MAE 212: SLAB ANALYSIS FOR FLAT ROLLING

N. Zabaras, Spring 2000

Figure 1: Schematic of flat rolling showing the neutral point, N .

XFx = 0 ) (�x + d�x) (h + dh)� �xh

� (2�pRd�) cos�

+ 2pRd� sin� (1)

� = entry

+ = exit

�x h+ �xdh+ hd�x + dhd�x| {z }zero

��xh� 2�pRd� cos�+ 2pRd� sin� = 0)

d(�xh)

d�= 2pR (�sin�� � cos�)

Small angles:sin� � �

cos� � 1

375 )

d (�xh)

d�= 2pR (��� �) (2)

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Figure 2: Stresses on an element in rolling: (a) entry zone and (b) exit zone.

For small angles take �z ' �p and for plane strain (�y = 0))

�x + p =2p3Y

| {z }

true anywhere inside the deformation zone

(Y = Yield Stress) (3)

h changes with � as follows:

Figure 3: Approximation of h in terms of �.

h = hf + 2 (R�) sin�

2

= hf + 2 (R�)�

2= hf +R�

2 (4)

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We assume that as the material advances inside the deformation zone, its hardening behavior issuch that: Y h = constant (so as h decreases, Y increases such that the product Y h remainsconstant!!! A ridiculous assumption that however is better than assuming that Y is constant insidethe deformation zone!).

Returning to the equilibrium equation with the above assumption, we can write:

d (�xh)

d�=

d�

2p3Y � p

�h

d�=

2p3

d (Y h)

d�| {z }zero

�d (ph)

d�

= �d (ph)

d�= �

d�p 1

YY h

�d�

= �d (p=Y )

d�Y h|{z}const

(5)

Finally:

�d�p

Y

�d�

Y h = 2pR (��� �))

�d�p

Y

�p

Y

= 2R��� �

hd�)

�d(

p

Y)

p

Y

= 2R ����hf+R�

2d� (6)

Let us integrate the above equation in the entry region from � = � to a general angle �. Similarcalculation can be applied to the exit region.

�Z

�

entry

d�p

Y

�p

Y

= 2R

Z�

entry

��+ �

hf +R�2d�) (7)

��lnp

Y� ln

p

Yjentry

�= �ln

�hf +R�2

�+ ln

�hf +R�2

�

+ 2R�1qhfR

0@tan�1

sR

h�� tan

�1

sR

h�

1A (8)

Note that in the last calculation we used the following integral formula:Z

dx

a2 + b2x2=

1

abarctan

bx

a; (arctan � tan

�1) (9)

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At the entry region using the yield condition, one can write the following:

p

Y|entry =

2Y√3− σx

Y|entry =

(2√3− σb

Yentry

)=

2√3

(1− σb

Y ′entry

)(10)

where Y ′entry =

2√3Yentry.

So returning to equation (8), we can write:

−lnp

Y+ ln

2√3

(1− σb

Y ′entry

)

= −ln

h︷ ︸︸ ︷hf + Rφ2

+ ln

ho︷ ︸︸ ︷hf + Rα2

+ 2Rµ1√hfR

tan−1

√R

hφ − tan−1

√R

hα

(11)

Define:

H = 2

√R

hf

tan−1

(√R

hf

φ

)

Ho = 2

√R

hf

tan−1

√

R

hf

entry︷︸︸︷α

(12)

Equation (11) is now simplified as:

−lnpY

ho

2√3

(1− σb

Y ′entry

)h= µ (H − Ho) ⇒

−lnpY ′ho

h(1− σb

Y ′entry

) = µ (H − Ho)

lnpY ′ho

h(1− σb

Y ′entry

) = µ (Ho − H) (13)

Finally, the following pressure distribution is derived in the entry region:

p

Y ′ =

(1− σb

Y ′entry

)h

ho

eµ(Ho−H) (14)

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where H and Ho are given by equation (12).

To derive the corresponding equation in the exit region, you can repeat the above calculationsby integrating equation (6) (with the bottom sign in ±) from angle φ to angle 0 (exit).

It is also possible to derive the distribution of p at the exit using equation (14) with somechanges! (

here ho → hf , Y ′entry → Y ′

exit

Ho → 0 (because α = φat the exit = 0)

)(15)

p

Y ′ =

(1− σf

Y ′exit

)h

hf

eµH (16)

Equations (14) and (16) define the complete pressure distribution in the deformation zone.

Calcuation of the Neutral Point

Equate the two pressure expressions from equations (14) and (16):(1− σb

Y ′entry

)h

ho

eµ(Ho−H) =

(1− σf

Y ′exit

)h

hf

eµH (17)

⇒ eµ(2H−Ho) =1− σb

Y ′entry

1− σf

Y ′exit

hf

ho

(18)

Simplifying for the case σb = σf = 0 leads to:

Hn =1

2

(Ho − 1

µln

ho

hf

)(19)

2

√R

hf

tan−1

(√R

hf

φn

)= Hn ⇒ (20)

φn =√

hf

Rtan

(√hf

RHn

2

)(21)

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