# ME322 HW 5 T131 solution ```King Fahd University of Petroleum &amp; Minerals
Mechanical Engineering Department
ME 322: MANUFACTURING PROCESSES
Semester (131)
Homework # 5
Assigned on: 15/December/2013
Due date: 22/ December/2013
Name: ________________________
ID #: _____________________
Section #: _________
Problems
Marks
1
10
2
10
3
10
4
10
5
10
Total
50
Problem 1
Calculate the force required in direct extrusion of 1100-O aluminum from a diameter of 15.24 cm
to 5.08 cm. Assume that the redundant work is 30% of the ideal work of deformation, and the
friction work is 25% of the total work of deformation.
Solution
2
2
The extrusion ratio is R = 15.24 /5.08 = 9, and thus the true strain is ε1= ln(9) = 2.20. For 1100-O
aluminum, we have from Table 2.3 on p. 37 K = 180 MPa, n = 0.20. Therefore average flow stress
Y = K ε n / n+ 1 = 175.6 Mpa
The ideal extrusion pressure
p = Y ln R = 175.6 ln 9 = 385.88 MPa
Ideal extrusion force
F = pA = 7039018N
Ftotal = Fideal + 0.3 F ideal + 0.25 Fideal = 10.9 MN
Problem 2
A planned extrusion operation involves steel at 800◦C, with an initial diameter of 100 mm and
final diameter of 20 mm. Two presses, one with a capacity of 20 MN and the other of 10 MN, are
available for this operation. Obviously, the larger press requires greater care and more expensive
tooling. Is the smaller press sufficient for this operation? If not, what recommendations would you
make to allow the use of the smaller press?
Solution
2
For steel k = 425 MPa (From Fig. 6.53 on p. 313). The initial and ﬁnal areas are 0.00785 m and
−4 2
3.14&times;10 m , respectively. From Eq. (6.62) on p. 313, the extrusion force required is
F = Aok ln (Ao/ Af )
=10.6MN
Thus, the smaller and easier to use press is not suitable for this operation, but it almost has
suﬃcient capacity. If the extrusion temperature can be increased or if friction can be reduced
suﬃciently, it may then be possible to use this machine
Problem 3
A linearly strain-hardening material with a true-stress-true-strain curve σ = 34473 +
172368ε kPa is being drawn into a wire. If the original diameter of the wire is 0.635 cm, what
is the minimum possible diameter at the exit of the die? Assume that there is no redundant
work and that the frictional work is 15% of the ideal work of deformation. (Hint: The yield
stress of the exiting wire is the point on the true-stress-true-strain curve that corresponds
to the total strain that the material has undergone.)
Solution
σd =
ε1
Y = 34473.8 + (34473.8 + 172369 ε1) / 2
= 34473.8 + 86184.5 ε1
The yield stress of exiting wire
Y1 = 34473.8 + 172368 ε1
Equating both expression
34473.8+172368.9 ε1 = 1.15 (34473.8 + 86184 ε1) ε1
ε1 = 1.56
ε1 = ln (Ao/Af ) = ln (0.635/Df)2
Df = 0.291 cm
Problem 4
Estimate the maximum bending force required for a 0.3175 cm thick and 30.48 cm wide Ti-5Al2.5Sn titanium alloy in a V -die with a width of 15.24 cm.
Solution
The bending force is calculated from Eq. (7.11). Note that Section 7.4.3 states that k
takes a range from 1.2 to 1.33 for a V-die, so an average value of k =1.265 will be used.
From Table 3.14, we ﬁnd that UTS = 860 MPa. Also, the problem statement gives us L
= 30.48 cm, T = 0.3175 cm, and W = 15.24 cm. Therefore, Eq. (7.11) gives
F max = k (UTS)LT2 / W = 21.93 kN
Problem 5
A straight bead is being formed on a 1-mmthick aluminum sheet in a 20-mm-diameter die
cavity, as shown in the accompanying figure. (See also Fig. 7.25a.) Let Y = 150 MPa. Considering
springback, calculate the outside diameter of the bead after it is formed and unloaded
from the die.
For this aluminum sheet, we have Y = 150 MPa and E = 70 GPa (see Table 2.1 on p. 32). Using
Eq. (7.10) on p. 364 for springback, and noting that the die has a diameter of 20 mm and the sheet
thickness is T = 1 mm, the initial bend radius is
Ri = 20/2 – 1 = 9 mm
RiY /ET = 0.0193
Ri / Rf = 4 (RiY /ET)3 – 3 (RiY/ ET) + 1
= 4(0.0193)3 − 3(0.0193) + 1 = 0.942
Rf = Ri / 0.942 = 9.55 mm
Hence, the ﬁnal outside diameter will be
OD = 2Rf +2T
= 2(9.55 mm) + 2(1 mm)
= 21.1 mm
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