Assignment #4
IE 337
Polymers, Polymer Processing and Metal Forming
Materials & Manufacturing Processes
Review Questions:
13.3 Viscosity is an important property of a polymer melt in plastics shaping processes.
Upon what three parameters does viscosity depend?
Answer. Viscosity of a polymer melt depends on (1) temperature and (2) shear rate. Also,
(3) the molecular weight of the polymer affects viscosity.
13.4 How does the viscosity of a polymer melt differ from most fluids that are Newtonian.
Answer. A polymer melt exhibits pseudoplasticity, which means that its value decreases
with increasing shear rate.
13.5 What does viscoelasticity mean, when applied to a polymer melt?
Answer. Viscoelasticity is a combination of viscous and elastic properties which cause
the melt to exhibit memory - the tendency to return to its previous shape, as exhibited by
die swell in extrusion.
Multiple Choice:
8.1 Of the three polymer types, which one is the most important commercially: (a)
thermoplastics, (b) thermosets, or (c) elastomers?
Answer. (a).
8.3 Which one of the three polymer types does not involve cross- linking: (a)
thermoplastics, (b) thermosets, or (c) elastomers?
Answer. (a).
8.5 Which one of the following is the chemical formula for the repeating unit in
polyethylene: (a) CH2, (b) C2H4, (c) C3H6, (d) C5H8, or (e) C8H8?
Answer. (b).
8.8 A copolymer is a mixture of the macromolecules of two different homopolymers: (a)
true or (b) false?
Answer. (b).
13.10 Use of a parison is associated with which one of the following plastic shaping
processes: (a) bi-injection molding, (b) blow molding, (c) compression molding, (d)
pressure thermoforming, or (e) sandwich molding?
Answer. (b).
13.11 A thermoforming mold with a convex form is called which one of the following
(may be more than one): (a) a die, (b) a negative mold, (c) a positive mold, or (d) a threeplate mold?
Answer. (c).
18.1 Which of the following are bulk deformation processes (three correct answers): (a)
bending, (b) deep drawing, (c) extrusion, (d) forging, (e) rolling, and (f) shearing?
Answer. (c), (d), and (e).
18.5 Hot working of metals refers to which one of the following temperature regions
relative to the melting point of the given metal on an absolute temperature scale: (a) room
temperature, (b) 0.2Tm, (c) 0.4Tm, or (d) 0.6Tm?
Answer. (d).
18.6 Which of the following are advantages and characteristics of hot working relative to
cold working (four correct answers): (a) fracture of workpart is less likely, (b) friction is
reduced, (c) increased strength properties, (d) isotropic mechanical properties, (e) less
overall energy is required, (f) lower deformation forces is required, and (g) more
significant shape changes are possible, and (h) strain-rate sensitivity is reduced?
Answer. (a), (d), (f), and (g).
18.7 Increasing strain rate tends to have which one of the following effects on flow stress
during hot forming of metal: (a) decreases flow stress, (b) has no effect, or (c) increases
flow stress?
Answer. (c).
20.1 Most sheet metalworking operations are performed as which of the following: (a)
cold working, (b) hot working, or (c) warm working?
Answer. (a).
20.4 A circular sheet metal slug produced in a hole punching operation will have the
same diameter as which of the following: (a) the die opening or (b) the punch?
Answer. (a).
20.7 Sheet metal bending involves which of the following stresses and strains (two
correct answers): (a) compressive, (b) shear, and (c) tensile?
Answer. (a) and (c).
Problems:
18.2 A metal has a flow curve with parameters: strength coefficient = 850 MPa and strainhardening exponent = 0.30. A tensile specimen of the metal with gage length = 100 mm is
stretched to a length = 157 mm. Determine the flow stress at the new length and the average flow
stress that the metal has been subjected to during the deformation.
Solution: ε = ln (157/100) = ln 1.57 = 0.451
0.30
Flow stress Y = 850(0.451)
f
= 669.4 MPa.
Average flow stress fY= 850(0.451)
0.30
/1.30 = 514.9 MPa.
18.15 A tensile test is performed to determine the parameters strength constant C and strain-rate
sensitivity exponent m in Eq. (18.4) for a certain metal. The temperature at which the test is
performed = 500°C. At a strain rate = 12/s, the stress is measured at 160 MPa; and at a strain rate
= 250/s, the stress = 300 MPa. (a) Determine C and m. (b) If the temperature were 600°C, what
changes would you expect in the values of C and m?
m
m
Solution: (a) Two equations: (1) 160 = C(12) and (2) 300 = C(250)
(1) ln 160 = ln C + m ln 12 or ln 160 - m ln 12 = ln C
(2) ln 300 = ln C + m ln 250 or ln 300 - m ln 250 = ln C
(1) and (2): ln 160 - m ln 12 = ln 300 - m ln 250
5.0752 – 2.4849 m = 5.7038 – 5.5215 m
(5.5215 – 2.4849)m = 5.7038 – 5.0752
3.0366 m = 0.6286
m = 0.2
(1) C = 160/(12)
0.207
= 160.1.6726 = 95.658
0.207
= 300/3.1361 = 95.660
(2) C = 300/(250)
Averaging these values, C = 95.7
20.4 A blanking die is to be designed to blank the part outline shown in Figure P20.4. The
material is 4 mm thick stainless steel (half hard). Determine the dimensions of the blanking punch
and the die opening.
Solution: From Table 20.1, A = 0.075. Thus, c = 0.075(4.0) = 0.30 mm
c
Blanking die: dimensions are the same as for the part in Figure P20.4.
Blanking punch: 85 mm length dimension = 85 - 2(0.3) = 84.4 mm
50 mm width dimension = 50 - 2(0.3) = 49.4 mm
Top and bottom 25 mm extension widths = 25 - 2(0.3) = 24.4 mm
The 25 mm inset dimension remains the same.
20.9 A bending operation is to be performed on 5.00 mm thick cold rolled steel. The part drawing
is given in Figure P20.9. Determine the blank size required.
Solution: From drawing, α’ = 40°, R = 8.50 mm
α = 180 - α’ = 140°.
A = 2π(α/360)(R + K t)
b
ba
R/t = (8.5)/(5.00) = 1.7, which is less than 2.0; therefore, K = 0.333
ba
A = 2π(140/360)(8.5 + 0.333 x 5.0) = 24.84 mm
b
Dimensions of starting blank: w = 35 mm, L = 58 + 24.84 + 46.5 = 129.34 mm
20.12 Determine the bending force required in Problem 20.9 if the bend is to be performed in a
V-die with a die opening dimension of 40 mm. The material has a tensile strength of 600 MPa
and a shear strength of 430 MPa.
Solution: For V-bending, K = 1.33.
2
bf
2
F = K (TS)wt /D = 1.33(600)(35)(5.0) /40 = 17,460 N
bf