The slab method model [Salimi and Kadkhodaei, 2004; Salunkhe, 2006] used in the present work is briefly explained here. Figure 6.1 shows an asymmetric rolling process. It is assumed that the speed of lower roll is higher than that of upper roll (Vl >Vu). Now, in an asymmetric rolling process, due to asymmetric conditions, two
neutral points will be generated i.e., neutral point at upper roll-work interface and neutral point at lower roll-work interface. This leads to form three zones in the roll gap. In zone I, the strip velocity is lower than the speeds of both the rolls and the frictional stresses on the upper and the lower surfaces are in the forward direction. In zone II, the strip velocity is more than the speed of the upper roll and less than the speed of the lower roll. This zone is known as cross shear zone. In this zone, the frictional stresses on the upper surface act in the backward direction and that on the lower surface act in the forward direction. In zone III, as the speeds of both the rolls are lower than the strip velocity, the frictional stresses at both the surfaces are in the backward direction.
Figure 6.1. The schematic diagram of asymmetric rolling
A small slab element of length dx and the strip thickness h at a distance of x is chosen. For the equilibrium, net forces along the rolling direction, net forces along the normal direction and the net moments of the forces about any chosen point on the slab are equated to zero. This leads to the following set of equations:
d d
( ) ( ) 0
2 d d
u l u l
x u l u l
u l eq
p p x h
F x
R R R x x
σ σ
σ σ τ τ
= + − + − + − + = , (6.1)
2 d
( ) 0
d
l u
y l u
eq l u
F x p p x h
R R R x
τ τ τ
τ
= + − + − + = , (6.2)
2
( ) ( )
2 2 2
d d
( 5 ) 0,
6 12 d d
l u
o u l u l
l u u
u l
u l
eq
p p
xh xh h
M h
R R R
xh h
R x x
τ σ σ τ τ
σ σ
σ σ
= + − + + + −
− + + − =
(6.3)
where Ruand Rl denote the radii of the upper and lower rolls and equivalent roll radius Reqis given by
2 u l ,
eq
u l
R R R
R R
= + (6.4)
Rearranging the terms in Eqs. (6.1 and 6.3), we get
1
d d 2 2 ( ) 2( )
d d
u l u l
u l u l
u l eq
p p
x x A
x x h R R hR h
σ + σ = + − σ σ+ − τ τ+ = , (6.5)
1
d d 12 6 6
( )
d d
6( ) 2 ( 5 ) .
l u l u
u l
l u u
u l u l
eq
p p
x x
x x h h R R hR
x B
h hR
σ σ τ σ σ
τ τ σ σ
− = + − + +
+ − − + =
(6.6)
The normal pressures at the upper and lower rolls are denoted by pu and pl. u and l are the interfacial shear stresses at the upper and the lower interfaces, respectively.
The average shear stress acting on the vertical surface of the material is . The longitudinal compressive stresses on the top and bottom rolls are denoted by σu and σl . For small bite angle, one can write [Salimi and Kadkhodaei, 2004]
2 2
2 ( ) 3
u pu σY u
σ = − −τ , (6.7)
and
2 2
2 ( ) 3
l pl σY l
σ = − −τ . (6.8) For most of the strain hardening materials, the flow stress σY can be given by (Eq.
3.8). Assumptions of volume constancy, plane strain and across the thickness uniform plastic strain in the thickness direction allow us to write the equivalent strain as
1 2
2 ln 3
h
ε = h . (6.9) It is to be noted that although the present model accounts for an internal shear stress, yet in Eq. (6.9) the shear strain is neglected in the computation of overall strain. It was found that for the cases studied in this thesis, the shear strain was limited to a maximum 20% of overall equivalent strain in the most severe cases. This introduces less than 5% error in the estimation of flow stress because of the lower value of exponent n. Hence, for the estimation of flow stress, the approximate equivalent strain expression given by Eq. (6.9) is justified. Thus, the equation for the flow stress becomes
( )
1 2 0
2 ln
1 3
Y Y
h n
h
σ = σ + b . (6.10)
(The above equation was used by Salunkhe [2006], but no justification for its suitability was provided.) As the strip thickness is a function of x, the above expression of flow stress is also a function of x. Differentiating Eq. (6.10) we obtain
1
d d
d d
Y h
x C x
σ = , (6.11)
where
( )
1 1 2
1 0
2 3 1 2 ln 3
n
Y
h n h
C σ bh b
−
= − + . (6.12)
Differentiating Eq. (6.7–6.8) with respect to x, and substituting these derivatives as well as Eq. (6.11) in Eqs. (6.2, 6.5 and 6.6), the following system of equations is obtained.
1 1 1
2 2
d ( ) d 3 4 d
d 2 d ( ) 3 3 2 d
u u u Y u
u
u u Y u eq
p A B x C
x R R x R x x
τ τ σ τ τ
σ τ
= − − − + −
− , (6.13)
1 1 1
2 2
d ( ) d 3 4 d
d 2 d ( ) 3 3 2 d
l l l Y l
l
l l Y l eq
p A B x C
x R R x R x x
τ τ σ τ τ
σ τ
= + − − + −
− , (6.14)
d 2
d
u l u l
u l eq
p p
x x
x h R R h R h
τ τ
τ = − + − − τ , (6.15)
where A B1, 1 and C are defined as per Eqs. (6.5–6.6 and 6.12) respectively. These 1 are system of three first order ordinary differential equations. A MATLAB function ODE45 is used to solve these differential equations. Solving these differential equations, the values of p pu, l andτ are obtained. The locations of two neutral points can be calculated using following relationship:
2
( ) 2( 1)
nu A nl eq A
x = V x +R h V − , (6.16) where, x xnu, nl are the distances of upper and lower neutral points from the centers of the roll, respectively and VA is the speed ratio (i.e. the ratio of surface velocities of the lower roll to that of upper roll).
During rolling process, the rolls get elastically deformed. This roll flattening effect is taken into consideration using Hitchcock’s formula [1935] that is discussed
in Eq. (3.17). It is to be mentioned that Hitchcock’s formula is incapable of estimating roll deformation for rolling of thin and hard strips particularly at low reduction. However, for the cases examined in the present thesis, Hitchcock’s formula is appropriate. This is supported by the work of Chandra and Dixit [2004], who have estimated the roll deformation by treating the roll as an elastic half space and using a theory of elasticity solution. The authors found that for moderate strip thickness, reduction, flow stress of the material and friction, their model provides almost same results as the model of Dixit and Dixit [1996], which uses Hitchcock’s model. A more accurate roll deformation model will be needed for the studying the rolling of thin and hard strips at low reduction.
6.2.1 Roll Force, Roll Torque and Strip Curvature
The roll force per unit width can be calculated as [Salimi and Kadkhodaei, 2004]
0 0
d and d
L L
u u u l l l
u l
x x
F p x F p x
R τ R τ
= + = + . (6.17)
where L is the contact length. The converged solution is used for calculating the strip curvature. Taking moment about the center of roll, the rolling torque per unit width is given by
0 0
d and d
L L
u u l l
T = xp x T = xp x, (6.18) The asymmetry in the rolling process leads to produce an undesirable curvature to the rolled strip. There are two different types of effects that contribute to this curvature. The first is due to the difference in the axial strains at the upper and the lower surfaces and the second is due to the difference in the shear strains at the upper and the lower surfaces. The total curvature will be equal to the summation of the curvatures due to these two effects.
6.2.1.1 Strip Curvature Due to Difference in Axial Strains
Figure 6.2 (a) shows the rolled strip with a curvature. This type of curvature (convex upward) is considered negative in the sign convention adopted. Let l1, l2, and l0 be
the lengths of the upper, lower and the neutral layer of the strip. From the geometry of the figure one can write,
( )
1 2
1 2
1 2 0 2
1 1 1
x x ,
l l
r = −h l− = − h ε −ε (6.19) where r1 is the radius of curvature due to difference in axial strains, h2 is the outlet
strip thickness and εx1 and εx2 are the axial strains at the upper and lower surfaces, respectively.
Figure 6.2. The strip curvature due to (a) difference in axial strains (b) difference in shear strains [Salunkhe, 2006]
6.2.1.2 Strip Curvature Due to Difference in Shear Strains
As the material undergoes deformation in the roll gap, the strip adopts a curvature due to the differential shear strains also. It is shown in Fig. 6.2 (b). The strip curvature due to difference in the shear strains is given by
1 2
0 2
(d d )
1 ,
L
xy xy
r L
ε − ε
= − (6.20) where the incremental shear strain at any point for the upper and lower surfaces is given by dεxy1 and dεxy2. Thus, the total curvature to the strip will be the summation
of the curvature due to the difference in the axial strains and curvature due to the difference in the shear strains. Thus, the resultant radius of curvature Rcurv is given by
1 2
1 2
curv . R r r
= r r
+ (6.21) The curvature of the strip is the reciprocal of the resultant radius of curvature (i.e.1/Rcurv).