ENGI 3941
Production Technology
Problem Set #1
Suggested Answers
Text References are to “Manufacturing Engineering and Technology”, Kalpakjian & Schmid,
6/e, 2010
Chapter 10: Fundamentals of Metal Casting
1. Text 10.15 Explain the effects of mold materials on fluid flow and heat transfer in
casting operations.
Answer
Heat transfer and fluid flow have a direct effect on the formation or suppression of defects in
metal casting. Defects such as porosity (due to either shrinkage or gas), hot tears, and misruns
(i.e., when the molten metal in a casting freezes before the mold is completely filled, shutting off
that portion of the mold) are all controlled by these factors. Furthermore, the grain structure
(hence properties such as strength and toughness of a metal casting) is dependent upon the rate
and direction of heat transfer.
2. Text 10.21 Why does porosity have detrimental effects on the mechanical properties of
castings? Would physical properties, such as thermal and electrical
conductivity, also be adversely affected by porosity? Explain.
Answer
Pores are, in effect, internal discontinuities that are prone to propagate under external stresses.
Thus, the toughness of a material, for example, will decrease as a result of porosity. Furthermore,
the presence of pores in a metal part under tension requires that the material around the pores
support a greater load than if no pores were present; thus the strength and elastic modulus are
also lowered. Considering thermal and electrical conductivity, porosity decreases both the
thermal and electrical conductivity because of the presence of a vacuum or air.
3. Text 10.41
A 100-mm (4-in.) thick square plate and a right circular cylinder with a
radius of 100 mm (4 in.) and a height of 25 mm have the same volume. If
each is to be cast with the use of a cylindrical riser, will each part require the
same-size riser to ensure proper feeding? Explain.
Answer
First note that it is important for the riser to solidify after the casting has solidified. A
casting that solidifies rapidly would be expected to require a smaller riser than one that
solidifies over a longer period of time. Let’s now calculate the relative solidification times.
For the cylindrical part, we have
V = πr2h = π(4)2(1) = 50.3 in3
And
A = 2πr2 + 2πrh = 2π(4)2 + 2π(4)(1) = 125.7 in2
Thus
tcylinder = C(50.3/125.7)2 = 0.160C
For a square plate with sides L and height h, we have
V = 50.3 = L2h = L2(4), or L = 3.55 in.
And
A = 2L2 + 4Lh = 2(3.55 in)2 + 4(3.55 in)(4 in) = 82.0 in2
Thus
tplate = C(50.3/82.0)2 = 0.376C
Therefore, the cylindrical casting will take less time to solidify and hence will require a smaller
riser.
Chapter 11: Metal Casting Processes
4. Text 11.16 Why are risers not as useful in die casting as they are in sand casting?
Answer
There are a number of reasons that risers are not as useful in die casting as they are in sand
casting. Recall that in sand casting, a riser is sized and located so that it provides molten
metal to the die cavity to compensate for metal shrinkage. In sand casting, the cooling rate
is relatively low, so that the cooling rate can be effectively manipulated by placement and
size of a riser. In die casting, it is essential that the cooling rate be high, or else the economic
justification for tooling and equipment cannot be made. Using risers would of course slow
the cooling time, and therefore they are economically undesirable. Further, the metals that
are used in die casting will therefore be ones that develop internal shrinkage porosity, but do
not separate from the mold wall, so that risers are not as necessary.
5. Text 11.35 What are the benefits to heating the mold in investment casting before pouring
in the molten metal? Are there any drawbacks? Explain.
Answer
The benefits to heating the mold include: Greater fluidity for detailed parts (in that the molten
metal will not solidify as quickly), a possible reduction in surface tension and in viscous friction
in the mold, and slower cooling. The main drawbacks to heating the mold are that the mold may
not have as high a strength at the elevated temperature, and the metal may be less viscid and
becomes turbulent as discussed in Chapter 10. Also, the solidification time will be larger with
increased mold preheat, and this can adversely affect production time and process economics as
a result.
6. Text 11.42 In sand casting, it is important that the cope-mold half be weighted down with
sufficient force to keep it from floating when the molten metal is poured in. For
the casting shown in Fig. P11.42, calculate the minimum amount of weight
necessary to keep the cope from floating up as the molten metal is poured in.
(Hint: The buoyancy force exerted by the molten metal on the cope is dependent
on the effective height of the metal head above the cope.)
Answer
The cope mold half must be heavy enough or be weighted sufficiently to keep it from floating
when the molten metal is poured into the mold. The buoyancy force, F, on the cope is exerted by
the metallostatic pressure (caused by the metal in the cope above the parting line) and can be
calculated using the formula
F = pA
where p is the pressure at the parting line and A is the projected area of the mold cavity.
The pressure is
p = wh = (0.26 lb/in3)(2.00 in.) = 0.52 psi
The projected mold-cavity area can be calculated from the dimensions given on the right figure
in the problem, and is found to be 10.63 in2. Thus the force F is
F = (0.52)(10.63) = 5.52 lb
Chapter 13: Rolling of Metals
7. Text 13.15 Explain whether it would be practical to apply the roller-leveling technique shown
in Fig. 13.7a to thick plates.
Answer
It is doubtful that the roller-leveling process, shown in Fig. 13.7 on p. 324, can be applied to
plates. In this process, the strip is flattened by repeatedly flexing it in opposite directions. To do
the same with a plate would require much higher forces in order to develop stresses that are of
the same magnitude at the plate surface as they are in sheet. Also, unless it is sufficiently ductile,
the plate may develop cracks if bent to small radii.
8. Text 13.24 Note in Fig. 13.3d that the driven rolls (powered rolls) are the third set from the
work roll. Why isn’t power supplied through the work roll itself? Is it even
possible? Explain.
Answer
We note in Fig. 13.3d on p. 321 that the diameter of the rolls increases as we move away from
the work (smallest) roll. The reason why power cannot be supplied through the work roll is that
the significant power required for this rolling operation will subject the work roll to a high
torque. Since its diameter is small, the torsional stresses on the roll would be too high; the roll
will either fracture or undergo permanent twist. With the setup shown in the figure, the power is
applied to a larger-diameter roll, which can support a large torque.
9. Text 13.29 In Example 13.1, calculate the roll force and the power for the case in which the
workpiece material is 1100-O aluminum and the roll radius, R, is 8 in.
Answer
As discussed in Example 13.1 on p. 320, the roll-strip contact length, L, is given by
or L = 0.105 ft.
L = (RΔh)1/2 = ((8)(1.00 − 0.8))1/2 = 1.265 in.
Referring to Fig. 2.6 on p. 63 we find that for 1100-O aluminum the yield stress is about 8,000
psi, and that at a true strain of 0.223, the true stress (flow stress) is about 16,000 psi. Thus the
average stress Yavg is 12,000 psi, and the roll force, F, is given by Eq. (13.2) on p. 319 as
F = LwYavg = (1.265)(9)(12, 000) = 136, 600 lb
and the power is given by Eq. (13.4) on page 319 as:
P=
2πFLN / 33, 000 hp
=
2π(136, 600)(0.105)(100) / 33, 000
=
273 hp