Formul me-3074683

1. Stress Analysis

Moment of Inertias

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

∫ 2xI = y dA ∫ 2

yI = x dA ∫ 2 2zJ = (x + y )dA

Shape Ix Iy J

Rectangle bh3/12 hb3/12 ( )2 2bhb +h

12

Triangle bh3/36 hb3/36 ÷ ÷

2 2h +bbh

18

Circle πd4/64 πd4/64 πd4/32

Stresses

Normal Stresses Shear Stresses

Axial

TensileF

σ =A

Torsional

τTr

=J

• τ 3= 16T/πd for solid circular beam

CompressionF

σ =A

Bending

b

Mcσ =

I

• b 3

32Mσ =

πd

for solid circular beam

Transverse (Flexural)

=τVQ

Ib,

-Q = A y

• maxτ = 4V/3A for

solid circular beam

• τmax = 2V/A for

hollow circular section

• τmax = 3V/2A for

rectangular beam

Principle stresses ÷

τ2

x y x y 21,2 xy

-σ + σ σ - σσ = + +

2 2 τ xy

x y

2tan2φ =

σ - σMax. and min. shear stresses

÷

τ τ2

x y 21,2 xy

- σ - σ= + +

2

Von-Mises stresses 2 21 1 2 2σ' = σ - σ σ + σ or τ2 2

x xyσ' = σ + 3 (for biaxial)

Stresses States

Triaxial stress state

2 311

σ + σσε = - μ

E E

1 322

σ + σσε = - μ

E E

3 1 22

σ σ + σε = - μ

E E

Stress in Cylinders

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

2 2 2 2 2i o o i

t 2 2

p a - p b - a b (p - p )/rσ =

b - a

2 2 2 2 2i o o i

r 2 2

p a - p b + a b (p - p )/rσ =

b - a

2i

l 2 2

p aσ =

b - a

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

÷

2 2i

t 2 2 2

a p bσ = 1+

b - a r

÷

2 2i

r 2 2 2

a p bσ = 1-

b - a r

At the inner surface;

r=a⇒ r iσ = -p2 2

t i 2 2

b + aσ = p

b - a

At the outer surface;

r=b⇒ rσ = 0 2

t i 2 2

2aσ = p

b - a

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

÷

2 2

t o 2 2 2

b aσ = -p 1+

b - a r

÷

2 2

r o 2 2 2

b aσ = -p 1-

b - a r

At the inner surface;

r=a⇒ rσ = 02

t o 2 2

2bσ = -p

b - a

At the outer surface;

r=b⇒ r oσ = -p2 2

t o 2 2

b + aσ = -p

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):

it

pdσ =

2ti

l

pdσ =

4t

Curved Members In Flexure:

•∫

Ar =

dA

ρ

•My

σ =Ae(r - y)

⇒ oo

o

Mcσ =

Aer, i

ii

Mcσ =

Aer

Press and Shrink Fit:

•2 2

it 2 2

b + aσ = -p

b - a•

2 2

ot 2 2

c + bσ = -p

c - b

• ÷

δ2 2

o o2 2

o

bp c + b= +μ

E c - b

• ÷

δ2 2

i i2 2i

bp b + a= - -μ

E b - a

• ÷ ÷

δ δ δ2 2 2 2

o i o i2 2 2 2io

= +bp c + b bp b + a

= - +μ - μE c - b E b - a

•( ) ( )

( )

δ2 2 2 2

o i 2 2 2

c - b b - aE if E = E = E; interface pressure = p =

b 2b c - a

2. Deflection Analysis

Fk =

y, k=spring constant

T GJk = =

θ l, k=Torsional spring rate

AEk = for tension or compression loading

l

Castigliano’s Theorem:

Strain Energy

Axial Load 2F L

U =2AE

Direct Shear Force2F L

U =2AG

Torsional Load2T L

U =2GJ

Bending Moment ∫2M

U = dx2EI

Flexural Shear ∫2CF

U = dx2GA

, C is constantC=1.2 for rectangular shapeC=1.11 CircularC=2.0 for thin walled tubular,

Buckling Consideration:

Slenderness ratio= ÷

l

k,

Ik =

A

÷

1

22π ECSy

l=

k

Euler column

U Total energy

F Force on the deflection point

θ Angular deflection

∂

∂y =

U

F

Tlθ =

GJ

•( )

⇒≥ ÷ ÷

2cr

21

Pl l Cπ ECritical Unit Load = =

k k A l/k or

2

2

Cπ EIP =cr l

Johnson's Column

• ⇒

÷ ÷ ÷ ÷ ÷

2 2ycr

y1

=SPl l 1 l

< Critital Unit Load = S - k k A 2π 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

3.Design For Static Strength

Ductile Materials

1. Max. Normal Stress Theory (MNST):

• If, 1 2 3σ > σ > σ

• y

1

Sn =

σ

3. Distortion Energy Theory

• If, 1 2 3σ > σ > σ

•2 2 2

1 2 2 3 3 1(σ - σ ) + (σ - σ ) + (σ - σ )σ' =

2

• For biaxial stress state;

2 2x xyσ' = σ + 3τ

• ySn =

σ'

2. Max. Shear Stress Theory (MSST):

• Yield strength in shear

(Ssy)=Sy/2

• ( )τ 1 3

max

σ - σ=

2

for biaxial stress state;

τmax x xy

1 2 2=σ + 4τ2

sy

max

Sn =

τ

Brittle Materials

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

• If, 1 2 3σ > σ > σ• ut

1

Sn =

σ

• If, 1 2 3σ > σ > σ

or uc

3

Sn =

σ •uc

3uc ut 1

ut 3

SS =

S - S σ-1

Sσ

• 3

3

Sn =

σ

2. The Column Mohr Theory (CMT) or Internal Friction Theory (IFT):

•uc

3uc 1

ut 3

SS =

S σ-1

Sσ

• 3

3

Sn =

σ

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:



• For Aliminiums:



• 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 Nb

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, α =1.11 if Sut ≤ 1520 MPa and α =1 if Sut ≥1520 MPa for ductile materials.

• kc : reliability factor

• kd : temperature effects, kd=1 if T ≤ 3500 and kd=0.5 if 3500<T ≤ 5000

• 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

uf

1n =

σ σ+

S Sa m

e y

1n =

σ σ+

S Sa m

f y

1n =

σ σ+

S S

• Fa=(Fmax-Fmin)/2 • Fm=(Fmax+Fmin)/2

6. Tolerances and Fits

Clearance fit:

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

Transition fit:

TF=Imax+Cmax Cmax=DU-dL Imax=dU-DL

Interference fit:

TF=Imax-Imin Imax=dU-DL Imin=dL-Du

Tolerances on the shaft, on the hole and on the fit

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

7. Design of Power Screws

÷

m mR

m

Fd L +πd μT =

2πd - μL

÷

m mL

m

Fdπd μ - LT =

2πd +μL

Considering μ = tanρ ;

( )mR

FdT = tanλ + ρ

2( )m

L

FdT = tanρ - λ

2

If ≥μ tanλ or µ >m

L

πd or ρ λ> or TL>0, then screw is self locking.

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

( ) c cmR

μ d FFdT = tanλ + ρ +

2 2( ) c cm

L

μ d FFdT = tanρ - λ +

2 2

Condition for self locking is : TL>0

Efficiency of screws: o

R R

T FLε = =

T 2πT when collar friction is negligible, we

obtain ε as, ( )tanλ

ε =tanλ + ρ

Thread Stresses:

• Bearing Stresses

( )b 2 2r

4pFσ =

πh d - dor b

m

Fpσ =

πd thwhere

pt =

2

• Shear Stresses

For Screw Thread For Nut Thread

τsr

2F=

πd hτn

2F=

πdh

• Bending Stresses

The maximum bending stress, mh

6Fσ =

πd

Stresses on the body of screw:

• Tensile or Compressive stresses

xt

Fσ =

A

2t

t

πdA =

4

r mt

d + dd =

2

• Shear Stresses

τ Rxy 3

t

16T=

πd

• Combined stresses:

Based on distortion energy theory;

τ2 2x xyσ' = σ + 3 yS

n =σ'

Based on maximum shear stress theory;

τ τ2 2x xy

1=σ + 4max

2 τsy

max

Sn =

8. Design of Bolted Joints

Feb and Fep are forces shared by the bolt and by the members respectively.

Fe=Feb+Fep Feb=CFeFep=(1-C)Fe stiffness ratio:b

b m

kC =

k k+

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

Stiffness of bolt: b bb

A Ek =

L Stiffness of members:

m 1 2 n

1 1 1 1= + +.......... +

k k k k

• Shigley and Mishke approach;

For cone angle of α 0= 30 ,

÷

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

For cone angle of α 0= 45 ,

( )( )

÷

i

i

i

i

πE dk =

5 2L + 0.5dln

2L + 2.5d

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

m

πEdk =

L + 0.5d2ln 5

L + 2.5d

• Wileman approach;

(B d/L)im 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;

÷ ÷

π d-B15 L

m eq2

π 1k = E d e

2 1- Bwhere

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

( )N n Ns

≤ ≤i

S AnF CnFp te e1- C F - n=load factor of safety

Critical load per bolt = ice

FF =

1- C

Dynamic Loading:

ea

t

CnFσ =

2Ai

m at

Fσ = + σ

As

a m

e u

1n =

σ σ+

S S

÷

t u e uimax

s e

A S CnF SF = - +1

n 2N S

where: ns is strength factor of safety and N is the number of bolts

Constraints:

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

• ÷

e uti t utmax

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 and bb

Dc

N

π=

9. Design of Riveted Joints

Shearing of Rivets:

τF

=A

, F=Force on each rivet

2πdA =

4

Secondary Shear Force:

∑

ii N 2

i1

MrF'' =

r

Bearing (compression) Failure:

σ F= -

A, A=td, t=thickness of the

plate

Plate Tension Failure:

σF

=A

, ( )A = w - Nd t

w=width of plate

N=number of rivets on the selected cross section

Primary Shear Force:

∑N

i1

FF' =

A

10. Design of Welded Joints

• Primary Shear Stress

τF

' =A

• uJ = 0.707hJ

• Secondary Shear Stress

τMr

'' =J

• uI = 0.707hI

• Bending Stress

Mc

σ =I

Table 9-3 Minimum weld-metal properties

AWS electrode

Number

Tensile Strength

Yield Strength

MPa

Percent

ElongationE60xx 420 340 17-25E70xx 480 390 22E80xx 530 460 19E90xx 620 530 14-17E100xx 690 600 13-16E120xx 830 740 14

Table 9-5 Fatigue-strength reduction factors

Type of Weld Kf

Reinforced butt weld 1.2Toe of transverse fillet weld 1.5End of parallel fillet weld 2.7T-butt joint with sharp corners 2.0

Table 9-1 Torsional Properties of Fillet Welds*

Weld Throat Area Location of GUnit 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.

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

Formul me-3074683

Engineering

Transcript of Formul me-3074683