Plastic- Case Study 4 Actual Case Study End Flashcards

1
Q

In cold rolling what does and does not affect the flow stress of the material?

A

Strain rate doesn’t affect the flow stress. But the material will work harden due to it being strained. Means the material exiting the rolls has a flow stress higher than the material entering the rolls

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2
Q

What is the mean flow stress of the material in cold rolling?

A

The average of the exit and entry flow stresses

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3
Q

What is rolling pressure and how is it distributed?

A

The rolling load divided by the area of contact. Roll pressure vs length of contact (L) graph starts at flow stress of material at entrance, almost straight diagonal up to neutral point (curved top) then back down to flow stress of material at exit (higher). Neutral point is a bit to the right of middle due to work hardening during rolling

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4
Q

What are back and front tensions?

A

Tension in the plane of the sheet applied to material as it either enters or exits the roll gap.
Back tension: produced by controlling the speed of the uncoiled relative to the roll speed.
Front tension: created by controlling he speed of the wind up reel.
They can both significantly reduce the rolling load (and pressure)

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5
Q

How does tension affect the roll pressure distribution?

A

Front tension: follows normal curve up from starting flow stress but peaks earlier so N shifted left, then comes back down to a lower value than normal exit flow stress.
Back tension: starts lower than normal starting flow stress and peaks later (to right) but still lower and joins normal curve to end flow stress.
Back tension about twice as effective as front tension in reducing rolling load.

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6
Q

Other effects of tension

A

Less wear on the rolls.
Improved flatness of the product across the width

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7
Q

What is mill spring and what is it determined by?

A

Where as the arterial is rolled the force exerted by the material causes the rolls to separate elastically, leading to an exit thickness greater than the original roll separation. This is mill spring.
Determined by the overall stiffness of the rolling mill

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8
Q

How do roll mill spring curves work?

A

Load vs plate thickness. See slide 30. Have one straight diagonal line up from x=roll gap for the elastic deformation of the rolling mill. Have a concave curve down for plastic deformation of plate and the rolling load as a function of final thickness hf (down to initial thickness h0). Plastic curve from F=χ.w.rt(R.(h0-hf)).σflow. Elastic line obtained from mill stiffness. Point of intersection of lines is final plate thickness.

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9
Q

Experimental steps to finding the mill stiffness

A

Take series of slabs of varying thickness.
Pass them through rolling mill with known roll gap.
Measure rolling load and final thickness of slabs.
Calculate mill spring for each slab (hf-roll gap).
Plot rolling load against mill spring.
Slope of line gives mill stiffness.
Emill=F/(hf-hmill) or F=Emill.(hf-hmill)

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10
Q

How to find the roll gap (hmill) for a particular hf

A

Equate both F=χ.w.rt(R.(h0-hf)).σflow
and F=Emill.(hf-hmill).
Rearrange to make hmill the subject
hmill=hf-χ.w.rt(R.(h0-hf)).σflow/Emill
Sub the numbers in from the case study
Mill stiffness seems to be 5
For 5mm thick plate need hmill at 4.72mm

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11
Q

Why do we get variation in the final thickness of the plate?

A

Looking at the rolling load equation, any variation in rolling load leads to a change in final thickness. If rolling load increases then mill spring will increase and so will final thickness. If friction increases increases then thickness will increases. Increased initial plate thickness or flow stress (both can vary according to remit by +/-0.5mm or +/-10MPa) increases final thickness.

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12
Q

Why will the level of friction vary?

A

In reality lubrication control is extremely difficult to maintain

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13
Q

Ideal situation for mill spring curve

A

Friction, flow stress and entrance thickness are all normal (140MPa and 160MPa and 6mm). Then the intercept of the two lines on the mill spring curve is at 5mm height and within acceptable region from 4.95 to 5.05mm. Elastic curve up from 4.72mm roll gap and plastic curve down to 6mm initial plate thickness.

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14
Q

Mill spring curved at the extremes

A

If entrance thickness is 5.5mm then plastic curve is left and lower so intersection is just less than 4.95mm so out out of specification.
If entrance thickness 6.5mm then plastic curve right and higher so intersection just over 5.05mm so out of specification.
Means roll mill control is required

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15
Q

Process control that could be added to the rolling operation

A

One method is to measure the thickness of the plate as it exits the roll. If thickness starts to increase then roll gap could be automatically reduced and vice versa.
Or could apply a front or back tension to plate during rolling. If thickness increased then tension could be increased (effect of decreasing rolling load and reducing mill spring) and vice versa.

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16
Q

What happens if reduction in height of plate is too small per pass?

A

Deformation is concentrated in the surface layers and a concave profile may develop. Can result in edged having to be trimmed which is expensive and wasteful. In extreme cases the residual stresses together with any voids or defects in mid-thickness plane can result in material splitting apart along the centre plane (alligatoring). This also happens for too large reduction in height.

17
Q

What happens when reduction in height per pass is too large?

A

Convex profile normally develops mainly due to friction at the roll/material interface reducing lateral spread. In this case the state of stress can lead to edge cracking. If this is only about 0.5mm in then can trim edges but if more than this then is a bigger problem

18
Q

How is the concept of rolling torque derived?

A

The rolling load can be though of as acting at a single point along the arc of contact. The location of this point with respect to the roll centres determines the torque and power required to produce the required reduction. Horizontal distance between centre of gravity of the pressure distribution from the exit (or roll centre) is a. Torque applied by the rolls is T=F x a. F is rolling load

19
Q

Describe the distance a for rolling torque

A

It is some fraction of the length of the arc of contact such that a=λL. The fraction λ is known as the lever arm and has a value of about 0.5 in hot rolling and 0.45 for most cold rolling

20
Q

Formula for power supplied to each roll

A

WsubR=2θT
This is power for the two rolls.
θ is angular velocity 2πN
N is number of revolutions per min.
WR=4πNT
T is roll torque

21
Q

Effect if friction at the roll bearings and equations t go with this

A

Work will be done against friction at the roll bearings. On each roll there will be 2 bearings each carrying load 1/2.F, giving frictional drag of 1/2.μB.F where μ sub B is coefficient of friction at bearing. For diameter d bearings the required torque is 1/4.μB.F.d.
For 2 rolls with 4 bearings the power required at the bearings WB is:
WB=θ.μB.F.d=2πN.μB.F.d J/min

22
Q

Formula for total power for two rolls with 4 bearings

A

Wtot=WR+WB
=4πNT+2πN.μB.F.d