Homework solutions for test 1
1. Calculate the volume change from fcc γ-iron to bcc α-iron. The
atom radius is R.
fcc
bcc
fcc metal:
The unit cell of an fcc metal is a = 2 2R
The volume of a unit cell V = a 3 = 16 2R 3
There are 4 atoms per unit cell. The special density of the fcc
4
4 ! " R3
3
= 0.74
structure is ! =
16 2R 3
The density of bcc structure can be calculated as ρ=0.68.
The volume change from fcc to bcc is
1 / !bcc !1 / ! fcc 1 / 0.68 !1 / 0.74
=
= 8.8%
1 / ! fcc
1 / 0.74
2. For homogeneous nucleation during solidification, derive the
"16!" 3Tm2 % 1
" 2! Tm % 1
critical nucleus size, r* = 2! / !GV = $
,
and
!G*
=
$
' 2.
2
# L '& !T
# 3L & !T
4
!G = " ! r 3!GV + 4! r 2"
3
!"G
= 0 yields r*=2γ/ΔGv
!r
!GV = !H " T !S
!H = L
!S = L / Tm
r
" 2! Tm % 1
r* = 2! / !GV r* = 2! / !GV = $
# L '& !T
#16!" 3Tm2 & 1
4
!G* = " ! r *3 !GV + 4! r *2 " = %
( 2
2
3
$ 3L ' !T
5.8 A flat plate is to be cast in an open mold whose bottom has a square
shape that is 200 mm by 200 mm. The mold is 40 mm deep. A total of 1,000,000
mm3 of molten aluminum is poured into the mold. Solidification volumetric
shrinkage is known to be 6.0%. Table 5.1 lists the linear shrinkage due to
thermal contraction after solidification to be 1.3%. If the availability of
molten metal in the mold allows the square shape of the cast plate to
maintain its 200 mm by 200 mm dimensions until solidification is completed,
determine the final dimensions of the plate.
Solution: The initial volume of liquid metal = 1,000,000 mm3. When
poured into the mold it takes the shape of the open mold, which is 200
mm by 200 mm square, or 40,000 mm2. The starting height of the molten
metal is 1,000,000 / 40,000 = 25 mm.
Volumetric solidification shrinkage is 6%, so when the aluminum has
solidified its volume = 1,000,000(0.94) = 940,000 mm3. Because its base
still measures 200 mm by 200 mm due to the flow of liquid metal before
solidification, its height has been reduced to 940,000 / 40,000 = 23.5
mm.
Thermal contraction causes a further shrinkage of 1.3%. Thus the final
dimensions of the plate are 200(0.987) by 200(0.987) by 23.5(0.987) =
197.40 mm by 197.40 mm by 23.195 mm.
5.9 In the casting of steel under certain mold conditions, the mold constant
in Chvorinov's rule is known to be 4.0 min/cm2, based on previous experience.
The casting is a flat plate whose length = 30 cm, width = 10 cm, and
thickness = 20 mm. Determine how long it will take for the casting to
solidify.
Solution: Volume V = 30 x 10 x 2 = 600 cm3
Area A = 2(30 x 10 + 30 x 2 + 10 x 2) = 760 cm2
Chvorinov’s rule: TTS = Cm (V/A)2 = 4(600/760)2 = 2.49 min
5.11 A disk-shaped part is to be cast out of aluminum. The diameter of the
disk = 500 mm and its thickness = 20 mm. If the mold constant = 2.0 sec/mm2
in Chvorinov's rule, how long will it take the casting to solidify?
Solution: Volume V = πD2t/4 = π(500)2(20)/4 = 3,926,991 mm3
Area A = 2πD2/4 + πDt = π(500)2/2 + π(500)(20) = 424,115 mm2
Chvorinov’s rule: TTS = Cm (V/A)2 = 2.0(3,926,991/424,115)2 = 171.5 s =
2.86 min
5.12 In casting experiments performed using a certain alloy and type of sand
mold, it took 155 sec for a cube-shaped casting to solidify. The cube was 50
mm on a side. (a) Determine the value of the mold constant in Chvorinov's
rule. (b) If the same alloy and mold type were used, find the total
solidification time for a cylindrical casting in which the diameter = 30 mm
and length = 50 mm.
Solution: (a) Volume V = (50)3 = 125,000 mm3
Area A = 6 x (50)2 = 15,000 mm2
(V/A) = 125,000/15,000 = 8.333 mm
Cm = TTS /(V/A)2 = 155/(8.333)2 = 2.232 s/mm2
(b) Cylindrical casting with D = 30 mm and L = 50 mm.
Volume V = πD2L/4 = π(30)2(50)/4 = 35,343 mm3
Area A = 2πD2/4 + πDL = π(30)2/2 + π(30)(50) = 6126 mm2
V/A = 35,343/6126 = 5.77
TTS = 2.232 (5.77)2 = 74.3 s = 1.24 min.
5.13 A steel casting has a cylindrical geometry with 4.0 in diameter and
weighs 20 lb. This casting takes 6.0 min to completely solidify. Another
cylindrical-shaped casting with the same diameter-to-length ratio weighs 12
lb. This casting is made of the same steel, and the same conditions of mold
and pouring were used. Determine: (a) the mold constant in Chvorinov's rule,
(b) the dimensions, and (c) the total solidification time of the lighter
casting. The density of steel = 490 lb/ft3.
Solution: (a) For steel, ρ = 490 lb/ft3 = 0.2836 lb/in3
Weight W = ρV, V = W/ρ = 20/0.2836 = 70.53 in3
Volume V = πD2L/4 = π(4)2L/4 = 4πL = 70.53 in3
Length L = 70.53/4π = 5.61 in
Area A = 2πD2/4 + πDL = 2π(4)2/4 + π(4)(5.61) = 95.63 in2
(V/A) = 70.53/95.63 = 0.7375
Cm = 6.0/(0.7353)2 = 11.03 min/in2
(b) Find dimensions of smaller cylindrical casting with same D/L ratio
and w = 12 lb.
Weight is proportional to volume: V = (12/20)(70.53) = 42.32 in3
D/L ratio = 4.0/5.61 = 0.713; thus L = 1.4025D
Volume V = πD2L/4 = π(D)2(1.4025D)/4 = 1.1015D3
D3 = (42.32 in3)/1.1015 = 38.42 in3
D = (38.42)0.333 = 3.374 in
L = 1.4025(3.374) = 4.732 in
(c) V = πD2L/4 = π(3.374)2(4.732)/4 = 42.32 in3
A = 2πD2/4 + πDL = 0.5π(3.374)2 + π(3.374)(4.732) = 68.04 in2
V/A = 42.32/68.04 = 0.622 in.
TTS = 11.03(.622)2 = 4.27 min.
6.1
Name the two basic categories of casting processes.
Answer. The two categories are (1) expendable mold processes, and (2)
permanent mold processes.
6.3
What is a chaplet?
Answer. Chaplets are metal supports of various designs used to hold the
core in place in the sand mold.
6.4
What properties determine the quality of a sand mold for sand casting?
Answer. The usual properties are (1) strength - ability to maintain
shape in the face of the flowing metal, (2) permeability - ability of
the mold to allow hot air and gases to escape from the cavity, (3)
thermal stability - ability to resist cracking and buckling when in
contact with the molten metal, (4) collapsibility - ability of the mold
to give way during shrinkage of the solidified casting, and (5)
reusability - can the sand be reused to make other molds?
6.6
What are the most common metals used in die casting?
Answer. Common die-casting metals include zinc, tin, lead, aluminum,
brass, and magnesium.
6.7 Which die-casting machines usually have a higher production rate,
cold-chamber or hot-chamber, and why?
Answer. Hot-chamber machines are faster because cold-chamber die
casting machines require molten metal to be ladled into the chamber
from an external source. Ladling takes more time than injecting the
molten metal into the die as in the hot-chamber operation.