1

B5.9EH2: Mechanical Engineering Science 8

Part 1: Mechanics of Materials

Topic 3: Instability, thermal stress, bolt loading and fatigue

Sub-sections:

3.1: Plastic instability

3.2: Plastic collapse in bending

Tutorial 3: Plastic deformation

3.3: Summary of mechanics of deflection of elastic structures

3.4: Principal stresses in plane stress

3.5: Yield criteria

3.6: Variation of stress

3.7: The basis of strain theory

3.8: Strain compatibility for small displacement theory

3.9: Activity 1, Thick cylinders

3.10: Plastic behaviour of a pressurised thick cylinder

Tutorial 4: Plasticity of thick cylinders

3.11: Metallic creep

3.12: Creep under non-uniaxial loading

3.13: Stress relaxation

Tutorial 5: Creep

Tutorial solutions

3.1: Plastic instability

Plastic instability can be understood in terms of a tensile test where, at the UTS, there is a

turning point in the load-extension (F-Δl) curve where the slope dF/d(Δl) changes from being

positive to being negative. Up to the turning point, each increment of extension requires more

work to be done whereas, after that point, the amount of load required to sustain further plastic

extension decreases (see below):

Tensile test curves for a material which exhibits plastic instability (left) and one which

exhibits ductile rupture before plastic instability (right). [Note: Curves are for engineering

stress and strain and so are of similar shape to load-extension curves]

2

The figure belowi shows some typical microscopical aspects of ductile failure for the example

of a tensile test specimen. The cup-and-cone type of failure is usual in ductile test specimens

because the planes of maximum shear stress are at 45° to the tensile axis, and the shear dimples

shown in the lower photograph are formed by the metal plastically deforming in approximately

micron-sized cells whose walls thin until slip bands entirely cross the wall and it parts. The

dimples in the centre are actually formed under the triaxial stresses present when necking of

the test piece occurs and arise from non-conformity with less favourably oriented grains or

inclusions (some of which can be seen in the high magnification photograph. These types of

dimples are not typical of all ductile fractures.

Because plastic collapse involves unconstrained deformation, there is usually a great deal of

distortion of any failed parts in service, particularly around the fracture surface. The

photographs below show examples of ductile failures of parts of a drilling system for a semi-

submersible rig, where the chain was being used to pull the drill string from the well. It is clear

that both the chain and pipe have been overloaded in tension because of the significant

reduction in cross-sectional area at the points of failure.

Although most tensile tests are carried out under displacement control, the point of plastic

instability can be understood best by a load control experiment where, for example, weights of

increasing magnitude are hung on the end of a specimen; the weight which takes the stress up

to the UTS will then cause the specimen to fail:

50mm OD rod 100mm OD pipe

10µm 5µm

3

Although models for uniaxial stress-strain curves exist, they are not very practicable for

carrying out advanced analyses, and it is normal to use one or other of two possible models;

perfectly plastic or elastic-linear strain-hardening (see belowii).

3.2: Plastic collapse in bending

If a beam is subjected to an increasing bending load, the outer fibres yield first leaving an

elastic coreiii:

Wei

gh

t

Extension

Weight at which extension

becomes unconstrained

4

In the elastic régime, the stress in any fibre a distance y from the neutral surface is given by

M y I so that the bending moment which will produce yielding at the outer fibres of the

beam is:

y y

IM

y

More generally, in the case of elastic-plastic bending, M y dA , so that, for the

particular case of a rectangular section b × d above, the elastic component can be written as

2

2 6e yM b d h , where My above corresponds to the case where h = 0. Within the plastic

part, the stress is independent of y and the resulting moment is p yM bh d h . Thus, the

total moment for a beam which may be anywhere from the elastic limit (h = 0) to fully plastic

(h = ½d) is:

2

1 2 16

y

ep

bd h hM

d d

As the radius of curvature is decreased, the bending moment increases, but this is only linear

in the elastic region, 3

12ee

EbdM

R , and the radius of curvature in the elastic-plastic region can

be written using the fact that 12

@y y d h . Since, from elastic bending theory,

Ey R

:

12

1 y

R E d h

Exercise A2.4 shows how there is a value of moment at which the curvature becomes

unconstrained, and this is known as the plastic moment at which the beam becomes a

mechanism, with a plastic hinge.

Cross-

section Stress distributions

Elastic limit Elastic-plastic Plastic

5

The moment at which the section becomes fully plastic (h = ½d) is given by 2

32

4p y y

bdM M , and values of Mp/My for other sections (known as shape factors) are

available in standard literature. Plastic collapse of pressure vessels can be similarly determined

as can bars under torsion, and standard solutions are available in mechanics compilations.

6

Tutorial 3: Plastic deformation

Exercise 3.1: An isotropic metal exhibits elastic-strain hardening plastic stress strain behaviour

as shown below:

The slope of line AB is given by the Young’s modulus, E, and the slope of BF is given by βE,

where β is the strain-hardening factor.

(a) By estimating the coordinates of points E, F, G and H for a specimen of length 3.5m and

cross-sectional area 1000mm2, determine β, E, σY and εY for the steel whose idealised stress-

strain curve is shown below:

(b) A pin-jointed structure of the type shown below has members made from the steel in Part

(a). If the cross-sectional area of each member is 645mm2, calculate the deflection u, for loads,

P, of 170kN, 270kN and 300kN.

σY

εY

7

(c) Plot a load-deflection graph up to 300kN, and include unloading from 300kN. What is the

residual strain after removal of the 300kN load?

Exercise 3.2: A suspended structure consists of a beam supported by five rods placed

symmetrically about the centre as shown below:

The members are each of cross-sectional are 100mm2, are made of steel (E = 200GPa) and can

be assumed to behave in an elastic-perfectly plastic fashion. The inner members (CD, FG and

HJ) yield at a stress of 500MPa and the outer ones (MN and RS) yield at 250MPa.

If the weight of the beam is negligible, find the load and the corresponding downward

displacement of the beam at which yield first occurs and at which all members have yielded.

Construct the load displacement diagram and determine the residual displacement and forces

in the members once a load above general yield has been removed.

8

3.3: Summary of mechanics of deflection of elastic structuresiv

The equations of elasticity for the general case of an anisotropic material (one which has no

planes of symmetry) can be written as:

where the Cij are the elastic coefficients, of which all 36 are different for truly anisotropic

materials. The σs and εs are the components of stress and strain, respectively. Figure 1 defines

the stress tensor, T, at a point in an arbitrarily-defined orthogonal co-ordinate system:

T =

xx xy xz

yx yy yz

zx zy zz

of which only 6 components are independent due to the principle of complementary shear

stresses, [σij = σji]. It should be noted that stresses and strains for which the subscripts are the

same, e.g. σxx, are direct stresses and strains, and those for which the subscripts are different,

e.g εyz, are shear stresses and strains. These equations only hold for materials which are linear

elastic (most metals, ceramics and polymers and composites over relatively small increments

of strain).

For isotropic materials, like metals, ceramics and polymers (although not generally

composites), whose elastic properties are the same in all directions (corresponding to an infinite

number of planes of symmetry), the stress-strain relations can be simplified considerably:

xz

xx

xz

yz

xy

zz

yy

xx

E

.

.

.

.

22100000

02

210000

002

21000

0001

0001

0001

211

xz

xx

xz

yz

xy

zz

yy

xx

CCC

CCC

CCC

.

.

.

.

...

......

......

......

...

...

662616

262212

161211

Figure 1

9

Furthermore, it is always possible to find a set of orthogonal co-ordinate axes 1, 2, 3, oriented

in such a way that the shear stress components are zero and these axes are known as the

principal axes, i.e.:

Txyz =

zzxyzx

yzyyyx

xzxyxx

T123 =

3

2

1

00

00

00

3.4: Principal stresses in plane stress

In the simple case of uniaxial tension, convention has dictated that the yield stress is defined

when a certain amount of plastic deformation has take place, and this will correspond to slip

having occurred to a greater or lesser in extent in a number of grains. The preceding analysis

is rather simple in that it does not acknowledge that there needs to be compatibility between

differently oriented grains, Figure 6v. Whereas there have been theoretical extensions to the

critical resolved shear stress which help to understand the effect, say, of grain size on strength,

a more pragmatic approach is needed to define yield in practical situations.

Figure 2 shows a general two dimensional state of stress (plane stress) with respect to axes x

and y and, given that σxy = σyx, this can be resolved onto a plane at some angle, θ to the axesvi

to give:

2sin2cos21

21

xyyyxxyyxx

and

2cos21

xyyyxx

In order to get a picture of a state of stress which acknowledges that it might be resolved in any

plane, these two equations can be combined to give:

22

4122

21

xyyyxxyyxx

which is the equation of a circle using axes of normal and shear stress, Figure 3.

z

x

y 2

3

1

10

Figure 2

Figure 3 is a representation of what is known as Mohr’s circle for two dimensional stress,

assumed to have been constructed for the plane upon which σxx, σxy and σyy are known.

The radius of the circle (see equation) is 22

41

xyyyxx and the co-ordinates of its centre

are 0,21

yyxx . Once the circle has been drawn, the normal and shear stress components

on any other plane oriented at an angle θ to the original can be found by rotating the plane

through an angle 2θ (note that the angle is 2θ because the sin and cosine functions operate on

this angle for the resolution of forces). It might also be noted that the complementary shear

stresses are always of the same magnitude and opposite sign because the circle centre is always

on the normal stress axis. It can also be seen that the principal stresses (defined as the plane

where there is no shear stress) are the maximum and minimum direct stresses than can be

σxx σxx

σyy

σxy

σyx

σxy

σyy σyx

θ

θ

σxy

σxx

σθ

τθ

σyy σyx

2φ σn σn

τs

Figure 3 Figure 4

τs

τmax τmax = 0.5 σ1

σxx σyy

σxy

σyx

2θ

σ1 σ2

σ1

11

experienced for a given stress state at a point, and that the maximum shear stresses are

encountered on a plane angled at 45° to the principal plane. Both of these observations mean

that the radius of Mohr’s circle, and hence the magnitude of the maximum shear stress can be

given in terms the principal stresses, 2

21max

.

Also, by corollary, the principal stresses and the principal plane can be found for any plane

stress situation as follows:

22

21

1 42

xyyyxx

yyxx

22

21

2 42

xyyyxx

yyxx

yyxx

xy

2tan 1

21

3.5: Yield criteria

Although it is convenient to measure the yield stress, σy, in uniaxial tension, it is relatively

uncommon for metals to be used in pure tension, so a means of determining the loads which

can be applied in more complex stress states is needed. There is no fundamental way of

determining yield criteria from first principles, but two theories (the Tresca and von Mises

criteria) have become well-accepted on the basis that they operate satisfactorily in practice. For

the output of finite element codes, stresses are often given as Tresca or von Mises stresses,

which can be directly compared with the uniaxial tensile yield stress.

The Tresca Criterion supposes that it is the maximum shear stress in any stress state which

causes yield when it reaches a critical value. Given that the maximum shear stress in uniaxial

tension is half the uniaxial tensile stress, and, for more complex stress states, it is half the

difference between the maximum and minimum principal stresses, the Tresca Criterion can be

written that yield will occur if:

y 31

The von Mises criterion depends on an (intuitive) understanding of the meaning of

“dilatational” and “deviatoric” components of the strain energy. The first of these is related to

change in volume and the second is related to change in shape. This is most easily visualized

without the complication of shear stresses, so imagine a three-dimensional element subjected

to principal stresses σ1, σ2 and σ3. The effect (strain) of these three stresses in an isotropic solid

can be divided into the effect brought about by the average of the three principal stresses

32131 acting in each principal direction, which will produce a dilatation, and the

deviatoric stresses:

ii

where i is 1, 2 or 3.

12

The proposition that yield was associated with the deviatoric contribution to the strain energy

was made independently by Hencky, Maxwell and von Mises, and it can be shown (relatively

simply, although it is a little involved) by comparing an arbitrary case with uniaxial tension

that this leads to the criterion:

22

13

2

32

2

21 2 y

Other yield criteria have been developed for non-isotropic materials but these will not be

covered here.

3.6: Variation of stress

In most applications, the stress is not the same at all points in a body, and it is necessary to be

able to describe the variation in a way that it can be used to determine stress distributions and

devise load-deflection characteristics.

Stress variation analysis relies simply on Newton’s Laws, and the following diagram shows

the most general case of a rectangular prismatic element of dimensions dx, dy, dz subject to

forces resulting from the stresses and inertial forces per unit volume (due to accelerations) Bx,

By, Bz acting along the three Cartesian axes.

For equilibrium in the x-direction (say):

0

yxxxxx xx yx yx

zxzx zx x

dx dydz dy dxdzx y

dz dxdy B dxdydzz

Dividing by the volume of the element gives: 0yxxx zx

xBx y z

13

By symmetry, the three equations of motion can be written:

0

0

0

yxxx zxx

xy yy zy

y

yzxz zzz

Bx y z

Bx y z

Bx y z

It is often of interest to write these equations in curvilinear co-ordinates, of which cylindrical

polar co-ordinates is one example:

Using the approximation that, as dθ → 0, sin½dθ → ½dθ and cos½dθ → 1, and dividing by

rdrdθdz, the above reduces to:

1

1 1 1 10

2 2

rr rr zr zrrr zr

rr r r

r drdr dz

r rdr dr z dz dz

d d Br r rd rd

Expanding, and neglecting second- and higher order terms in the differentials gives:

θ

dθ

Bθ

Br

σrr

σθθ

σθr

σrθ

r

dr

Taking a cylindrical segment of

dimensions dr, rdθ, dz as shown

above, the equilibrium along the

radial centre-line can be written

(note that the inertial force, and

stresses in the z-plane or z-direction;

Bz, σzz, σrz, σzr, σzθ, σθz, are not shown

on the diagram for clarity):

14

10r rrrr zr

rBr r z r

Similarly, resolving in the other two dimensions gives:

210

10

r z r

zrz zz rzz

Br r z r

Br r z r

3.7: The basis of strain theory

Whereas the theory of stress is based on Newton’s Laws, the theory of strain is based purely

on geometry (deformation). When a body (Region R) is deformed (Region R*) the particle P:

(x, y, z) passes to P*: (x*, y*, z*) and an adjacent particle Q: (x + dx, y + dy, z + dz) passes to

Q*: (x* + dx*, y* + dy*, z*+ dz*) so that the line element PQ = ds passes to the line element

P*Q* = ds*:

The deformation can be described by the equations:

, ,

, ,

, ,

x x x y z

y y x y z

z z x y z

where the functions x*, y* and z* need to be continuous (and therefore differentiable) unless a

discontinuity (fracture) has occurred.

15

The engineering strain of the line element PQ is defined as:

E

ds ds

ds

and this is the most commonly used kind of strain.

The total differential of the above equation is:

x x xdx dx dy dz

x y z

y y ydy dx dy dz

x y z

z z zdz dx dy dz

x y z

…….[2.1.7.19]

and, using the (x, y, z) components (u, v, w) of the displacement from P to P*:

x x u

y y v

z z w

Also, from simple geometry:

2 2 2 2

2 2 2 2

ds dx dy dz

ds dx dy dz

The above equations can be combined1 to determine the “magnification factor”, from which

the engineering strain can be found:

2 2 2 21

22 2 2E E xx yy zz xy xz yzM l m n lm l n mn

and the finite strain-displacement relations are:

1 If you have any sense, you will take this as read.

16

2 2 2

2 2 2

2 2 2

1

2

1

2

1

2

1

2

xx

yy

zz

xy yx

u u v w

x x x x

v u v w

y y y y

w u v w

z z z z

v u u u v v

x y x y x y

1

2

1

2

xz zx

yz zy

w w

x y

w u u u v v w w

x z x z x z x z

w v u u v v w w

y z y z y z y z

For small displacements, the quadratic terms in the differentials can be neglected and these

reduce to the small-displacement relations:

1

2

1

2

1

2

xx

yy

zz

xy yx

xz zx

yz zy

u

x

v

y

w

z

v u

x y

w u

x z

w v

y z

17

3.8: Strain compatibility for small displacement theory

The small displacement relations:

1

2

1

2

1

2

xx

yy

zz

xy yx

xz zx

yz zy

u

x

v

y

w

z

v u

x y

w u

x z

w v

y z

can be used to find the strain if the three displacements are chosen arbitrarily, but the reverse

is not true, i.e. the integration of the above equations requires the use of some boundary

conditions.

For the example of x-y plane strain, which can be defined as the displacement components u

and v being functions of x and y only, and the component w being constant, so that the strain

displacement relations become:

xx

u

x

; yy

v

y

; 1

2xy yx

v u

x y

0xz yz zz

Differentiating these twice:

2 3

2 2

xx u

y x y

;

2 3

2 2

yy v

x x y

;

2 3 3

2 2

2 xy v u

x y x y x y

and adding the right hand sides of the first two gives the right hand side of the third, so we can

write a differential equation in strain only, which is the strain compatibility relation for plane

strain:

2 22

2 2

2yy xyxx

y x x y

Elimination of (u, v, w) from the other small displacement relations in a similar way yields a

more general set of compatibility equations.

The strain - small-displacement relations can also be written in cylindrical polar co-ordinates:

18

1

12

2

12

rr

zz

r r

rz rz

z z

u

r

u v

r r

w

z

u v v

r r r

u w

z r

v w

z r

3.9: Activity 1, Thick cylinders: A long, hollow cylinder is subjected to an internal pressure,

pi, and an external pressure, po. Consider a section away from the ends and write the equations

of equilibrium for the element in the plane shown below, given that the circumferential stress,

σθθ is constant around the circumference at a given value of r.

Thus, show that:

0rrrr

r r

ro

ri pi

po

19

and, since the axial stress is constant:

0zz

z

Use the small-displacement equations to show that:

rr

u

r

; u

r and

zz

w

z

Use the above, along with the strain-stress equations with x = r, y = θ and z = z to show that:

rrrr zz

u

r E E

……………(1)

rr zz

u

r E E

……………(2)

zzzz rr

w

z E E

……………(3)

Differentiate equation (2) with respect to r and substitute for u

r

and

u

r from (1) and (2) to

show that:

1 zz rr

rrr r r r

……………(4)

Noting that εzz is constant, differentiate (3) with respect to r to show that:

zz rr

r r r

……………(5)

Substitute this into (4), and use the equation of equilibrium above to show that:

21 0rr

r r

……………(6)

which, along with (5) shows that 0zz

r

, i.e. that the axial stress is constant through the wall

thickness.

Integrating (6):

. 2rr const A ……………(7)

and using 0rrrr

r r

to eliminate gives:

2 20rr rr A

r r

Multiplying by r2:

20

22 2 0rrrrAr r r

r

Recognising the last two terms as a differential:

22 0rrAr rr

which is easily integrated to give: 2 2

rrAr r B , i.e.

2rr

BA

r and, using (7),

2

BA

r

Now, apply the boundary conditions that the radial stress is equal to the internal pressure at the

inner radius, and that it is negative (compressive). Similarly, the radial stress at the outer surface

is equal to the external pressure, again negative. Show that: 2 2

2 2

i i o o

o i

p r p rA

r r

and

2 2

2 2

i o i o

o i

p p r rB

r r

and therefore that: 2 2

2

2 2 2

11 1

1

o irr i o

r rp p k

k r r

and

2 2

2 2 2

11 1

1

o ii o

r rp p k

k r r

where o

i

rk

r

Also show that, when the external pressure is negligible, these equations reduce to:

2

2 21

1

i orr

p r

k r

and

2

2 21

1

i op r

k r

There are three possible cases of axial stress:

1. The ends are free to move as in a hydraulic cylinder: Here 0zz , i.e. we have plane stress.

Use Equation (3) to show that the axial strain: 2

2

1

izz

p

E k

2. The ends are capped, but the cylinder is free to change in length, as in a portable cylinder:

Here, if the external pressure is negligible, the force exerted by the pressure on the end cover

must equal the force due to the axial stress integrated across the (annular) wall of the vessel:

2 2 2

zz o i i ir r p r

21

Using this, and Equation (3), show that the axial stress and the axial strain are given by:

2 1

izz

p

k

and

2

1 2

1

i

zz

p

E k

3. The ends are capped and are restrained from movement by rigid end supports, as in some

designs of fixed cylinder: Here 0zz , i.e. we have plane strain. Again using Equation (3) and

the expressions for rr and above, show that:

2

2

1

izz

p

k

The three stresses above are, in fact, the principal stresses in the cylinder. Show that the internal

pressure to cause yielding (if the external pressure is negligible) is given by:

2

, 2

1

2i Y Y

kp

k

for all three end conditions using the Tresca Criterion.

Also show that, using the von Mises Criterion, the pressure to cause yielding is:

2

, 2

1

3i Y Y

kp

k

, for condition 1.

2

,4

1

3 1i Y Y

kp

k

, for condition 2, and

2

,4 2

1

3 4 4 1i Y Y

kp

k

, for condition 3.

Given that the thick cylinder is more accurate than the thin cylinder approximation, calculate

the error that a designer would incur in applying the Tresca Criterion as a function of ro/ri.

Express the error in terms of the pressure to cause yielding thus:

thick thin

y y

thick

y

p p

p

Estimate the error between the Tresca and von Mises criteria for each of the thick and thin

cylinders, assuming the Tresca Criterion to be the “correct” one.

22

3.10: Plastic behaviour of a pressurised thick cylinder

Here we will consider a thick cylinder made from an elastic-perfectly plastic material under

internal pressure only. From Activity 1, two of the principal stresses are: 2

2 21

1

i orr

p r

k r

and

2

2 21

1

i op r

k r

where o

i

rk

r and the maximum shear stress at any point in the plane of the cross-section is:

2

max 22 1

rr i op r

k r

irrespective of the end condition because rr zz

The maximum shear stress in the cross-section is at the inside wall so, as the pressure is

increased, the yielded zone boundary will spread from the bore outwards.

At any pressure between that for first yield and for general yield, the vessel can be regarded as

a compound cylinder with the outer cylinder made from elastic material. From Activity 1, when

the external pressure is negligible, the stresses in the elastic part ( ar ) can be obtained in

terms of pressure at the yielded interface, i.e. ap :

2

2 21

1

a orr

p r

k r

and

2

2 21

1

a op r

k r

The stresses at the elastic-plastic interface r a are therefore:

2 2

2 2 21a o

rr

o

p a r

r a a

and

2 2

2 2 21a o

o

p a r

r a a

so that: 2

max 2 2

a o

o

p r

r a

, and, according to the Tresca Criterion, this is equal to half of the yield

stress since yield has just occurred at r = a. Thus:

2 2

22

Ya o

o

p r ar

……….(1)

Thus, the stresses in the elastic part of the cylinder are given by: 22

, 2 21

2

oYrr

o

ra

r r

Within the plastic zone, 0rrrr

r r

[Activity 1], and, using a perfectly plastic

model, the maximum shear stress is everywhere equal to half the yield stress, so:

max2 0rr rr rrrr Yr r r

r r r

23

and: rr Y

r r

Integrating: lnrr Y r C , and substituting the boundary condition:

rr ap at r a

gives

lnrr Y a

ap

r

Substituting (1) for pa gives the radial stress distribution in the plastic zone ( r a ):

2 2

2ln

2

Yrr Y o

o

ar a

r r

,

and the corresponding hoop stress is:

2 2

21 ln

2

YY rr Y o

o

ar a

r r

The internal pressure to cause yielding to a radius a is found by noting that, at the internal

surface, ir r and rr ip , therefore:

2

( ) 2ln 1

2

Yi e p Y

i o

a ap

r r

Thus, the internal pressure to cause yielding right through the wall ( oa r ) is:

( ) ln oi p Y

i

rp

r

and the internal pressure to cause first yield at inner surface ( ia r ) is:

2

21

2

iYiY

o

rp

r

(c.f. Activity 1, approached from elastic side)

24

Tutorial 4: Plasticity of thick cylinders

Exercise 3.3: Calculate the internal pressure required to cause yielding at the inner surface,

one third of the way through the wall and completely through the wall of a hydraulic cylinder

of bore 150mm and wall thickness 30mm, made from steel with a yield stress of 250MPa.

Sketch the stress distribution across the wall of the cylinder in each case.

Exercise 3.4: In order to produce a compressive hoop stress (autofrettage) at the inner wall of

a cylindrical pressure vessel to resist stress corrosion cracking, it is necessary to bring 20% of

the wall thickness into the plastic range. The external diameter of the vessel is 800mm, the

internal diameter is 320mm and the yield stress is 400MPa. Calculate the internal pressure

required to cause this amount of plastic deformation and find the corresponding hoop stresses

at the bore, at the elastic/plastic interface and at the outer surface. Determine the pressure

required to cause yield through the entire wall.

25

3.11: Metallic creep

Creep in metals is a visco-plastic process (as opposed to a visco-elastic process, as in

polymers). Metallic creep only occurs at relatively high homologous temperatures, i.e.

temperatures which are a significant fraction of the melting point. This means that, for low

melting alloys, such as solders, creep must be considered at normal operating temperatures,

whereas, for alloys like steels, it only becomes a consideration at temperatures of above about

500ºC.

Three distinct mechanisms of metallic creep have been identified; dislocation climb, vacancy

migration and grain boundary processes. In normal plastic deformation, dislocations slide along

slip planes:

and slip generally becomes easier as the temperature increases. However, the fact that the yield

stress becomes lower at higher temperatures is not creep. Dislocation creep involves

dislocations overcoming the normal obstacles to their movement (such as small particles or

other dislocations) by a process of climb:

Climb requires the atoms around the

dislocation line to move lattice position

under the driving force of the stress,

which they can only do by diffusion.

Since diffusion is a thermally activated

process, this means that creep is also a

thermally activated process. Creep can

occur by other diffusional processes

under the driving force of an applied

stress.

26

For example, vacancies (empty lattice sites) can “move” by adjacent atoms moving into the

vacancies, the vacancies moving into areas where there is a Poisson contraction making the

elastic strain plastic. Similar effects can occur due to grain boundaries rearranging under the

applied stress (often called grain boundary sliding.

The involvement of a thermally-activated process means that creep strain is not only a function

of applied stress, but also depends on time and temperature:

c f g T h t

Typically a creep curve shows a characteristic time-dependence with three phases known as

primary, secondary and tertiary creep, respectively:

The time evolution of the creep curve is often approximated by a function: 1

3 3

c t t t

but, for design purposes, it is normal to focus on the secondary (linear) portion of the curve, so

that the creep strain at any time is approximated by an effectively instantaneous value plus an

amount which increases linearly with time:

It has been observed empirically that the steady-state creep rate shows a power law dependence

with stress and also, since creep is thermally activated, the creep rate shows an Arrhenius

dependence with temperature so that:

expm m

s

QA B

RT

Time

Increasing stress or

temperature

Primary

Secondary or steady-

state Tertiary

Rupture t0

0 0

n

s st t t

27

3.12: Creep under non-uniaxial loading

Creep analysis of a component subjected to bending can be analysed making similar

assumptions to plastic deformation during bending, i.e. that plane sections remain plane, that

axial fibres parallel to the neutral plane experience only an axial direct stress and creep

behaviour is the same in tension and compression. Given this, the strain can be expressed in

purely geometrical terms: y

R and, considering secondary creep at a constant temperature:

mB . Thus:

m

yd

RB

dt , which can be integrated by separating variables: my

B tR

.

Rearranging: 1

my

RBt

Taking a rectangular bar b d , the internal-external equilibrium can be written by:

12 2 2

1

1

1

0 0 0

22 2

d d dm

m

m

y bM by dy by dy y dy

RBt RBt

which can be integrated to give the moment-curvature relationship:

1

1

22

2 1 2

m

m

b m dM

mRBt

From the stress-curvature relationship, 1 11

m mRBt y

, so, substituting and rearranging, gives

the relationship for stress across the cross-section:

1 1

2 1 2

3

mMy m y

I m d

which can be compared (see tutorial) with the elastic case where My I

When the loading is multi-axial, it is necessary to use three laws of plasticity: that the principal

stresses and strains are coincident, that the volume remains constant and that the maximum

shear stresses and strains are proportional. Using this, the three principal creep rates can be

written:

1

1 1 2 3

1

2 2 3 1

1

3 3 1 2

1

2

1

2

1

2

m

m

m

B

B

B

Where the equivalent uniaxial von Mises stress is:

28

2 2 2

1 2 2 3 3 1

1

2

3.13: Stress relaxation

Stress relaxation is the complement of creep where a component is held at constant strain, and

the stress reduces with time. If a component is subjected to an initial stress in the elastic range:

0 E , and the strain is held constant, then:

0 c

T

E

Differentiating with respect to time:

0 10 cd d d

dt dt E dt

i.e.

1c

d

E dt

Considering only steady-state creep:

1m dB

E dt

Separating variables:

1m

ddt

EB

and integrating from 0t , 0 to t t , t :

0

1 1

0

1 1 1 1 1

1

t

m m m

t

dt

EB EB m

29

Tutorial 5: Creep

Exercise 3.5: Some measurements have been made of the steady state creep rate of a nickel

based superalloy and it has been found that the stress exponent, m, is 5, the activation energy,

QC , is 120,000 J/mol and the pre-exponential constant, K2 , is 1.31x10-8 (the constants K2 and

n are in MPa, % and hour units).

The alloy is to be used to manufacture turbine blades with a maximum service temperature of

800C. It is expected that the blades will experience a tensile load of 5kN at operating speed

and the end clearance allows for a maximum creep strain of 2% over the design life of 10,000

operating hours. Calculate the minimum required cross-sectional area of the blade. What would

be the creep life if the temperature were to be increased to 850C?

Exercise 3.6: Assume the cross-section calculated above to be approximately rectangular with

breadth, b, three times the depth, d. If the blades are 0.5m long, and can be regarded as

cantilevers, calculate the stress across the cross-section at the blade root for an effective end-

load on the cantilever of 1kN: (i) at normal ambient temperature (no creep), (ii) at 800C and

(iii) at 850C.

Exercise 3.7: A cylindrical alloy steel pipe of outer diameter 100 mm and wall thickness 3 mm

requires to have a life of 10,000 hours at 400C. If the creep strain is not to exceed 0.5%, m =

3 and B = 1.45 × 10-19 hr-1 MPa-1, find the allowable pressure.

Exercise 3.8: The bolts holding down the flange on a pressure vessel have been tightened to a

stress of 400 MPa at 400°C. Assuming the flange to be rigid, find the remaining stress in the

bolts after 100,000 hours. Also find the time it will take for the stress to relax to 80% of its

original pretension.

For the bolting alloy, E = 200GPa, the stress exponent, m, is 6.3, the activation energy, QC , is

220,000 J/mol and the pre-exponential constant, K2 , is 1.31×10-8 (the constants K2 and m are

in MPa, % and hour units)

30

Tutorial Solutions

Exercise 3.1: An isotropic metal exhibits elastic-strain hardening plastic stress strain behaviour

as shown below:

The slope of line AB is given by the Young’s modulus, E, and the slope of BF is given by βE,

where β is the strain-hardening factor.

(a) By estimating the coordinates of points E, F, G and H, determine β, E, σY and εY for the

steel whose idealised stress-strain curve is shown below:

Point Load (kN) Deflection (mm)

O 0 0

E 170 2.9

F 265 4.5

G 270 6.5

H 300 13

σY

εY

31

Thus, 3

3170 10 3500205 10

1000 2.9E MPa

,

3270 10270

1000Y MPa

,

4.50.13%

3500Y ;

3

3300 270 10 3500

16.15 1013 6.5 1000

E MPa

, so 16.15

0.079205

Thus, the stress-strain relationships are: 3205 10 ; 0.13%

3266.5 16.15 10 ; 0.13%

(b) A pin-jointed structure of the type shown below has members made from the steel in Part

(a). If the cross-sectional area of each member is 645mm2, calculate the deflection u, for loads,

P, of 170kN, 270kN and 300kN.

From equilibrium at joint D: 0 2 cosxF F P and cos 0.8 , so:

105.25,168.75,187.51.6

PF kN

The stress is given by:

163.2,261.6,290.7F

MPaA

The first two of these are below yield, so the strain:

4 3

37.96 10 ,1.276 10

205 10

For the stress beyond yield, the strain is:

3

3

266.51.498 10

16.15 10

The elongation in the direction of the rod is therefore:

2.39,3.83,4.49L L mm

and so:

cos 1.91,3.06,3.59u L mm

(c) Plot a load-deflection graph up to 300kN, and include unloading from 300kN. What is the

residual strain after removal of the 300kN load?

To plot the load-deflection graph we need the yield deflection and load, most easily found from

the yield strain: cos 3.12Y Yu L mm and the yield stress: 1.6 279Y YP A kN

So:

P (kN) 0 170 270 279 300

u (mm) 0 1.91 3.06 3.12 3.59

32

The residual deflection can be calculated by extrapolating the dotted line from 3.59, 300)

parallel to the elastic loading line. The gradient can be found from any two points on the elastic

line, e.g. zero and yield: 279

tan3.12

. Thus:

300 3003.12 3.35

tan 279u mm

and so the residual deflection is 3.59-3.35 = 0.24mm

Exercise 3.2: A suspended structure consists of a beam supported by five rods placed

symmetrically about the centre as shown below:

The members are each of cross-sectional are 100mm2, are made of steel (E = 200GPa) and can

be assumed to behave in an elastic-perfectly plastic fashion. The inner members (CD, FG and

HJ) yield at a stress, i , of 500MPa and the outer ones (MN and RS) yield ( o )at 250MPa.

If the weight of the beam is negligible, find the load and the corresponding downward

displacement of the beam at which yield first occurs and at which all members have yielded.

Construct the load displacement diagram and determine the residual displacement and forces

in the members once a load above general yield has been removed.

The members will all have the same elongation, , (i.e. parallel loading) and so the strains in

the members are:

1

MN FG HJ RSl

and

2

CDl

0

50

100

150

200

250

300

350

0 1 2 3 4

Lo

ad

(P

), k

N

Deflection (u), mm

Plastic

Elastic

u’

33

The stress-strain relationships for each of the members are:

MN RS MN RSE E o and o

E

MN RS o o

E

FG HJ FG HJE E i and i

E

FG HJ i i

E

CD CDE i and i

E

CD i i

E

Yield will first occur in members MN and RS when their stress is o and when the load is yP

. The strain in these members can be determined as:

oMN RS

E

and hence the deflection at first yield: 1 1y MN RSl l

The stresses in the remaining members are therefore:

, , , ,

1

y

FG y HJ y FG y HJ y oE E El

and

1, ,

2 2

y

CD y CD y o

lE E

l l

so, the total load,

1, , , , ,

2

4 100 4.5 250 112.5y MN y RS y FG y HJ y CD y o

lP A A kN

l

and the deflection:

1 3

2501000 1.25

200 10

oy l mm

E

As the deflection increases, the members FG and HJ will yield next at a deflection:

2 1 , 2 1 , 2 1i

y HJ y FG yl l lE

At this deflection, the stress in members MN and RS is still o and that in members FG and

HJ is i . The stress in member CD is:

2 1, 2 , 2

2 2

y

CD y CD y i

lE E

l l

34

so, the total load,

12 , 2 , 2 , 2 , 2 , 2

2

2 2

100 2 250 2.5 500 175

y MN y RS y FG y HJ y CD y o i i

lP A A

l

kN

and the deflection:

2 1 3

5001000 2.5

200 10

iy l mm

E

The last member to yield will be CD, and this will occur at a deflection:

3 2 3

5002000 5

200 10

iy l mm

E

The corresponding force will be:

3 , 3 , 3 , 3 , 3 , 3 2 3

100 2 250 3 500 200

y MN y RS y FG y HJ y CD y o iP A A

kN

Thus, load displacement curve:

Exercise 3.3: Calculate the internal pressure required to cause yielding at the inner surface,

one third of the way through the wall and completely through the wall of a hydraulic cylinder

of bore 150mm and wall thickness 30mm, made from steel with a yield stress of 250MPa.

Sketch the stress distribution across the wall of the cylinder in each case.

Only one equation is needed here: 2

( ) 2ln 1

2

Yi e p Y

i o

a ap

r r

0

50

100

150

200

250

0 1 2 3 4 5 6

Lo

ad

(kN

)

Displacement (mm)

Load

Members FG

and HJ yield

Members MN

and RS yield

Member CD

yields

Unload

35

Where, for first yield, ia r , for one third through the wall 30

75 853 3

o ii

r ra r mm

,

and, for fully plastic oa r . So, the pressure for first yield is:

2 2

2 2

250 751 1 61.2

2 2 105

iYY

o

rp MPa

r

for one-third through the wall:

13

2 2

2 2

85 250 85ln 1 250ln 1

2 75 2 105

250 0.125+125 0.345=31.25+43.1

=74.4MPa

YY

i o

a ap

r r

for full plasticity:

105ln 250ln 250 0.336=84MPa

75

op Y

i

rp

r

For the stress distributions, the elastic stress equations (Activity 1) are needed for the wall

beyond the yielded zone boundary:

When the external pressure is negligible, and putting the internal radius at the plastic zone

boundary, the elastic equations are:

2 2

2 2 21a o

rr

o

p a r

r a r

and

2 2

2 2 21a o

o

p a r

r a r

where

2 2

22

Ya o

o

p r ar

So 2

2

2 21

2

oYrr

o

ra

r r

and

22

2 21

2

oY

o

ra

r r

When r = a:

2 2

2 2

2 2 2 2

22

2 2

1 12 2

22

o oY Yrr

o o

oYY

o

r ra a

r a r a

ra

r a

For the wall inside the yielded zone boundary:

2 2

2ln

2

Yrr Y o

o

ar a

r r

and 2 2

21 ln

2

YY o

o

ar a

r r

36

For first yield (internal pressure 61.2MPa), ia r so only the elastic equations are required:

2 4

2 2 2

61.2 105 1.102 101 63.75 1

1051

75

rrr r

and 2 4

2 2 2

61.2 105 1.102 101 63.75 1

1051

75

r r

The plastic equations:

2 2

2

75 250 75250ln 105 75 250ln 61.22

2 105rr

r r

and

75250 1 ln 61.22

r

only hold for r = 75mm

For internal pressure 74.4MPa 85a mm so the plastic equations are required for

75 85mm r mm :

2 2

2

85 250 85250ln 105 85 250ln 43.1

2 105rr

r r

and

85250 1 ln 43.1

r

For 85 105mm r mm

22 2 2

, 2 2 2 2

250 85 1051 1

2 2 105

oYrr

o

ra

r r r

For complete yield, oa r so

105250lnrr

r

and

105250 1 ln

r

These are shown plotted below:

37

Note that, when the wall is plastic, the difference between hoop and radial stress is equal to the

yield stress (because we have used the Tresca Criterion to define yield). Also, there is a

discontinuity at the elastic/plastic boundary where the slope of the hoop stress with radius

changes sense. The radial stress always becomes less negative, from the inner value (equal to

the internal pressure) to the outer value (equal to the external pressure, here negligible).

Exercise 3.5: Some measurements have been made of the steady state creep rate of a nickel

based superalloy and it has been found that the stress exponent, n, is 5, the activation energy,

QC , is 120,000 J/mol and the pre-exponential constant, K2 , is 1.31x10-8 (the constants K2 and

n are in MPa, % and hour units).

The alloy is to be used to manufacture turbine blades with a maximum service temperature of

800C. It is expected that the blades will experience a tensile load of 5kN at operating speed

and the end clearance allows for a maximum creep strain of 2% over the design life of 10,000

operating hours. Calculate the minimum required cross-sectional area of the blade. What would

be the creep life if the temperature were to be increased to 850C?

You are given the data to allow you to write a specific version of the steady-state creep

equation:

8 5 1200001.31 10 exp

8.34s

T

When T = 1073 K, the maximum allowable creep strain rate is 2/10000 %hr-1. Inserting these

values into the above equation, you can solve for:

14 5

15

10

21.966 10

10000

2

1.966 10

-150-100-50

050

100150200250300

75 85 95 105

Str

es

s (

MP

a)

Radius (mm)

Hoop stress(a=ri)Radial stress(a=ri)Hoop stress(a=1/3)Radial stress(a=1/3)

38

Thus, the allowable stress is about 100 MPa. From the applied load of 5 kN, you can now find

the required XSA of 50 mm2. This is the minimum, since a smaller XSA would result in a

higher stress and therefore greater creep rate.

If the temperature is now increased to 1123 K and the stress is kept at 100.3 MPa, the creep

rate becomes:

8 5 41200001.31 10 100.3 exp 3.625 10

8.34 1123s

So, if the creep rate is 3.625 x 10-4 %/hr, 2% strain will arise in 5517 hr, which is the new creep

life.

Exercise 3.6: Assume the cross-section calculated above to be approximately rectangular with

breadth, b, three times the depth, d. If the blades are 0.5m long, and can be regarded as

cantilevers, calculate the stress across the cross-section at the blade root for an effective end-

load on the cantilever of 1kN: (i) at normal ambient temperature (no creep), (ii) at 800C and

(iii) at 850C.

From 2.3.1, 5m , 14

800 1.966 10B and 14

850 3.572 10B .

The bending moment: 51000 500 5 10M F L Nmm .

The XSA: 2 23 50A b d d mm , and so 4.08d mm

Thus, the stress across the section is: 0.8

0.2 4 0.2

3 3.2

12 12 21.838 1.019 10

15c

M y My y y MPa

bd d d

The elastic stress would be:

3

3 4

124 7.218 10e

M My y yMPa

bd d

Plotting these:

39

Note that the effect of creep is to relax the stress in the outer fibres of the beam and to increase

the stress towards the neutral axis.

Exercise 3.7: A cylindrical alloy steel pipe of outer diameter 100 mm and wall thickness 3 mm

requires to have a life of 10,000 hours at 400C. If the creep strain is not to exceed 0.5%, m =

3 and B = 1.45 × 10-19 hr-1 MPa-1, find the allowable pressure.

Assuming thin cylinder, 1 22 and 3 0

2 2 2 2 2 2

1 2 2 3 3 1 1 1 1 1 1

1 1 30.5 0.5

22 2

1

1 3

1 1 2 3 1 1 1

1

2 2 3 1

1

1 3

3 3 1 2 1 1 1

1 30.75 0.5625

2 2

10

2

1 30.75 0.5625

2 2

m

m

m

m

m

B B

B

B B

So, design is constrained by the hoop strain, ε1, and the hoop stress 12

pD

h . The total hoop

strain at time, t, is:

3

1 1 0.56252

pDt B t

h

, so the maximum allowable pressure is:

13 319

2 2 3 0.005510

0.5625 100 0.5625 1.45 10 100000

hp MPa

D B t

Exercise 3.8: The bolts holding down the flange on a pressure vessel have been tightened to a

stress of 400 MPa at 400°C. Assuming the flange to be rigid, find the remaining stress in the

-20000

-15000

-10000

-5000

0

5000

10000

15000

20000

-2.1 -1.1 -0.1 0.9 1.9

Str

es

s (

MP

a)

Distance from neutral surface (mm)

Creep stress

Elastic stress

40

bolts after 100,000 hours. Also find the time it will take for the stress to relax to 80% of its

original pretension.

For the bolting alloy, E = 200GPa, the stress exponent, m, is 6.3, the activation energy, QC , is

220,000 J/mol and the pre-exponential constant, K2 , is 1.31×10-8 (the constants K2 and m are

in MPa, % and hour units)

At T = 673 K the steady-state creep equation becomes:

8 4.3 25 4.32200001.31 10 exp 1.10 10

8.314 673s

So 251.10 10B

in hr, % and MPa units.

Stress after 100,000hr is given by solving:

1 1 3 25 5.3 5.3

0

14

5.3 5.3

1 1 1 1 1 1 1 1100000

1 200 10 1.10 10 5.3 400

1 11.166 10

400

m m

t t

t

tEB m

which gives:

14 14

5.3 5.3

1 11.166 10 2.784 10

400 t

so the relaxed stress is 361MPa

Relaxation time is given by:

1 1 3 25 5.3 5.3

0

18

5.3 5.3

1 1 1 1 1 1 1 1

1 200 10 1.1 10 5.3 320 400

1 18.576 10 314,000

320 400

m m

t

tEB m

hr

i Photographs from Engel L and Kilingele H An Atlas of Metal Damage, Hanser Verlag 1981. ii Diagram from Boresi, Schmidt and Sidebottom Advanced Mechanics of Materials, Wiley, 1983 iii Diagram from Benham P P, Crawford R J and Armstrong C G Mechanics of Engineering Materials, Longman, London, 1996 iv Most essential mechanics of solids texts, such as Benham P P, Crawford R J and Armstrong C G Mechanics of Engineering Materials,

Longman, London, 1996 cover the material in this section. v Photograph from Callister W D Materials Science and Engineering, Wiley, 2000. vi This resolution requires each stress to be considered as a force, and includes resolving the plane upon which it is acting