Hinch, E. J.

English

Mathematics in Industry Information Service Eprints Archive

Friction welding
Rolls Royce
1 The problem
A novel welding process is investigated in which two work-pieces are vibrated
in a sliding motion while being pressed together. The sliding generates heat
which softens a thin layer of material, which can then deform in plastic flow.
Applying the pressure produces a squeeze-film flow, which removes impurities
on the original surface, and which also plays a controlling role by trapping
heat in the thin layer, thereby determining its thickness.
2 Simplifications
In this first study we make a number of simplifications. First, we examine
a two-dimensional problem, appropriate for work-pieces long in the third
direction. Second, we will only study the steady state. Third, we assume
that the displacement of the sliding motion is small compared to the width,
so that we can ignore end-effects and ignore any variations in the applied
pressure while the load is held constant. Fourth, we take the variation in
time of the sliding motion to be a square wave. Of course the real sliding
is more nearly sinusoidal, but the constant magnitude of the shear-rate in
a square wave allows us to ignore fluctuations in the heat generated. Fifth,
we assume the the squeezing velocities are smaller than the sliding velocities.
This means that heat is generated only by the sliding motion. Further, the
plastic flow which has a nonlinear rheology can be linearised around the
base sliding motion, thus yielding a simpler decoupled problem for the small
squeeze flow. Sixth, we ignore inertia in the flow in the thin softened layer.
3 Rheology
The plastic flow in the softened layer is predominantly a shearing motion.
We assume that the shear rate 'Y is related to the shear stress (J by a power
The power law index a is positive and less than unity. For metals it is
typically i. The temperature dependency of the material modulus K,(T)
has a rapidly varying exponential factor corresponding to a high activation
temperature Ta and a more slowly varying linear factor which vanishes at a
melting temperature Tm. It is convenient for the asymptotic analysis later
to refer the two factors to the lower melting temperature. Thus we take
4 Governing equations
Let x be measured in the sliding direction, and y across the softened layer,
with the origin at the centre. We take the width of the work-piece to be 2L,
so that the ends of the flow region are x = ±L.
The temperature T(x, y, t) is governed by the heat equation with advec-
tion, diffusion across the thin layer, and heat generation. Thus with density
p, specific heat cp and thermal conductivity k
The plastic flow (u, v) in the thin softened layer, a combination of the slid-
ing and squeezing motions, we take to governed by the standard lubrication
approximation in which inertia is ignored. Thus the flow is predominantly
a shearing flow, with shear rate -r = 8u/8y. This shearing produces a shear
stress (J given by equation (1). Any variations of the shear stress across the
layer are generated by a pressure gradient along the layer
8p 8(J
0= --+-.8x 8y
au av _ 0ax + ay - .
where 2Uo is the imposed sliding velocity of one work-piece over the other,
o the frequency, made into a square wave by the application of the function
sign(.), and V is the constant velocity at which the work-pieces approach,
caused by the applied pressure. It is our goal to predict this approach velocity.
At the ends of the layer, where the squeeze flow escapes, the pressure us set
to atmospheric
It is convenient to express the externally imposed squeezing load as an aver-
age pressure P,
1 1L2L pdx = P.
-L
The problem is to predict the squeezing velocity V, the thickness of the
softened layer, its temperature and the difference of temperature across it.
The given parameters are the sliding motion Uo and 0, the squeezing pressure
P, the size of the work-piece L, the ambient temperature To, and the material
properties p, cp, k, K,(T) and a. As a concrete example consider: Uo =
0.22 ms-\ n = 188 s-\ P = 1.57 X 108 Pa, L = 6 X 10-3 m, p = 4420 kgm-3,
cp = 750 m2 S-2 OK-I, k = 15 kg mS-3 oK-I, a = 0.25, K,m = 5.4 X 108 Pa SO.25,
Tm = 1350 OK, and Ta = 8400 OK.
This is a first attempt to find the scalings of the unknowns. Refinements
will emerge later in the asymptotic analysis. The non-dimensionalisation has
been designed not to depend heavily on the particular rheology.
It is clear that distance along the layer x should be scaled with the half-
width L, the velocities along the layer u scaled with the imposed sliding Un,
and so that time t is scaled with L/Uo. Further the pressure p should be
scaled with the imposed average pressure P. We chose to scale temperature
T with the melting temperature Tm.
The scaling for the thickness H of the softened layer is not obvious. If
we suppose temporarily that we knew H, then we would scale the shear-rate
'Y with Uo/H, the velocities v across the layer with UoH/L, and the shear
stresses (J with PH / L. With these estimates we can balance heat diffusion
and heat generation
kTm PHUo
H2 LH'
As we shall see later, this scaling is false because the variation of the tem-
perature across the softened layer is much smaller than Tm.
The above scalings produce seven non-dimensional groups. First, a non-
dimensional frequency
which is 0(1) and does not enter the subsequent analysis. Second, the aspect
ratio of the layer
which we have assumed to be small, and as such also does not enter the
subsequent analysis. Third and fourth are two ratios of temperature
This is a usefully large 0(1) parameter. Sixth, the power law index a. Sev-
enth, a parameter which measures the strength of the material compared
with the applied forces. l.From the scaling of the shear stress PH / L and the
scale of the shear-rate Uo/ H, we can construct a scaling for the modulus of
the plastic flow PH/L(Uo/HYt. Comparing this measure to the modulus of
the material Km, we obtain the non-dimensional group
Km(L/ p)(1-a)/2ucil+3a)/2
(kTm)(1+a)/2
For the example data, w = 5.1, HI L = 0.31, Ta = 6.2, To = 0.22, Pe = 28,
a = iand lC= 36.
Henceforth we shall work with non-dimensional variables, for which we
will use the same symbols. The governing equations (1 - 9) are little changed
by this non-dimensionalisation: Km, Tm and TalTm in (2) are replaced by lC,
1 and Ta, PCp and k in (3) are replaced by Pe and 1, Uo and n in (7) are
replaced by 1 and w, and Land P in (8) and (9) are replaced by 1.
In a later section we shall make an asymptotic theory for strong materials
lC»1.
6 Sliding dominating squeezing
In this section we present the scheme for solving the equations when the
sliding is larger than squeezing. The difficulty to be overcome is the nonlinear
rheology, which couples the two motions. Assuming that one dominates
enables us to decouple the two, and to obtain for the smaller a problem
linearised around the larger.
The dominant sliding motion is driven by an externally applied shear
stress, say S sign( cos wt). A simple solution can be constructed with the
sliding velocity and the temperature fields both independent of position x
along th softened layer. With temperature only a function of position y
across the layer, the modulus K(T) only varies across the layer. Hence the
shear-rate 1only varies across the layer. And finally we conclude that the
heat generated also only varies across the layer, which is consistent with the
original assumption that the temperature varies only across the layer. With
the sliding motion independent of position along the layer, there is no need
for any pressure gradient to keep the net flux constant. Hence the shear
stress is constant across the layer.
The problem for the sliding motion ii(y) sign( cos wt) therefore reduces to
The unknown shear stress S is fixed by the boundary conditions from (7)
ii -+ ±1 as y -+ ±oo.
By symmetry ii(O) = O.
We consider the squeeze flow to produce small changes in the shear-rate,
from 1 to 1+ 61. Linearising the rheology for the small changes, the change
in shear-rate produces a change in shear stress, from a to a + 6a, given by
6a = J-l61, with J-l= aK(Tha-1•
We evaluate this effective viscosity J.L using the dominant sliding motion for
the shear-rate, i.e., 'Y = Bu/By. Thus using (11)
J.L= 0'.5(0.-1)/0. fi,1/0.
The squeezing flow is therefore governed by a linear lubrication problem with
a viscosity J.L which only varies across the softened layer.
Now the squeezing motion brings the work-pieces together at the approach
velocity ±V, which is independent of position along the softened layer be-
cause the work-pieces are solid. This will produce a flux of softened material
along the layer proportional to distance x from the centre. We therefore seek
a simple solution to the squeeze flow which varies across the layer and is
proportional to distance along the layer, xu(y). This is possible, because the
effective viscosity J.L is independent of position along the layer. To drive the
flux velocity proportional to x a pressure gradient is required, also propor-
tional to x. Thus the pressure varies quadratically along the layer, vanishing
at the ends by (8). Given the load (9) and the non-dimensionalisation, the
pressure must be
Integrating this from y = 0 where au/By vanishes by the symmetry of the
squeeze flow, and then substituting the expression (12) for the effective vis-
cosity, we obtain
The unknown value u(O) is selected so as to satisfy this condition.
i,From the squeeze velocity along the layer, xu(y) and the mass conser-
vation (6), we obtain a equation for squeeze velocity across the layer
Bv
By = -u.
This is to be integrate from the centre line where by symmetry of the squeeze
flow
Finally we can return to the non-dimensionalised version of heat equation
(3), now with expressions for the heat generated by the sliding motion and
with advection form the squeezing motion.
When the sliding dominates the squeezing, we needs to solve equations
(11), (13), (14) and (15), a coupled system of ordinary differential equations,
to obtain the profiles of the sliding and squeezing motion, u and u, the
velocity of approach to the work-pieces V, and the temperature profile T.
The condition that the sliding dominates the squeezing is that u ~ O(u) = 1.
7 Scaling for hard materials, J( » 1
When the material is hard, in the sense that its modulus of the plastic flow
/'i,m is much greater than the scaling modulus PH/L(Uo/HYY.; the material
becomes sufficiently soft to flow only very close to the melting temperature.
At such temperatures, the variation in the modulus comes from the linear
factor in (2) while the exponential factor varies little. Thus in the softened
layer we can approximate the non-dimensional version of (2) by
We shall find that the difference between the temperature on the centreline
and the melting temperature
With the linear variation (16) of the modulus, its value will be double
the centreline value K!:::..T where the temperature is twice as far below the
melting temperature
At this point, the factor K,-l/a in the equations (11), (13) and (15) has
decreased by a factor of 16 when a = h and so beyond this point the
sliding, squeezing and heating become negligible compared with that on the
centreline. The temperature variation across the softened layer is thus merely
!:::..T. We shall find the contrast between this small temperature difference and
the large difference between ambient and melting explains why the softened
layer is very thin for hard materials.
Vve now seek the new scalings of the variables of interest in the limit of
hard materials K » 1, working in terms of the non-dimensional problem
in §5. Let h be the unknown (non-dimensional) thickness of the softened
layer. The shear-rate in the sliding motion (11) is then au/ay = O(l/h).
Thus the effective viscosity (12) is f-l = O(aK,h1-a). The squeeze flow velocity
u = O(h2 / f-l) produces an approach velocity V = O(h3 / f-l) = O(a-1K,-lh2+a).
We substitute these estimates into the heat equation (15), being careful to
distinguish between the temperature difference 1-To of the material advected
towards the softened layer and the temperature difference !:::..T which diffuses
across the layer. Thus
V !:::..T
Pe(l - To)a-1K,-lhl+a = Peh(l - To) = h2 = a-y = K,h-(l+a).
Finally recalling the temperature dependence of the modulus (16), we have
where f32 = Pe(l- To) / a. We will return is §9 to discuss the consequences of
these scalings on the original dimensional variables of interest. We comment
here that harder materials have thinner layers and slower approach velocities.
Vve now see that for hard materials (K » 1) that the modulus in the
softened layer is indeed small (K, « 1) because the temperature on the cen-
treline is close to melting (!:::..T « 1). The thickness of the softened layer
is also much smaller than our earlier estimate H in (10) by the new factor
h « 1. The squeezing flow is however 0(1), as JC -+ 00. However there is a
useful factor (Pe(1 - 7O)a)1/2 to keep it smaller than the sliding motion.
In making the above estimates of the terms in the heat equation, we
used the small temperature differences !:1T in the diffusion across the layer
and the large temperature difference 1 - 70 in the advection towards the
layer. The latter is necessary so that the rate of heat generation is equal
to the specific heat required to raise the temperature from ambient to near
melting. The former gives the temperature differences necessary to diffuse
heat out of the softened layer. A consequence of using these two scales is
that within the softened layer, where one should use the smaller scale, the
advection term is negligibly small, by a factor !:1T/(1 - 70). Thus within
the softened layer, heat generation just balances diffusion. The advection
becomes important on a much longer length scale, O(h(1 - 70)/ !:1T), which
in fact is large 0(JC1/(1-0)). In this outer thermal boundary layer, there is no
heat generation and the advection velocity is constant. The solution there,
tending to the ambient 70 at great distances and tending to melting towards
the softened layer y -+ 0+, is therefore
T = 70 + (1 - 7O)e-PelVyl.
aTI- = -(1 - 7O)Pe V.ay 0+
We examine the details within the thin softened layer using the non-dimensionalisation
of §5, and the rescalings just found (17) for the hard material limit. We intro-
duce 'fJ as the resealed distance across the layer in y = 'fJh, and 0 the resealed
temperature within the layer in T = 1 - O!:1T. The modulus thus becomes
K, = {3JC-(l+o)/(l-o)O. We now find the resealed versions of the governing
equations (11), (13) and (15).
The sliding velocity u is imposed and so is not resealed. The shearing
stress in the sliding motion is however resealed to s in S = s{3JC-1/(1-0). (The
shearing stress S is in fact small, because in the limit for hard materials the
modulus is smaller than the shear rate is large.) The sliding equation (11)
therefore becomes
au _ 1/00-1/0
a'fJ - s .
This is to be integrated from u(O) = 0 with unknown s selected so that u ---+ 1
as 'T/ ---+ 00.
The squeezing velocity along the thins softened layer 71 is resealed to 71*
in 71= U*(aj3t1• The squeezing equation (13) becomes
:* = _3s(1-O:)/O:'T/()-1/o:.
The unknown 71*(0) is selected so that 71* ---+ 0 as 'T/---+ 00.
The squeezing velocity across the thins softened layer v is resealed to v*
in v = v*(aj3)-lJ(-l/(l-o:). The squeezing equation (14) becomes
ov* -*
O'T/ =-U,
which is integrated from v*(O) = 0 to find the resealed velocity of approach
of the work-pieces V* in v* ---+ - V* as 'T/ ---+ 00.
As explained at the end of the previous section, the advection term in
the heat equation (15) drops out in the thin softened layer. In terms of the
resealed variables, the equation becomes
02() = s(1+o:)/o:()-l/a
O'T/2
The matching condition (18) with the outer thermal boundary layer becomes,
in these resealed variables
O() V*O'T/ ---+ as 'T/ ---+ 00.
The resealed heat equation (22) is integrated from 'T/ = 0 where ()T/ = 0, and
with the unknown ()(O) selected so as to satisfy the matching condition (23).
Numerical solution of the equations (19), (20), (21) and (22) for the case
a = t are shown in figure 1. The values for the shooting parameters used
are s = 1.425, 71(0)= 1.046, ()(O) = 1.245. These give the scaled velocity of
approach of the work-pieces V* = 1.425.
Note that the integrals implied in the above differential equations con-
verge if a < ~.
9 Conclusions for hard materials
Now assemble the above numerical results for a = t, the rescalings of §7 and
the original non-dimensionalisation of §5.
ut-uo ~.__.
v·····
lh .
._-------------------------------------------------
Figure 1: The profiles of the sliding velocity it (ut), squeezing velocity long
the layer 11 (uo), squeezing velocity across the layer v (v), and temperature
() (th), as functions of distance TJ across the thin layer.
The shear stress for the sliding motion was non-dimensionalised by PH / L.
For hard materials it was resealed by /3K-4/3• The numerical value of the
resealed solution was s = 1.425. Thus
(j = 2.850(1 - To/Tm)1/2 p1/2C~/2T~1/6 k4/3 ",-4/3U;;5/3 p1/2 L -1.
For the example data, the value is 5.46 x 106 Pa.
The squeezing motion along the thin layer was non-dimensionalised by
Uo. For hard materials it was resealed by (a/3t1• The numerical value of
the resealed solution was 11(0)= 1.046. Thus
11(0)= 2.092(1- To/Tm)-1/2p-1/2c;1/2T~1/2UoP1/2.
For the example data, the value is 9.84 x 10-2m S-1.
The squeezing motion across the thin layer was non-dimensionalised by
UoH/L. For hard materials it was resealed by (a/3t1K-4/3. The numerical
value of the resealed solution was V* = 1.425. Thus
v = 2.850(1- To/Tm)-1/2p-1/2c;1/2T:,/6k4/3",-4/3U;;2/3p1/2L-1.
For the example data, the value is 3.50 x 10-4m S-1.
The thickness of the thin softened layer was non-dimensionalised by H =J kJu~' For hard materials it was resealed by 1(-4/3. Thus
For the example data, the value is 1.56 x 10-5 m.
The thermal boundary layer on the other hand was resealed by (0.(3)1(4/3 Pe-1.
The numerical value of the resealed solution was l/V* = 0.702. Thus its
thickness is
0.351(1 - To/Tm)-1/2 p-l/2C;1/2T~5/6k-l/3 ",4/3Ug/3 p-1/2 L.
For the example data, the value is 1.30 x 10-2 m.
Contributions from: Julian Spence, Sam Howison, John Ockendon, Jon
Chapman, Stephen Cowley, Gail Duursma, Julie, Guneratne, Bruce Gillies,
Stefan Llewellyn-Smith, David Parker.


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