Analysis of Rolling Theory by Analog Computer — Karman's Differential Equation —
HITACHI Analog-Hybrid Computer
Technical Information Series No.8
Analysis of Rolling Theory by Analog Computer
— Karman's Differential Equation —
1968
Hitachi, Ltd.
printed in Japan
Analysis of Rolling Theory by Analog Computer
(Karman's Differential Equation)
§1. Introduction
Since Karman's differential equation cannot be solved strictly analytically, Tselikov,
Trinks, Nadai, Hill and others obtained its solution through approximation or construction
by various means, which, however, require considerable labor. The analysis by analog
computer presented here is an automatically converging calculation system using an auto-
matic programming system, markedly saving calculation time, e.g. determination of a
pressure distribution curve, roll radius or projected contact length etc. takes only 1 ~ 2
minutes. Besides, in the logical operation and control of analog computer, reverse time
operation system is adopted, widely differing from the conventional boundary value problem.
The outline of this analysis will be described below.
§2. Outlines of Rolling
Assume that a plate of thickness h; is rolled to thickness h,. While circumferential
speed of roll, v, is constant everywhere, material speed at roll exit v', is always faster
than that at roll entrance v',, since the material is elongated by rolling. In Fig. 1, there-
fore, material speed of hatched area A is slower than v, and material is drawn in toward
roll exit by frictional force against the roll face, while material speed of hatched area B
is faster than v, preventing flow of material due to frictional force against the roll face.
At the mid-point between A and B, e.g., Y in Fig. 1, material speed is equal to v, elimi-
nating mutual slipping. Hence point Y is called neutral point or watershed, and ¢, roll
angle at neutral point.
The absolute value of frictional force between roll
and material is represented by a single-peaked curve
with maximum at neutral point, as illustrated in Fig. 1.
When horizontal pressure q is applied to a block of
material which is deformed by vertical pressure p as
shown in Fig.2, deformation requires vertical pressure
ptq to overcome the effect of horizontal pressure q.
Accordingly, distribution of vertical pressure takes
thatch-shaped curve as shown in Fig. 1.
-
§3. Derivation of Equation
eo Ns Projected Contact Length
Frictional force at
roll face
If the roll is assumed not to deform in the
course of rolling, projected contact length L (to be
represented by X hereinafter) is given by Eq. (1),
Distribution of vertical
Pressure between
roll and material
exactly, or by Eq. (2), in approximation. Fig. 1. Rolling Process
2
X, = / RAh - Bie (1)
4 P p+q
X, =. RAh (2) ee cel
In case that the segment of roll in contact with —+ —¢
material is elastically deformed in the course of roll- —
ing, such as cold rolling of thin plate, the actual | | | I
projected contact length is longer than X,, as given by p ptq
Hitchcock's formula, Eq. (3) F
= 2 = ' Fig. 2. Deformation under
- /[8R (1 -7?)p'm 8R (1 -7*)p'm g. 2.
sat i xE | +Rah+ xE (3) Plane Stress
a
Or, more frequently, using modified roll radius R' caused by elastic deformation, which
can be represented by Eq. (4) according to Hitachicock, X ' may be represented by Eq. (5)
in place of Eq. (3).
pile 16:(1-=72): “p (4)
R TE bAh
Xt= [R'Ah (5)
where +: - Poisson's ratio
E: Young's modulus
b: Width of material
pm: mean rolling pressure
Ps rolling load
Rolling load P is represented by Eq. (6) as a function of rolling pressure p as in
Eq. (6).
x"
P = bf pdx (6)
fe)
3-2. Karman's Differential Equation
In discussion of stress equilibrium at a minute segment bounded by sections parallel
to roll shaft, as the hatched area in Fig.3, Karman derived a differential equation in the
following way, assuming 9 hypotheses given below.
1. Transverse expansion of material can
be neglected.
2. Coefficient of friction «is uniform
everywhere.
3. Additional shearing strain does not
occur.
4. Elastic deformation does not occur.
5. Material is plastic-rigid. hy h+dh h < he
dx-=)
6. Since two-dimensional constraint
strain resistance (two-dimensional
yielding stress S) is 1.15 times
simple compressive strain resistance <——— FE Eke
(So), the following formula holds be-
tween maximum and minimum
principal stress, S, and S;, respective- Fig.3 Pressure Equilibrium
ly. _ in Rolling
S=1.15S)=Si -S;
7. Strain resistance S is uniform throughout arc of contact.
8. Roll speed is constant.
9. Horizontal stress q of material distributes uniformly in the direction of thickness.
That is, in discussing the equilibrium of horizontal stress at the hatched area in Fig. 3,
transverse extension can be assumed always 1, to omit calculation.
Horizontal stress 4Q acting on the small portion of material through the cross-
section is: ;
AQ = (h+dh)(q+dq) - h-q
dx F
Pr ae r *sing
Hence, 4Q = hq + dh-q + dq-h+ dh-dq - hq
= dh-q + dq-h
d(h-q) (7)
Since horizontal component AQ of stress
acting on the small portion of material through the
surface is the sum of horizontal component of
pressure vertical to roll face pr and of horizontal
h+dh q+dq—™
component of frictional force pr acting to the sur- oe
face of material, horizontal stress on the one side
of material 4 Q/2 is given by (8), (8') or (8"). a2
aQ dx dx
5 7 ese ) wine = # (Pray) cos 6 (8)
Fig.4 Details of Stress
=?,(tang - tan f) dx (8!) Equilibrium
where # = tané.
While Fig. 4 gives stress equilibrium at the
entrance of roll, the direction of frictional force is Ap=APreosé nee
reversed at the exit of roll, thus giving Eq. (8")
from Eq. (8'). Pr
A
= ?; (tan oytan f) dx (8") yi = :
2255)
Furthermore, putting compressive force and iy
compressive stress acting on material in dx from the HII |
roll surface in Fig.5 as Apr and pr, respectively, Pp
the following relation holds: dx
dx
Fig.5 Relation between p
If vertical component of Apr is put as and pr
Ap = Apr cos @ and compressive stress in the vertical
direction as p
Ap = Apr cos 6 = p.dx (10)
Accordingly, the following relation always holds from Eq. (9) and (10),
p=pr y (11)
Hence, p and pr will not be distinguished in the subsequent discussion.
From Eqs. (7) and (8'), Karman's differential equation is derived as an equation for
equilibrium of AQ as below:
d (39) = p(tano Ftan 1) dx (12)
where minus sign refers to entrance side and plus sign to exit side.
Since vertical compressive stress p is sum of horizontal compressive stress and of
stress due to two-dimensional constraint strain resistance,
q+k=P (13)
where k is a function of h, as represented below
k=k@) (14)
1
From Fig. 6, R ae
R'cos @
Be Bet R' - R'cos@
ie h 4/2
hd [_h/2) | ha/2
hence=—>4 RV = RPS ei bait ert) |
Do R Sir ;
Fe tee iez ioe
sB+R'-R! (1-32)")
2
a Fig.6 Deformation Process
And tan @ = x of Material
R' cos@
xX
i ri (16)
1x
TP siege
R 2R!
As boundary conditions, tension to material at entrance and exit, ,, and o2, should satisfy
the following relations:
where X = 0, Ps Io kyi=" g22 BP Cr)
k = ke
h = he P
where X=X,', P=1.15k: - o=B Ph he vw
k = ki (18) ac ve
NS 7
~“ 7
h =h, Sie?
p+
Pz 1
§ 4. Operation of Analog Computer ‘
xX
0 1
4-1. Construction of Automatic Program .
Since pressure distribution p is as illustrated
in Fig. 7, for projected contact length X, of roll without Fig.7. Pressure
elastic strain, p” value is calculated starting from the distribution
given initial value p, in the direction of Xi: ~ 0, and the
value p' at X= 0 is memorized.
Then at X = 0, Giving initial values p, for p*
and p' for p’, calculating simultaneously in the direction 0 >X:, to obtain pressure distribu-
tion curve p as shown by thick lines in Fig.7. At the same time, determining rolling load
P, radius R! of roll with elastic strain is obtained from Eq. (4). From R! and Eq.(5), new
projected contact length X' is determined, and with this the above calculation is repeated in
the same way. By repeating these trial calculations, X', R', p and P converge to finite
values, which are solutions of Karman's differential equation.
A flow chart for the automatic program and time chart for its control process are
given below.
& in(p-10) = plian 6 <a)
dx
Calculator of p™
Calculator of
independent
variable x
0O>x1' or x;" 70
2 th(p-k)) = p(tang +n)
dx
Calculator of p*
Summation of
p- and p*,
calculator of
pressure
distribution
p (pressure
distribution)
x,' (projected
contact
length)
Calculation of
X,', calculator
R! (Roll radius
with elastic
strain)
Calculation of
P, calculator
of rolling
load
load)
Calculation of
R', calculator
of elastic
Strain of roll
of projected
contact length
P (rolling
Fig. 8 Automatic Program System
Calculate HOLD Calculate
xX - lat Rest Rest
calculator X,' +0 atX=0 0- x; e e
Calculate HOLD Calculate
= in reversed at in normal
js time from x =0 time from Beet nee
K=X! xX =0
Calculate
4 in normal
P' -calculator Rest Rest timesfrom Rest Rest
xX =0
p-calculator Rest Rest Calculate Rest Rest
P-calculator Rest Rest Calculate HOLD Rest
R'-calculator Rest Rest Calculate HOLD Rest
X,'-calculator HOLD HOLD HOLD Calculate HOLD
Fig.9 Control Process at the Program Controller
Composition of Analog Computer Circuit
h
(1) Function from for k = k ra
1
While function of k may take various forms, this calculation adopted that given
in Fig. 10. k
kg/mm?
(2) Synthesis circuit for p~ and pt
The circuit shown in Fig. 12 is used
for synthesis, i.e., in general, taking
lower segments of two intersecting curves
as illustrated in Fig. 11. 20
(3) Numerical values ; ‘ ; ; bh
0.2 04 0.6 08 1.0 4x
hi 1.0 mm Fig.10 Function form
of K
he 0.7mm
Bh 0¢3
b 1,000 mm P-
R 270 mm se
8 kg/mm?
oo 12kg/mm? Fig.11 Synthesis of
p” and p*
u 0.07
E 2.1 x 10‘kg/mm’*
v 0.3
(4) Block diagram of
analog computer
Fig.12 Circuit for
Synthesis
Block diagram of the analog computer
is given in Fig.13, in which there are five
integrators controlled separately by the sequence shown in Fig. 14.
Control time | 3 ~5 sec. | X1' sec/mm |3 ~ 5 sec| X:' sec/mm | 3 ~5 sec
etree 2 | RESET | COMPUTE] HOLD | COMPUTE | RESET
soa. in| RESET | RESET RESET | COMPUTE | RESET
Eh OR in) RESET | RESET RESET | COMPUTE | HOLD
Aa awn in| HOLD HOLD HOLD | HOLD COMPUTE
oe OFF OFF ON ON OFF
Fig.14 Operation control sequence
for program controller
et 34
nt 00r
(74 -4guny)OOt
(4) 008
9s
(77-9 ues) 4%
I
'
(7749 wy) g~ y I
I
Zz. "os\ [euB}s jorjU09
gau+ “Zot T mi9L 34S Z°ON = of ot
g 49{]01qU0D wesbosg 4
d
I guryt-= Sf
ul "Wa {t)}—
T Zz \ NSA
(:U/e X7- a 00F"0
ENSURE eT Add+
Xo.
As2° Riches oset enna
JL 9 I
is ie x 008 (6)
Aga, 1) 008 vos o=¢6 ¢ na: aA
td (ea-1)9t'°) 4 OF uz 00S I Wa+ W I ‘ya (
(7-9 ua) or (Se wr 00% 9% }
(9) ext Wry 7 1x x soo NO
¥ <i 1G) ac Re
Tt 008"0 ° = Rick! em
* yo0s I 00s — +
or I ee
a Om
4” u Aqu-
atta *y [euBys jo1;U09
Yr , WAL US FON
I 25 6 y Aa[[4WOD wessorg
I ost"0
I ws YX 00S (or 4
LL ie!
S$ Aaa ba
I v Ȣ fe
N 00s
I (*9-"¥ST°0)' 4 T-
dnosb_y
ZAG I 440
Wg, 008 v = om 06 | 7 rat Ty fi
— da
fas (4-4) 008. (4-4) 208 aa ne
Calculating conditions
R=203mm
h, =3.63mm
Rolling prossure h, =2.54mm
(kg/mm? )
wu =0.224
t,=tr=0
60 4 ————— _ Solution obtained by eigital
computer
If \ == ----- Solution obtained by analog
if Computer
i ——-——-- Strain resistance Value
40 - y, \
se
+
L.
5.0 10.0 15.0
Fig 15 Exampls of digital and analog computation 7
Rolling pressure
(kg/mm*)
60 =
Calculation conditions
R =203mm
h ,=3.63mm
h 2=2.54mm
“=0.224
t,=t,=5.0kg/mm*
Solution obtained by digital computer
———-——-— Solution obtained by analog computer
—— -———- Strain resistance value
te
5.0
Fig 16 Examples of digital and analog computation @)
10.0 15.0
(mm)
Calculating conditions R=203mm
h, =3.63mm
h, =2.54mm
we =0.224
Rolling pressure
Pp t, =t¢ =10kg/mm?
(kg/mm? ) S =20kg/mm?
—_ Solution obtained by
30-4 digital computer
p=. an Le Solution obtained by
analog computer
— - — - Strain resistance value
>
—+>
4
5.0 10.0 . 15.0
Fig 17 Examples of digital and analog computation () X (mm)
=H6—
Rolling pressure
| p
(kg/mm? )
30 - / \
Calculating conditions R=203mm
h, =3.63mm
h, =2.54mm
ft =0.112
107
tp =tr= 0
Solution obtained by
digital computer
-----—-- Solution obtained by
analog computer
——— Strain resistance value
~~
+
5.0 10.0 15.0
X (
Fig 18 Examples of digital and analog computation 4 an)
-ll-
Calculating conditions R=203mm
h, =3.63mm
h, =2.54mm
uw =0.112
Rolling pressure
t, =te =5.0kg/mm?
(kg mm?) fp
ees Solution obtained by
digital computer
Se Solution obtained by
30- analog computer
— -— - Strain resistance value
10-
5.0 10.0 15.0
Fig 19 Examples of digital and analog computation %, i
-12-
Calculating conditions R=203mm
Rolling pressure h, =3.63mm
h, =2.54mm
100-
# =0.112
S =80.0( 7 +0.00817)°3
aN
(kefmm®.) Solution obtained by
digital computer
Solution obtained by
analog computer
Strain resistance value
50
0 ~ T 7
5.0 10.0 15.0
———> X (mn)
Fig 20 Examples of digital and analog computation 6
SAVY