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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
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