Roller Deflection
Roller Deflection
by
J. J. Shelton
ABSTRACT
A coordinate system with its origin at the center of the roller, and with the
origin moving as the roller deflects, is used for solution of the fourth order differential
equation of beam mechanics. The result is a continuous polynomial in terms of x, the
distance from the center. The form of the equation is adaptable to further analyses
such as finding the profile of a nip roller, predicting wrinkling of a web, or predicting
the natural frequency of a live-shaft roller as demonstrated in this paper.
NOl.vIENCLATURE
398
w width of web
w weight of roller per unit length
x,y coordinates of roller
Ii deflection
liR deflection of roller because of its own bending
liRmax bending deflection of the roller at its center
Ii, deflection of roller because of deflection of stub shafts
p density of material in roller shell
Oln natural frequency (radians/second)
DEFLECTION
A roller is nearly always symmetrical about its center, with the maximum-width
web approximately centered on the roller. The tension distribution is genera11y
unknown, but an assumption of a uniformly distributed tension is usually satisfactory
for roller design. The load on a roller is usually the vector sum of the tensile forces in
the entering and exiting web spans (equal for an idler with the tension in its steady
state) and the weight of the roller. Additional loads may come from liquid inside the
roller or from a nip roller.
The coordinate system and sign conventions are shown in Figure 1, which
illustrates a live shaft roller. The governing equation([]], page 2) for any span of the
roller with continuous loading q and constant stiffness is:
(I)
where positive loading is upward. The loadings shown in Figures 1, 2, and 4 are
negative. Other relationships are
EI d3y =N (2)
dx3
2
ctx2y -- M .
EI d (3)
399
bearing, but the contribution of this weight to deflection is usually negligible. A
separate solution of equation (I) is required for each section between discontinuities of
loading or stiffness. Separate solutions for tensile forces and weight forces are
generally more useful than a combined solution.
The general solution of the differential equation for the roller loaded by the
tensile load f is
Y = _f_
24EI
x4+Ctx 3 +Cox 2 +C3x+C4
- (4)
for a positive (upward) loading. The first term would be negative for a downward
loading. The constants Ct through C4 are determined by substitution of boundary
conditions into equation (4) and its derivatives.
Boundary conditions at the center of the roller are that the deflection (as defined
by the moving coordinate system) and the slope are zero. At the edge of the web, the
moment and shear are obtained from a free body diagram:
Yo =0, (5a)
and
Ely'"(W/2) = - fW
2
(5d)
( L - W)
2
2 _f 4
x - 24El x ' (6)
tW
Y' = 4EI (7)
( L-~) (8)
and
f
y'" ::::: -EIX. (9)
Evaluation of equations (6) and (7) at x = W/2 results in equations for the
deflection of the center relative to the edge of the web and the slope of the roller at the
edge of the web, results useful in analyzing distortion of the web:
400
2 2 3 3
Y(W/2) = 48EI
3
tWL [3 (w) /LR) 7 (w) /I,R) ]
2 LR \ L -8 LR \ L '
where the function inside the brackets is plotted as f3 in Figure 3, and
, tW3 ( L 2)
y (W/ 2) = SEI W - 3 (11)
The deflection curve of the roller beyond the edge of the web (W/2 < x < LR/2)
is similar to equation (4) except that the first term does not exist, because there is no
distributed force fin this span. The constants C1 through C4 assume different values,
based on the boundary conditions
4
tw (-_~wL
= 48EI _7- ) (12a)
Y(W/2) , 8
Y'(W/2) = fW3
SEI ( WL - 13) (12b)
EIY"(LR/2) tW
=4 (L - L R ) (12c)
and
EIY'" (LR/2) tW ·
=-2 (12d)
The resulting equations for deflection and its derivatives for this span between
the edge of the web and the end of the roller face are
(13)
(14)
(15)
and
y"'=-~r· (16)
Evaluation of equations (13) and (14) at x = LR/2 results in equations for the
deflection of the center relative to the end of the roller face and the slope of the roller
at the end of the roller face:
401
and
(18)
The total deflection caused by the resultant uniformly distributed tensile force f
is the deflection of equation (17), plus the deflection caused by tl1e angle of the stub
shafts as given by equation (18) multiplied by (L-LR)/2, plus the deflection of the stub
shafts as cantilever beams. The latter det1ection is equal to fW/2, the force on the end
of the cantilever (at the center of the bearing), multiplied by the cube of the length of
the cantilever, (L-LR)/2, and divided by 3 E 5 l 5 :
3
tW cL-LR)
Y(U2) = Y(LR/2) + (L-LR)Y (LRi2/2 + 6Esls ~
0
(19)
fWL3 [(LR)
3 l(W) 2 (LR) 2 l(W) 3(LR) 3 (LR) (LR) 2]
Y(U2) = 48EI L -2 LR L + 8 LR L + 3 L - 3 L
(20)
The weight of the stub shafts and the heads are neglected in this analysis. This
assumption is valid for most rigidly mounted rollers even if the numerical comparison
of weights does not seem to justify the assumption, because the energy associated with
deflection near the support bearings is low in comparison to the energy at points near
the center of the roller.
The weight per unit length of the roller, w, can be substituted for f and LR for W
in equations (5a) through (9), resulting in the equation for deflection of the center of
the roller relative to its end because of its weight:
(21)
(22)
402
The total deflection caused by the weight of the roller is the deflection of
equation (21) plus the deflection caused by the angle of the stub shafts as given by
equation (22) multiplied by (L-LR)/2, plus the cantilever deflection of the stub shafts.
The first two of these components of deflection are dependent on the stiffness EI of the
roller, while the third is dependent on the stiffness E,Is of the shafts. After
combination of the first two components, collection of terms, and algebraic
manipulation, these components of deflection of the center of the roller relative to the
bearings caused by ro1Ier weight can be expressed as
(23)
Equation (23) is the deflection of the center of the live shaft roller shown in Figure
2(B) because of the weight of the roller, if the usually negligible deflection caused by
the weight of the stub shafts is neglected, and assuming the shafts to be infinitely stiff.
The third component of deflection of the roller because of its weight is the
deflection of the entire roller as if it were a rigid body supported on its cantilevered
stub shafts. The force on the end of the cantilever (at the center of the bearing) is
wLR/2, and the length of the shaft is (L-LR)/2. The well-known equation for the
deflection caused by compliance of the shafts therefore becomes
(24)
For a dead shaft roller supported on bearings at the ends of the roller (LR =L)
as shown in Figure 2(A), equation (23) simplifies to
which shows the total deflection caused by the uniformly distributed weight to be 5/8
as great as if the total weight wL were concentrated at the center.
403
EI t·
f I +EI
ss
4 ] ' (26)
where
(l.7)
and
(28)
For the usual dead shaft roller with L, the spacing of the bearings, equal to LR,
the face length of the roller, f1 simplifies to
(29)
The minimum value of this special case of f1 is 5/8 for a web width equal to the face
length. The maximum value is unity for the hypothetical case of a concentrated
tension in the center of the web, for which case the deflection is given by the factor
outside the brackets in equation (26).
For the live shaft roller of Figure I, the deflection of the center relative to the
end of the roller, equation (17), can be written as
(30)
where
(31)
For a dead shaft roller with L = LR, f2 of equation (31) becomes identical to the same
special case of f1 in equation (29).
(32)
where
(33)
404
In equation (32), w is the weight per unit length of the roller shell including the cover;
therefore, WLR is the total weight of the roller shell. The weights of the heads (hubs)
and stub shafts are neglected, as explained previously. The function fJa is the same as
fJ in equation (27) and Figure 3 ifW/LR = 1.
(34)
where f4 is given by equation (28) and is plotted in Figure 3. Therefore, the deflection
(because of weight) of the center of the roller relative to the bearings is
(35)
Equations (32) and (34) are used for calculation of the natural frequency in the
second section of this paper.
The profile of a nip roller is correct for only one level of loading, which must
generally be determined by experience or experimentation. Some nip processes,
notably calenders, extend the range of satisfactory operation by imposing a moment on
the ends of one of the nip rollers by means of hydraulic cylinders. However, such
correction of the profi1e is nut exact, bt!cause the terms of the polynomial of the
variable x are not the same for the desired uniform distribution as for the pure moment
on the ends. Another level of compensation for this discrepancy is sometimes
implemented in the form of an "M" profile of the roller of the order of 0.001 inch (0.03
mm), but again such a profile is only exact for one value of loading and roller-bending
moment. The basic crowning profile or compensation for the mismatch of
polynomials can be determined by the methods of this report.
In the above arrangement, the usual practice is to significantly wrap the smaller,
moving, crowned nip roller with the web. The advantage of this arrangement is that
the nip roller turns if the nip is open, so that there is no acceleration required of the nip
roller when the nip closes. Disadvantages compared to slightly (JO degrees or so)
405
wrapping the larger roller are: (]) Tension may significantly deflect the small roller,
causing the crowning to be incorrect except at one value of tension. (2) Alignment of
the moving roller is difficult, especially in the open-nip condition. (3) Crowning of the
roller which is wrapped by the web tends to create wrinkles. (4) Wrinkling is less
likely in a web feeding on to a large, low-friction roller than a small, high-friction
roller.
The assumption throughout this paper of simple supports (no constraint imposed
by the bearings against angular deflection) might be questioned by one unfamiliar with
web handling machinery. Total constraint could be approached only by heavily
preloaded bearings, such as double tapered roller bearings, at each support. The
turning torque of such bearings would usually be prohibitive for an idler roller;
furthermore, one end of a roller must generally be allowed to float axially to allow for
differential expansion of the roller and its mounting. Constraint of this floating
bearing would be extremely complicateU. Therefuri::, musl support bearings for ro11ers
are completely self aligning within the small angles of misalignment encountered in
operation.
The basic beam theory behind this analysis of deflection does not consider the
component of deflection caused by shear stresses; therefore, the prediction of
deflection will be low for a roller which is large and short, such as in calenders and
throughout many metals processing lines. The complexity of incorporating the effect
of shear stresses is rarely justified unless the diameter of the roller is greater than 20%
of its length, if the roller shell is metal. (Hopkins [3] lists design factors and equations
which show the deflection from shear stresses to be 6 percent of the deflection from
beam bending for a uniformly distributed load on a simply supported thin-walled metal
cylinder if DmlLR is 0.1, 24 percent for DmiLR of 0.2, and equal bending and shear
deflections ifDmiLR is 0.41.) Roller shells of fiber composites, however, have much
greater relative shear deflections because of low moduli in shear. The deflection
because of shear stresses in a solid metal shaft is rarely worth consideration, because if
the shaft is short enough for this deflection to be relatively significant, it is likely to be
so short that the total shaft deflection is negligible in comparison to the roller
406
deflection. (From Hopkins [3], design factors and equations show that the deflection
from shear stresses in a round cantilevered solid metal shaft to be 14 percent of the
deflection from beam bending if D/1..R is 0.5, and 25 percent for D/1..R of 0.67.)
The analysis of a live shaft roller does not consider angular deflection of the
stub shafts in addition to the angle of the ends of the roller shell as caused by
flexibility of the heads (hubs) of the roller. Heads are usually designed to be quite
stiff, by such means as thick heads, double heads, or stiff flanges with large bolt circles
on removable stub shafts. If the deflection of the heads is significant, this effect could
be readily incorporated into the equations for deflection. (See [4], Case 21, page 368,
for the angular deflection of a flat circular plate with a fixed edge, loaded by a central
moment.)
The first equation is o=5 FLR 3/384 EI, where F is specified as the total load
(pounds). This equation is for a dead shaft roller with the load uniformly distributed
over the face and with the bearings spaced at the ends of the roller shell. The
calculated deflection is 0.014 inch.
The size of the 15-inch-long stub shafts was not mentioned, but if they were
four-inch diameter solid steel, equation (20) gives a deflection of 0.0586 in., more than
triple the higher deflection from the manufacturer's equations.
cycles/second. (36)
407
If tR, the thickness of the wall of the roller, is small in comparison to the mean
diameter Dm, I= (rr/8) Dm 3 tR and A= rrDmtR, so that equation (36) becomes
(37)
Because E/p is the same, for practical purposes, for carbon steel as for aluminum, the
natural frequencies of a thin-walled aluminum roller and a steel roller of the same
length and diameter are equal. The natural frequency is also unaffected by the
thickness of the wall, for a given mean diameter.
The "exact" deflection curve for a live shaft roller as previously derived proved
too unwieldy for a closed-form solution by Rayleigh's method and a practical
presentation of the results.
The deflection curve (because of weight) of the roller body extended to include
the stub shafts at the angle nf the ends of the roller shell (as if the shafts were infinitely
stiff) is assumed to be a parabola:
(38)
Note that the coordinates shown in Figure 4 are different from those used for the
derivations of deflection.
The total deflection 8 is the sum of the above roller deflection OR and the
cantilever deflection 85 of the stub shafts, as shown in Figure 4:
(39)
The deflections ORmax and 8,; can be calculated with equations (32) and (34).
408
The natural frequency by Rayleigh's method is:
fL/2
O
wodx
COn2 ;::::; (40)
The masses of the stub shafts are neglected; however, all points on the roller
undergo the full amount of the deflection of the stub shafts.
Integrating:
4
w [ ( 0Rmax+lis) X --,BRmaxX 3]L/2
3L- o
2 2
8 16 o ]L/
m [ ( ORmax+lis) x- ( ORmax+li,) CL28Rmax)x 3 + SL4 0Rmax-x5
0
Substituting limits:
m = wig, the L's in the numerator and denominator cancel, and fractions are cleared by
multiplying the numerator and denominator by 30:
(41)
cps. (42a)
409
The deflections 6Rmax and Ss (because of weight) are calculated from equations (32)
and (34).
fn - r:i--
=3.50 "\J~ cps. (43)
Equation (43) can be checked against the "exact" derivation of equation (36) in English
units for a uniformly distributed mass between bearings (dead shaft roller) for which
6Rmax =5wL4t384EI, and the difference is shown to be 1.0 percent:
This result is the same as that obtained from the analysis of a single mass on a
massless spring. Confirmation of equation (42a) is therefore excellent for the two
extreme cases. Another special case is if 65 =6Rmax (8Rmax in inches):
If the overall deflection S of the center of the roller because of its weight is
determined experimentally witl1out a knowledge of the components 6Rmax and 65 , the
natural frequency can vary by a factor no greater than 1.12 by an incorrect assumption
of the components of the deflection. This factor is equation (43) for zero shaft
deflection divided by equation (44) for zero roller deflection, with 8 substituted for
6Rmax and I\;, respectively.
410
highly deflected state. The internal damping of the material in the usual vibration
problem is absent, but the web (when present) would help to limit the deflection.
The above cyclical deflection type of vibration would occur at the frequency of
roller rotation in the common case of an eccentric bore of a roller, in which a curved
tube is machined only on the outside to make the outside cylindrical. If a flattened
tube were machined only on the outside, vibration would occur at twice the frequency
of rotation. (Two maximum and two minimum moments of inertia would occur per
revolution.) Evidently, this latter condition has been labeled a "half-critical" vibration,
because a resonance was observed at one half of the calculated or experimentally
determined critical frequency. The term "half critical" is a misnomer, because the
problem arises from a disturbance at twice the frequency of rotation instead of a
resonance at half the calculated natural frequency.
The natural frequency of a given roller can be easily related to the velocity of
the web. Simplified equations which have been used in industry include
(a)Vcr=49.1D/3
1n,and(b)Vcr=55.3D/li 1n ,
for an answer in fpm if D is the outside diameter in inches and Ii is the total deflection
(because of weight) at the center in inches. The constants in these equations are Sn
multiples of those in equations (44) and (43), respectively, for conversion from inches
to feet and seconds to minutes, and to lineal travel per revolution. Equation (a) is for
behavior as if all of the deflection occurred in the stub shafts, with no bending of the
roller, or as if all of the mass were concentrated in the center. Equation (b) is for an
evenly distributed mass as in a dead shaft roller, or for negligible deflection of the stub
shafts. The equations, however, differ by a factor of only 1.12 if Ii is the total
deflection at the center of the roller.
This study shows that simple approximations commonly used in calculating the
natural frequency or critical velocity are reasonably accurate if the total deflection of
the center of the roller under the influence of its own weight is correct.
Example of Critical Velocity. The roller which was used as an example for
deflection has a shell weight of 3865 lbr, or w = 20.4 lbr/in. The deflection because of
weight is 0.00237 inch according to the manufacturer's first equation, and 0.00327 inch
according to the second. The critical velocity according to the more conservative
industry equation (a) above, and using the deflection of 0.00327 inch, is 20,600 fpm.
If the deflection of 0.00237 inch is used in equation (b), the critical velocity is
calculated as 27,300 fpm.
By the method of this paper, for the above roller, equation (32) gives 0Rmax =
0.00414 inch, while equation (34) gives Os= 0.00576 inch. From equation (42a), fn =
411
33.6 cps. The critical velocity is 12,700 fpm, or 62 percent and 47 percent of the
above two calculations which utilized less accurate methods.
This example shows that critical velocity calculations may be quite inaccurate if
common equations for deflection are used.
ACKNOWLEDGMENTS
REFERENCES
-
I LOAD PER UNIT LENGTH
!SHOWN NEGATIVEJ
iwy----J SIGNS:
_,,/ •SLOPE
+ CURVATURE
' " -
/+CHANGE IN
- X /' CURVAT.URE
LR
q-.Giil, + LOADING
M
L ( c::JJU• MOMENT
NI Cl I•. ~~~"Al
412
I-FORCE PER
UNIT LENGTH
w-WEIGHT PER
UNIT LENGTH
STIFFNESS EI
BEARING BETWEEN
NONROTATING SHAFT
i-----L -----i AND ROLLER SHELL
LR -----{,--I-FORCE PER
W UNIT LENGTH
STIFFNESS w-WEIGHT PER
Esls UNIT LENGTH
STIFFNESS El
BEA.'!NG BETWEEN
I Bl LIVE SHAFT ROTATING SHAFT
AND SUPPORT
STRUCTURE
413
NUMBERS ON CURVES ARE W/LR
!WEB WIDTH/ROLLER FACE)
1.0
JI-. -0~4 I
/: _,
0.9
, ~
/ 1/ ,....
.,, _, - .8
'
0.6
- I /
- t...
,
/
~i -
<ii 0.8 / 1.0
/ ")(
"'w
..J
~
5(::; ,, I/
5 0.7 ~o v,
~
.,, K
co = 0.4-
z 11 0.6 ,, 0
w
::. 0.6 2.
VJ/ "' .,,1.0,7
§
V
""" .// ,, .
f f2- /
2i 0.5
~
/
~ 0.4
LL
O.!!,
--
\ /
5 / /
/ Q.6.,,
,-
t3w 0.3 " J , ,,,..
iiw ' / ~
_,,.
C 0.2 0.4-
~
Q / ""
-
-
3 1..,.,-
~ .,, I,..,' /
0.1 _ _,,, _,, ..... ,___. ..... -~ ,__
o.g.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
LR/L (ROLLER FACE/BEARING SPACING)
STIFFNESS El
J-o----LR-----o-1
,___ _ _ _ _ L - - - - - . . . . . ! STIFFNESS
414
Shelton, J.J.
Deflection and Critical Velocity of Rollers
6/2 l/95 Session 6 9:50 - JO: 15 a.m.
Question - Your comment as to the Jack of available published formulas, may be correct;
I'm sure suppliers of equipment other than Beloit have more than adequate formulas but
they just don't publish them.
Answer - One of the critical-velocity formulas reportedly came indirectly from Beloit.
Question - We do have formulas which more than adequately handle the situation which
you described. A case which you did not cover is a roller with a massive head joining the
shaft and the shell. Depending on the design of the roll and the dimensions of the
assembly, sometimes the mass of these heads has a profound effect on the natural
frequency. When you start going through your energy calculations for computing natural
frequency, this is not insignificant.
Answer- Yes, I am sure that there are rollers with massive heads and Jong stub shafts for
which the mass of the hubs would be significant. The deflection of the stub shafts,
because of the weight of these hubs, would add to the deflection as caused by the weight
of the roller shell as discussed in the paper.
Question - You just basically go through the procedure, plugging in numbers for
deflections into the natural frequency formula?
Answer - I am not answering all your questions, but believe that this paper contains
information which you will find useful. If you are trying to fine tune your calculation of
natural frequency, this equation derived from Rayleigh's method wilJ be useful. If you
don't need great accuracy, you are not for off with published equations for natural
frequency, if you know the deflection of the roller accurately. I have seen some very
poor published equations for deflection of rollers.
Question - You talk about a stub shaft. You mean the shaft ends at the beginning of the
roller, that is not a true shaft, upon which the roller is mounted?
Answer - The shaft in a live-shaft roller usually does not extend through the roller.
Answer - That is why I call it a "stub shaft". A dead shaft roller may need to have its
shaft analyzed separately. You don't want it flopping around.
Question - The shaft of a dead-shaft roller would be a beam with two points of loading
and two supports?
Answer - Yes.
Thank you.
415