A tension test uses a test specimen that has a measurement length of 50 mm.

and an area of 200 mm2During the test, the specimen fails under a load of 98000 N.
The corresponding measurement length is 50.23 mm. This is 0.2% of the deformation point.
The maximum load of 168000 N is reached with a measuring length of 64.2 mm. Determine a)
a) the yield strength, b) the modulus of elasticity, and c) the tensile strength.

a) O = F/A = 98,000 N/0.0002 m2= 490 000 000 Pa = 490 MPa

b) O = E/e E = O/e = 490/0.0026 = 188 461.5385

e = (50.23–50/50)–0.0002 = 0.0026

c) O = F/A = 168,000/200 = 840 MPa

2. In problem 1, the fracture occurs at a measuring length of 67.3 mm. a) Determine the
percent elongation, b) if the specimen necks when the area is 92 mm2, determine the
percentage reduction of the area.
a) Percent elongation

34.6%

b) Percentage reduction of the area

92 x 100 / 200 = 46 %

3. The specimen in a tensile test has a gauge length of 2.0 in and an area of 0.5
in2During the test, the specimen fails under a load of 32,000 lb. The measurement length
the corresponding is 2.0083 in. This is 0.2% of the deformation point. The maximum load of 60
000 lb is reached with a measurement length of 2.60 in. Determine a) the resistance of
deformation, b) the modulus of elasticity, and c) tensile strength.

a) O = F/A 32,000 lb264,000 lb/in2

b) O = E/e E = O/e = 64,000 lb / 0.00415 = 15,421,686.75

e = 2.0083–2 / 2 = 0.00415

c) O = F/A 60,000 lb / 0.5 in2120,000 lb/in2

In problem 3, the fracture occurs when the measurement length is 2.92 in, a) determine the
percentage elongation, b) if the specimen is constricted to an area of 0.25 in2determine the
percentage reduction of the area.
a) Percent elongation

0.92 x 100 / 2 = 46 %

b) Percentage reduction of the area

0.25 x 100 / 0.5 = 50 %

5. During a tension test where the initial measurement length is 125.0 mm and the area
the cross-section is 62.5 mm, the following data of force and length are collected
medición: 1)17 793 N a 125.23 mm, 2)23 042 N a 131.25 mm, 3) 27 579 N a 1410.05 mm, 4) 28 913
N at 147.01 mm, 5) 27 578 N at 153.00 mm, y 6) 20 462 N at 160.10 mm. The maximum load is 28
913 N, and the last data point occurred just before the failure. a) Plot the curve
engineering stress-strain. Determine b) the strain resistance, c) the modulus
of elasticity, d) the tensile strength.
a) O = F/A 20 462 N / 62.5 mm2= 327.392 MPa

b) O = E/e E = O/e = 327.392 /0.2808 = 1165.92 MPa

e = (160.10–125 /125) = 0.2808

c) O = F/A 28 913 / 62.5 mm2= 462.608 MPa

6. A plate 42 mm thick made of low carbon steel is reduced to 34.0 mm in one step.
from laminating. As the thickness is reduced, the plate thickens by 4%. The yield strength of
the steel plate has a strength of 174 MPa and the tensile strength is 290 MPa. The entry speed
The plate speed is 15.0 m/min. The radius of the reel is 325 mm and the rotational speed is 49.0.
rev/min. Determine a) the minimum required coefficient of friction that would make this operation work
possible laminating, b) the exit speed of the plate and c) the forward sliding.
(a) dmax=μ2R
Since d = to- tf= 42-34 = 8.0 mm
μ2= 8/325 = 0.0246
μ= (0.0246) 0.5 = 0.157
(b) towovo= tfwfvf
wf= 1.04 wo
42 (wo) (15) = 34 (1.04wo) vf
vf= 42 (wo) (15) / 34 (1.04wo17.8 m / min
(c) vr= π = π R2N (0.325) 2 (49,0) = 16.26 m / min
 s = (vf - vr) / vr = (17.8 to 16.26) / 16.26 = 0.0947

A board that is 2.0 in thick has a width of 10 in and a length of 12.0 ft. The thickness is reduced by
three steps of hot rolling. Each step reduces the plate by 75% of its previous thickness. For
this metal and this reduction is expected to see a widening of 3% at each step. If the speed of
The entry of the sheet at the first step is 40 ft/min, and the speed of the rollers is the same.
para los tres pasos, determine a) la longitud y b) la velocidad de salida de la plancha después de la
final reduction.
(a) After three passes, tf=(0.75)(0.75)(0.75)(2.0) = 0.844 in
wf =(1.03) (1.03) (1.03) (10.0) = 10.927 in
towoLoInvadnilppue;ltaseprovdieavadeitloxratntsaleft.wfLf
(2,0) (10,0) (12 x 12) = (0.844) (10.927) Lf
Lf(2,0) (10,0) (12 x 12) / (0,844) (10,927) = 312.3 in = 26.025 ft
(b) Considering that the speed of the roller is the same in the three positions and that
towovo=fwfvf,
Step 1: VF = (2.0) (10.0) (40) / (0.75 x 2.0) (1.03 x 10.0) = 51.78 feet / min
Step 2: VF = (0.75 x 2.0) (1.03 x 10.0) (40) / (0.752 x 2.0) (1.032 x 10.0) = 51.78 feet/min
Step 3: VF = (0.752 x 2.0) (1.032 x 10.0) (40) / (0.753 x 2.0) (1.033 x 10.0) = 51.78 feet / min
8. A series of cold rolling operations is used to reduce the thickness of a plate from 50 to
25 mm in a reversible 2-roller mill. The diameter of the roller is 700 mm and the coefficient
The friction between the rollers and the work is 0.15. The specification is that the draft should be equal in
the minimum number of passes required
7.875 mm
Minimum number of passes = (a - tf) / dmax = (50-25) / 7.875 = 3.17 → 4 passes
(b) Project per pass d = (50-25) / 4 = 6.25 mm
9. In problem 19.3, suppose that a percentage reduction is specified equally at each step in
lugar del draft: a) ¿cuál es el número mínimo de pases requerido? b) ¿cuál es el draft para cada
step?
a) dmax = μ2 = R (0.15) 2 (350) = 7.875 mm
This results in a maximum possible reduction x = 7.875 / 50 = 0.1575
Let x = reduction fraction per pass, and n = number of passes. The number of passes
It must be an integer. To reduce from a = 50 mm to 25 mm, a = n passes, the relationship
the following must be satisfied:
50 (1 - x) n = 25
 (1 - x) n = 25/50 = 0.5
(1 - x) = 0.51 / n
Try n = 4: (1 - x) = (0.5) 1/4 = 0.8409
x = 1 and 0.8409 = 0.1591, which exceeds the maximum possible reduction of 0.1575.
Treat n = 5: (1 - x) = (0.5) 1/5 = 0.87055
x = 1 a 0.87055 = 0.12945, which is within the maximum possible reduction of 0.1575.
(b) Step 1: d = 50 (0.12945) = 6.47 mm, tf = 50 to 6.47 = 43.53 mm
Step 2: d = 43.53 (0.12945) = 5.63 mm, tf = 43.53 to 5.63 = 37.89 mm
Step 3: d = 37.89 (0.12945) = 4.91 mm, tf = 37.89 to 4.91 = 32.98 mm
Step 4: d = 32.98 (0.12945) = 4.27 mm, tf = 32.98 to 4.27 = 28.71 mm
Step 5: d = 28.71 (0.12945) = 3.71 mm, tf = 28.71 at 3.71 = 25.00 mm
10. A continuous hot rolling mill has two frames. The thickness of the initial plate is 25
mm and the width is 300 mm. The final thickness will be 13 mm and the radius of each frame is 250 mm.
The rotation speed of the first frame is 20 rev/min. Drafts will be produced in each frame.
6 mm thick. The plate is wide enough in relation to its thickness so that it does not occur
an increase in width. Under the assumption that the forward slip is equal in
the speed vr in each frame
determine also the exit speed at each roller frame if the entry speed to
The first frame is 26 m/min.

(a) Let a = thickness of the sheet entering in stand 1. a = 25 mm. Let t1 = exit of
thickness of the sheet in stand 1 and the thickness to enter in stand 2. t1 = 25-6 = 19 mm.
Let t2 = leaving the thickness of the sheet at stand 2. t2 = 19-6 = 13 mm.
Deje vo = speed of the plate entering stand 1.
Let v1 = speed of the plate exiting at stand 1 and entering speed at stand 2.
Leave v2 = speed of the plate to leave at stand 2.
Leave speed VR1 = roll in stand 1. VR1 = πDNr = π(2 x 250) (3.10) (20) = 31.42 m/min
speed vr2 = roll in stand 2. vr2 = ?
Forward sliding s = (vf - vr) / vr
svr = vf - vr
(1 + s) vr = vf
At stand 1, (1 + s) VR1 = v1 (Eq. 1)
At stand 2, (1 + s) VR2 = v2 (Eq. 2)
 For constant volume, towovo = t1w1v1 = t2w2v2
Since there is no change in the width, wo = w1 = w2
Therefore, Tovo = t1v1 = t2v2
1.0vo = 0.75v1 = 0.50v2
v2 = 1.5v1 (Eq. 3)
(Combining equations 2 and 3), (1 + s) vr2 = v2 = 1.5v1
(Substituting (Eq. 1), (1 + s) VR2 = 1.5 (1 + s) VR1, therefore VR2 = 1.5 VR1
1.5 (31.42) = 47.1 m/min
(b) 25vo = 19v1
v1 = 25 (26) / 19 = 34.2 m / min
(Ec. 1): (1 + s) VR1 = v1
(1 + s) (31.4) = 34.2
(1 + s) = 34.2 / 31.4 = 1.089
s = 0.089
34.2 m/min, calculated previously in (b)
v2 = 1.5v1 = 1.5 (34.2) = 51.3 m/min
11. A hot rolling mill has eight stands. The dimensions of the initial sheet
They are 3.0 in thick, 15.0 in wide, and 10.0 ft long. The final thickness will be 0.3 in.
diameter of the roller in each frame of 36 in and the rotational speed in frame number 1 of
30 rev/min. It has been observed that the speed of the sheet entering frame number 1 is
240 ft/min. Assume that there is no widening of the sheet during the sequence of
laminated. The percentage reduction in thickness is equal in each frame and it is assumed that the
forward sliding will be equal in each frame. Determine a) the percentage reduction in
each frame, b) the rotation speed of the rollers in frames two to eight, c) the
forward sliding, d) what is the draft in frames one and eight and e) what is the length
What is the exit speed of the final strip coming out of frame eight?

(a) Reduce from a = 3.0 to tf = 0.3 in more than 8 stands, 3.0 (1 - x) 8 = 0.3
(1 - x) 8 = 0.3 / 3.0 = 0.10
(1 - x) = (0.10) 1/8 = 0.74989
x = 1 a 0.74989 = r = 0.2501 = 25.01% in each stand.
Sliding advance s = (vf - vr) / vr
svr = vf - vr
vf
At stand 1: (1 + s) VR1 = v1, where the speed VR1 = roll, speed v1 = slab output.
At booth 2: (1 + s) VR2 = v2, where the speed vr2 = roll, speed v2 = slab output.
At the stand 8: (1 + s) VR8 = v8, where the speed VR8 = roll, speed v8 = slab output.
For constant volume, towovo = t1w1v1 = t2w2v2 = ... = t8w8v8
Since there is no change in width, wo = w1 = w2 = ... w8
Therefore, Tovo = t1v1 = t2v2 =... = t8v8
a = 3.0
3Vo = 3 (1 - r) v1 = 3 (1 - r) = 2v2. . . 3 (1 - r) 8v8, where r = 0.2501 as determined in the
part (a).
Since s is a constant, vr1: vr2:...: VR8 = v1: v2:...: v8
Given that Nr1 = 30 rev / min, VR1 = πDNr1 = (2π x 18/12) (30) = 282.78 feet / min
In general Nr = (30 / 282.78) = 0.10609vr
Nr2 = 0.10609 x 282.78 / (1-r) = 0.10609 x 282.78 / (1-0.2501) = 40 rev / min
Nr3 = 0.10609 x 282.78 / (1-r) 2 = 53.3 rev / min
Nr4 = 0.10609 x 282.78 / (1-r) 3 = 71.1 rev / min
Nr5 = 0.10609 x 282.78 / (1-r) 4 = 94.9 rev / min
NR6 = 0.10609 x 282.78 / (1-r) = 5 rev 126.9.3 / min
Nr7 = 0.10609 x 282.78 / (1-r) 6 = 168.5 rev / min
224.9 rev/min
(c) Given vo = 240 feet/min
v1 = 240 / (1-r) = 240 / 0.74989 = 320 ft / min
v2 = 320 / 0.74989 = 426.8 m / min
(1 + s) VR1 = v1
(1 + s) (282.78) = 320
(1 + s) = 320 / 282.78 = 1.132 s = 0.132
Consult with support 2: given v2 = 426.8 ft/min from above
Nr2 = 0.10609vr2
Reorganize, VR2 = Nr2 / 0.10609 = 9.426Nr2 = 0.426 (40) = 377.04 feet/min
(1 + s) (377.04) = 426.8
(1 + s) = 426.8 / 377.14 = 1.132 s = 0.132, as before
(d) Project in stand 1 d1 = 3.0 (0.2501) = 0.7503 in
Project at stand 8 d8 = 3.0 (1-0.2501) 7 (0.2501) = 0.10006 in
(e) Duration of the final strip Lf = L8
towoLo = t8w8L8
Given that wo = w8, Tolo = t8L8
3.0 (10 feet) = 0.3L8 L8 = 100 ft
t8w8v8
Tovo = t8v8
v8 = 240 (3 / 0.3) = 2,400 ft / min
12. A plate 250 mm wide and 25 mm thick is reduced in a single pass in a two-roller mill.
rollers to a thickness of 20 mm. The roller has a radius of 50 mm and its speed is 30 m/min.
The working material has a strength coefficient of 240 MPa and an exponent of
hardening due to deformation of 0.2. Determine a) the rolling force, b) the moment of
twisting of rolling and c) the power required to carry out this operation.

(a) Project of d = 25-20 = 5 mm,

Contact with length L = (500 x 5) 0.5 = 50 mm

Real deformation ε = ln (25/20) = ln 1.25 = 0.223

Y f = 240 (0.223) 0.20 / 1.20 = 148.1 MPa

Rolling force F = 148.1 (250) (50) = 1,851,829 N

(b) by T = 0.5 (1,851,829) (50 x 10^-3) = 46.296 Nm

(c) N = (30 m / min) / (2π x 0.500) = 9.55 rev / min = 0.159 rev / s

Power P = 2π (0.159) (1.851,829) (50 x 3.10) = 92.591 Nm/s = 92.591 W

13. Solve problem 19.7 using a roller radius of 250 mm.

(a) Project of d = 25-20 = 5 mm,

Contact length L = (250 x 5) 0.5 = 35.35 mm
Deformación real ε = ln (25/20) = ln 1,25 = 0,223
Y f = 240 (0.223) 0.20 / 1.20 = 148.1 MPa
Rolling force F = 148.1 (250) (35.35) = 1311095 N
(b) by T = 0.5 (1,311,095) (35.35 x 3.10) = 23,174 Nm
(c) N = (30 m / min) / (2π x 0.250) = 19.1 rev / min = 0.318 rev / s
Power P = 2π (0.318) (1.311,095) (35.35 x 03.10) = 92.604 Nm / s = 92.604 W

14. Solve problem 19.7 assuming a combined roller mill whose working rollers
they have a radius of 50 mm. Compare the results with the two previous problems and note the
important effect of the roller radius on the force, torque, and power.
Project of d = 25-20 = 5 mm,
Contact with length L = (50 x 5) 0.5 = 15.81 mm
Real deformation ε = ln (25/20) = ln 1.25 = 0.223
Y f = 240 (0.223) 0.20 / 1.20 = 148.1 MPa
Rolling force F = 148.1 (250) (15.81) = 585,417 N
(b) by T = 0.5 (585.417) (15.81 x 3.10) = 4.628 Nm
(c) N = (30 m / min) / (2πx 0.050) = 95.5 rev / min = 1.592 rev / s
Power P = 2π (1.592) (585.417) (15.81 x 10-3) = 92.554 Nm / s = 92.554 W
A 4.50 in thick plate that is 9 in wide and 24 in long will be reduced in a single
Pass through a two-roll mill to a thickness of 3.87 in. The roll rotates at a speed of 5.50.
rev/min and has a radius of 17.0 in. The work material has a resistance coefficient equal to
30,000 lb/in2 and a strain hardening exponent of 0.15. Determine a) the force
of the laminate, b) the torsion moment of the laminate and c) the power required to perform this
operation.
Project of d = 4.50-3.87 = 0.63 in,

Contact with length L = (17.0 x 0.63) = 0.5 3.27 in

Real deformation ε = ln (4.5 / 3.87) = ln 1.16 = 0.1508

Y f = 30,000 (0.1508) 0.15 / 1.15 = 19.642 pounds / in²

Running force F = S f wL = 16.414 (9.0) (3.27) = 483,000 pounds

(b) by T = 0.5FL = 0.5 (483,000) (3.27) = 789,700 N-cm.

(c) N = 5.50 rev / min

Power P = 2π(5.50) (483,000) (3.27) = 54,580,500 in-lb / min

HP = (54580500 in-lb / min) / (396 000) = 138 hp

16. A one-step rolling operation reduces a plate from 20 mm thick to 18 mm.

The initial plate has a width of 200 mm. The radius of the roller is 250 mm and the rotation speed
it is 12 rev/min. The work material has a resistance coefficient of 600 MPa and a
strain hardening exponent of 0.22. Determine a) the rolling force, b) the
twisting moment of rolling and c) the power required for this operation.
(a) Project of d = 20-18 = 2.0 mm,

Contact with length L = (250 x 2) 0.5 = 11.18 mm = 0.0112 m

Real deformation ε = ln (20/18) = ln 1.111 = 0.1054

Y f = 600 (0.1054) 0.22 / 1.22 = 300 MPa

Rolling force F = 300 (0.0112) (0.2) = 0.672 MN = 672000 N

(b) by T = 0.5 (672,000) (0.0112) = 3,720 N-m

(c) Considering that N = 12 rev/min

Power P = 2π (12/60) (672,000) (0.0112) = 37.697 W

A hot rolling mill has rolls with a diameter of 24 in. It can exert a
maximum force of 40,000 pounds. The mill has a maximum power of 100 hp. It is desired to reduce
a plate of 1.5 in thick to the maximum possible draft in one pass. The initial plate is 10 in
width. The hot material has a resistance coefficient of 20,000 lb/in2 and an exponent of
strain hardening of 0
associated and c) the maximum speed of the rollers for this operation.
(a) Assumption: the maximum possible project is determined by the capacity of the mill's force and
do not roll due to the coefficient of friction between the rollers and the work.

Project of d = 1.5 - tf

Contact with length L = (12d) 0.5

Y f = 20,000 (ε) 0 / 1.0 = 20,000 pounds / inch2

Force F = 20,000 (10) (12d) 0.5 = 400,000 (the limiting force of the rolling mill)

(12d) 0.5 = 400.000 / 200.000 = 2.0

12 d = 2.02 = 4

d = 4/12 = 0.333 in

(b) The true strain ε = ln (1.5 / tf)

tf = a -d = 1.5 a 0.333 = 1.167 in

ε = ln (1.5 / 1.167) = ln 1.285 = 0.251

(c) Taking into account the maximum possible power HP = 100 hp = 100 x 396,000 (in-lb / min) / CV =
39600000 InlB /

min

Contact with length L = (12 x 0.333) 0.5 = 2.0 in

P = 2πN (400,000) (2,0) = 5,026,548N in-lb / min

5,026,548 N = 39,600,000

N = 7.88 rev / min

vr = 2πRN = 2π(12/12) (7.88) = 49.5 m / min

18. Solve problem 19.12, except that the operation is hot rolling over the
the crystallization temperature and the strain hardening exponent is 0.18.
Assume that the resistance coefficient remains at a value of 20,000 lb/in2.
(a) Asunción (like in the previous problem): maximum possible project is determined by
the capacity of the rolling mill force and not by the coefficient of friction between the rollers and
I work.

Project of d = 1.5 - tf

Contact with length L = (12d) 0.5

ε = ln (1.5 / tf)

Y f = 20,000 (ε) 0.18 / 1.18 = 16.949ε.18

F=Y

(10) (12d) = 0.5 34.641Y f (d) = 0.5 400,000 (as indicated)

Y f (d) 0.5 = 400.000 / 34.641 = 11.547

Now use trial and error for values of Y and yd that fit into this equation.

d = 0.3 m, tf = 1.5 - 0.3 = 1.2 m

ε = ln (1.5 / 1.2) = ln 1.25 = 0.223

Y f = 16,949 (0.223) 0.18 = 13.134 pounds / inch2.

(d) 0.5 = 11.547 / 13.134 = 0.8791

d = 0.773, which is not equal to the initial test value of d = 0.3

Treat d = 0.5 in, tf = 1.5 - 0.5 = 1.0 in

ε = ln (1.5 / 1.0) = ln 1.50 = 0.4055

Y f = 16 949 (0.4055) 0.18 = 14.538 pounds / inch2.

(d) 0.5 = 11.547 / 14.538 = 0.7942

d = 0.631, which is not equal to the test value of d = 0.5

Treat d = 0.6, tf = 1.5 - 0.6 = 0.9

ε = ln (1.5 / 0.9) = 0.5108

Y f = 16,949 (0.5108) 0.18 = 15.120 pounds/inch2.

(d) 0.5 = 11.547 / 15.120 = 0.7637

d = 0.583, which is too much compared to d = 0.6

Trate d = 0,58 en, tf = 1,5-0,58 = 0.92 en

ε = ln (1.5 / 0.92) = ln 1.579 = 0.489

Y f = 16,949 (0.489) 0.18 = 15.007 pounds / inch2.

(d) 0.5 = 11.547 / 15.007 = 0.769

d = 0.592, which is close, but still above the test value of d = 0.55

Treat d = 0.585 in, tf = 1.50 at 0.585 = 0.915 in

ε = ln (1.5 / 0.915) = 0.494

Y f = 16,949 (0.494) 0.18 = 15.036 pounds / inch2.

(d) 0.5 = 11.547 / 15.036 = 0.768

d = 0.590, which is close, but still above the test value of d = 0.585.

Trate d = 0.588 in, tf = 1.50 at 0.588 = 0.912 in

ε = ln (1.5 / 0.912) = 0.498

Y f = 16,949 (0.498) 0.18 = 15.053 pounds / inch².

(d) 0.5 = 11.547 / 15.053 = 0.767

d = 0.588, which is almost the same as the test value of d = 0.588.

(b) The true strain ε = ln (1.5 / 0.912) = 0.498

(c) Taking into account the maximum possible power HP = 100 hp = 100 x 396,000 (in-lb / min) / CV

= 39600000 in-lb / min

Contact with length L = (12 x 0.588) 0.5 = 2.66 in

P = 2πN (400,000) (2.66) = 6,685,000 N in-lb / min

6,486,000N = 39,600,000

N = 5.92 rev / min

vr = 2πRN = 2π (12/12) (5.92) = 37.2 m/min

19.A cylindrical piece is pressed in an open die. The initial diameter is 45 mm and the height
The initial is 40 mm. The height after forging is 25 mm. The coefficient of friction in the
the punch-work interface is 0.20. The work material has a yield curve defined by
a resistance coefficient of 600 MPa and a strain hardening exponent of
0.12. Determine the instantaneous force in the operation: a) at the moment when the point is reached
of fluency (fluency at a deformation of 0.002). b) if h = 35 mm and c) if h = 25 mm.
(a) V = πD2L / 4 = π (45) 2 (40) / 4 = 63 617 mm3

Given ε = 0.002, Yf = 600 (0.002) 0.12 = 284.6 MPa, y h = 40 at 40 (0.002) = 39.92

A = V / h = 63617 / 39.92 = 1594 mm2

Kf = 1 + 0.4 (0.2) (45) /39.92 = 1.09

1.09 (284.6) (1594) = 494,400 N

(b) Given h = 35, ε = ln (40/35) = ln 1.143 = 0.1335

Yf = 600 (0.1335) 0.12 = 471.2 MPa

V = 63.617 mm3 from part (a) above.

A h = 35, A = V / h = 63617/35 = 1818 mm2

Corresponding D = 48.1 mm (from A = πD² / 4)

Kf = 1 + 0.4 (0.2) (48.1) / 35 = 1.110

F = 1.110 (471.2) (1818) = 950 700 N

(c) Given h = 25, ε = ln (40/25) = ln 1.6 = 0.4700

Yf = 600 (0.470) 0.12 = 548.0 MPa

V = 63.617 mm3 from the previous part (a).

A h = 25, A = V / h = 63.617 / 25 = 2545 mm2

Corresponding D = 56.9 mm (from A = πD² / 4)

Kf = 1 + 0.4 (0.2) (56.9) / 25 = 1.182

F = 1,182 (548.0) (2545) = N 1649000

20. A cylindrical piece is cold worked in an open die with D = 2.5 in and h = 2.5 in and a height
end of 1.5 in. The coefficient of friction at the die-work interface is 0.10. The material of
Work has a yield curve defined by K = 40,000 lb/in2 and n = 0.15. Determine the force.
snapshot in operation: a) when the yield point is reached (yield at deformation of
0.002), b) h = 2.3 in, c) h = 1.9 in and d) h = 1.5 in.
(a) V = πD2L / 4 = π (2.5) 2 (2.5) / 4 = 12.273 in³

Given ε = 0.002, Yf = 40.000 (0.002) 0.15 = 15748 pounds/in2 and h = 02/05 to 02/05 (0.002) = 2.495

A = V / h = 12.273 / 2.495 = 4.92 in²

Kf = 1 + 0.4 (0.1) (2.5) / 2.495 = 1.04

F = 1.04 (15.748) (4.92) = 80.579 pounds

(b) Given h = 2.3, ε = ln (2.5 / 2.3) = ln 1.087 = 0.0834

Yf = 40000 (0.0834) 0.15 = 27.556 pounds / inch2

V = 12.273 IN3 from part (a) above.

A h = 2.3, A = V / h = 12.273 / 2.3 = 5.34 in²

Corresponding D = 2.61 (from A = πD2 / 4)

Kf = 1 + 0.4 (0.1) (2.61) / 2.3 = 1.045

F = 1,045 (27,556) (4.34) = 153822 pounds

(c) Given h = 1.9, ε = ln(2.5 / 1.9) = ln 1.316 = 0.274

Yf = 40,000 (0.274) 0.15 = 32.948 pounds / inch2

V = 12.273 IN3 from part (a) above.

A h = 1.9, A = V / h = 12.273 / 1.9 = 6.46 in²

Corresponding D = 2.87 (from A = πD² / 4)

Kf = 1 + 0.4 (0.1) (2.87) / 1.9 = 1.060

F = 1,060 (32,948) (6.46) = 225.695 pounds

(d) Given h = 1.5, ε = ln(2.5 / 1.5) = ln 1.667 = 0.511

40,000 (0.511) 0.15 = 36.166 pounds/inch2

V = 12.273 IN3 from part (a) above.

h = 1.5, A = V / h = 12.273 / 1.5 = 8.182 in2

Corresponding D = 3.23 (from A = πD² / 4)

 Kf = 1 + 0.4 (0.1) (3.23) / 1.5 = 1.086

F = 1,086 (36,166) (8,182) = 321,379 pounds

21. A workpiece has a diameter of 2.5 in and a height of 4.0 in. It is resized to a height of
2.75 in. The coefficient of friction at the die-work interface = 0.10. The work material has
a yield curve with a resistance coefficient of 25,000 lb/in2 and an exponent of
hardening by deformation of 0.22. Build a graph of force against height of the work.
Volume of the cylinder V = πD2L / 4 = π (2.5) 2 (4.0) / 4 = 19.635 in3

Vamos a calcular la fuerza F en valores seleccionados de altura h: h = (a) 4,0, (b) 3,75, (c) 3,5, (d)
3.25, (e)

3.0, (f) 2.75, and (g) 2.5. These values can be used to develop the plot.

A h = 4.0, we assume that creep has just occurred and the height has not changed significantly. use
ε = 0.002 (the approximate yield point of metal).

In ε = 0.002, Yf = 25,000 (0.002) 0.22 = 6.370 pounds/in2

Adjustment of the height of this strain, h = 4.0 to 4.0 (0.002) = 3.992

A = V / h = 19.635 / 3.992 = 4.92 in²

Kf = 1 + 0.4 (0.1) (2.5) / 3.992 = 1.025

F = 1,025 (6,370) (4.92) = 32,125 pounds

A h = 3,75, ε = ln (4,0 / 3,75) = ln 1.0667 = 0.0645

Yf = 25000 (0.0645) 0.22 = 13.680 pounds / inch2

V = 19.635 in3 calculated previously.

A h = 3.75, A = V / h = 19.635 / 3.75 = 5.236 in2

Corresponding D = 2.582 (from A = πD² / 4)

Kf = 1 + 0.4 (0.1) (2.582) / 3.75 = 1.028

F = 1,028 (13,680) (5,236) = 73,601 pounds

A h = 3.5, ε = ln (4,0 / 3,5) = ln 1,143 = 0,1335

Yf = 25000 (0.1335) 0.22 = 16.053 pounds / inch2

A h = 3.5, A = V / h = 19.635 / 3.5 = 5.61 in²

Corresponding D = 2.673 (from A = πD² / 4)

Kf = 1 + 0.4 (0.1) (2.673) / 3.5 = 1.031

F = 1,031 (16,053) (5.61) = 92,808 pounds

 A h = 3,25, ε = ln (4,0 / 3,25) = ln 1,231 = 0,2076

Yf = 25000 (0.2076) 0.22 = 17.691 pounds / in2

A h = 3.25, A = V / h = 19.635 / 3.25 = 6.042 in2

Corresponding D = 2.774 (from A = πD² / 4)

Kf = 1 + 0.4 (0.1) (2.774) / 3.25 = 1.034

F = 1,034 (17,691) (6,042) = 110.538 pounds

A h = 3.0, ε = ln (4,0 / 3,0) = ln 1,333 =0,2874

Yf = 25000 (0.2874) 0.22 = 19.006 pounds/inch2

A h = 3.0, A = V / h = 19.635 / 3.0 = 6.545 in2

Corresponding D = 2.887 (from A = πD2 / 4)

Kf = 1 + 0.4 (0.1) (2.887) /3.0 = 1.038

F = 1.038 (19,006) (6,545) = 129.182 pounds

A h = 2.75, ε 0_5_3_ _l_ = ln (4.0 / 2.75) = ln 1.4545 = 0.3747

Yf = 25000 (0.3747) 0.22 = 20144 pounds/inch2

V = 19.635 in3 calculated earlier.

A h = 2.75, A = V / h = 19.635 / 2.75 = 7.140 in2

Corresponding D = 3.015 (from A = πD2 / 4)

Kf = 1 + 0.4 (0.1) (3.015) / 2.75 = 1.044

F = 1,044 (20,144) (7,140) = 150,136 pounds

A cold heading operation is performed to produce the head of a steel nail.

The resistance coefficient of steel is 600 MPa and the hardening exponent by
deformation is 0.22. The coefficient of friction at the die-work interface = 0.14. The wire
the nail is made of 5.00 mm in diameter. The head has a diameter of 9.5 mm and a
thickness of 1.6 mm. The final length of the nail is 120 mm. a) What length of wire is required?
Projecting out of the mold to provide sufficient material volume for this operation.
recalcado?, b) calculate the minimum force that the punch must apply to form the head in this
open die operation.
(a) Volume of nails head V = πDf

2HF / 4 = π (9.5) 2 (1.6) / 4 = 113.4 mm3.

Ao = πDo

2/4 = π (4.75) 2/4 = 19.6 mm²

 ho = V / Ao = 113.4 / 19.6 = 5.78 mm

(b)ε = ln(5.78 / 1.6) = ln 3.61 = 1.2837

Yf = 600 (1.2837) 0.22 = 634 MPa

Af = π (9.5) 2/4 = 70.9 mm2

Kf = 1 + 0.4 (0.14) (9.5 / 1.6) = 1.33

F = 1.33 (634) (70.9) = 59.886 N

23. Get a large common nail with a flat head. Measure the diameter of the head and its
thickness as well as the diameter of the nail shank, a) what length of material should be projected outside
from the die to provide enough material to produce the nail?, b) using the values
appropriate for the resistance coefficient and the deformation hardening exponent
from the metal with which the nail is produced (table 3.4), calculate the maximum force in the operation of
header to form the head.

24.A hot re-pressing operation is carried out in an open die. The initial size of the piece
It has Do = 25 mm and ho = 50 mm. The piece is reshaped to a diameter = 50 mm. At this high
temperature, the work metal flows at 85 MPa (n = 0). The friction coefficient at the interface
the final height of the piece
operation.
(a) V = πDo

2HO / 4 = π (25) 2 (50) / 4 = 24,544 mm3.

Af = πDf

2/4 = π (50) 2/4 = 1963.5 mm2.

hf = V / Af = 24.544 / 1963.5 = 12.5 mm.

(b) ε = ln (50 / 12.5) = ln 4 = 1.3863

Yf = 85 (1.3863) 0 = 85 MPa

The force is maximum at the highest value of the area, Af = 1963.5 mm2

D = (4 x 1963.5 / π) = 0.5 50 mm

Kf = 1 + 0.4 (0.4) (50 / 12.5) = 1.64

F = 1.64 (85) (1,963.5) = 273.712 N

25. A forging hydraulic press is capable of exerting a maximum force of 1,000,000 N. A piece
The cylindrical piece is cold-deformed. The initial piece has a diameter of 30 mm and a height of 30 mm.
The metal flow curve is defined by K = 400 MPa and n = 0.2. Determine the maximum reduction in
height to which the piece can be compressed with the press, if the friction coefficient is 0.1.
Work volume V = πDo

2HO / 4 = π (30) 2 (30) / 4 = 21,206 mm3.

Final area Af = 21.206 / hf

ε = ln (30 / hf)

Yf = 400ε0.2 = 400 (ln 30 / hf) 0.2)

Kf = 1 + 0.4μ (Df / hf) = 1 + 0.4 (0.1) (Df / hf)

Forge strength F = KfYfAf = (1 + 0.04Df / hf) (400 (ln 30 / hf) 0.2) (21.206 / hf)

Requires a trial and error solution to find the value of hf that will match the force of
1,000,000 N.

(1) Handle hf = 20 mm

Af = 21.206 / 20 = 1.060.3 mm2

ε = ln (30/20) = ln 1.5 = 0.405

Yf = 400 (0.405) 0.2 = 333.9 MPa

36.7 mm

Kf = 1 + 0.04 (36.7 / 20) = 1.073

1,073 (333.9) (1,060.3) = 380.050 N

Too low. Try a smaller value of hf to increase F.

(2) Try hf = 10 mm.

Af = 21.206 / 10 = 2.120.6 mm2

ε = ln (30/10) = ln 3.0 = 1.099

Yf = 400 (1.099) 0.2 = 407.6 MPa

Df = (4 x 2120.6 / π) 0.5 = 51.96 mm

Kf = 1 + 0.04 (51.96 / 10) = 1.208

F = 1,208 (407.6) (2,120.6) = N 1,043,998

A little high. You have to try a value of hf between 10 and 20, closer to 10.

Handle hf = 11 mm
 Af = 21.206 / 11 = 1,927.8 mm2

ε = ln(30/11) = ln = 2.7273 1.003

Yf = 400 (1.003) ^ 0.2 = 400.3 MPa

Df = (4 x 1927.8 / π) 0.5 = 49.54 mm

Kf = 1 + 0.04 (51.12 / 11) = 1.18

1.18 (400.3) (1,927.8) = 910.653 N

By linear interpolation, try hf = 10 + (44/133) = 10.33 mm

Af = 21206 / 10.33 = 2052.8 mm²

ε = ln (30 / 10.33) = ln = 2.9042 1.066

Yf = 400 (1.066) 0.2 = 405.16 MPa

51.12 mm

Kf = 1 + 0.04 (51.12 / 10.33) = 1.198

F = 1,198 (405.16) (2,052.8) = 996.364 N

(5) In a new linear interpolation, try hf = 10 + (44/48) (0.33) = 10.30

Af = 21206 / 10.30 = 2058.8 mm2

ε = ln (30 / 10.30) = ln 2.913 = 1.069

Yf = 400 (1.069) 0.2 = 405.38 MPa

Df = (4 x 2058.8 / π) ^ 0.5 = 51.2 mm

Kf = 1 + 0.04 (51.2 / 10.3) = 1.199

F = 1,199 (405.38) (2,058.8) = N 1000553

Maximum reduction height = 30.0 - 10.3 = 19.7 mm

26. A piece is designed to be hot forged in a stamping die. The projected area of the piece,
including the burr, is 16 in2. After cutting, the piece will have a projected area of 10 in2.
The geometric configuration of the part is complex. The material flows at 10,000 lb/in² when heated.
and does not tend to harden due to deformation. At room temperature, the material flows at 25,000.
lb/in2. Determine the maximum force required to perform the forging operation.

Since the work material does not have a tendency to work harden, n = 0.

From Table 21.1, choose Kf = 8.0.

F = 8.0 (10000) (16) = £1,280,000

27.A blank is designed for hot forging in a stamping die. The projected area of the piece
It is 6,500 mm2. The design of the die will cause the formation of burrs during forging, thus
that the area, including the burr, will be 9,000 mm2. The shape of the piece is complex. When heated
the working material flows at 75 MPa and does not tend to harden by deformation. Determine the
maximum force required to perform the operation.
Since the working material does not tend to work harden, n = 0.

From Table 21.1, choose Kf = 8.0.

F = 8.0 (75) (9,000) = 5,400,000 N.