22

x y

xy

Von-Mises stresses 2 2

1 1 2 2' or 2 2

' 3x xy (for biaxial)

Stress States

Triaxial stress state

2 311

E E

1 322

E E

3 1 22

E E

Stress in Cylinders

Thick-Walled (t/r>1/20) Wessels (internally and externally pressurized cyclinders):

2 2 2 2 2

2 2

( ) /i o o it

p a p b a b p p r

b a

2 2 2 2 2

2 2

( ) /i o o ir

p a p b a b p p r

b a

2

2 2

il

p a

b a

Page 2: Me307 machine elements formula sheet

If the external pressure is zero (po=0);

2 2

2 2 21i

t

a p b

b a r

2 2

2 2 21i

r

a p b

b a r

r=a r ip

2 2

2 2t i

b ap

b a

r=b 0r

2

2 2

2t i

ap

b a

If the internal pressure is zero (pi=0);

2 2

2 2 21t o

b ap

b a r

2 2

2 2 21r o

b ap

b a r

r=a 0r

2

2 2

2t o

bp

b a

r=b r op

2 2

2 2t o

b ap

b a

a=inside radius of the cylinder b=outside radius of the cylinder pi=internal pressure po=external pressure

Thin-Walled Wessels(t/r<1/20):

2

it

pd

t

4

il

pd

t

Curved Members In Flexure:

A

rdA

( )

My

Ae r y

o

o

o

Mc

Aer , i

i

i

Mc

Aer

Press and Shrink Fit:

2 2

2 2it

b ap

b a

2 2

2 2ot

c bp

c b

2 2

2 2o o

o

bp c b

E c b

2 2

2 2i i

i

bp b a

E b a

2 2 2 2

2 2 2 2o i o i

io

bp c b bp b a

E c b E b a

2 2 2 2

2 2 2; interface pressure

2o i

c b b aE if E E E = p =

b b c a

Page 3: Me307 machine elements formula sheet

2. Deflection Analysis

Fk

y , k=spring constant

T GJk

l ,k=Torsional spring rate for tension or compression loading

AEk

l

Castigliano’s Theorem:

Strain Energy

Axial Load

2

2

F LU

AE Direct Shear Force

2

2

F LU

AG

Torsional Load

2

2

T LU

GJ Bending Moment

2

2

MU dx

EI

Flexural Shear

2

2

CFU dx

GA , C is constant

Buckling Consideration:

Slenderness ratio=l

k

, I

kA

1/2

1

l 2 EC

k Sy

2

2

1

Critical Unit Load = Euler Column/

crPl l C E

k k A l k

; 2

2Pcr

C EI

l

2 2

1

1Critital Unit Load Johnson's Column

2

ycry=

SPl l l= S

k k A CE k

1. Both ends are rounded-simply supported C=1

2. Both ends are fixed C=4

3. One end fixed, one end rounded and guided C=2

4. One end fixed, one end free C=1/4

U Total energy

F Force on the deflection point

Angular deflection

Uy

F

Tl

GJ

Page 4: Me307 machine elements formula sheet

3.Design For Static Strength

Ductile Materials

1. Max. Normal Stress Theory (MNST):

If, 1 2 3

1

ySn

3. Distortion Energy Theory

If, 1 2 3

2 2 2

1 2 2 3 3 1( ) ( ) ( )'

2

For baxial stress state;

2 2' 3x xy

1

ySn

2. Max. Shear Stress Theory (MSST):

Yield strength in shear (Ssy)=Sy/2

1 3

max2

, for biaxial stress state;

max

1 2 24

2x xy

max

sySn

Brittle Materials

1. Max. Normal Stress Theory (MNST): 3. The Modified Mohr Theory (MMT)

If, 1 2 3 1

utSn

or

3

ucSn

If, 1 2 3

31

3

1

uc

uc ut

ut

SS

S S

S

3

3

Sn

2. The Column Mohr Theory (CMT) or Internal

Friction Theory (IFT):

31

3

1

uc

uc

ut

SS

S

S

3

3

Sn

Page 5: Me307 machine elements formula sheet

5. Design for Fatigue Strength

Endurance limit for test specimen (Se’);

For ductile materials:

Se’=0.5 Sut if Sut<1400 MPa

Se’=700 MPa if Sut 1400 MPa

For irons:

Se’=0.4 Sut if Sut<400 MPa

Se’=160 MPa if Sut 400 MPa

For Aliminiums:

Se’=0.4 Sut if Sut<330 MPa

Se’=130 MPa if Sut 330 MPa

For copper alloys:

Se’ 0.4 Sut if Sut<280 MPa

Se’ 100 MPa if Sut 280 MPa

Se = ka kb kc kd ke Se’

Sf=10c N

b

u

e

0.8S1b log

3 S

2

u

e

0.8Sc log

S

ka= surface factor, ka=aSutb

Surface Finish Factor a Factor b

Ground 1.58 -0.065

Machined or Cold Drawn 4.51 -0.265

Hot Rolled 57.7 -0.718

As Forged 272 -0.995

kb= size factor;

kb=1 if d 8 mm and kb= 1.189d-0.097

if 8 mm<d 250 mm for bending & torsional loading.

For non-rotating element, 0.097

b eqk 1.189d deq=0.37d

For pure axial loading, kb=1 and Se’=0.45Sut

For combined loading, =1.11 if Sut 1520 MPa and =1 if Sut 1520 MPa for ductile materials.

kc=reliability factor

kd=temperature effects, kd=1 if T 3500 and kd=0.5 if 350

0<T 500

0

ke=stress concentration factor, ke=1/Kf Kf=1+q(Kt-1)

Kt=geometric stress concentration factor, q=notch sensitivity.

Modified Goodman Soderberg

Infinite Life Finite Life Infinite Life Finite Life

a m

e u

1n =

σ σ+

S S

a m

f u

1n =

σ σ+

S S

a m

e y

1n =

σ σ+

S S

a m

f y

1n =

σ σ+

S S

Fa=(Fmax-Fmin)/2 Fm=(Fmax+Fmin)/2

Page 6: Me307 machine elements formula sheet

6. Tolerances and Fits

TF=Cmax-Cmin dL=DU-c Cmax=DU-dL Cmin=DL-dU

TF=Imax+Cmax dU=dL+TS Imax=dU-DL Imin=dL-Du

TF=Imax-Imin dU=DL+Imax

TS=dU-dL TH=DU-DL TF=TH+TS

7. Design of Power Screws

m mR

m

Fd L dT

2 d L

m m

L

m

Fd d LT

2 d L

Or considering tan ;

mR

FdT tan

2 m

R

FdT tan

2

If the friction between the stationary member and the collar of the screw is taken into consideration;

c cmR

d FFdT tan

2 2

c cm

R

d FFdT tan

2 2

o

R R

T FL

T 2 T

when collar friction is negligible, we obtain as,

tan

tan

If tan or m

L

dthen screw is self locking.

Bearing Stresses

b 2 2

r

4pF

h d d

b

m

Fp

d th

pt

2

Shear Stresses

For Screw Thread For Nut Thread

s

r

2F

d h

n

2F

dh

Bending Stresses

The maximum bending stress, m

6F

d Np

N=h/p

Page 7: Me307 machine elements formula sheet

Tensile or Compressive stresses

x

t

F

A

2

tt

dA

4

r m

t

d dd

2

Combined Stresses

Rxy 3

t

16T

d

Based on distortion energy theory;

Rxy 3

t

16T

d

2 2

x xy' 3 yS

n'

Based on maximum shear stres theory;

2 2

max x xy

14

2

sy

max

Sn

8. Design of Bolted Joints

Fe=Feb+Fep Feb=CFe Fep=(1-C)Fe b

b m

kC =

k k

Fb=Fi+CFe Fm=Fi-(1-C)Fe

b bb

A Ek

L

m 1 2 n

1 1 1 1..........

k k k k

ib

b

F

k i

m

m

F

k

Shigley and Mishke approach;

For cone angle of 030 ,

ii

i

i

1.813E dk

1.15L 0.5dln 5

1.15L 2.5d

m 1 2 n

1 1 1 1..........

k k k k

If L1=L2=L/2 and materials are same, m

1.813Edk

2.885L 2.5d2ln

0.577L 2.5d

Page 8: Me307 machine elements formula sheet

For cone angle of 045 ,

ii

i

i

E dk

5 2L 0.5dln

2L 2.5d

If L1=L2=L/2 and materials are same, m

Edk

L 0.5d2ln 5

L 2.5d

Wileman approach;

i(B d/ L)

m ik EdA e

Where Ai and Bi are constants related to the material. For Steel Ai=0.78715 and Bi=0.62873, for

Aliminium Ai=0.79670 and Bi=0.63816, for Gray cast iron Ai=0.77871 and Bi=0.61616.

Filiz approach;

1

dB

5 L

m eq

2

1k E d e

2 1 B

1 2eq

1 2

E EE

E E

2

1

0.1dB

L

8

11

2

LB 1

L

Static loading;

b y tF S A or b p tF S A p yS 0.85S mF 0

e i p t e1 C nF F S A CnF n=load factor of safety

Critical load= ice

FF

1 C

Dynamic Loading:

ea

t

CnF

2A i

m a

t

F

A s

a m

e u

1n

S S

t u e ui

s e

A S CnF SF 1

n 2 S

Fi=the maximum value of preload for there is no fatigue failure.

Limitations:

p i p0.6F F 0.9F where p t pF A S

e utimax t ut

e

cF n SF A S 1

2N S

e ei t p

F cF(1 c) F A S

N N b3.5d c 10d b

180c

N

Page 9: Me307 machine elements formula sheet

9. Design of Riveted Joints

Shearing of Rivets:

F

A , F=Force on each rivet

2d

A4

Secondary Shear Force

ii N

2

i

1

MrF ''

r

Bearing (compression) Failure:

F

A , A=td, t=thickness of the plate

Plate Tension Failure:

F

A , A w Nd t

w=width of plate

N=number of rivets on the

selected cross section

Primary Shear Force

N

i

1

FF'

A

10. Design of Welded Joints

Primary Shear Stress

F

'A

uJ 0.707hJ

Secondary Shear Stress

Mr

''J

uI 0.707hI

Bending Stress

Mc

I

Page 10: Me307 machine elements formula sheet

Table 9-3 Minimum weld-metal properties

AWS electrode

Number

n

Tensile Strength

MPa

Yield Strength

MPa

Percent

Elongation

E60xx 420 340 17-25 E70xx 480 390 22

E80xx 530 460 19

E90xx 620 530 14-17

E100xx 690 600 13-16

E120xx 830 740 14

Table 9-5 Fatigue-strength reduction factors

Type of Weld Kf

Reinforced butt weld 1.2 Toe of transverse fillet weld 1.5

End of parallel fillet weld 2.7

T-butt joint with sharp corners 2.0

Page 11: Me307 machine elements formula sheet

Table 9-1 Torsional Properties of Fillet Welds*

Weld Throat Area Location of G Unit Polar Moment of

Inertia

*G is centroid of weld group; h is weld size; plane of torque couple is in the plane of the paper; all

welds are of the same size.

Page 12: Me307 machine elements formula sheet

Table 9-2 Bending Properties of Fillet Welds*

Weld Throat Area Location of G Unit Moment of Inertia

*Iu, unit moment of inertia, is taken about a horizontal axis through G, the centroid of the weld

group; h is weld size; the plane of the bending couple is normal to the paper; all welds are of the

same size

Page 13: Me307 machine elements formula sheet

Table A3-8 Stress concentration factors for round shaft with

shoulder fillet in tension

d

r

D

.

o= F/A, where A= d2/4

D/d =1,02 D/d =1,05 D/d =1,1 D/d=1,5

r/d Kt Kt Kt Kt

0,025 1,800 - - -

0,028 1,728 - 2,200 -

0,031 1,678 2,000 2,125 -

0,037 1,610 1,868 2,020 -

0,044 1,550 1,778 1,938 2,522

0,050 1,508 1,714 1,866 2,400

0,062 1,452 1,626 1,766 2,235

0,075 1,408 1,550 1,684 2,086

0,088 1,370 1,502 1,624 1,970

0,100 1,336 1,457 1,568 1,893

0,125 1,286 1,400 1,496 1,760

0,150 1,254 1,364 1,452 1,662

0,175 1,230 1,340 1,400 1,600

0,200 1,220 1,314 1,372 1,546

0,250 1,216 1,292 1,342 1,508

0,275 1,200 1,270 1,325 1,480

0,300 1,200 1,250 1,296 1,452

* Adopted from Ref. [12]

Page 14: Me307 machine elements formula sheet

Table A3-9 Stress concentration factors for round shaft with shoulder fillet

in torsion

d

r

DT T

.

o= Tc/J, where c=d/2 and J=d4/32

D/d =1,09 D/d =1,20 D/d =1,33 D/d =2,0

r/d Kt Kt Kt Kt

0,009 - - - -

0,012 1,800 2,300 - 2,600

0,030 1,566 2,040 2,144 2,288

0,025 1,472 1,894 2,020 2,122

0,033 1,384 1,761 1,878 1,966

0,042 1,322 1,644 1,755 1,828

0,050 1,283 1,576 1,677 1,750

0,062 1,244 1,500 1,600 1,644

0,075 1,206 1,434 1,516 1,572

0,087 1,184 1,378 1,458 1,510

0,100 1,166 1,342 1,412 1,466

0,125 1,144 1,275 1,344 1,400

0,150 1,122 1,220 1,294 1,344

0,200 1,110 1,160 1,220 1,266

0,250 1,100 1,130 1,178 1,222

0,300 1,100 1,120 1,160 1,200

* Adopted from Ref. [12]

Page 15: Me307 machine elements formula sheet

Table A3-10 Stress Concentration factors for round shaft with shoulder

fillet in bending

d

r

DM M

.

o= Mc/I, where c=d/2 and I=d4/64

D/d =1,02 D/d =1,05 D/d =1,1 D/d =1,5 D/d =3

r/d Kt Kt Kt Kt Kt

0,012 2,290 2,553 2,700 - -

0,017 2,120 2,378 2,500 3,000 -

0,021 2,000 2,240 2,366 2,774 3,000

0,025 1,926 2,134 2,260 2,600 2,862

0,036 1,760 1,936 2,046 2,310 2,600

0,050 1,644 1,782 1,865 2,060 2,310

0,062 1,574 1,700 1,750 1,925 2,140

0,075 1,518 1,628 1,688 1,800 1,986

0,087 1,472 1,563 1,630 1,728 1,880

0,100 1,440 1,534 1,580 1,660 1,804

0,125 1,380 1,468 1,500 1,584 1,684

0,150 1,330 1,412 1,450 1,510 1,584

0,175 1,297 1,358 1,400 1,450 1,510

0,200 1,264 1,336 1,360 1,400 1,457

0,225 1,242 1,308 - - 1,410

0,250 1,225 1,286 - - 1,374

0,275 1,210 1,264 - - 1,340

0,300 1,200 1,242 - - 1,320

* Adopted from Ref. [12]