IE 215 Solutions for Problems due Oct 17, 2008 (Ch 19)
19.15 A cylindrical workpart with
D
= 2.5 in and
h
= 2.5 in is upset forged in an open die to a height = 1.5
in. Coefficient of friction at the die-work interface = 0.10. The work material has a flow curve defined by:
K
= 40,000 lb/in
2
and
n
= 0.15. Determine the instantaneous force in the operation (a) just as the yield point is
reached (yield at strain = 0.002), (b) at height
h
= 2.3 in, (c)
h
= 1.9 in, and (d)
h
= 1.5 in.
Solution:
(a)
V
=
π
D
2
L
/4 =
π
(2.5)
2
(2.5)/4 = 12.273 in
3
Given
ε
= 0.002,
Y
f
= 40,000(0.002)
0.15
= 15,748 lb/in
2
and
h
= 2.5 - 2.5(0.002) = 2.495
A
=
V/h
= 12.273/2.495 = 4.92 in
2
K
f
= 1 + 0.4(0.1)(2.5)/2.495 = 1.04
F
= 1.04(15,748)(4.92) =
80,579 lb
(b) Given
h
= 2.3,
= ln(2.5/2.3) = ln 1.087 = 0.0834
Y
f
= 40,000(0.0834)
0.15
= 27,556 lb/in
2
V
= 12.273 in
3
from part (a) above.
At
h
= 2.3,
A
=
V/h
= 12.273/2.3 = 5.34 in
2
Corresponding
D
= 2.61 (from
A
=
π
D
2
/4)
K
f
= 1 + 0.4(0.1)(2.61)/2.3 = 1.045
F
= 1.045(27,556)(4.34) =
153,822 lb
(c) Given
h
= 1.9,
= ln(2.5/1.9) = ln 1.316 = 0.274
Y
f
= 40,000(0.274)
0.15
= 32,948 lb/in
2
V
= 12.273 in
3
from part (a) above.
At
h
= 1.9,
A
=
V/h
= 12.273/1.9 = 6.46 in
2
Corresponding
D
= 2.87 (from
A
=
π
D
2
/4)
K
f
= 1 + 0.4(0.1)(2.87)/1.9 = 1.060
F
= 1.060(32,948)(6.46) =
225,695 lb
(d) Given
h
= 1.5,
= ln(2.5/1.5) = ln 1.667 = 0.511
Y
f
= 40,000(0.511)
0.15
= 36,166 lb/in
2
V
= 12.273 in
3
from part (a) above.
At
h
= 1.5,
A
=
V/h
= 12.273/1.5 = 8.182 in
2
Corresponding
D
= 3.23 (from
A
=
π
D
2
/4)
K
f
= 1 + 0.4(0.1)(3.23)/1.5 = 1.086
F
= 1.086(36,166)(8.182) =
321,379 lb
19.17 A cold heading operation is performed to produce the head on a steel nail. The strength coefficient for