Forming Processes - ROLLING - Numerical Problems with Solutions (solved)

FORMING PROCESSES

Example 3.1 (ROLLING)

A strip with a cross-section of 150 mm × 6 mm is being rolled with 20% reduction in area, using 400-mm-diameter steel rolls. Before and after rolling, the shear yield stress of the material is 0.35 kN/mm² and 0.4 kN/mm², respectively. Calculate (i) the final strip thickness, (ii) the average shear yield stress during the process, (iii) the angle subtended by the deformation zone at the roll centre, and (iv) the location of the neutral point θ_n. Assume the coefficient of friction to be 0.1

SOLUTION

(i) As no widening is considered during rolling, 20% reduction in the area implies a longitudinal strain of 0.2 with consequent 20% reduction in the thickness. Therefore, the final strip thickness is given as

(ii) The average shear yield stress during the process is taken to be the arithmetic mean of the initial and the final values of the yield stress. So,

(iii) From figure, it is clear that

Substituting the values, we get

(iv) To determine θ_n, first λ_n has to be calculated. Since nothing has been mentioned regarding the forward and the back tension, σ_xi and σ_xt will be assumed zero. Hence,

where

Substituting the values of R, t_f, and θ_i in the expression for λ_i, we get λ_i = 5.99.

Now using this value of λ_i, we find that the value of λ_n is 1.88

θ_n can be expressed as

Hence,

EXAMPLE 3.2 (ROLLING)

Assuming the speed of rolling to be 30 m/min, determine (i) the roll separating force and (ii) the power required in the rolling process describes in Example 3.1

SOLUTION

From above equation, we see that to find out the roll separating force F, we need to know the pressure p (as a function of θ), θ_n and θ_i. In Example 3.1, we have already calculated θ_n and θ_i. For the numerical integration of the above equation, we employ Simpson’s rule, using four divisions after θ_n and eight divisions before θ_n. Thus,

With these intervals, the pressure at different station points can be computed, using following equations.

Next, we compile the following data. After the neutral point:

Station Point	θ (rad)	(mm)		pafter (kN/mm2)
1	0.00000	2.40000	0.000	0.75
2	0.00575	2.40331	0.479	0.788
3	0.0115	2.4132	0.958	0.830
4	0.01725	2.4298	1.432	0.876
5	0.023	2.453	1.88	0.925

Before the neutral point:

Station Point	θ (rad)	(mm)		Pbefore (kN/mm2)
1	0.023	2.453	1.88	0.925
2	0.02981	2.489	2.454	0.887
3	0.03662	2.534	2.997	0.855
4	0.04343	2.589	3.529	0.828
5	0.05024	2.65	4.049	0.804
6	0.05705	2.725	4.555	0.786
7	0.06386	2.81	5.048	0.772
8	0.07067	2.899	5.525	0.759
9	0.0775	3.000	5.99	0.75

Now, from equation (1), the roll separating force per unit width is

(i) Since the width of the strip is 150 mm, the roll separating force is

(ii) The driving torque per unit width for each roll can be computed, as

Taking the values before and after the neutral point and using Simpson’s rule, we find

So, the total driving torque for each roll is 100.8 × 150 N-m = 15210 N-m

Now, the total power required to drive the two rolls is 2 × 15, 120 × ω W, ω being the roll speed in rad/sec. The rolling speed (same as peripheral speed of rolls) is given to be 30 m/min. So,

Thus, the total power (2P_R) = 75.6 kW.

The power required in the rolling process is 75.6 kW

EXAMPLE 3.3

If the diameter of the bearings in Examples 3.1 and 3.2 is 150 mm and the coefficient of friction (µ_b) is 0.005, estimate the required mill power.

SOLUTION

Substituting the values of µ_b, d_b, ω, and F in following equation, we obtain

So, the mill power is given as