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

FORMING PROCESSES (Extrusion)

EXAMPLE 3.11 (EXTRUSION)

Estimate the maximum force required for extruding a cylindrical Aluminium billet of 50 mm diameter and 75 mm length to a final diameter of 10 mm. The average tensile yield stress for Aluminium is 170 N/mm². What percent of the total power input will be lost in friction at the start of the operation?

FORMULAE

SOLUTION

The ram force is obviously maximum just after the start of the extrusion process as l is maximum. For the given dimensions and assuming a 45°-dead zone, the value of l at the beginning is given as

Next, the value of µ is determined by solving equation (1) through a trial and error method. Thus, we get µ = 0.15. Now, σ_x|_BB is found out from equation (2) with this value of µ. Hence,

Finally, using equation (3), we have

If the velocity of the ram is V_B mm/sec, then the frictional power loss (using equations (4) and (5)) is given as

Now, the total input power is

Therefore, the power loss in friction is

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

FORMING PROCESSES (Drawing & Deep Drawing)

EXAMPLE 3.7 (DRAWING)

A steel wire is drawn from an initial diameter of 12.7 mm to a final diameter of 10.2 mm at a speed of 90 m/min. The half-cone angle of the die is 6° and the coefficient of friction at the job-die interface is 0.1. A tensile test on the original steel specimen gives a tensile yield stress 207 N/mm². A similar specimen shows a tensile yield stress of 414 N/mm² at a strain of 0.5. Assuming a linear stress-strain relationship for the material, determine the drawing power and the maximum possible reduction with the same die. No back tension is applied.

FORMULAE

SOLUTION

To start with, we determine the average tensile yield stress of the material during the process. For the given operation, the strain is

Since the stress-strain relationship is linear, the value of the tensile yield stress at the end of the operation is given by

So, the average yield stress is

From the given data,

For the strain hardening a material, the value of σ_xf, strictly speaking, can go up to the value of the tensile yield stress at the end of the operation (σ_Yf) which may be considerably greater than the average value σ_Y. So, when (σ_xf)_max is put equal to σ_Yf,

Now, from equation (1) with F_b = 0,

Using equations (2) and (3), the drawing power we obtain is

It is interesting to note that the drawing stress is more than the tensile yield stress of the original material. To find out the maximum allowable reduction, we use equation (4) with F_b = 0 and ϕ = 2. Hence,

If we use equation (5), then D_max = 0.652

EXAMPLE 3.8 (DEEP DRAWING)

A cold rolled steel cup with an inside radius 30 mm and a thickness 3 mm is to be drawn from a blank of radius 40 mm. The shear yield stress and the maximum allowable stress of the material can be taken as 210 N/mm² and 600 N/mm², respectively.

(i)                Determine the drawing force, assuming that the coefficient  of friction µ = 0.1 and β = 0.05

(ii)             Determine the minimum possible radius of the cup which can be drawn from the given blank without causing a fracture.

FORMULAE

SOLUTION

(i) We first calculate the blank holding force F_h from the given data as

Next, we find the value of σ_r at r → r_d by using equation (1). Thus,

Now, using equation (2), we get

It should be noted that this is much less than the fracture strength though r_j – r_p = 10 mm = 3.33t, i.e., very close to the limit set by the condition of plastic buckling. From equation (3), the drawing force is given as

(ii) In this case, σ_z = 600 N/mm². From equation (2), we have

Using equation (1) and taking F_h = 52778 N, we get

Again, it is interesting to note that r_j – r_p = 30.8 mm >> 4t, and this goes much beyond the limit set by the plastic buckling condition.

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

FORMING PROCESSES (Forging of Strip & Forging of Disc)

EXAMPLE 3.4 (FORGING OF A STRIP)

A strip of lead with initial dimensions 24 mm × 24 mm × 150 mm is forged between two flat dies to a final size of 6 mm × 96 mm × 150 mm. If the coefficient of friction between the job and the die is 0.25, determine the maximum forging force. The average yield stress of lead in tension is 7 N/mm².

SOLUTION

First, let us determine the shear yield stress K for lead by using following equation. Thus,

To find p, the value of x_s is required. From following equation, we have

Now, from the following equations, the expressions for the pressures p₁ and p₂ (for the non-sticking and the sticking zones, respectively) can be found out. Thus,

Using following equation, the force per unit length we get is

Since the length of the strip is 150 mm, the total forging force is

EXAMPLE 3.5

Solve Example 3.4 when the coefficient of friction µ = 0.08

SOLUTION

Using the following equation, we obtain

Since, x_s is more than l, the entire zone is non-sticking, and, as a result, the expression for the pressure throughout the contact surface is given by following equation. Thus,

So,

The corresponding value of the total forging force is

EXAMPLE 3.6 (DISC FORGING)

A circular disc of lead of radius 150 mm and thickness 50 mm is reduced to a thickness of 25 mm by open die forging. If the coefficient of friction between the jib and the die is 0.25, determine the maximum forging force. The average shear yield stress of lead can be taken as 4 N/mm².

FORMULAE:

SOLUTION

Since the volume remains constant, the final radius of the disc is

R = √2 × 150 mm = 212.1 mm.

Þ R = 212.1 mm

Now, using equation (1)

The expressions for p₁ and p₂ using equations (2) and (3) are

Substituting the values of K, h, R, µ and r_s in equation (4), the total forging force we obtain is

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