SME
Manufacturing Processes · Problem type
Can one rolling pass take this draft?
Use this for “A 25 mm slab is rolled to 20 mm with 250 mm rolls and μ = 0.15. Can one pass do it?”
Diagram
L
=
√
Rd
=
√
250 mm × 5 mm
L = 35.36 mm
Bite
angle
α
=
cos⁻¹(1
−
d
2R
)
=
8.11°.
d
≤
dₘₐₓ
=
μ
2
R
=
5.625
mm:
the
rolls
bite,
one
pass
works.
Drawn to scale near the gap; the roll centres are far above and below.
This is an example. Type your own values to start a new problem.
d (mm)
=
t₀ (mm)
−
tf (mm)
dₘₐₓ (mm)
=
μ
2
×
R (mm)
ε
True strain
calculated
L
Contact length
calculated
P
One pass?
worked out (not typed)
nₚ
Passes needed
calculated
Formulas
nₚ
=
⌈
d
dₘₐₓ
⌉
1
=
⌈
5
5.625
⌉
d
=
t₀
−
tf
5 = 25 − 20
dₘₐₓ
=
μ
2
×
R
5.625
=
0.15
2
×
250
ε
=
ln(
t₀
tf
)
0.2231
=
ln(
25
20
)
L
=
√
R
×
d
35.3553
=
√
250 × 5
P
:
one
pass
works
if
d
≤
dₘₐₓ,
else
too
much
draft
5 ≤ 5.625, so one pass works
Assumptions
•
Flat rolling of a wide slab between two equal rolls; the width hardly changes.
•
The rolls can pull the slab in only while the draft is at most μ²R (friction at the entry).
•
The true strain is ln(t₀ ÷ tf); the contact length is about √(Rd) for a draft small next to R.
Step-by-step
WE KNOW
t₀ = 25 mm, tf = 20 mm, R = 250 mm, μ = 0.15
FIND
draft (d), largest draft (dₘₐₓ), true strain (ε), contact length (L), one pass? (P), passes needed (nₚ)
Step 1 · Find draft (d)
Use d = t₀ − tf
The draft is the thickness the pass takes off.
d
=
25
−
20
d
=
5
mm
Step 2 · Find largest draft (dₘₐₓ)
Use dₘₐₓ = μ² × R
Friction can drag the slab into the gap only up to this draft.
dₘₐₓ
=
0.15
2
×
250
dₘₐₓ = 0.0225 × 250
dₘₐₓ = 5.625 mm
Step 3 · Find true strain (ε)
Use ε = ln(t₀ ÷ tf)
True strain adds up each small thinning: the log of the thickness ratio.
ε
=
ln(
25
20
)
ε
=
ln(1.25)
ε
=
0.2231
Step 4 · Find contact length (L)
Use L = √(R × d)
The
arc
of
roll
touching
the
slab,
for
a
draft
much
smaller
than
R
.
L
=
√
250 × 5
L
=
√
1
,250
L
=
35.3553
mm
Step 5 · Find one pass? (P)
Use P: one pass works if d ≤ dₘₐₓ, else too much draft
The draft is within what friction can pull in, so the rolls bite.
5 ≤ 5.625: one pass works
P
=
one
pass
works
if
d
≤
dₘₐₓ
P
=
one
pass
works
if
5
≤
5.625
P
=
one
pass
works
Step 6 · Find passes needed (nₚ)
Use nₚ = ⌈d ÷ dₘₐₓ⌉
Each pass takes at most dₘₐₓ, so the draft over dₘₐₓ, rounded up, is the passes needed.
nₚ
=
⌈
5
5.625
⌉
nₚ = 1
Check
Put the numbers back in. Both sides should match.
1
=
⌈
5
5.625
⌉
✓
5 = 25 − 20 ✓
5.625
=
0.15
2
×
250
✓
0.2231
=
ln(
25
20
)
✓
35.3553
=
√
250 × 5
✓
5 ≤ 5.625, so one pass works ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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