Lectures 1-4 Hand out 2012.pdf

8/10/2019 Lectures 1-4 Hand out 2012.pdf

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Lectures 1-4 Hand out 2012.doc 1

MEC302 The Integrity of Materials and Components

Introduction to the Module

Two definitions of Integrityare:

1. an unimpaired condition

2. the quality of being whole and complete

In Engineering we need to design materials, components, structures, objects, systems etc.

that have no defects and that are structurally sound, i.e are unimpairedand wholeand

complete. However if they do contain defects through a material fault or due to operating

conditions then we need to know about failure so that it can be avoided. So this module

considers failure of components, how to identify it, and how to prevent it.

We will begin by studying the stress analysis of structures which have axi-symmetric

stress distributions, such as rotating shafts and pressure vessels. We will study thestresses in these components, consider how they might fail, and look at ways to strengthen

their design.

Then we will move on to how to identify defects in components using non destructive

examinationand consider how we might choose the most appropriate technique for a

particular component.

A set of lectures on fracturewill consider how to assess the integrityof a component

under load which contains a known defect.

We will then consider the collapse of cracked structuresand how to assess the failure ofsuch structures using R6 Failure Assessment Diagrams. This technique was developed in

the nuclear industry for fail safe design.

Failure due to contact loads, friction and wearwill be addressed and a study will be

undertaken of how to prevent such failure.

In week 7 a case study will be set as a learning assignmentwhich brings together stress

analysis, defect sizing and fracture mechanics assessments of a component. Students willsubmit their solutions to this problem in Week 8 and feedbackwill be given on their

understanding in Week 9.

After looking at static loading of defects we will then consider fatigue loading.

All through the course we will consider the same case studies on rotating shafts and

pressure vessels, but using different approaches and loading.


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Lectures 1-4 Axisymmetric Stress Distributions

Background reading: Mechanics of Engineering Materials, 2ndEd. P P Benham, R J

Crawford and C G Armstrong, Longman 1996, pp 367-412

Several important design problems are solved by using the results of this analysis Thick discs subjected to rotation

e.g. turbine discs , gas turbines

Thick cylinders subjected to pressure loadinge.g. gas cylinders, pressure vessels, gun barrels

Also power generating equipment, rotors, alternators, and thermal strains in pipes all

covered by this theory.

We can derive a general theory to cover all cases then modify it for specific cases to suit a

specific problem.

General Equations of Equilibrium

For an axisymmetric stress-strain system, the stresses on a typical element of material will

be as shown, the radial, hoop and axial stresses all vary with radius.

r = radial stress

= hoop stress

RB= body force in units

force/unit volume e.g. due toacceleration

Figure 1Stresses on a typical element of material under an axisymmetric stress-strain

system in the radial and hoop directions

In order to choose an appropriate material for a design we need to determine the radial

stress, r , the hoop stress, and the axial stress, Aand compare these with the properties

of the materials available. First we must consider radial equilibrium of forces.

d

2

d

2

d

r

drdr

dr

r

RBdr

x

y

r


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dr

dr

dr

r

r

A

A

Axial direction

RB

r = radial stress

= hoop stress

A= axial stress

Figure 2Stresses on a typical element of material under an axisymmetric stress-strain

system in the radial, hoop and axial directions

Radial equilibrium of forces (consider unit thickness)

02

sin2...

ddrdrdrdrRddrrdr

dr

drB

rr

Assume:22

sin dd and neglect second order terms

We also assume that the stress field is axisymmetric and stresses vary only with radius.

hence on simplifying we get

0.. rRdr

dr B

rr

(1)

From Equation (1) we can see that we have two unknowns, and r, so we need another

equation in order to solve (1).

Therefore we now consider how a typical element of material deforms such that the whole

material remains continuous, i.e., we must ensure Compatibility of Displacements. Wedo this by considering the geometry of a typical displacement or deformation and using the

elastic Hookes Lawrelationships.

Compatible deformation

We have a disc which moves radially outwards due to rotational forces

r

dru

(u + du)

dr + du

Let :u = Radial displacement at radius, r

w = axial displacement at radius, r

(out of the page)

= strain with suffix denoting

direction

= Poissons ratio

Figure 3Deformation radially


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Considering the displacement geometry we have

dr

du

dr

drdudrr

radial strain

dA

dwA axial strain

ru

rrur

2

22 hoop strain

If we want to calculate radial displacement, we cannot obtain this from radial strain, but it

can be easily obtained from hoop strain.Now we use Hookes Law to express the strains in terms of stresses:

ArrEdr

du

1 (2)

rAEr

u

1 (3)

rAAEdA

dw 1 (4)

Using the straindisplacement relationships in equations (2,3,4) gives

dr

dErrR

dr

dr ABr

...

1.1

(5)

Note: The introduction of expressions for RBand Aenables (1) and (5) to be solvedsimultaneously.

Stresses due to the rotation of thick discs of uniform thickness .

Consider an element of unit axial thickness.

dr

r

centrifugal force Let the specific weight of the material =

Centrifugal force = 2r x (volume of

element)

runitvolume

forceRB2

Figure 4 Centrifugal force due to rotation

We must now make an assumption regarding A. For a thick disc we assume that the plane

sections remain plane. This implies that the axial strain Ais constant with radius.

i.e. A= constant

0dr

d A

Thus equation (1) can now be written as:


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0... 22 rdr

dr rr

(6)

and equation (5) as

22 ..1 rdr

dr r (7)

These lead to:

22

2

22

2

18

21

2

18

23

2

rr

BA

rr

BAr

(8)

These equations (8) solve the problem once we put in boundary conditions to find A/2 and

B. The boundary conditions will depend on the component to be studied and severalexamples follow.

******************************

Note: For all problems in this course we will assume that (8) is appropriate.

In practise, if the disc is thin, then we do not assume that 0dr

d A but assume that A=0.

This leads to

222

22

2

8

31

2

8

3

2

rr

BA

rr

BAr

(8a)


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Case Study On A Rotating Shaft

Find the maximum stresses in the shaft in Figure 5

Ro

rRo

r

Ro= 0

Figure 5A long shaft of radius, Rorotating at radians/second

To find A/2 and B we must specify appropriate boundary conditions:

i.e. rat r= Rowill be zero i.e. Ro= 0.

This is the only boundary stress we can specify in this problem. But we need twoconditions since we have two unknowns.

Examine equations (8). Both these equations include

2

r

Bas a term and we have material r

= 0.

If B 0, then 2r

B, which says that we will have infinite stress at the centre of the shaft,

which is impossible. Therefore, for a solid shaftB must equal 0.

Equations (8) become

22

22

18

21

2

1823

2

rA

rAr

We now use the condition r= 0 at r= Ro

22

18

23

20 oR

A

22

1823

2 oR

A

222

222

23

21

18

23

18

23

rR

rR

o

or

(9)

At the centre of the shaft r = 0,

22

18

23or R

which are the maximum stresses in the shaft.

****************************


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Case Study On A Hollow Rotating Shaft, or Hollow Cylinder

Derive an expression for the radial and hoop stresses for the hollow rotating shaft in

Figure 6

Ro

R1

r

R1

Ro

Figure 6A long hollow shaft of outer radius, Roand inner radius R1, rotating at

radians/second

The boundary conditionshere are that at r = R1, and r = Ro, r= 0, that is:

R1 = 0, and Ro = 0

The expression for r in equations (8) becomes

22

2

2

1

2

2

1

1

18

23

20

18

23

20

o

o

Ro

R

RR

BA

RR

BA

solving these simultaneously

22

1

2

22

1

2

18

23

18

23

2

o

o

RRB

RRA

2

2

22

122

1

2

2

2

22

122

1

2

23

21

18

23

18

23

rr

RRRR

rr

RRRR

oo

oor

(10)


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Rotating Components Example 1

(a) A thinsteel disc is rotated at 6000 r.p.m. Plot the radial and hoop stresses with

increasing radius if = 8000 kg/m3, Inner radius Ri= 0.075 m, Outer radius Ro= 0.3 m

(b) If this were a long cylinder, how would the stresses differ?

(Hint: Which equations should be used? Equations (8) or (8a)?)

***************************************************************************

Rotating Components Example 2

A rotor in the form of a long hollow cylinder is shrunk onto a shaft such that the interface

radial pressure due to shrinkage is 50MN/m2.

(a) Find the speed of rotation when the shaft ceases to drive the rotor

(b) Find the maximum stress in the cylinder at this speed

= 8000 kg/m3, E = 21 x 1010N/m2, = 0.3

Inner diameter di= 100 mm, Outer diameter do= 600 mm

Note: The radial stress due to shrinkage is compressive (i.e.ve)

The radial stress due to rotation is tensile (i.e. +ve)

***********************************************************************


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Lam Equations - Thick cylinders subjected to internal and external pressure

The same theory as has been developed for rotating thick walled cylinders may be used for

thick cylinders subjected to internal and external pressure, but now the body forces are

zero (since = 0).

Equations (8) lead to:

Radial stress:222 r

BC

r

BAr

Lam Equations

Hoop stress:222 r

BC

r

BA

Axial Stress: 0A for open ends

CE

E

AA

rAA

2

for closed ends, i.e. independent of r

As before, B and C are found from boundary conditions.

Note: rand both increaseas r decreases. So the maximum stresses always occur on

the inner surfaces in this case.

We can use the Lam Equationsis terms of the diameter, provided we are consistent.

i.e.

2

2

**

**

d

BC

dBCr

and remember that the constants C* and B* will be different from C and B.

o

2max

ro

ro

r

ri

r

Figure 7Stress distribution of radial and hoop

stress in a thick walled cylinder subject to

internal pressure

Hoop stresses are positive

Radial stresses are negative

(for internal pressure conditions)

Since the problem is axisymmetric

the hoop and radial stresses are

principal stresses.

i.e. 1= 2= r

22

r21max

The maximum shear stress in

the cylinder is at the inside

maxr

2

)i()i(


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Yielding in a cylinder

Let k = ro/rifor an internally pressurized cylinder with pressure P

Using the Lam Equations

2

2

r

BC

r

BCr

with the boundary conditions:

At r = ro, r= 20or

BC

At r = ri, r= 2ir

BCP

We obtain the expressions:

1

1

1

1

2

2

2

2

k

r

rP

k

r

rP

o

o

r

12 2

2

max

k

Pkr at r = ri

Since yield will commence in the bore, using Trescas criterion

2

2

2

max

11

2

12

kP

k

Pk

y

y

************************************************************************

Pressurised Cylinders Example 1

A cylinder where ri = 0.5 rohas an internal pressure, P and closed ends. Plot the stress

distribution along the radius and find the maximum shear stress.

************************************************************************

Pressurised CylindersExample 2

A pipe of 100 mm ID is subjected to internal water pressure of 10 MN/m2. If the pipe

material has a safe tensile stress of 20 MN/m2, and a safe shear stress of 40 MN/m

2, find

the OD of the pipe.

************************************************************************


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Turbine discs of variable thickness

Turbine discs are seldom flat, but they thicker near the shaft. Optimum weight is attained

if the stresses rand are equal, and uniform at all radii.

For varying thickness, we require a new equilibrium equation, as thickness, t varies with

radius, r.

0.... 22 rttrtdr

dt

rr

may be derived from the free body diagram.

If r= f(r), the solution gives:

ro

rtt

225.0exp

and this is an ideal shape to be used in turbines.

Uniform thickness cylinders

In order to strengthen a pressurized cylinder one would automatically make the walls

thicker. However for uniform thickness cylinders, above thickness, wopt ,we get very little

reduction in ior maxfor massive increases in wall thickness.

For internal pressure, is tensile, i.e. positive. If we can induce an initial negative, i.e.

compressive hoop stress, the initial pressurisation of the cylinder would be employed in

overcoming the negative initial .

In practice to contain a higher pressure one can:

(a)shrink fitone or more cylinders around the pressure containing cylinder (e.g.hydraulic cylinders, gun barrels etc.) to form a compound cylinder to induce

compression

(b)wind wire on a cylinder under tension to induce compression at the bore.(c)overstrain once with a pressure beyond yield, to leave compressive residual stress

distribution near the bore.

imax

Cylinder wall thickness

Wopt

Thickness, t


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Shrink Fit Example 1 - Comparison of stresses in a single cylinder withthose in a compound cylinder under the same internal pressure.

(a) Single Cylinder

2r

d

BA

2d

BA

Internal Pressure = 45 MPa

d = 0.1r

= -45 -45 = A + 100B

d = 0.2 r = 0 0 = A +

25B

} B = -0.6

A = 15

2r d

6.015

2d

6.015

Stresses are max at d = 0.1:

Max. Tensile Stress, 751.0

6.015

2 MPa

Max. Shear Stress, 601.0

6.0

2 2r

MPa

d (m) )MPa(r )MPa( )MPa( 0.1 -45 75 60

0.15 -11.7 41.7 26.7

0.2 0 30 15

(b) Compound Cylinder - An outer cylinder shrunk onto an inner cylinder.(Both cylinders are made of the same material)

Shrinkage pressure between cylinders = 7 MPa

Internal pressure = 45 MPa

Resultant Stress = Stress due to shrinkage+ stress due to internalpressure.

0.1 m

0.2 m

0.2 m

0.15 m


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Due to shrinkage pressure:Inner cylinder, B100A00,1.0d r

B9

400A77,15.0d r

}6.12A

126.0B

r = -12.6 + 2d126.0

= -12.6 - 2d

126.0

d, m MPa,r MPa,

0.1 0 -25.2

0.15 -7 -18.2

Outer

Cyl inder

''r B

9

400A77,15.0d

''r 25A00,2.0d B

}

9A

36.0B

'

'

2r d

36.09

2d

36.09

d, m MPa,r MPa,

0.15 -7 25

0.2 0 18

Resultant Stress

d, mDue to

shrinkagepressure

Due to internalpressure

Resultant

r r r

InnerCylinder

0.1 0 -25.2 -45 75 60 -45 49.8 47.4

0.15 -7 -18.2 -11.7 41.7 26.7 -18.7 23.5 21.1

OuterCylinder

0.15 -7 25 -11.7 41.7 26.7 -18.7 66.7 42.7

0.2 0 18 0 30 15 0 48 24

Single

Cylinder

Compound

Cylinder

Stress Distribution

-6 0

-4 0

-2 0

0

20

40

60

80

100

0.1 0 .15 0 .2

d, mm

S

tress,

MPa


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Diametral interference required to produce a shrunk fi t

In order to produce ashrink fit, the outer diameter of the inner cylinder must be slightly

greater than the inner diameter of the outer cylinder. These dimensions need to be

accurately machined in order to obtain the required interface pressure due to shrinkage.

Ro Ri

R = initial radial difference

+ve R

-ve RFinal position o

common radius

outer Ri= change in radius ofinner cylinder

Ro= change in radius of

outer cylinder

DDDorRRR ioio for interference allowance.

General case for cylinders of different materials

I

R1 R2 R3

II

Hookes Law

rE

1 (1)

Assuming A= 0.

Hoop strain

DD

D D

DD

lengthntialcircumfereoriginal

lengthntialcircumfereinchange

Diameter change D = D (2)


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At the interface diameter, D2:

(I)cylinderinnerfor

(II)cylinderouterfor

222

22

22

2

22

I

r

IIII

II

r

IIIIIIII

E

DDD

E

DDD

Now r2I

= r2II

= r2Diametral interference allowance = D2= D2

II- D2I

This gives us the general expression:

222

22

II

II

I

I

rI

I

II

II

EEEEDD

(3)

If both cylinders are the same material this expression reduces to:

2222 IIIE

DD (4)

Shrink Fit Example 2

Look at the case in Shrink Fit Example 1

Consider the effects due to shrinkage only. What is the diametral interference allowance?

Case of sleeve on solid shaft

Let the interference pressure = -P

For the sleeve then the procedure is as before.

For the shaft:

2

2

d

BC

d

BCr

Since there is material at r = 0, then B = 0, so that we do not have r= = at r = 0.

B = 0 for the shaft

r= = C = (-P)

i.e. in the shaft we have a uniform compressive stress (-P), everywhere


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Lectures 1-4 Hand out 2012 doc 16

Summary of Axisymmetric Stresses

Derived expressions for radial and hoop stresses in rotating thick cylinders

22

2

22

2

18

21

2

18

23

2

rr

BA

r

r

BAr

(8)

Solution methoduse boundary conditions to find A and Bsubstitute back intoequations.

Rules:

Solid Shafts: B = 0

Hollow shafts r= 0 at inner and outer radii

Long shafts give similar solution to thin discs

Lams EquationsThick cylinders under pressure

Compound Cylindersto reduce stress at bore

Diametral Interferencehow to design to shrink fit

Documents

Lectures 1-4 Hand out 2012.pdf