Transcript Survey of Rolling Processes

Rolling
flat rolling
Shape Rolling
Note appearance of surfaces
Hot rolling is used to reduce
forces and promote ductility
material savings, ideal grain
orientation, strengthening through cold
working, fatigue resistance
Ring Rolling
Thread Rolling
Rolled Products Made of Steel
Frame, drive, jacks, and rolls comprise
a mill or stand which are connected
together via conveyors
Compressive
Stress
Rolling Mill Designs
3-high reversing mill.
2-high mill
4-high mill
6-high or cluster mill
d t o t f
where d = draft; to = starting thickness; and tf = final
thickness
Reduction = draft expressed as a fraction of starting
stock thickness:
d
r 
to
where r = reduction
Tandem Rolling Mill
Note how workpiece
thickness decreases as the
material passes through the
roll gap
- We need to use average
flow stress
Note how workpiece
velocity must increase while
roll velocity will be constant
We need to
calculate force and
power needed to
turn the rolls
Roll speeds (𝜋𝐷𝑁) must account for increased work velocity
Assume no deformation perpendicular to rolling direction (plane strain)
Ainlin  Aout lout
lin
l
 Aout out
t
t
Ain vin  Aout vout
Ain
Ain
hin w vout


Aout hout w vin
Volume in = Volume Out
∴ 𝑣𝑜𝑢𝑡 > 𝑣𝑖𝑛
At point where friction changes direction, this is
termed the no-slip point
At the no-slip point, the roll pressure distribution is
at a maximum
We can consider that the average roll pressure is the
same as the average flow stress for rolling
We can develop an estimate of the force and power for a single stand in cold
rolling as follows:
K
Yf 
1 n
n
For an annealed material:
We can also calculate the average flow stress at the i th stand by:
 i 1
Yf 
 K 
n
d
Note average yield stress will increase as the
next stand in tandem cold rolling.
i
 i 1   i
In cold rolling, the average flow stress will increase due to work hardening .
In hot rolling we often assume that flow stress is constant.
Mill Load
Force = Average Yield Stress * area
Change in Width of workpiece = 0 (plane strain)
Assume rolling is compression between 2 inclined plates
L
Roll radius is large compared to arc contact length
We can reasonably approximate the length, L, as being a straight line due to roll
flattening where 𝐿 = 𝑅(𝑡𝑜 − 𝑡𝑓)
Force F, on each roll, will be
𝑌𝑓 𝑤𝐿
F
Based on the force equilibrium, we
can see that a torque is generated
on the roll
When no front or back pull is
applied, F is midway along the
contact length, L
L
𝑇=𝐹
F
𝐿
2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
𝑃 = 𝐹(𝑙𝑏𝑠) 2(inches) 𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑁
Hp =
𝑙𝑏
𝑃 (𝑓𝑡𝑚𝑖𝑛)
𝑙𝑏
33000(𝑓𝑡𝑚𝑖𝑛)
𝑟𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠
𝑚𝑖𝑛𝑢𝑡𝑒
*hp would need to double hp for 2 rolls
𝐿
2
Power is more
useful for the
drive motor
We would like to roll a workpiece using the minimum number of stands due to the
equipment and operations costs
i.e. maximize the reduction taken at each stand
Clearly, mill load is one factor
Rolling requires that we have sufficient friction to pull the workpiece into the roll gap,
this represents a second limiting factor as to the maximum reduction that can be taken
Consider an element at the entry to the roll gap
- Roll force (normal)
- Friction force (parallel)
Minimum condition for feasible
rolling
q
FN
R
Fnsinq = Ffcosq
m tanq
R-d/2
Successive Application of
Pythagoreans Theorem Yields
dmax=m2R
q
d/2
Ff
Suggests max roughness and roll
sizes
  10 0.25 ksi
Annealed
D = 40” diameter
W = 24”
What is the rolling load at each
stand?
.050”
.041”
.045”
Second Stand
Force F, on each roll, will be 𝑌𝑓 𝑤𝐿
𝐿=
 i1
𝑅(𝑡𝑜 − 𝑡𝑓)
K
Yf 
1 n
n
First Stand
Yf 
 K
n
d
i
 i 1   i
Annealed
.050”
.045”
.041”
10  0.2 
1.25
 .05 
10
ln


n
K
.045

Yf 
 
1 n
1.25
First Stand
 i 1
Second Stand
Yf 


0.25
 4.6 ksi
Note work hardening from first stand
K nd
i
 i 1   i
10  0.2   10  0.1
K


1

n



  2 1 
1.25 0.2  0.1
n 1
1.25
1.25
 6 ksi
Now find contact areas, w is constant in flat rolling
Second Stand
First Stand
𝐿=
𝑅(𝑡𝑜 − 𝑡𝑓)
𝐿=
𝑅(𝑡𝑜 − 𝑡𝑓)
𝐿 = 20 .05 − .045 = 0.316
Area, A = 0.316*24 = 7.58
𝐿 = 20 .045 − .041 = 0.283
Area, A = 0.288*24 = 6.91in2
Load = 4.6*7.58 = 34.9 kips
Load = 6*6.91 = 41.46 kips
Uneven thickness across
workpiece width (crowning)
Minimum thickness achievable due to
roll flattening
note aluminum foil is 0.2 mm thick
No-load
Loaded
Rolling Thin Sheet
Backer Rolls
Work Roll
12 High Sendzimir Mill Stand
Used for rolling very thin sheet material