SOLUTIONS MANUAL FOR
Product Design for
Manufacture and
Assembly,
Third Edition
by
Geoffrey Boothroyd
Peter Dewhurst
Winston A. Knight
SOLUTIONS MANUAL FOR
Product Design
for Manufacture and
Assembly,
Third Edition
by
Geoffrey Boothroyd
Peter Dewhurst
Winston A. Knight
Boca Raton London New York
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Chapter 1 Introduction
1. For each example in Figure 1.17, determine the theoretical minimum part count. Also,
establish the part count when practical considerations are taken into account. In example
b, assume both box and cover are die castings and, in example c, assume that the purpose
is to cover the hole in the base plate.
Figure 1.17
Solution:
a) The two L shaped brackets do not fulfill any of the three
minimum part criteria, as they do not move relative to the base
part, can be made of same material as the base part and they do
not need to be separate to assemble the base part into a fixture.
In general screws and fasteners do not meet the minimum part
criteria because methods of assembly that do not require
separate fasteners are usually possible. Consequently, the
theoretical minimum of parts for this assembly is one and the item could be made as a
one piece part.
b) Assuming that a separate cover is required to allow access to the
inside of the base part then the cover part is required for assembly
purposes. Usually screws do not meet the minimum part criteria,
but assuming that the cover may need to be easily removed then
attachment by screws may be necessary. However, this could be
achieved with only one screw rather than three. A conservative
theoretical minimum number of parts is three.
Note: If easy removal of the cover is not necessary then the screw could be eliminated by
using a snap fit cover for example.
c) Assuming that the small plate attached with screws and
fasteners is needed to cover up the hole in the base plate, then the
theoretical minimum number of parts is two, as the fasteners do
not meet the minimum part criteria. A two part assembly can be
achieved by using a snap fit cover for the hole in the base plate.
d) For this terminal block the base must be an insulation material and the terminals must
be isolated from each other, therefore the theoretical minimum number of parts is the
same as shown. Simplification may be possible by using another form of termination
rather than screws which are inherently time consuming to assemble.
2. Estimate the theoretical minimum part count for the Latch Mechanism shown in Figure
1.18.
Figure 1.18
Solution: This product includes several subassemblies which must be separate from each
other as they move during operation of the latch. The lock is presumably a purchased
item and its configuration is fixed. This means that the attachment parts (014) and (015)
will be necessary parts even though they are fasteners. The larger nut (016) could be
replaced with a different fastening part which may be quicker to assemble.
i) Sub 1 – Handle
This whole subassembly
moves together during
operation of the latch and
could all be made of the same
material. The two screw
fasteners (021) attach the sub
to the base part and allow it to
slide relative to the base part.
At least one part is needed for assembly purposes in this case. The theoretical minimum
number of parts is thus two.
ii) Sub 2 – Latch Sub
This whole subassembly
moves as one piece
during operation of the
latch mechanism and is
activated by the handle
sub. The individual items
do not need to move
relative to each other. All
items could be made of
the same material. The
various fasteners simply
hold the three sheet metal
parts together. The
theoretical minimum
number of parts for this
subassembly is one and a more complex single sheet metal part could replace the three
separate sheet metal parts.
iii) Sub 3 – Lock sub
The items (020) and (034) simply
provide a pivot on which the sheet
metal part (036) rotates during
action of the lock. These could be
replaced by a single post perhaps
riveted to the base part of the
assembly. As discussed above, the
large nut (016) could be replaced
by a snap on part to hold the lock
in place.
iv) Sub 4 – Body
The items (033) and (031) do not move
relative to each other and could be made of
the same material. However, they must be
separate for assembly purposes. The various
fasteners (032), (030), (002) simply provide a
post, which could be a single item perhaps
riveted to the sheet metal part (033)
Incorporating all these possible simplifications gives a possible redesign of this
subassembly as shown below. Most of the parts that could be eliminated are the separate
fasteners. Some of the proposed sheet metal parts are more complex and these would
need to be economically justified. In practice replacing several sheet metal parts with one
more complex part is most often less expensive, in particular when the eliminated
assembly processes are taken into account.
3. Figure 1.3 shows a design recommendation which is shown to be misleading under a
wide range of conditions. List the circumstances where the design recommendation might
lead to a lower total manufacturing cost.
Wrong
Right
Solution: Left to the individual student. Things to be taken into consideration include the
quantities to be produced, assembly labor costs, tool manufacturing cost rates, material
costs and so on.
4. Perform a literature search to uncover reports where the applications of DFA or DFMA
type studies have resulted in product simplification or reductions in manufacturing costs.
For each report give the reference and a brief summary of the findings.
Solution: Left to the individual student. Numerous articles describing case studies of the
application of DFMA have been published in the past 25 years. These typically have
appeared in various magazines, including: Industry Week, Machine Design, Assembly,
Design News, Manufacturing Engineering, Quality, American Machinist, Appliance
Design and similar journals. Case studies can also be found on the Boothroyd Dewhurst
website at www.DFMA.com
Chapter 2 Selection of Materials and Processes
1. Figure 2.24 shows the side view of a modern hollow golf driver head. The preferred
weight of a driver head is 200 g and the volume is 460 cc; the latter value is the
maximum allowable by the United States Golf Association. To achieve these two design
specifications modern driver heads are constructed as hollow shells. Most commonly the
face is manufactured separately, as shown in section on the right of the figure, and either
welded or bonded to the hollow shell body. For highest ball speed off the face after
impact, the face of a modern driver is designed to act as a stiff diaphragm spring. Use the
material data in Table 2.5 and the diaphragm spring maximum performance parameter in
Table 2.6 to determine candidate metal alloys for the face, which also possess high
strength needed for golf ball impact. Since it is difficult to make modern large driver
heads within the 200 g target weight, repeat the calculations using the derived parameter
for best diaphragm spring property per weight. Do your selections agree with the
manufacturers’ material of choice for driver heads?
Figure 2.24
Solution: From Table 2.6, the criteria of merit for diaphragm springs are
for best
performance and
for best performance per unit weight. Comparing metal
alloys from Table 2.5, which can be manufactured in thin-wall parts gives the following:
Beryllium copper
0.287
3.48 x 10-5
Titanium
0.257
5.41 x 10-5
Alloy steel (high strength)
0.248
3.15 x 10-5
Magnesium
0.080
4.44 x 10-5
Almost all modern golf driver heads are made from high-strength titanium alloys which
corresponds to the ranking in the third column. The earliest metal drivers were much
smaller in volume (< 300 cc compared to the current USGA 460 cc limit). Material
weight was not an issue for these smaller heads and the material of choice was alloy steel.
The second column shows little discrimination between alloy steel and the much more
expensive choices of titanium or beryllium copper.
2. Figure 2.25 shows a support platform for a precision electrical instrument. The
platform is 100 mm high, and the platform base and top have outer dimensions 75x75
mm. The square cutout in the top plate has dimensions 50 x 50 mm. The platform is to
be made from an electrically conductive metal.
Use the procedures in Sections 2.3 and 2.4 to identify all candidate primary processes
from the list in Fig. 2.3. The top and bottom surface of the platform must be flat and
parallel. Any other surface on the part may have slight taper or draft, if required, for
particular primary process choices. Some of the features on the finished part may be
assigned to secondary machining processes to increase the list of candidate primary
processes. For these cases add an ordered list of the required secondary processes to the
primary one, to create a simple production plan. When considering machining from stock
as a possible option, also consider if a primary process could be used to eliminate the
need for machining some of the required main features.
Figure 2.25
Solution: The platform has the following shape attributes:
Depress:
UniWall:
UniSect:
AxisRot:
RegXSec:
CaptCav:
Enclosed:
No Draft:
Yes Yes (two directions: vertically and between the legs)
Yes
Yes
(I – section across width)
No
No
No
No
No
With reference to Table 2.2, and excluding processes which cannot process conductive
material, gives candidate primary processes;
Sand casting
Investment casting
Die casting
Closed die forging
Machining from stock
Note that because of the depression (through hole) across the width, closed die forging is
unlikely to be cost effective: see suffix ‘a’ in column 2 of Table 2.2.
The direction of mold opening for sand casting or die casting would be in the vertical
direction (to produce the required flat and parallel top and bottom faces), with side cores
(sand casting) or side die action (die casting) to form the side depressions and the main
slot between the legs. For investment casting two dies would likely be used for the wax
pattern pieces; one for the top plate and one for the T-section legs. All three casting
processes could produce the screw clearance holes. However, sand casting would be
unlikely to provide precise enough surfaces for instrument mounting. The production
processes list would be:
Primary Process
Secondary Processes
Investment casting
Die casting
Sand casting
none
trim flash (see Ch. 10)
mill top and bottom faces
For machining from stock with significant production volume, consider investment in a
die plate for hot extrusion of the required I-section shape. The process list for this might
be:
1. Cut to length; 2. Mill large center ‘slot’ across width; 3. Mill rectangular hole in face;
4. Drill holes
3. A version of the support platform in Fig. 2.25 is required for aerospace use, for which
its electrical conductivity should be as high as possible combined with minimum weight.
Assume that all of the section thicknesses should have the same value, h, so that the part
volume can be expressed approximately in the form
, where C0 is a constant.
Using electrical resistivity,
, and density,
, determine the derived
parameter which represents electrical current flow/ weight. Use this parameter to
compare different possible materials. Refer to materials handbooks for resistivity and
density data, or use web material databases such as www.matweb.com. Since alloying
elements can have a significant effect on electrical resistivity, compare only pure metals
in this exercise as a starting point for investigation of candidate alloys. (Hint: assume an
applied voltage, v, and conductive paths up through the legs of width, w, and length, L
are fixed by the design. Also assume that leg thickness h changes with change of material
to obtain the required level of conductivity. Write an expression for current flow, I, in
terms of these parameters and the variable thickness, h. Use this with an approximate
expression of part weight to obtain the required result.)
Solution:
Consider the conductivity path up one of the legs. I = current flow in amps.
From Ohm’s law I = v/R,
(1)
where v is the voltage (given) and R is the resistance of the conductor. From the
definition of resistivity,
, ohms
(2)
Conductor volume is:
and weight,
(3)
Substituting (2) into (1) gives:
(4)
and eliminating h between (3) and (4) gives:
Hence for best performance per unit weight we require:
A comparison of likely pure metal candidates is:
Metal
ρ (g/cc)
γ (ohm – cm)
Copper
Aluminum
Magnesium
Silver
Gold
8.83
2.70
1.74
10.49
19.32
1.7 x 10-6
2.7 x 10-6
1.74 x 10-6
1.6 x 10-6
2.2 x 10-6
6.7 x 104
1.3 x 105
1.3 x 105
5.9 x 104
2.4 x 104
Thus it appears that aluminum and magnesium alloys are equally likely candidates.
4. Figure 2.26 illustrates the outer housing of rotor assembly, of which a production
volume of 100,000 is required. The assembly is designed to spin at high speed during
which the housing is subjected to high tensile stresses. Preliminary designs calculations
suggest that the housing could be made of aluminum alloy with a wall thickness in the
range of 2.0 to 3.0 mm depending on the selected alloy. The wall thickness may be
different for other candidate materials but low strength materials requiring an excessively
thick wall should be avoided. A good surface finish of approximately 50 in is required.
The part is 13 cm high. The large diameter is 20 cm which steps down to 17.5 cm at the
midpoint. The bottom of the housing is open and the top has large hole 12.5 cm diameter
in the center surrounded by twelve 1.5 cm diameter holes.
Figure 2.26
Use the procedures in Sections 2.3 and 2.4 to identify all candidate primary processes, for
manufacture of the housing, from the list in Fig. 2.3. The inner and outer main surfaces
may have slight taper or draft, if required, for particular primary process choices. Some
of the features on the finished part may be assigned to secondary processes. For these
cases add an ordered list of the required secondary processes to the primary one, to create
a simple production plan.
(Hint: First eliminate processes from Table 2.2 on the basis of shape attributes. This will
leave a relatively large set of candidate processes. Next review these candidates in Table
2.1 to check the requirements of part size, surface finish, process limitations and
associated materials against the design requirements.)
Solution: The rotor housing has the following shape attributes:
Depress:
UniWall:
UniSect
AxisRot:
RegXSec:
CaptCav:
Enclosed:
No Draft:
Yes No
Yes
Yes
Yes
No
No
No
No
Using Table 2.2 only eliminates:
Blow molding (extrusion)
Blow molding (injection)
Rotational molding
Hot extrusion
Rotary swaging
The requirement for strength equivalent to 2 to 3 mm thick aluminum alloy eliminates:
Structural foam molding
The requirement for a surface finish of approximately 50 µin (referring to Table 2.1)
eliminates:
Sand casting
Hot forging
Reference to the ‘Process limitations’ column and ‘Part size’ columns of Table 2.1
eliminates:
Cold heading
Impact extrusion
Hot forging
Pressing and sintering
The remaining candidate processes are:
Investment casting
Impact extrusion
Injection molding
Sheet metal stamping
Metal spinning
Machining
ECM
EDM
Note: Wire EDM is excluded since it can only generate near 2-D profiles.
Since 100,000 are required machining, ECM and EDM will certainly be non-competitive
with respect to the forming and casting processes. Possible manufacturing sequences are:
Primary processes
Secondary processes
Injection molding
Die casting
Investment casting
Sheet metal stamping
Metal spinning
None
Trim flash
None
Punch holes
Punch holes
5. Saturn Automobile Corporations is one of only a very few car manufacturers to
replace some of the commercial quality steel body panels with injection molded ones.
The material they chose to use is glass-reinforced polycarbonate, blended with ABS for
improved mold flow characteristics. The elastic modulus of this blended reinforced
thermoplastic is E=5 GN/m2, and the yield stress is Yt=80 MN/m2. The steel panels had a
nominal thickness of 1mm and corresponding material properties of E=200 GN/m2, and
Yt=300 MN/m2.
Use an appropriate derived parameter to investigate Saturn’s marketing claim that the
thermoplastic panels are more ‘ding’ resistant. Determine the wall thicknesses Saturn
would have needed to use to obtain the same panel stiffness as the sheet steel ones being
replaced.
(Hint: ‘ding resistance’ is determined by the diaphragm spring quality of the panels, so
use the appropriate derived parameter for comparison. Refer to Section 2.5.3 for help
with the last part of this problem.)
Solution:
The ‘ding resistance’ is governed by the diaphragm spring quality of the panels.
. Comparison of the two
From Table 2.6 this is defined by the derived parameter
materials gives:
Material
Yt
E
Commercial quality steel
300
200
26
PC/ABS glass reinforced
80
5
143
These values amply support Saturn’s claims.
From section 2.5.3, the thickness, h, of the material B to replace material A for equivalent
panel (bending) stiffness is
6. Review the equations relating the maximum center load which can supported by a
simply-supported beam with length L, width w, thickness h, and yield strength in tension
Yt. If length and width are fixed by the design, show that two beams A and B (or
equivalently two plates) will have the same load carrying capability if
or equivalently
Use this relationship as an approximate test of the requirement for equal or improved
bending strength of the Saturn panels, described in Problem 5, compared to the sheet steel
ones.
Solution: From any strength of materials text, for a simply supported beam of length, L,
width, w, and thickness, h, the maximum stress produced by a center load F is:
, where the second moment of area
Therefore
For maximum load, Fmax, corresponding to maximum tensile stress, Yt,, this gives:
Assuming the right-hand side of this equation is fixed by the design specification, then
for alternate materials A and B to satisfy the same specifications we require:
or
For substitution PC/ABS (glass reinforced) with 1mm thick commercial quality steel
panels requires:
7. Suggested class project: Study Table 2.1, which covers the more common
manufacturing processes in consumer product production. Identify a product
manufacturing process not included in Table 2.1. Complete an entry for your process in
Tables 2.1 and 2.2. Provide references for your process information.
Solution: Left to individual student
8. Construct the Excel spreadsheet illustrated in Table 2.5. Use this to explore the best
material choices for each of the 18 criteria given in Table 2.6.
Solution: Left to individual student
Chapter 3 Product Design for Manual Assembly
1. Estimate the manual assembly time and design efficiency of the piston assembly
shown in Figure 3.56. Suggest a redesign of this assembly to reduce the number of parts
and increase the assembly efficiency. Estimate the manual assembly time and design
efficiency of the redesigned assembly. The assembly sequence of the piston assembly is
as follows:
a) The base is placed in a fixture
b) The piston is inserted into the base. The piston must be carefully aligned
because the piston rod does not engage the hole in the bottom of the base
before it must be released.
c) The piston stop is inserted into the cylinder
d) The spring is inserted into the cylinder and tangles severely with others in
bulk
e) The cover is placed onto the base and held in place to maintain alignment.
f) The two screws are inserted and driven home to secure the cover.
Figure 3
4
5
6
7
8
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling
time per part, s
Two digit manual
insertion code
1
2
3
4
5
6
1
1
1
1
1
2
30
10
10
05
23
11
1.95
1.5
1.5
1.84
2.36
1.8
00
10
00
00
08
39
Figure for theoretical
minimum number of
parts
3
Total operation time
(2)*{(4) + (6)}
2
Manual insertion
time per part, s
1
1.5
4.0
1.5
1.5
6.5
8.0
3.45
5.50
3.00
3.34
8.86
16.60*
1
1
1
1
0
0
40.75
4
TM
NM
Pneumatic Piston
Main body
Piston
Piston stop
Spring
Cover
Screw
Design efficiency = 3NM/TM = 0.29
Being fairly conservative the theoretical minimum number of parts is 4. The main body
is the base part. The piston moves relative to the body. The spring probably needs to be a
different material. The piston stop needs to be separate to allow the piston to be
assembled, but could be combined with the cover. The combined cover piston stop can
be snap fitted to the base thus eliminating the screws. A possible redesign of this
assembly is shown below, together with its DFA worksheet.
Note: This solution is somewhat conservative. The main function of the cover is to
provide something for the spring to push against. The cover does not seal the cylinder.
There is a central hole in the original design. It is probably possible to reduce the number
of parts to three, by using a spring with a larger end coil that snap fits into a groove in the
top of the cylinder.
4
5
6
7
8
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling time
per part, s
Two digit manual
insertion code
1
2
3
4
1
1
1
1
30
10
05
10
1.95
1.5
1.84
1.5
00
00
00
30
Figure for theoretical
minimum number of
parts
3
Total operation time
(2)*{(4) + (6)}
2
Manual insertion time
per part, s
1
1.5
1.5
1.5
2.0
3.45
3.00
3.34
3.50
1
1
1
1
13.29
4
TM
NM
Pneumatic Piston
Main body
Piston
Spring
Cover and stop
Design efficiency = 3NM/TM = 0.90
2. Estimate the manual assembly time and design efficiency for the shaft assembly shown
in Figure 3. 57. The assembly procedure is as follows:
a) The retainer bush is placed in the work fixture.
b) The shaft is inserted in the bush.
c) The securing screw is inserted into the bush and is difficult to align and position.
d) The spring tangles severely and, when separated from others, is dropped onto the
shaft.
e) The washer is inserted onto the shaft and then held down while the snap fastener
is positioned using a standard grasping tool.
Figure 3.57
Solution:
The worksheet for this assembly is shown below. The theoretical minimum number of
parts is 3. The bush is the base part. The shaft does not move relative to the bush so could
be eliminated as a separate part by combining with the bush to form a headed shaft. The
screw would then not be needed. The washer moves relative to the shaft and therefore
must be separate. The snap is a fastener and could be eliminated by riveting over the end
of the shaft.
4
5
6
7
8
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling time
per part, s
Two digit manual
insertion code
1
2
3
4
5
6
1
1
1
1
1
1
21
10
13
80
03
68
2.1
1.5
2.06
4.1
1.69
8.0
00
00
39
00
06
31
Figure for theoretical
minimum number of
parts
3
Total operation time
(2)*{(4) + (6)}
2
Manual insertion time
per part, s
1
1.5
1.5
8.0
1.5
5.5
5.0
3.6
3.0
10.06
5.6
7.19
13.00
1
0
0
1
1
0
42.45
3
TM
NM
Shaft Assembly
Bush
Shaft
Screw
Spring
Washer
Snap
Design efficiency = 3NM/TM = 0.21
3. Analyze the terminal block shown in Figure 3.58 for manual assembly. Give the
estimated assembly time and the design efficiency.
Figure 3.58
Solution:
Manual assembly worksheet is below. The theoretical number of parts is 7, as indicated
by the notes in Figure 3.58. Note that this number could be reduced by 2 if the leads
could be snap connected to the terminals.
Figure for theoretical
minimum number of
parts
8
Total operation time
(2)*{(4) + (6)}
7
Manual insertion
time per part, s
6
Two digit manual
insertion code
5
Manual handling
time per part, s
4
Two digit manual
handling code
3
Number of repeat
operations, Rp
2
Part ID number
1
1
1
30
1.95
00
1.5
3.45
1
Base
2
1
33
2.51
08
6.5
9.01
0
Nameplate
3
2
68
8.00
39
8.00
32
0
S.T. Screw
-
-
-
-
98
4.50
4.50
-
Reorientation
4
1
21
2.10
08
6.50
8.60
1
Rectifier
5
2
68
8.00
49
10.50
37.00
0
Screw
6
1
03
1.69
40
4.50
6.19
1
Fuse
7
2
83
5.60
00
1.50
14.20
2
Lead
8
2
41
6.85
00
1.50
16.70
0
Washer
9
2
68
8.00
39
8.00
32.00
2
Nut
163.65
7
TM
NM
Terminal Block
Design efficiency = 3NM/TM =0.128
4. Figure 3.59 shows an exploded view of a diaphragm assembly from a gas flow meter.
Determine a suitable assembly sequence for this product and carry out an analysis for
manual assembly. Estimate the assembly time, together with the design efficiency.
Develop a redesign for this assembly with reduced part count and reduced assembly time.
Estimate the assembly time and design efficiency for manual assembly of the redesigned
assembly.
Figure 3.59
Solution:
There are several possible assembly sequences for this assembly.
If it is assumed that the assembly is carried out by hand without the use of fixtures then a
possible sequence is as follows:
1) The plate is picked up and placed in a simple holding device.
2) The bearing housing is place in position aligned with the holes in the plate and held in
place (Nothing to fix the alignment).
3) The screws are inserted through the holes in the bearing housing and plate and held in
place.
4) The washers are placed onto the screws that protrude through the plate.
5) The nuts are screwed onto the screws locking everything in place.
The manual assembly worksheet for this sequence is shown below.
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling
time per part, s
Two digit manual
insertion code
Manual insertion
time per part, s
Total operation time
(2)*{(4) + (6)}
Figure for theoretical
minimum number of
parts
1
2
3
4
5
6
7
8
1
1
13
2.06
00
1.50
3.56
1
Plate
2
1
03
1.80
08
6.50
8.00
1
Bearing housing
3
2
11
1.80
06
5.50
14.60
0
Screw
4
2
03
1.69
00
1.50
6.38
0
Washer
5
2
01
1.43
39
8.00
18.83
0
Nut
51.67
2
TM
NM
Manual assembly without fixture
Design efficiency = 3NM/TM =0.116
A simple assembly fixture could be built and a possible assembly sequence is then as
follows:
1) The nuts are placed in suitable recesses in the fixture.
2) The washers are placed over the nuts into suitable recesses in the fixture.
3) The plate is positioned over the holes in the washers and held in place (nothing to fix
the alignment).
4) The bearing is placed over the holes in the plate and held in place (nothing to fix the
alignment).
5) The screws are screwed into the nuts locking everything in place.
The manual assembly worksheet for this sequence is shown below. Note that the
assembly design efficiency has increased because the estimated assembly time is reduced.
This indicates that this efficiency is dependent on the assembly methods assumed.
The theoretical minimum number of parts for this assembly is two – the plate and bearing
housing. A possible redesign of this assembly is shown in the figure below and the
assembly sequence is as follows:
1) The bearing housing is loaded into a fixture on a riveting machine.
2) The plate is placed over the peg on the bearing housing and riveted in place.
The manual assembly worksheet for this redesigned assembly is shown below. The
design efficiency of this assembly is increased to 0.528 (52.8%).
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling time
per part, s
Two digit manual
insertion code
Manual insertion time
per part, s
Total operation time
(2)*{(4) + (6)}
Figure for theoretical
minimum number of
parts
1
2
3
4
5
6
7
8
1
2
01
1.43
00
1.50
3.86
1
Nut
2
2
03
1.69
00
1.50
6.38
0
Washer
3
1
13
2.06
08
6.50
8.56
0
Plate
4
1
20
1.80
08
6.50
8.30
1
Bearing housing
5
2
11
1.80
39
8.00
19.60
0
Screw
46.7
2
TM
NM
Manual assembly with fixture
Design efficiency = 3NM/TM = 0.128
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling time
per part, s
Two digit manual
insertion code
Manual insertion time
per part, s
Total operation time
(2)*{(4) + (6)}
Figure for theoretical
minimum number of
parts
1
2
3
4
5
6
7
8
1
1
20
1.80
00
1.50
3.30
1
Bearing housing
2
1
03
1.69
37
7.00
8.69
1
Plate
11.99
2
TM
NM
Redesigned assembly
Design efficiency = 3NM/TM = 0.50
5. Analyze the gear-box assembly shown in Figure 3.60 for manual assembly. Obtain the
estimated assembly time, assuming that none of the parts are easy to align and position.
Also obtain the theoretical minimum number of parts, assuming that the springs and the
earth lead must be different materials from the remaining parts. Suggest a redesign for the
gear box, assuming that two screws only are needed to secure the upper cover, and
estimate the assembly cost for the redesign. Assume a manual labor rate of $36 per hour.
Figure 3.60
Solution:
The manual assembly worksheet for the gear box is shown below. The theoretical
minimum number of parts is 9. The lower cover and casing can be combined and this will
eliminate one gasket, four washers and four screws. The name plate information can be
engraved onto the casing. The upper gasket could be eliminated by using a liquid gasket
bead applied to the upper casing surface. Gears A and C could be combined together, but
this would be possible if the gears could be assembled the other way up. All screws and
washers are technically unnecessary, but the upper cover may need screws so that
disassembly for servicing is possible. If this is the case then consideration to using screws
with captive washers should be given. The worksheet for a redesigned gear box that
incorporates these suggestions is shown after the worksheet for the current design.
With these changes the estimated assembly time is reduced from 287 seconds to 104
seconds. The corresponding change in design efficiency is an increase to 26% from 9.4%.
Figure for theoretical
minimum number of
parts
8
Total operation time
(2)*{(4) + (6)}
7
Manual insertion
time per part, s
6
Two digit manual
insertion code
5
Manual handling
time per part, s
4
Two digit manual
handling code
3
Number of repeat
operations, Rp
2
Part ID number
1
1
1
00
1.13
00
1.5
2.63
1
Casing
-
1
-
-
-
14
14
-
Apply adhesive
2
98
00
8
02
2.5
10.5
0
Gasket
3
1
00
1.13
08
6.5
7.63
0
Lower cover
4
4
03
1.69
08
6.5
32.76
0
Washer
5
4
01
1.43
38
6
29.72
0
Screw
-
1
-
-
-
9
9
-
Reorientation
6
1
33
2.51
35
7
9.51
0
Name plate (rivet)
-
1
-
-
-
9
9
-
Reorientation
7
1
33
2.51
96
12
14,51
1
Leaf spring (spot weld)
-
1
-
-
-
14
14
-
Apply adhesive
8
1
98
8
02
2.5
10.5
0
Gasket
9
1
00
1.13
00
1.5
2.63
1
Gear B
Gear Box
Figure for theoretical
minimum number of
parts
8
Total operation time
(2)*{(4) + (6)}
7
Manual insertion time
per part, s
6
Two digit manual
insertion code
5
Manual handling time
per part, s
4
Two digit manual
handling code
3
Number of repeat
operations, Rp
2
Part ID number
1
10
1
00
1.13
00
1.5
2.63
1
Gear C
11
1
80
4.1
02
2.5
6.6
1
Spring
12
1
00
1.13
06
5.5
6.63
1
Gear D
13
1
00
1.13
3.1
5
6.13
0
Gear A
-
1
-
-
-
20
20
-
Apply grease
14
1
00
1.13
08
6.5
6.63
1
Upper cover
15
4
03
1.69
08
6.5
32.76
0
Washer
16
4
01
1.43
38
6
29.72
0
Screw
17
1
23
2.36
00
1.5
2.86
1
Lever
18
1
01
1.43
38
6
7.43
0
Screw
19
1
83
5.6
95
8
13.6
1
Earth lead
287.23
9
TM
NM
Gear box (continued)
Design efficiency = 3NM/TM = 0.094
Figure for theoretical
minimum number of
parts
8
Total operation time
(2)*{(4) + (6)}
7
Manual insertion
time per part, s
6
Two digit manual
insertion code
5
Manual handling
time per part, s
4
Two digit manual
handling code
3
Number of repeat
operations, Rp
2
Part ID number
1
1
7
9
1
1
1
00
33
00
1.13
2.51
1.13
00
96
00
1.5
12
1.5
2.63
14.51
2.63
1
1
1
Casing
Leaf spring (spot weld)
Gear B
10
11
12
-
1
1
1
1
00
80
00
-
1.13
4.1
1.13
-
00
02
06
-
1.5
2.5
5.5
14
2.63
6.6
6.63
14
1
1
1
-
Gear A and C Combined
Spring
Gear D
Apply gasket bead
14
16
17
19
1
4
1
1
00
01
23
83
1.13
1.43
2.36
5.6
08
38
32
95
6.5
6
2
8
6.63
29.72
4.36
13.6
103.94
1
0
1
1
9
Upper cover
Screw
Lever
Earth lead
TM
NM
Gear box redesign
Design efficiency = 3NM/TM = 0.26
6. A design concept consists of a cylindrical stepped base machined from nylon, a
diaphragm disk and a clamping ring as shown in Fig. 3.61a. For ease of assembly, and to
reduce manufacturing costs, you propose to combine the diaphragm and clamping ring
into a single aluminum alloy diecasting as shown in Fig. 3.61b. For the diecast diaphragm
design, assume the coefficient of friction between the diaphragm and the nylon base
during assembly is 0.15
(a) Would you expect jamming problems during assembly with this value of friction and
the given diameter and overall disk thickness dimension of 15 mm.
(b) If you determine that jamming would occur, to what value would you increase the
overall thickness, from the initial value of 15 mm, to stop the jamming condition?
Alternatively, if you determine that jamming would not occur, how much smaller could
you make the overall thickness and still avoid jamming during assembly?
Solution:
Clamping ring
Diaphragm disk
Base
a)
80 mm
t
b)
15 mm
79.4 mm
Figure 3.61
Solution: a) Diaphragm will jam is
.
, l = 15/80 = 0.1875. Therefore, l2 = 0.0352 and
= 0.0374. Therefore the part will tend to jam.
b) To avoid jamming requires l2 > 0.0374 or l > 0.1935 or L = 80l > 15.84 mm.
7. The final assembly of a new high-performance aircraft engine turbocharger involves
182 assembly operations. The assembly time has been estimated, using the DFA
standard data times, to take 1700 seconds.
(a) Using this information, determine the probability that any given one of the completed
turbochargers will contain an assembly error. Assume that with attention to quality being
the central issue in aircraft manufacture, the assembly workers perform at 2x the quality
level of the benchmark Motorola workers; i.e. make assembly mistakes at only half the
rate.
(b) You have an order for the new turbocharger from Piper Aircraft Company, to supply
a batch of 24 immediately, and then a second batch of 24 in 3 month’s time. Assuming
the same assembly workers will build both batches, estimate the total time for assembly
of the first batch and then the total time for the second batch. Use an 85 percent learning
curve as the basis for your calculations.
Solution: a)
s;
Therefore,
b)
.
and
Therefore, total time for batch
.
hrs and
s.
Therefore total time for second batch =
hrs
8. Your company has just designed a new aircraft fuel pump with higher output than a
previous model. You decide that the previous model is similar enough to the new one to
use previous assembly times for estimating purposes. You need to produce a quotation to
Boeing for delivery of a first test set of 24 pumps. As part of the quote you need to
estimate the average assembly time for the first 24 units. A group of technicians, not
production workers, will assemble the batch of 24 prototype units.
a) Assume you have found from records that the total time to assemble the first six units
of the previous model was 122 hours. Use this data to calculate your estimate with a 85%
learning curve.
b) Assume that times were never recorded for the first assemblies of the previous model.
However, you know that currently after assembling 280 of the previous model the
production assembly time per unit is 6.5 hours. Use this information instead of that given
in (a) to determine your assembly estimate with a 85% learning curve.
(c) Assume that both sets of data (a and b). Use both sets of data to estimate the exponent
b of the learning curve and the percent learning curve value (r%) which best fit the data.
Use your calculated values of b and r% to determine an improved assembly time
estimate.
Solution: a)
and
hrs
hrs and assembly estimate is:
Therefore,
24 x 14.7 = 353 hrs.
b)
Therefore,
hrs.
hrs and assembly estimate = 24 x 15.1 363 hrs.
c) Using both sets of data gives:
(1)
and
(2)
Dividing (2) by (1) gives:
which gives
and therefore
From (1),
and estimated assembly
hrs
hrs, which gives
Chapter 4 Electrical Connections and Wire Harness Assembly
1. Complete all of the partially completed worksheets for the sample analysis in this
chapter.
Solution:
I.D.
of
leg
preparation
code
number of
different
termination
styles (2)
number of
conductor
ends to be
prepared
preparation
time per
conductor
end, s
total
preparation
time, s
4
setup time
per group of
identical
terminations
s
5
1
A, B
D, H
2
3
6
(3x5+4x6)
01
3
10
2.2
3.9
45.6
C, F
02
3
2
6.2
18.2
42.6
E
25
1
6
25.6
21.4
154.0
G
03
1
2
2.2
6.1
14.4
H
21
1
6
18.6
5.5
51.6
308.2
Wire/cable preparation
insertion
time per
item, s
4
connector
hardware
assembly
time (3)
seconds
5
I.D.
of leg
insertion
code
1
2
number of
items
to be
inserted
3
total
insertion
time, s
A
0
2
2.5
-
5.0
B
0
3
2.5
-
7.5
C
0
1
2.5
-
2.5
D
0
3
2.5
82.5
E
0
1
2.5
82.5
85.0
F
0
1
2.5
-
2.5
G
0
2
2.5
-
5.0
H
0
7
2.5
-
17.5
(3 x 4) +
5
90.0
215.0
Assembly – Insertion
additional
operation
code
1
number
of setups
number of
operations
constant
time, s
4
time per
operation,
s
5
total
operation
time, s
(2x4+3x5)
2
3
2
1
24
1.9
13.7
330.7
3
1
4
3.1
2.3
12.3
4
1
1
3.1
8.0
11.1
354.1
Additional Operations – A
preparation
worksheet
total time
seconds
total time
minutes
1
308.2
5.1
2
60.4
1.0
3
215.0
3.6
4
115.7
1.9
5
27.7
0.5
6
27.0
0.5
7
125.8
2.1
8
61.4
1.0
9
354.1
5.9
10
20.4
0.3
1315.7
21.9
assembly
assembly
or
installation
installation
additional
operations
Wire Harness Summary Data
2. Obtain the assembly time for the harness shown in Figure 4.52 below
Figure 5.52
Solution:
I.D.
of
leg
preparation
code
number of
different
termination
styles (2)
number of
conductor
ends to be
prepared
preparation
time per
conductor
end, s
total
preparation
time, s
4
setup time
per group of
identical
terminations
s
5
1
A, B
D, E
2
3
6
(3x5+4x6)
01
3
10
2.2
3.9
45.6
C, F
02
1
2
6.2
18.2
42.6
G
03
1
2
2.2
6.1
14.4
102.6
Wire/ Cable Preparation
I.D. of
wire or
cable
1
w1, w2,
w3, w4,
w6, w7
Wire/cable
handling
code
2
number of
wires/cables
laid
3
total length
of all cable
paths, ft (2)
4
basic
handling
time, s
5
additional
handling
time per
foot, s
6
0
6
29
2.7
0.8
39.4
w5
1
1
5
4.7
1.2
10.7
total
handling
time, s
(3x5)+(4x6)
50.1
Wire/ Cable Handling
I.D.
of leg
insertion
code
number of
items
to be
insertion
time per
item, s
connector
hardware
assembly
time (3)
total
insertion
time, s
1
2
3
4
5
(3 x 4) +
5
A
00
2
1.9
-
3.8
B
00
3
1.9
-
5.7
C
00
1
1.9
-
1.9
D
00
3
1.0
-
5.7
E
00
1
1.9
-
1.9
F
00
1
1.9
-
1.9
G
00
2
1.9
-
3.8
24.7
Assembly – Insertion
1
wire
dressing
code
2
number of
items to
be dressed
3
time per
wire end,
s
4
total
dressing
time, s
(3 x 4)
A,G
10
4
2.3
9.2
B,D
10
6
9.5
13.8
I.D. of leg
23.0
Dressing
additional
operation
code
1
number
of setups
number of
operations
constant
time, s
4
time per
operation,
s
5
total
operation
time, s
(2x4+3x5)
2
3
2
1
12
1.9
13.7
166.3
3
1
4
3.1
2.3
12.3
4
1
1
3.1
8.0
11.1
189.7
Additional Operations – A
additional
operation
code
number of
operations
constant
time, s
time
per ft, s
total
operation
time, s
2
length for
all
operations,
ft
3
1
4
5
(2x4+3x5)
4
1
2
8.0
6.2
20.4
20.4
Additional Operations – B
preparation
worksheet
total time
seconds
total time
minutes
1
102.6
1.7
2
50.1
0.8
3
24.7
0.4
4
23.0
0.4
5
-
-
6
-
-
7
-
-
8
-
-
9
189.7
3.2
10
20.4
0.3
410.5
6.8
assembly
assembly
or
installation
installation
additional
operations
Wire Harness Summary Data
Chapter 5 Design for High-Speed Automatic Assembly and Robot Assembly
1. Eight different small parts are shown in Figure 5.24. Using the classification charts
shown in Figures 5.3, 5.4, 5.5 and 5.8, determine the feeding and orienting 5 digit codes
for these parts.
Figure 5.24
Solution:
1) Code number 60000
Digit 1 – The envelope is a rectangular prism. A/B = 30/10 = 3; A/C = 30/6 = 5. The first
digit is 6
Digit 2 – The part is 1800 about all three axes and therefore the second digit is 0.
Digit 3 - Because three adjacent surfaces of the envelope have significantly different
dimensions, the third digit is 0. The remaining features are not needed for orienting
purposes.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
2) Code number 07400
Digit 1 – The envelope is the cylinder that encloses the part excluding the two
projections. L/D = 7/23 = 0.3. The first digit is 0.
Digit 2 – This part has a feature on one end only and hence is not alpha symmetric. The
feature causing alpha asymmetry is not beta symmetric and the second digit is 7.
Digit 3 – The part has beta asymmetric projections on both the side and end surfaces so
the third digit is 4.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
3) Code number 22000
Digit 1 – The envelope is the cylinder that completely encloses the part. L/D = 30/12 =
2.5. The first digit is 2.
Digit 2 – The part has a beta symmetric step. It cannot be fed in a slot because the center
of mass is not within the shank. The second digit is 2. The concentric blind hole in the
rounded end surface is not needed for orienting.
Digit 3 – The part will be fed end to end and hence the third digit is 0.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
4) Code number 84000
Digit 1 – The envelope is the square prism that encloses the part without the rib. A/B =
20/20 = 1 and A/C = 20/6 = 3.33. The first digit is 8.
Digit 2 – The part has no symmetry that defines the orientation, which is defined by one
main feature (the rib). The second digit is 4.
Digit 3 – The orientation is defined by a step parallel to the X axis. The third digit is 0.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
5) Code number 86600
Digit 1 – The envelope is a square prism. A/B = 30.5/11 = 2.77 and A/C = 30.5/11 =
2.77. The first digit is 8.
Digit 2 – The part has no symmetry that defines the orientation, which is defined by two
main features (the ribs) and at least one is a step. The second digit is 6
Digit 3 – The orientation is defined by the hole in one end that cannot be seen in
silhouette. The third digit is 6.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
6) Code number 66500
Digit 1 – The envelope is a rectangular prism. A/B = 30/21 = 1.5 and A/C = 30/6 = 5. The
first digit is 6.
Digit 2 – The part has no symmetry that defines the orientation, which is defined by two
main features and at least one is a step. The second digit is 6
Digit 3 – The orientation is defined by the groove that is parallel to the Z axis. The third
digit is 5.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
7) Code number 84300
Digit 1 – The envelope is a rectangular prism. A/B = 21/14.5 = 1.5 and A/C = 21/11 =
2.1. The first digit is 8.
Digit 2 – The part has no symmetry that defines the orientation, which is defined by one
main feature (the groove). The second digit is 4.
Digit 3 – The orientation is defined by a groove parallel to the X axis. The third digit is 3.
Digits 4 and 5 – There are no additional features that will result in additional feeding and
orienting costs so the fourth and fifth digits are both 0.
8) Code number 00504
Digit 1 – The envelope is a cylinder. L/D = 3.5/18 = 0.19. The first digit is 0.
Digit 2 – The part is alpha symmetric. The second digit is 0.
Digit 3 – The slot needs to be oriented and is beta asymmetric. The slot passes through
the part and can be seen in end view. The third digit is 5.
Digit 4 and 5 – The part will tangle with other parts but not severely. The part is small
and not abrasive. The part does not tend to overlap during feeding and is not delicate or
flexible. The fourth digit is 0. The part will tangle but not severely. The part is not light
or sticky. The fifth digit is 4.
2. The part shown in Figure 5.25 is to be delivered to a workhead on an automatic
assembly machine at a rate of 13 per minute. Estimate the cost of feeding and orienting
each part assuming that the cost of using a standard feeder for one second is 0.05 cents.
Figure 5.25
Solution: 5 digit code:
First digit – the part is rotational and the envelope has an L/D ratio of 1.58. The first digit
value is 2.
Second digit – because the part is rotational, the chart in Fig. 5.4 is used. The second digit
is the row number on this chart. Since the part has a β symmetric chamfer on its external
surface and does not satisfy the descriptions for rows) and 1, then the second digit value
is 2.
Third digit – The third digit is the column number in Fig. 5.4. The step which cuts into
both the side and end surfaces is the main feature causing beta asymmetry and so the
third digit is 4,
Fourth and Fifth digits – This part presents no additional handling difficulties so the
fourth and fifth digits are zero.
The five digit code is therefore “22400”. From the chart the orienting efficiency, E is 0.37
and the partial relative feeder cost, Cr, is 1.5. The additional relative feeder cost is 0. The
maximum feed rate obtainable from a standard feeder is,
Fm = 1500 x 0.37/20 = 27.75 parts/ minute.
The total relative feeder cost is, Cr = 1.5.
Since the required feed rate of 13 parts/minute is less than Fm, the cost of feeding and
orienting each part is:
Cf = (60/13) x 1.5 x 0.05 = 0.35 c.
3. The final operation in the automatic assembly of a small mechanism, involves the
insertion of an adjusting screw as shown in Figure 5.26. The screw is inserted with a
straight-line motion from the side and has a pilot point which makes it easy to align and
position in the hole. The assemblies are required at a rate of four per minute, estimate the
cost of automatic insertion of the screw, assuming that the cost rate of using a standard
workhead for 1 s is 0.06 cents.
Figure 5.26
Solution:
The first digit of the automatic insertion code is the row number of the chart in Fig. 5.9
and the second digit is the column number. The screw insertion takes place with straight
line motion not from vertically above and so the first digit is 4. The operation is screwing
with an easy to align pilot point screw so the second digit value is 8. For a part with an
insertion code of 48 the relative workhead cost, Wc, is 1.3 and thus the cost of inserting
each screw is given by:
Ci = (60/4) x 1.3 x 0.06 = 1.17 c.
4. The part shown in Fig. 5.27 is a coupling in a motorcycle gearbox, which is to be
assembled automatically at a rate of five per minute. The features of this part which can
be used for automatic orienting are the symmetric groove in the cylindrical body of the
part, and the grooves across the end faces. Estimate the following items for automatic
handling of this part.
(i) The maximum feed rate which would be obtained from a standard vibratory bowl
feeder capable of delivering 25 mm symmetrical cubes at a rate of 1 per second.
(ii) The estimated automatic handling cost per part assuming 0.03 cents/ sec as the rate
for a standard feeder.
iii) The approximate cost of a feeder to deliver the part at the required rate of five per
minute.
iv) The estimated insertion cost per part, assuming 0.06 cents/ sec as the rate for a
standard pick and place workhead. The part is easy to assemble and is inserted directly
from above.
The gearbox designer decides that the slots must be moved to the center position
in the end faces. If this were the only design change, what effect would it have on
automatic handling of the part? If this design change is essential, what other changes
would you recommend to improve the design for automatic handling?
Figure 5.27
Solution: This part is rotational and the length to width ratio is 30/18 = 1.67. The first
digit of the Feed Code is 2. The part is not alpha symmetric and the feature that can be
used for end to end orientation is the beta symmetric groove in the side surface of the
cylinder. The second classification digit is 4. The features that can be used for orienting
the part around the principal axis are the beta asymmetric grooves in the end surfaces.
The third classification digit is 6. There are no other feeding difficulties so the complete
classification code is 24600. From the chart for rotational parts the orienting efficiency,
E, is 0.5 and the relative feeder cost, Cr, is 2.
i) The maximum federate for this part is given by:
per minute.
ii) The feeding cost is given by:
cents
iii) The cost of feeding is also given by Equation (8.1):
, where Rf is the
operating cost in cents per second of the feeding equipment. Thus from the result of ii)
The operating cost is given by Equation (8.2):
or
Assuming a payback period Pb, of 30 months, number of shifts, Sn of 2 and an overhead
ratio, E0 of 2:
iv) The insertion code for this part is 00. Thus the relative feeder cost, Wc, is 1. The cost
of insertion is 0.06(60/5) = 0.72 cents.
The part has 360 degree beta symmetry because of the hole in the one end face. The
asymmetric grooves in the end faces of the part enable to part to be fed automatically and
delivered with the hole in the correct position. If the designer makes the grooves in the
end face symmetric this will not be possible and automatic feeding will unlikely to be
possible. If the use of such symmetric grooves is necessary a solution is to, if possible,
have a second hole in the end face such that the part has 180 degree beta symmetry. In
this case automatic feeding will be possible.
5. Figure 5.28 shows a small three part assembly that is to be assembled automatically.
Carry out an analysis of this assembly for high speed automatic assembly and from this
determine the estimated assembly cost. The assemblies are required at a rate of 10 per
minute. The cost of using a standard feeder for 1 s is 0.03 cents and the cost of using a
standard workhead for 1 s is 0.06 cents. The screw has a pilot point so is easy to align and
position into the screw hole.
Figure 5.28
Solution: The automatic assembly worksheet for this assembly is shown below. The cost
of assembly is 1.96 cents.
No. of repeats
Feed code
Orienting Efficiency, E
Relative feeder cost, Cr
Maximum feed rate/ min.,
Fm
Feeding cost, Cf
Insertion code
Relative workhead cost,
Wc
Insertion cost, Ci
Assembly cost , Ca
Minimum part count
Required assembly rate, Fr (parts/min): 10
Item or operation number
Name of Assembly: Three Part Assy.
1
1
76100
0.1
2
7.5
0.48
00
1
0.36
0.84
1
Base
2
1
64100
0.15
1
18.75
0.18
06
1.3
0.65
0.65
1
Clip
3
1
21000
0.9
1
270
0.18
38
0.8
0.47
0.47
0
Screw
Totals
1.96
2
Total
Item name
1. Fm = 1500 x 0.1/20 = 7.5; Cf = 0.03(60/7.5) x 2 = 0.48; Ci = 0.06(60/10) x 1 = 0.36; Ca = (0.48 + 0.36) = 0.84
2. Fm = 1500 x 0.15/12 = 18.75; Cf = 0.03(60/10) x 1 = 0.18; Ci = 0.06(60/10) x 1.3 = 0.47; Ca = (0.18 + 0.47) = 0.65
4. Fm = 1500 x 0.9/5 = 270; Cf = 0.03(60/10) x 1 = 0.18; Ci = 0.06(60/30) x 0.8 = 0.29; Ca = (0.18 + 0.29) = 0.47
6. Figure 3.59 shows an exploded view of the diaphragm assembly for a gas flow meter.
Carry out an analysis for high speed automatic assembly, including determination of the
theoretical minimum number of parts. The assembly sequence begins with the insertion
of the nuts into a suitably designed work carrier and ends with the insertion of the two
screws. The cost of using a standard feeder for 1 s is 0.03 cents and the cost of using a
standard workhead for 1 s is 0.06 cents. From the analysis determine the assembly cost
for each assembly. The required rate of assembly is 30 per minute.
Note: The diaphragm plate may be difficult or impossible to feed and orient
automatically. If it is decided that this part must be assembled manually, use the data in
Chapter 3 to estimate the assembly cost of this item, assuming a manual labor rate of
$36/hr.
Figure 3.59
Solution:
The automatic assembly worksheet for this assembly is shown below. The diaphragm
plate is uneconomic to feed automatically and must be inserted manual. The cost of
manual assembly is obtained from the manual assembly worksheet also shown below.
The labor cost rate of $36/hour is equivalent to 1 cent per second.
The total assembly cost is 10.31 cents and the theoretical minimum number of parts is 2.
No. of repeats
Feed code
Orienting Efficiency, E
Relative feeder cost, Cr
Maximum feed rate/ min.,
Fm
Feeding cost, Cf, cents
Insertion code
Relative workhead cost, Wc
Insertion cost, Ci, cents
Assembly cost , Ca, cents
Minimum part count
Required assembly rate, Fr (parts/min): 30
Item or operation number
Name of Assembly: Diaphragm Assy.
1
2
10000
0.7
1
131
0.06
00
1
0.12
0.36
1
Nut
2
2
00000
0.7
1
88
0.06
00
1
0.12
0.36
0
Washer
3
1
00800
3.42
8.56
1
Plate
4
1
72500
5
2
21000
Total
Manual assembly required
0.2
5
0.9
Item name
1
7.8
0.23
08
2
0.12
0.47
0
Bearing
1
84
0.06
39
1.8
0.12
0.56
0
Screw
Totals
10.31
1. Fm = 1500 x 0.7/8 = 131; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
2. Fm = 1500 x 0.7/12 = 87.5; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
4. Fm = 1500 x 0.25/48 = 7.8; Cf = 0.03(60/7.8) x 1 = 0.23; Ci = 0.06(60/30) x 2 = 0.24; Ca = (0.23 + 0.24) = 0.47
5. Fm = 1500 x 0.9/16= 84; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1.8 = 0.22; Ca = 2(0.06 + 0.22) = 0.56
1
2
3
4
5
6
7
8
Part ID number
Number of repeat
operations, Rp
Two digit manual
handling code
Manual handling time
per part, s
Two digit manual
insertion code
Manual insertion time
per part, s
Total operation time
(2)*{(4) + (6)}
Figure for theoretical
minimum number of
parts
Manual Assembly Worksheet for Plate
Assembly Labor Rate: 0.4 cents /s.
3
1
13
2.06
08
6.5
8.56
Plate
7. For the diaphragm assembly used in the previous problem, suggest changes to the
diaphragm plate that will allow easier feeding and orienting automatically. Carry out an
analysis for high speed automatic assembly with the redesigned diaphragm plate.
Estimate the assembly cost per assembly and compare this with cost in Problem 6
Solution: Features can be added to the plate that will enable easy automatic feeding. Two
possibilities are shown below, together with the automatic assembly worksheet for the
redesigned assembly.
Redesign A
Redesign B
No. of repeats
Feed code
Orienting Efficiency, E
Relative feeder cost, Cr
Maximum feed rate/ min.,
Fm
Feeding cost, Cf, cents
Insertion code
Relative workhead cost, Wc
Insertion cost, Ci, cents
Assembly cost , Ca, cents
Minimum part count
Required assembly rate, Fr (parts/min): 30
Item or operation number
Name of Assembly: Plate Design B
1
2
10000
0.7
1
131
0.06
00
1
0.12
0.36
1
Nut
2
2
00000
0.7
1
88
0.06
00
1
0.12
0.36
0
Washer
3
1
07600
1
2.5
0.72
00
1
0.12
0.84
1
Plate
4
1
72500
1
7.8
0.23
08
2
0.12
0.47
0
Bearing
5
2
21000
0.1
0.2
5
0.9
1
84
0.06
39
1.8
0.12
0.56
0
Screw
Totals
2.59
2
Total
Item name
1. Fm = 1500 x 0.7/8 = 131; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
2. Fm = 1500 x 0.7/12 = 87.5; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
3. Fm = 1500 x 0.1/60 = 2.5; Cf = 0.03(60/2.5) x 1 = 0.72; Ci = 0.06(60/30) x 1 = 0.12; Ca = (0.72 + 0.12) = 0.84
4. Fm = 1500 x 0.25/48 = 7.8; Cf = 0.03(60/7.8) x 1 = 0.23; Ci = 0.06(60/30) x 2 = 0.24; Ca = (0.23 + 0.24) = 0.47
5. Fm = 1500 x 0.9/16= 84; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1.8 = 0.22; Ca = 2(0.06 + 0.22) = 0.56
No. of repeats
Feed code
Orienting Efficiency, E
Relative feeder cost, Cr
Maximum feed rate/ min., Fm
Feeding cost, Cf, cents
Insertion code
Relative workhead cost, Wc
Insertion cost, Ci, cents
Assembly cost , Ca, cents
Minimum part count
Required assembly rate, Fr (parts/min): 30
Item or operation number
Name of Assembly: Plate Design B .
1
2
10000
0.7
1
131
0.06
00
1
0.12
0.36
1
Nut
2
2
00000
0.7
1
88
0.06
00
1
0.12
0.36
0
Washer
3
1
60000
1
20
0.09
00
1
0.12
0.21
1
Plate
4
1
72500
1
7.8
0.23
08
2
0.12
0.47
0
Bearing
5
2
21000
0.8
0.2
5
0.9
1
84
0.06
39
1.8
0.12
0.56
0
Screw
Totals
1.96
Total
Item name
1. Fm = 1500 x 0.7/8 = 131; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
2. Fm = 1500 x 0.7/12 = 87.5; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1 = 0.12; Ca = 2(0.06 + 0.12) = 0.36
3. Fm = 1500 x 0.8/60 = 20; Cf = 0.03(60/20) x 1 = 0.09; Ci = 0.06(60/30) x 1 = 0.12; Ca = (0.09 + 0.12) = 0.21
4. Fm = 1500 x 0.25/48 = 7.8; Cf = 0.03(60/7.8) x 1 = 0.23; Ci = 0.06(60/30) x 2 = 0.24; Ca = (0.23 + 0.24) = 0.47
5. Fm = 1500 x 0.9/16= 84; Cf = 0.03(60/30) x 1 = 0.06; Ci = 0.06(60/30) x 1.8 = 0.22; Ca = 2(0.06 + 0.22) = 0.56
8. Develop a redesign of the diaphragm assembly in Problem 6 that uses less parts and
that is suitable for high speed automatic assembly. Estimate the assembly cost for this
new design and compare this cost to costs in Problems 6 and 7.
Solution: The theoretical number of parts for the diaphragm assembly is 2 and a redesign
that has this number of parts, which are also easy to feed automatically, is shown below,
together with the associated automatic assembly worksheet.
The assembly sequence is as follows:
1. Insert bearing housing into work carrier.
2. Place plate over post on bearing housing.
3. Rivet post to complete the assembly.
The total assembly cost for this design is 0.57 cents, using the same assumptions as in
Problem 6.
1
00000
0.7
1
3
Total
Minimum part count
2
15.6
3
17.5
0
Assembly cost , Ca
1
Insertion cost, Ci
0.5
Relative workhead
cost, Wc
Relative feeder
cost, Cr
82000
Insertion code
Orienting
Efficiency, E
1
Feeding cost, Cf
Feed code
1
Maximum feed
rate/ min., Fm
No. of repeats
Required assembly rate, Fr (parts/min): 30
Item or operation
number
Name of Assembly: Diaphragm Redesign
0.12
00
1
0.12
0.24
1
0.1
00
1
0.12
0.22
1
91
0.9
0.11
0.11
-
Totals
0.57
2
Item name
Bearing housing
Plate
Riveting operation
1. Fm = 1500 x 0.5/48 = 15.63; Cf = 0.03(60/15.63) x 1 = 0.12; Ci = 0.06(60/30) x 1 = 0.12; Ca = (0.12 + 0.12) = 0.24
2. Fm = 1500 x 0.7/60 = 17.5; Cf = 0.03(60/17.5) x 1 = 0.10; Ci = 0.06(60/30) x 1 = 0.12; Ca = (0.10 + 0.12) = 0.22
3.
Ci = 0.06(60/30) 0.9 = 0.11; Ca = 0.11
9. The box shown in Fig. 5.29 is to be assembled at a rate of 40 per minute on a highspeed rotary indexing machine. The clearance between the cover and the walls of the
base is sufficient for the cover to be easy to align and insert. In addition, the screw has a
pilot point for ease of alignment. The base and cover are both symmetrical about the
vertical axis.
Complete a Design for Automatic Assembly Worksheet and hence estimate the
assembly cost and the assembly efficiency for this design. The cost of using a standard
feeder for 1 s is 0.03 cents and the cost of using a standard workhead for 1 s is 0.06 cents.
Note that all three parts can be oriented with a single feature visible in the external
silhouette, and that the relative workhead cost for a single-spindle vertical screwdriver
head is 0.9.
A second version of the box assembly is proposed in which the boss in the
molded base and the hole in the cover are offset as shown in Fig. 5.30. Would this offset
make any difference to the efficiency of automatic assembly?
Figure 5.29
Figure 5.30
Solution: a)
Relative feeder cost, Cr
Maximum feed rate/ min., Fm
Feeding cost, Cf
Relative workhead cost, Wc
Insertion cost, Ci
Assembly cost , Ca
Minimum part count
1
0.25
1
7.5
0.24
1
0.9
0.33
1
Base
2
1
0.4
1
15
0.12
1
0.9
0.21
1
Cover
3
1
0.7
1
66
0.045
0.9
0.081
0.126
1
Screw
Total
3
Totals
0.666
2
Insertion code
Orienting Efficiency, E
1
Feed code
No. of repeats
Required assembly rate, Fr (parts/min): 40
Item or operation number
Name of Assembly:
Item name
1. Fm = 1500 x 0.25/50 =7.5; Cf = 0.03(60/7.5) = 0.24; Ci = 0.06(60/40) = 0.09
2. Fm = 1500 x 0.4/40 =15; Cf = 0.03(60/15) = 0.12; Ci = 0.09
3. Fm = 1500 x 0.7/16 =66; Cf = 0.03(60/40) = 0.045; Ci = 0.06(60/40)0.9 = 0.081
Efficiency =
b) Base and cover will not possess features visible in the outer silhouette which can be used for orienting. Automatic assembly is
unlikely to be economic
Chapter 6 Printed Circuit Board Design for Manufacture and Assembly
1. Consideration is being given to the purchase of an automatic DIP insertion machine
that would be fully occupied inserting 26 identical 14-pin DIPs into a PCB. At present,
the DIPs are inserted manually. Estimate the breakeven total batch size for the boards if
one setup is used for the batch. Include the costs of rework and the cost (150 cents) for a
replacement component when rework is carried out. The assembly worker rate is $36/h,
including overheads.
Solution: Note: Assume batch size and life volume are the same.
Manual insertion time per component = 6 + 0.5 x 14 = 13 s.
Faults per insertion = 0.005
Rework time per component = 0.005(15 + 6 + 0.5 x 14 + 2.9 + 3 x 14) = 0.36 s
Total manual time = 13.36 s
Total manual cost = 13.36 x 36/36 + 0.005 x 150 = 14.11 cents/component
Auto insertion cost = 0.8 cents/component
Rework time/component = 0.002(15 + 6 + 0.5 x 14 + 2.9 + 3 x 14) = 0.14 s
Rework cost per component = 0.14 x 36/36 + 0.002 x 150 = 0.44 cents
Programming cost per component = 180/BS = 180/BS cents
Setup cost/component = 160/(26 x BS)cents
Total automatic cost = 0.8 + 0.44 + 180/BS + 160/(26 x BS) = 1.24 + 186.2/BS
cents/component
At break even manual cost = automatic cost or 14.11 = 1.24 + 186.2/BS or
BS ~ 15
2. A PCB has 25 axial leaded components of 15 types, 32 radial components (2 leads) of
4 types, 2 radial components (3 leads) of 1 type, 14 DIPs (16 pins) of 10 types, and one
connector (SIP) with 20 pins.
All the components are inserted manually, but the company is interested in the probable
savings if a radial inserter for two lead radials and a DIP inserter are available.
Estimate the total cost of manual insertion and the savings to be obtained through the use
of auto inserters. Assume a total life volume (total batch size) of 50 and an assembly
worker rate of $45/h. Neglect the cost of replacement components.
Solution: $45/hours is equivalent to 1.25 cents/s.
For automatic assembly the setup cost per component type is $1.6 and the programming
cost per part is $1.8. The worksheet for this problem is shown below and the total cost for
manual assembly is $14.30. The total saving from automatic insertion of the two lead
radials is $5.494 - $1.366 = $4.128. The total saving from automatic insertion of the DIPs
is $2.344 - $0.964 = $1.38. The total saving is $5.508.
Cost per part, $
Total insertion, $
4Ave. no. of faults
Removal time, s
Insertion time, s
Soldering time, s
Nt
Cc
Ti
Ci
Cti
Af
Trm
Ti
Trs
Crw
Axial
2
25
15
0
19
0.238
5.938
0.005
10
19
9
0.0594
5.997
Manual
Radial
2
32
4
0
13.6
0.17
5.440
0.005
10
8
8.9
0.0538
5.494
Manual
Radial
3
2
1
0
15.4
0.193
0.385
0.005
12
8
11.9
0.004
0.389
Manual
DIP
16
14
10
0
13
0.163
2.275
0.005
15
13
50.9
0.069
2.344
Manual
Connector
20
1
1
0
16
0.2
0.200
0.005
15
16
62.9
0.0059
0.206
Manual
Cst
Cpr
Total cost, $
Ave. Component
Cost, $
Insertion time, s
Rp
Setup cost, $
No. of part types
nl
Name or
operation
Total rework, $
No. of parts
Programming cost, $
Rework
No. of leads
Insertion
Cin
Total
14.430
Radial
2
32
4
0
0.002
0.064
0.002
10
8
8.9
0.0215
0.1280
1.152
1.366
DIP
16
14
10
0
0.008
0.112
0.002
15
13
50.9
0.0276
0.3200
0.504
0.964
Total
Description
Auto Insert
Auto Insert
3. A PCB has 28 14-pin DIPs of 6 types and 32 axial components of 10 types inserted
automatically. Manual insertion of 6 can-type components with three leads each.
Estimate the assembly cost per board if the batch size is 200. Include the cost of rework.
Assume an assembly worker rate of $30/h and an average component replacement cost of
$1.00.
Solution: The basic costs for this case are:
Labor, $/hr
Set up per part type, $
Program cost per part, $
Batch size
Life Volume
The worksheet for this problem is shown below.
The total assembly cost is $1.979
30
1.6
1.8
200
200
Cost per part, $
Total insertion, $
Ave. no. of faults
4
Removal time, s
Insertion time, s
Soldering time, s
Total rework, $
Setup cost, $
Programming cost, $
Total cost, $
Af
Trm
Ti
Trs
Crw
Cst
Cpr
Cin
1
0.002
0.064
0.002
10
19
9
0.0223
0.0800
0.288
0.454
Auto Insert
6
1
0.008
0.224
0.002
15
13
44.9
0.036
0.0480
0.252
0.560
Auto Insert
1
1
0.158
0.950
0.005
12
15.4
11.9
0.0148
0.965
Manual
No. of part types
Cti
No. of parts
Ci
No. of leads
Cc
nl
Rp
Nt
Axial
2
32
10
DIP
14
28
Can
3
6
Name or
operation
Rework
Ave. Component
Cost, $
Insertion time, s
Insertion
Ti
19
Total
Total
1.979
Description
4. A small video board is double sided and has the following main parameters.
Board length
Board width
Board thickness
Copper weight
Number of holes
Minimum trace width
Minimum conductor spacing
6.25 in.
2.5 in.
0.062 in.
0.5 oz
502
0.015 in.
0.006 in.
Table 6.13 Video Board Design Parameters
The board is made from FR-4 material and has gold plated edge connectors. It requires
solder masking and legend printing on both sides.
a) The cost of a simple double side board in FR-4 costs $0.04/ in2. Estimate the
manufacturing cost of the bare board, excluding hole drilling, legend printing, solder
masking and final testing.
b) During manufacture of the boards the dimensions of the standard panel size used are
length 24 inches and width 18 inches. The separation of the boards in the panel is 0.1
inches and a clearance of 0.2 inches at the edge of the board must be used. Determine the
number of boards per panel.
c) The maximum stack height of panels for hole drilling is 0.5 inches. The entry and exit
films have the same thickness as the boards. Assuming that it takes 2 s to load and unload
each panel and film and that each hole has a drilling time of 2 s, estimate the cost per
board for hole drilling. The operating cost of the drilling machine including labor is $60
per hr.
d) The operating cost rate for solder masking is $50 / hr and the masking operation time
is 60 s per side. If the panel loading and unloading time is 2 s, estimate the cost per board
for solder masking.
e) The operating cost rate for legend printing is $60 / hr and the masking operation time
is 30 s per side. If the panel loading and unloading time is 2 s, estimate the cost per board
for legend printing.
Solution: a) From Equation 6.1,
and
from this the board factor = 0.96 + 0.1 + 0.05 + 0.05 = 1.16. The board cost is 0.04 x 1.16
x 6.25 x 2.5 = $0.73.
b) If the boards are arranged in the panel with the board length parallel to the panel length
then the number of boards along the panel length is given by Equation 6.3:
and the number of boards along the panel width is given by Equation 6.5:
The number of boards per panel is 18. In this case if the boards are arranged with the
board length parallel to the panel width, Equations 6.4 and 6.6 give the same number of
boards per panel but 2 x 9.
c) The number of panels in the stack for drilling is given by Equation 6.7:
From this the panel loading time is (6 + 2)2 = 24 s. The drilling cost per board is:
60 (24 + 502 x 2 x 18)/ (3600 x 6 x 18) = $2.79
d) The cost of solder masking is given by Equation 6.10:
per board.
e) The legend printing cost per board is given by Equation 6.10:
per board.
5. The video board in Problem 4 has a mixture of through hole and surface mount
components on the upper side only. The parts list for the board is given in the following
table.
Package style
DIP
DIP
SIP
SIP
SIP
Radial
Can
PTH connector
PTH connector
PTH connector
PTH connector
PTH connector
PTH connector
PTH connector
Chip
Quad flat pack
Repeat
count
3
7
4
1
1
3
1
3
1
1
1
1
1
1
25
1
No. of
leads
20
14
8
6
5
2
2
10
3
2
24
40
34
25
2
100
No. of
different types
2
4
3
1
1
1
1
1
1
1
1
1
1
1
7
1
Insertion
method
Auto
Auto
Auto
Auto
Auto
Auto
Manual
Auto
Auto
Auto
Robot
Robot
Robot
Robot
Auto
Robot
Average part
cost, $
0.1
0.1
0.05
0.05
0.05
0.05
0.1
0.05
0.05
0.05
0.1
0.15
0.1
0.1
0.01
2.00
Table 6.14 Parts List for Video Board
a) What will be the probable assembly sequence for this board?
b) Using the data in Section 6.7, estimate the cost of assembling this board excluding
soldering operations. Assume that all components that are reworked are replaced with
new components. A copy of the worksheet in Table 6.11 can be used for this problem.
c) The can component on line 7 of the table requires the component to be bent over and
the can case soldered to the board by hand. What would be the savings if this component
could be replaced by a similar component that could be automatically inserted?
Solution:
a) The probable assembly sequence for this board will be:
1. Apply solder paste 2. Place surface mount components 3. Reflow solder 4. Insert and
clinch through hole components 5. Wave solder 6. Clean
b) The cost parameters for this assembly are:
Labor, $/hr
Set up per part, $
Program cost per part, $
Batch size
Life Volume
36
1.6
1.8
2000
10000
The worksheet for this case is shown below. The time to manual insert the can
component is 13 seconds, this should be increased to allow for bending over the can and
manually soldering the case to the board, say 20 seconds. The total cost of assembly
component placement is $0.992.
c) The total cost to manually insert the can component is $.203. Assuming the same
component cost of $0.1 for the alternative auto-insertable part, then the cost becomes
$0.14, made up of operation cost of $0.012, rework cost of $0.0009, setup cost of
$0.0008 and programming cost of $0.0002. The cost saving is therefore $0.203 - $0.14 =
$0.063.
Total insertion, $
Ave. no. of faults
4
Removal time, s
Insertion time, s
Soldering time, s
Total rework, $
Setup cost, $
Programming cost, $
Total cost, $
Af
Trm
Ti
Trs
Crw
Cst
Cpr
Cin
0.1
0.008
0.024
0.002
15
16
62.9
0.0058
0.0016
0.0005
0.032
Auto Insert
4
0.1
0.008
0.056
0.002
15
13
44.9
0.0104
0.0032
0.0013
0.071
Auto Insert
4
3
0.05
0.008
0.032
0.002
10
8
26.9
0.0037
0.0024
0.0007
0.039
Auto Insert
6
1
1
0.05
0.008
0.008
0.002
10
8
20.9
0.0009
0.0008
0.0002
0.010
Auto Insert
SIP
5
1
1
0.05
0.008
0.008
0.002
16
8
17.9
0.0009
0.0008
0.0002
0.010
Auto Insert
Radial
2
3
1
0.05
0.012
0.036
0.002
10
13.6
8.9
0.0021
0.0008
0.0005
0.039
Auto Insert
Radial
2
1
1
0.1
0.200
0.200
0.005
10
13.6
8.9
0.0021
0.0008
0.0002
0.203
Manual
PTH Connector
10
3
1
0.05
0.008
0.024
0.002
15
11
32.9
0.0036
0.0008
0.0005
0.029
Auto Insert
PTH Connector
3
3
1
0.05
0.008
0.024
0.002
15
7.5
11.9
0.0022
0.0008
0.0005
0.028
Auto Insert
PTH Connector
2
3
1
0.05
0.008
0.024
0.002
15
7
8.9
0.002
0.0008
0.0005
0.027
Auto Insert
PTH Connector
24
1
1
0.1
0.05
0.050
0.002
15
18
74.9
0.0024
0.0008
0.0002
0.053
Robot insert
PTH Connector
40
2
1
0.15
0.05
0.100
0.002
15
26
123
0.0069
0.0008
0.0004
0.108
Robot insert
PTH Connector
34
3
2
0.1
0.05
0.150
0.002
15
23
105
0.0088
0.0016
0.0005
0.161
Robot insert
PTH Connector
25
1
1
0.1
0.05
0.050
0.002
15
18.5
77.9
0.0024
0.0008
0.0002
0.053
Robot insert
Chip
2
25
7
0.01
0.002
0.050
0.001
10
10
9
0.0073
0.0056
0.0045
0.067
Auto Insert
Quad Flat pack
100
1
1
2
0.05
0.050
0.002
15
20
303
0.0108
0.0008
0.0002
0.062
Robot insert
Total
0.992
Ave. Component
Cost, $
Cti
No. of part types
Ci
No. of parts
Ti
No. of leads
Cost per part, $
Rework
Insertion time, s
Insertion
nl
Rp
Nt
Cc
DIP
20
3
2
DIP
14
7
SIP
8
SIP
Name or
operation
20
Description
6. A small control board for a PC has mainly surface mount components and two through
hole (PTH) connectors mounted on the upper side of the board. The parts list for this
board is given in the table below.
a) What will be the probable assembly sequence for this board?
b) Using the data in Section 6.7, estimate the cost of assembling this board excluding
soldering operations. Assume that all components that are reworked are replaced with
new components. A copy of the worksheet in Table 6.11 can be used for this problem.
c) A variant of this board has three 20 pin SOICs on the underside of the board, with an
average cost of $0.5 each. What is the probable assembly sequence for this variant of the
board and the additional cost of assembly?
Package style
SOIC
SOIC
Flat pack
SOT
Quad flat pack
Quad flat pack
Quad flat pack
Chip
PTH connector
PTH connector
Repeat
No. of
No. of
Insertion
count
leads
different types
method
1
40
1
Auto
1
50
1
Auto
1
8
1
Auto
3
4
3
Auto
1
156
1
Robot
1
80
1
Robot
1
44
1
Robot
122
2
80
Auto
1
4
1
Auto
1
53
1
Robot
Table 6.15 Parts List for Small PC Board
Average part
cost, $
1.0
1.0
0.5
0.1
3.00
2.00
1.50
0.01
0.1
0.5
Solution:
a) The usual sequence for a board with mixed through hole and surface mount parts on
the top side is:
1. Apply solder paste 2. Place surface mount components 3. Reflow solder 4. Insert and
clinch through hole components 5. Wave solder 6. Clean
However, because there are only two through hole components it should be possible to
reflow solder these and the most probable sequence becomes:
1. Apply solder paste 2. Place surface mount components 3. Insert and clinch through
hole components 4. Reflow solder 5. Clean
b) The basic cost parameters for this case are:
Labor, $/hr
Set up per part, $
Program cost per part, $
Batch size
Life Volume
36
1.6
1.8
2000
10000
The worksheet for this assembly is given below and the component placement costs are
$0.689.
c) If three 20 pin SOICs are added to the underside of the board the assembly sequence
will probably become:
1. Apply solder paste to bottom side 2. Place surface mount components with adhesive 3.
Cure adhesive 4. Invert board 5. Apply solder paste to upper side of the board 6. Insert
and clinch through hole components 7. Reflow solder 8. Clean
The additional placement cost for these additional components is $0.026 made up of
$0.018 operation cost, $0.0069 rework cost, $0.0008 setup cost and $0.0002
programming costs. There are, of course, additional costs for the bottom side solder paste
operation in the assembly sequence.
Cti
Af
SOIC
40
1
1
1
-
0.005
0.005
0.002
SOIC
50
1
1
1
-
0.005
0.005
Flat pack
8
1
1
0.05
-
0.002
SOT
4
3
3
0.1
-
Quad Flat pack
156
1
1
Quad Flat pack
80
1
Quad Flat pack
44
Chip
Total cost, $
Ci
Programming cost, $
Ave. no. of faults
4
Ti
Setup cost, $
Total insertion, $
Cc
Total rework, $
Cost per part, $
Nt
Soldering time, s
Ave. Component Cost,
$
Insertion time, s
Rp
Insertion time, s
No. of part types
nl
Removal time, s
No. of parts
Rework
No. of leads
Insertion
Tr
Ti
Trs
Crw
Cst
Cpr
Cin
15
10
123
0.005
0.0008
0.0002
0.011
Auto Insert
0.002
15
10
153
0.0056
0.0008
0.0002
0.012
Auto Insert
0.002
0.002
15
10
26.9
0.0011
0.0008
0.0002
0.004
Auto Insert
0.002
0.006
0.001
12
10
14.9
0.0012
0.0024
0.0005
0.010
Auto Insert
3
0.05
0.050
0.002
15
20
471
0.0161
0.0008
0.0002
0.067
Robot Insertion
1
2
0.05
0.050
0.002
15
20
243
0.0096
0.0008
0.0002
0.061
Robot Insertion
1
1
1.5
0.05
0.050
0.002
15
20
135
0.0064
0.0008
0.0002
0.057
Robot Insertion
2
122
80
0.01
0.002
0.244
0.002
10
10
9
0.0708
0.0640
0.022
0.401
Auto Insert
PTH Connector
4
1
1
0.1
0.008
0.008
0.002
15
8
14.9
0.001
0.0008
0.0002
0.010
Auto Insert
PTH Connector
53
1
1
0.5
0.05
0.050
0.002
15
32.5
162
0.0052
0.0008
0.0002
0.056
Robot Insertion
Total
0.689
Name or
operation
m
Description
Chapter 7 Design for Machining
1. Two thousand bars 80 mm in diameter and 300 mm long must he turned down to 65
mm diameter for 150 mm of their length. The surface-finish and accuracy requirements
are such that a heavy roughing cut (removing most of the material) followed by a lightfinishing cut are needed. The roughing cut is to be taken at maximum power. The lightfinishing cut is to be taken at a feed of 0.13 mm, a cutting speed of 1.5 m/s, and at
maximum power.
If the lathe has a 2-kW motor and an efficiency of 50 percent, calculate the total
production time in kiloseconds (ks) for the batch of work. Assume that the specific
cutting energy for the work material is 2.73 GJ/m3, the time taken to return the tool to the
beginning of the cut is 15 s, and the time taken to load and unload a workpiece is 120 s.
Solution:
Power available,
Also
m3/s
and therefore
For finish cut
Volume V to be removed in rough cut:
m3
s
s
Total production time = 2000(120 + 2 x 15 + 699) = 1698 ks
Alternatively:
Total volume removed = 256 x 10-6 m3
m3/s
Therefore
s
2. A workpiece in the form of a bar 100 mm in diameter is to be turned down to 70 mm
diameter for 50 mm of its length. A roughing cut using maximum power and a depth of
cut of 12 mm is to be followed by a finishing cut using a feed of 0.1 mm and a cutting
speed of 1.5 m/s.
It takes 20 s to load and unload the workpiece and 30 s to set the cutting
conditions, set the tool at the beginning of the cut and engage the feed.
The specific cutting energy for the material is 2.3 GJ/m3 and the lathe has a 3-kW motor
and a 70 percent efficiency. Estimate:
a. The machining time for the rough cut
b. The machining time for the finish cut
c. The total production time for each workpiece
Solution:
(a) Rough cut:
w
m/s
m3
Volume to be removed
ks = 190 s
(b) Finish cut:
f = 0.1 mm, ap = 3 mm, v = 1.5 m/s
s
(c)
s
3. A 1.5-m-diameter disc with a 600-mm-diameter hole in the center is to be faced,
starting at the outside, on a vertical-boring machine. The rotational frequency of the table
is 0.5 s-1, the feed is 0.25 mm, and the back engagement (depth of cut) is 6 mm. The
specific cutting energy for the work material and the particular cutting conditions is 3.5
GJ/m. Calculate:
a. The machining time, in kiloseconds
b. The power consumption (in kW) at the beginning of the operation
c. The power consumption just before the end of the operation
Solution:
r1=750mm
r2=300mm
(a)
Radial distance moved by tool = 750 – 300 =450 mm
Therefore number of revolutions of workpiece = 450/ 0.25
s
(b)
W
(c)
4. In a drilling operation using a twist drill, the rotational frequency of a drill is 5 s-1, the
feed 0.25 mm, the major cutting-edge angle 60 deg, and the drill diameter 12 mm.
Assuming that the specific cutting energy for the work material is 2 GJ/m3, calculate
a. The maximum metal removal rate
b. The undeformed chip thickness
c. The drill torque, in Newton-meters (N-m)
Solution:
µm3/s
(a)
(b)
(c) Torque,
mm
and
Therefore
Nm
5. In a slab-milling operation, the width of the medium carbon steel workpiece is 75 mm,
its length is 200 mm, and a 5-mm layer is to be removed in one pass.
a. What feed speed (mm/s) could be used if the power available for cutting is 3 kW and
the specific cutting energy for the work material is 3.6 GJ/m3?
b. If the cutter diameter is 100 mm and the ideal surface roughness is 1.5 micrometers
(µm), what should be the rotational frequency of the cutter?
c. What is the cutting speed’?
d. What is the machining time?
Solution:
(a)
Therefore
m/s = 2.2 mm/s
(b) Feed per revolution =
Roughness =
Therefore nt = 1.45 s-1.
(c)
mm/s
(d)
50
s
45
5
200
6. In a slab-milling operation, the cutter has 20 teeth and is 100 mm in diameter. The
rotational frequency of the cutter is 5 s-1, the work-piece feed speed is 1.3 mm/s. the
working engagement (depth of cut) is 6 mm. and the back engagement (width of the
workpiece) is 50 mm. The relationship between the maximum undeformed chip thickness
and the specific cutting energy ps. in gigajoules per cubic meter (GJ/m3), for the
work material is:
Estimate:
a. The maximum metal removal rate
b. The maximum power, in kilowatts (kW), required at the cutter
Solution:
µm
J/m3
µm3/s
(a)
(b)
kW
7. In a slab-milling operation, the length of the workpiece is 150 mm, its width is 50 mm,
and a layer 10 mm in thickness is to be removed from its upper surface. The diameter of
the cutter is 40 mm and it has 10 teeth. The workpiece is of medium carbon steel, the feed
speed selected is 2 mm/s. and the cutter speed is 2.5 rev/s. The specific cutting energy of
the material, ps, is 4 GJ/m3.
Estimate:
a. The power required (in kW)
b. The machining time for the operation
Solution:
mm
m3/s
(a)
kw
20
17.3
10
10
Effective,
mm
150
(b)
s
8. In a face-milling operation the back engagement (depth of cut) is 5 mm, the work feed
speed is 0.65 mm/s. and the working engagement (width of the workpiece) is 50 mm. The
cutter has a diameter of 100 mm and has 20 teeth. If the cutting speed is to be 1 rn/s.
calculate:
a. The rotational frequency of the cutter
b. The maximum metal removal rate
c. The time taken to machine 1000 workpieces of length 150 mm if it takes 180 s to load
and unload a workpiece and return the cutter to the beginning of the cut
Solution:
s-1
Therefore,
(a)
µm3/s
(b)
(c)
Effective length of workpiece = lw+dt = 150 + 100 = 250 mm.
s
Production time =100(385 + 180) = 565 ks.
(d)
mm
9. In a finish-surface-grinding operation on a horizontal-spindle surface grinder, the
length of the workpiece is 100 mm and its width 50 mm. The cross-feed is applied every
stroke of the worktable and is set at 0.25 mm, the back engagement (depth of cut) is 0.1
mm. and the maximum traverse speed is 250 mm/s. The frequency of worktable
reciprocation is 1-1. Calculate:
a. The machining time
b. The maximum metal removal rate
c. The maximum power consumption in watts (W) if the specific cutting energy for the
conditions employed is 25 GJ/m3
d. The maximum tangential force on the grinding wheel if its diameter is 150 mm and
has a rotational frequency of 60 s-1.
Solution:
(a) Cutting stroke frequency = 2 s-1.
Number of strokes required = 50/0.25 = 200
Therefore tm = 200/2 =100 s.
µm3/s.
(b)
(c)
(d)
W
, where F = tangential force.
Therefore,
N
10. A batch of 10,000 rectangular blocks 50 x 25 x 25 mm is to be slab-milled on all
faces. The material can be either mild steel costing $4,900/m3 or aluminum costing
$6,100/m3. Estimate the total cost of the batch of components in each material if the
average nonproductive time for each face = 60 s, the tool-changing time = 600 s, the cost
of a sharp tool = $20.00, the machine and operator rate = $0.01/s, the tool-life index n =
0.125, the cutting speed for 60 s of tool life is 1 m/s for mild steel and 4 m/s for
aluminum, the feed per revolution of the cutter = 1.25 mm, and the working engagement
(depth of cut) = 2 mm. (Assume that the distance traveled by the 50-mm-diameter, 38mm-wide cutter for each face of the workpiece is 5 mm greater than the length of the
face.)
Solution:
Volume per component =
Equivalent length of workpiece:
mm
Steel
Aluminum
0.625
2.5
3.98
15.9
1.0
4.0
280
70
140
562
(s)
644
431
($)
2.71
1.76
0.15
$28,600
0.19
$19,500
(m/s)
(s-1)
(m/s)
(s)
Cost of material ($)
Total for 10,000 parts
11. Aluminum rod is available in diameters of 1-mm increments to 25-mm diameter and
in 2-mm increments from 26- to 50-mm diameter. In the design of a particular shaft it has
been determined that its finished diameter D should be within the range
where L is the length of the shaft. If an allowance of 2mm in diameter is to be allowed for
machining, what finished diameter should be chosen for the shaft?
Solution:
Upper limit
Lower limit
32.99
31.57
27.80
25.62
31.97
28.97
27.56
27.43
L
230
240
250
254
Alternatives:
Finished diameter
Rod diameter
Length
Volume (mm)3 x 103
28
30
246
174
30
32
236
190
32
34
230
209
Therefore select 30 mm diameter stock.
12. Which of the two designs for a cylinder assembly shown in Fig. 7.54 would be
preferable from the manufacturer’s point of view? Also, list other manufacturing and
assembly aspects of both designs that could be improved. (Note that the omission of
details of the bolting arrangements for the cylinder ends is not to be considered an error.)
Solution:
Option (b) is preferable because the external groove is easier to machine than the internal
groove in (a).
Other features:
(1) Internal corners not radiused and external corners not chamfered to suit
(2) Stud not seated properly in piston
13. The end covers shown in Fig. 7.55 are to be milled from 10-mm-thick aluminum
plate. Which of these two designs is preferable from the manufacturer’s point of view?
Also, what other aspects of the designs should be changed?
Solution:
The design requiring the 10 mm diameter end mill is preferable because the machining
time for the pocket will be the shortest. Also the internal corner on the outer surface
should be rounded to 5 mm radius.
14. A designer has specified a surface finish on turned shafts of 0.4-pm arithmetical mean
when a surface finish of 1.6 pm would suffice. Estimate the cost of this mistake when
2000 shafts are to be produced with a tool with a rounded corner if the machining time
per component = 600 s, the number of components produced between tool regrinds = 4,
the cost of a sharp tool = $2.00, the machine and operator rate = $00033/s, the toolchanging time = 120 s, and the nonproductive time per component = 240 s.
Solution:
and therefore
(1) 0.4 µm finish:
s
Total cost = 2000 x 3.37 = $6740
(2) 1.6 µm finish:
s
s
Total cost =2000 x 2.08 = $4160
Cost of mistake = 6740 – 4160 = $2580
15. The part shown in Fig. 7.56 is to be produced on a CNC turret lathe from free
machining low carbon steel. Using the worksheet in Table 7.9 and the data in this
Chapter, estimate the processing time and costs for machining a batch of 500 parts.
Assume that 0.015 in. of material is faced off from the end of the part prior to drilling the
through hole. The center drilling operation prior to drilling the through hole can be
assumed to have a machining time of 5 s. The turning tool has a disposable carbide insert
and the drill is made from high speed steel.
2.000
2.000
dia.
1.000
dia.
0.500 dia.
through hole
2.500
Workpiece material:
Free machining steel
2.000 in. diameter bar stock
Dimensions in inches
Figure 7.56
Solution:
Allowing 0.05 in for facing the end of the part, the weight of the workpiece is given by:
lb.
From Table 13.10 the cost of bar steel is $0.51 per lb and hence the material cost of the
component is
.
The calculations for the operation times are given in the worksheet below. From
Figure 13.26 the estimated power for a CNC Lathe of the appropriate size is 5 hp. The
remaining data is obtained from the Tables in the Chapter.
The total time per part is 111.43 + 7.5 + 23.3 = 142.23 s. The setup time per part
is
s.
Thus the total time per part including setup is 142.23 + 6.84 = 149.07 s.
From Figure 13.27 the capital cost of the CNC lathe is about $80,000. Thus
assuming 5 years depreciation period and 1 shift working, the amortization cost of the
machine is
per hr. Therefore, assuming an operator rate of
$30 per hour and 100 percent overhead on the amortization cost, the cost rate for the
machine and operator is $30 + $16 = $46 per hr.
Therefore the machining cost of the part is
The total part cost is $1.16 + $1.88 = $3.04
.
D
Center drill
H
0.15
Drill
H
0.15
Totals
0.95
-
6.47
3.32
60.94
62.37
5
1.27
70.2
-
6.28
5.37
7.14
8.49
1.1
5
1.04
70.2
-
1.57
1.34
1.78
3.13
-
-
-
-
-
-
-
-
5.00
0.95
5
29.46
15.34
-
3.93
15.37
29.49
32.44
Total
111.43
lw
da
db
vm
ps
Pm
23.3
1.5
2.0
2.0
1.03
4.62
1.1
5
1.5
2.0
1.03
1.00
0.10
1.1
1.5
0.05
2.0
0
0.08
1.5
-
-
-
-
1.5
2.5
0.5
0
0.49
7.5
Time corrected for extra tool
travel (s)
117
tpt
23.3
Time corrected for tool wear
(s)
60.94
tl
* H = high-speed steel, C = carbide (brazed), D = carbide disposable insert
Machining time recommended
conditions (s)
area
D
Face end
Milling feed speed
Finish
0.65
Rate of surface generation
D
Am
Available power
Rough turn
vf
Specific cutting energy
CNC Lathe
vf
volume
operation
dimension
Machine
tool
dimension
Batch size (thousands): 0.5
dimension
Workpiece weight: 2.267
Tool positioning time (s)
3
Load and unload time (s)
Density: 0.283 lb/in
Tool type (HCD)*
Material: F.M. Carbon Steel
Setup time per batch (hr)
Part Name: Problem 7.15
Machining time max power (s)
Machining Cost Analysis Worksheet
tm
16. The aluminum alloy part shown in Fig. 7.57 is to be produced on a small machining
center from a piece of bright drawn rectangular bar of cross section 3 in. by 2 in. The
machining operations required are as follows:
1. Finish face mill the top surface to reduce the part height to 1.75 in. with a brazed
carbide face milling cutter.
2. Rough and finish the two 1 in. by 1.25 in. steps in the parts with a high speed steel end
milling cutter.
3. Finish mill the 0.5 in. by 0.25 in. in the top surface of the part with a high speed end
milling cutter.
4. Drill six 0.5 in. diameter holes with a high speed steel drill.
Using the worksheet in Fig.7.9 and the data in this chapter, estimate the processing time
and costs for a batch of 200 parts.
Figure 7.57
Solution:
The volume of the workpiece is 36 in3 and as the density of the work material is 0.1
lb.in3, the workpiece weight is 3.6 lb. From the data in Table 7.10, the material cost is
$1.93/ lb and the work piece cost is 3.6 x 1.93 = $6.95. Assuming that the workpiece is
held in a vise, then, from Table 7.11, the workpiece loading and unloading time is 18.6 s.
Based on the workpiece weight of 3.6 lb, the probable horse power available on the
machining center is about 5 hp (Figure 7.26). From Figure 13.27 the capital cost of the
machine is about $30,000. Thus assuming 5 years depreciation period and 1 shift
working, the amortization cost of the machine is $30,000/(5 x 50 x 5 x 8) = $3 per hr.
Therefore, assuming an operator rate of $30 per hour and 100 percent overhead on the
amortization cost, the cost rate for the machine and operator is $30 + $6 = $36 per hr.
The remaining data is obtained from the Tables in the Chapter.
It is assumed that the diameter of the face mill for the first operation is 4 inches. The
diameter of the end mill for the second operation is 1.5 inches and 0.015 inches is left on
both surfaces for the finishing operation three. The corrections for over travel and tool
approach in Table 7.5 are applied to the length of as opposed to the machining time for
milling operations and the corrected values are shown in the lw column of the worksheet.
The worksheet for machining this workpiece is shown below. The total estimated
machining time is 175.1 s, taking into account the repeat operations for the two side steps
and the 6 holes. The load/unload time is 18.6 s, giving a total time to machine each part
of 193.7 s.
The setup time for the machine is 0.9 hours, which for a batch size of 200 gives:
0.9 x 3600/200 = 16.2 s per part
The total cost of the workpiece is the sum of the material cost and the machining cost, or:
$6.95 + (193.7 + 16.2) 36/3600 = $9.049
Time corrected for extra tool
travel (s)
Time corrected for tool wear
(s)
area
19.44
216.5
-
2.80
19.44
19.44
31.54
20.4
-
22.3
34.17
34.17
5
0.86
20.4
-
22.3
29.74
29.74
20.4
-
19.32
25.77
25.77
-
0.196
0.15
0.23
0.34
Total
175.1#
tl
tpt
lw
da
db
vm
ps
Pm
18.6
8
10.1
-
-
4.5
0.36
5
8
7.55
-
-
7.3
0.36
5
7.55
-
-
0.2
0.36
Face mill
D
0.75
End mill * 2
H
0.05
End mill * 2
H
End mill
H
0.05
8
6.57
-
-
0.75
0.36
5
3.24
Drill * 6
H
0.05
8
0.5
0.5
0
0.10
0.18
5
0.21
Totals
0.90
18.6
Machining time recommended
conditions (s)
Milling feed speed
Machining time max power (s)
Am
Available power
vf
Specific cutting energy
Rate of surface generation
#
vf
volume
M/c center
operation
dimension
Machine
tool
dimension
Batch size (thousands): 0.2
dimension
Workpiece weight: 3.6
Tool positioning time (s)
3
Load and unload time (s)
Density: 0.1 lb/in
Tool type (HCD)*
Material: Al. alloy
Setup time per batch (hr)
Part Name: Problem 7.16
79.8
32
Total accounts for repeat operations * H = high-speed steel, C = carbide (brazed), D = carbide disposable insert
tm
Chapter 8 Design for Injection Molding
1. Figure 8.18 illustrates an annular ring component with eight hollow cylindrical bosses.
There are no hidden features on the reverse side. The outer and inner diameters of the
ring are 200 and 160 mm respectively. The main wall is 3 mm thick, and the boss
projections are 12 mm outer diameter, 6 mm inner diameter and 30 mm high. The
material proposed for the design is high-density polyethylene, which is a very
inexpensive polymer. However the poor material mechanical properties give rise to the
substantial wall thickness.
(a) Estimate the material cost for the part, assuming the runner system will be reused at
the machine (use the data in Tables 8.1 and 8.5).
(b) Estimate the appropriate machine size from the list in Table 8.4.
(c) Estimate the cycle time (use the data in Table 8.5).
(d) Estimate the process cost per part using the machine rate in Table 8.4. Assume in this
calculation that the molding shop has 85% plant efficiency; i.e. machines are typically
stopped 15% of the time for batch setup, adjustments and miscellaneous stoppages.
Increase the process cost appropriately to account for this.
Figure 8.18
= 77cm3.
Solution: a) Part volume,
From Table 8.2, percent addition for runners is 17. The shot size Vs is;
cm3.
Projected area, including runners is:
cm2.
Part weight is (from Table 8.5)
.
From Table 8.1 the material cost is $0.9 per kg and hence part material cost,
b) Separating force from Section 8.6 and Table 8.5 is:
From Table 8.4 the appropriate machine is 800 kN which also has sufficient shot size and
stroke.
c) From Equations 8.3, 8.7 and 8.9 and Tables 8.4 and 8.5
=1+1.75(3.3)(7.0+5)/
Therefore total cycle time t = 0.9 + 54 + 4.5 = 59.5 s.
d) From Table 8.4 machine rate is $33/hr. Therefore the processing cost with 85%
efficiency is:
(59.5/3600)(30)/0.85 = $0.64
Piece part cost = 0.64 + 0.07 = $0.71
2. Continuing from Problem 1, for the ring design in Fig. 8.18, you propose a design
change to use polyethylene terephthalate (PET) with 30% glass fill. PET is used for soda
bottles, and you have located a quality recycled material supplier who can deliver the
glass filled PET for $2.25/ kg. Using the data in Table 8.1, calculate the main wall
thickness required to give the same wall bending stiffness as for the high-density
polyethylene part in the original design proposal. Decrease the wall thickness of the
bosses in the same proportion.
Recalculate the material and process cost for this proposed redesign, using the same steps
as in problem 1. Is your redesign idea a sound economic proposal? If the recycled
material was unavailable and you had to pay the full supplier price of $3.74/ kg would the
change still be worthwhile?
(Hint: recall from Chapter 3 that for equivalent bending stiffness the wall thickness
should change in proportion to the cube root of the modulus)
Solution: Wall thickness for equivalent stiffness is, using modulus values from Table
8.1, :
The boss wall thickness, decreased in proportion, is:
The boss inner diameter is this
=12 -2(1.5) = 9 mm.
With these changes the part volume will have decreased in the same proportion. Thus:
cm3
Percentage runner system has increased to 20% (Table 8.2) and so total projected area is
now:
cm2.
The separating for f using PET (30%) glass is:
.
This means that a larger machine is needed at 1100 kN from Table 8.4.
The part weight with the new volume and higher material density is
The material cost using $2.25 per kg is Cp = (0.0597)(2.25) = $0.13.
The shot size Vs = 38.3 x 1.24 = 48 cm3. Using the PET processing properties and data
for the 1100 kN machine gives:
= 6.7 s
Therefore the total cycle time, t =12.1 s.
Using the increased machine rate of $36/hr and 85% efficiency gives:
Process cost = (12.11/3600)(36)/0.85 =$0.14.
The piece part cost is 0.13 + 0.14 = $0.27 (only 38% of the polypropylene part). If the
material cost was $3.74/kg the material cost per part would be:
Cp = 0.13(3.74/2.25) = $0.22.
The piece part cost would become 0.22 + 0.14 =$0.36 still approximately half of the pp
part.
3. A cable company is considering the possibility of changing their current die cast utility
pole mounted distribution box to an injection molded one. The motivation is to eliminate
the costs of trimming, plating (for corrosion resistance) and machining of screw threaded
holes for securing. In addition, the new design will have snap-fit hinges and an integral
snap-fit latch plate for self securing. The specifications of the box cover are estimated
from concept sketches, and calculations of equivalent strength and stiffness of the die cast
model. The cover is to be molded using polypropylene (PP) with 40% talc fill. The global
specifications are estimated to be:
Weight = 170g
Envelope width = 15 cm
Projected Area = 322 cm2
Envelope length = 25 cm
Envelope Depth = 4 cm
Wall thickness = 2.54 mm
The appearance level (Table 8.6) is to be standard-opaque and the tolerance should be
level 3 (Table 8.7). In addition, from the sketches, the complexity has been estimated as
shown in the table below, which gives a breakdown of the number of surface patches on
the inner and outer surfaces of the main wall, and on the separate features which make up
the cover design. The complexity numbers are calculated in the last column. Note that for
the repeated features the number of patches on a single feature are multiplied by the
number of repeats, r, raised to the 85% progress curve index.
Initially 50,000 covers are required, which can be made most economically in a single
cavity mold.
(a) Estimate the sum of the material and process costs (known as the piece part cost) for
this component. Use the material data and the machine specifications from Chapter 8.
Select the appropriate machine from Table 8.4. Allow for 15% downtime and assume the
runners are re-used at the machine.
(b) Estimate the cost of a single-cavity mold for the Verizon cover. The design requires
no mold mechanisms. Compare your estimated number of manufacturing hours with a
quick estimate from Fig. 8.10.
Name of feature
Repeats
(r)
Outer surface
Side projections
Latch plate &gussets
Hinges
Surface
Patches
(Ns)
29
6
11
12
Inner surface
Flange/ edge
Side projections
Latch plate underside
Hinges
8
15
10
3
8
1
1
2
1
2
1
2
1
2
Complexity
X
(0.1Ns*r0.766)
2.90
1.02
1.10
2.04
Xo = 7.06
0.80
1.50
1.70
0.30
1.36
Xi = 5.66
Solution:
a) Part volume = 170/1.22 = 139 cm3 and percent runner = 12.7.
Projected area,
cm2
Separating force,
A 5000 kN machine should be used from Table 8.4 and using data for this machine:
Fill time,
Cooling time,
Reset time,
Therefore the cycle time, t = 0.48 + 12.44 + 5.6 = 18.5 s and therefore the process cost
with a 5000 kN machine is (18.5/3600)(74)/0.85 = $0.45.
Material cost at $1.17/kg is (139)(1.17)/1000 = $0.20 per part. Therefore the piece part
cost = 0.45 + 0.20 = $0.65 per part.
b) Mold base plate area = (15 + 15)(25 + 15) = 1200 cm2, (allowing 7.5 cm between
cavity and plate edges).
Total plate thickness = D + 15 = 19 cm. From equation 8.11, the estimate of equivalent
hours for the mold base and its custom work is:
From the sum of Equations 8.13 and 8.19 the basic number of hours for the cavity, core
and ejection system is:
With the given total complexity values the additional hours for the particular part
geometry are:
Additional hours for improved appearance are (Table 8.6) = 0.15(112 +147) = 39 hrs.
Added hours for tighter tolerance are (Table 8.7) = 0.10(147) = 15 hrs.
Total manufacturing hours is (140 + 112 + 147 + 29 + 15) = 453 hrs.
Therefore estimated mold cost is :
From Figure 8.10, X =12.7 and Ap = 363 cm2 gives around 450 hours.
4. A particular manufacturing company uses a simple internal cost accounting method in
which all of the machines of a particular type are given the same hourly rate. In addition,
in the company’s internal tool making division no account is made for the savings in
making multiple identical items of tooling. Under this costing system all of the different
injection molding machines of different sizes are assigned an hourly rate of
$/hr,
and the cost of an n-cavity mold is n times that of a single-cavity mold (i.e. m = 1).
By making appropriate changes to equations in Section 8.10, show that this cost
accounting system would lead to the conclusion that the optimum number of cavities for
a particular part would be obtained when the mold cost per part was equal to the process
cost per part.
Solution: The equations simplify as follows:
Eq. (8.12)
Eq. (8.25)
Cr = k1
Cn = nC1
Eq. (8.20)
Squaring both sides of (3) gives:
(1)
(2)
(3)
or
(4)
Substituting (1) and (2) into (4) gives:
The right hand side equals (total number of cycles)(machine rate)(cycle time) = total
processing cost.
The left hand side is the cost of the multi-cavity mold.
5. The base part of the piston assembly illustrated in Fig. 8.16 is shown in Fig. 8.19. The
part outer dimensions are: length 76 mm, width 64 mm, height 30 mm. The part volume
is 29 cm3 and it is to be molded from ABS. The maximum wall thickness is 3.0 mm.
You have obtained a quote of $8,000 for a single cavity mold for the part. However,
large quantities of the part are required justifying the use of a 4-cavity mold. For the
proposed 4-cavity mold production estimate the following:
(a) The appropriate machine size from the selection in Table 8.4, making an approximate
allowance for the projected area of the runner system as well as the four cavities.
(b) The likely molding machine cycle time.
(c) The processing cost per part, assuming 85% plant efficiency and the machine rate in
Table 8.4.
(d) The likely 4-cavity mold cost assuming an 85% progress curve in manufacture of the
four cavities.
Figure 8.19
Solution: a) given: part volume, Vp = 29 cm3.
From Table 8.2, percent runner = 28 and therefore projected area is:
cm2.
Separating force for 4 cavities and pi = 1000 bars:
Therefore use 1600 kN machine.
b) Using data from the tables for the machine and ABS gives:
Cycle time, t = 1.35 + 15.69 + 4.22 = 21.27 s
c) Process cost =
d) Estimated 4-cavity mold cost is:
6. Figure 8.20 shows two views of a hose clamp which is to be injection molded using
30% glass filled polycarbonate; see Table 8.5. The envelope size of the part is 60x80 mm
and the wall thickness is 3.0 mm. A total of 4 million clamps will be required over the
mold amortization period of 30 months. You have already estimated the cycle time to be
19.6 seconds for a single cavity mold operation on a 300 kN machine, and have obtained
a quote of $7,000 for a single cavity mold. Assume that the injection molding machine
rate can be represented for your company by
$/hr (F = clamp force, kN)
(a) Using the 85% learning curve, determine the optimum number of cavities for the
mold which should be used to produce this part.
(b) Since this is a simple geometry to mold you decide to experiment with a lower mold
temperature of 800C. Assuming negligible effect on fill time and no effect on reset time,
by how much would this reduce the cycle time? (note that h = 3 mm)
(c) If you adopt the lower mold temperature for processing what would then be the
optimum number of cavities in the mold?
Figure 8.20
Solution: a) From Eq. (8.28)
Therefore optimum number of cavities is 16
b) Using the data for glass reinforced polycarbonate in Table 8.5 gives:
If the mold temperature is reduced from 102 0C to 80 0C, then cooling time changes to :
If the fill time and reset time remain unchanged then the cycle time will reduce by
(14.05 - 11.56) = 2.5 s. The cycle time becomes 19.6 – 2.5 = 17.1 s.
c) The optimum number of cavities then becomes:
and therefore around 14 cavities.
7. Saturn Corporation uses injection moldings for many of the body panels, including the
doors. In doing so they replaced the traditional low alloy steel sheet metal panels which
were nominally 1 mm thick. As previously described in the Chapter 2 problems, the
material they chose to use is 30% glass-reinforced polycarbonate, blended with ABS for
improved mold flow characteristics. The elastic modulus of this blended reinforced
thermoplastic is E=5 GN/m2, and the yield stress is Yt=80 MN/m2. The steel panels had a
nominal thickness of 1mm and corresponding material properties of E=200 GN/m2, and
Yt=300 MN/m2.
(a) Estimate the wall thicknesses Saturn would have needed to use to obtain the same
panel bending stiffness and strength as the sheet steel ones being replaced (hint: review
the information on material selection and derived parameter in Chapter 2).
(b) Using the processing parameters for this polymer blend in Table 8.5, determine the
cooling time for the panels during injection molding.
(c) Estimate the projected area of a Saturn automobile front door (don’t include the
window cutouts!). Determine the likely separating force and hence the likely machine
size. Extrapolate the data in Table 8.4 to estimate a likely hourly operating rate for the
machine. Your answer is probably conservative since these were custom made machines,
much larger than any standard machines available at the time.
(d) The Saturn roof panels were still manufactured as sheet metal formed parts. Can you
offer any explanation why these were not changed to injection moldings?
Solution:
a) Required wall thickness for equivalent stiffness:
and for equivalent bending strength:
Therefore use h = 3.4 mm.
b) From Table 8.5:
c) Measurements of a typical front door gives Ap approximately equals 7000 cm2. This
gives a separating force of:
Machines of 50 MN would be appropriate. Regression analysis of the data in Table 8.4
provides a good fit with the simple relationship:
A likely economic rate for the 50 MN machine would be:
d) A typical roof panel of a Saturn vehicle has Ap approximately equal to 16,000 cm2 and
required an almost certainly uneconomic machine with a clamp force capacity greater
than 100 MN.
Chapter 9 Design for Sheet Metalworking
In all of the questions below use the data from the Tables for material costs, material
properties, machine performance and machine hourly rates. Assume a rate of $40/hr for
all die making cost estimates.
1. For the part shown in Fig. 9.25, whose dimensions are given in mm, and which is
made from 12 gage commercial quality low-carbon steel, determine the following:
(a) The likely cost of a progressive die assuming a total production volume of 150,000
parts;
(b) The required press force for the external shearing, internal hole punching and the
bending operation;
(c) The press cycle time and the processing cost per part assuming 15 percent downtime
for batch setup and miscellaneous stoppages;
(d) The manufactured cost per part, including material cost, but neglecting the small
return on manufactured scrap for the low carbon steel. Assume the die is amortized over
the total production volume.
Figure 9.25
Solution:
a) Blanking die: Approximate part outer diameter is
assuming ¾ circular profile.
The plan area of the part is given by L = 11.5 cm and W = 10 cm. Allowing 5 cm space
around the part the required usable die area is Au = (21.5)(20) = 430 cm2.
The equivalent tool making hours for the die set from Eq. 9.2 is
The perimeter complexity value is:
Thus from Eq. 9.4 the basic die set manufacturing points are:
.
From Fig. 9.9, the area correction factor for LW = 115 cm2 is fLW = 1.6. For 150,000 parts
from 16 gage C. Q. low carbon steel, the required die plate thickness is (Eq. 9.5):
From Eq. 9.5, the die plate factor fp = 1.0. Thus the total points for a blanking die are:
.
Bending die: With one bend line of length 3 cm, the total manufacturing points for the
bending die are (Eq. 9.15):
Finally applying the approximate estimating procedure in Section 9.2.7 (Eq. 9.19) and
assuming $40/hr for die making gives:
Progressive die cost = 2(57.6 + 41.5 +27.3)40 = $10,000
b) Outside perimeter: Press force (Eq. 9.20),
Internal hole:
Bend: lb = 3 cm and using Eq. 9.24 gives:
and
.
c) The required press force = 170 + 65 +2 = 237 kN. Using the 500 kN press from Table
9.4 gives 60/90 =0.67 s cycle time at an hourly rate of $76/hr. With 15% down time the
processing cost per part is:
=1.7 c /part.
d) Assuming 2h = 5.24 mm clearance between each part and the strip edges, then material
used per part has a volume:
3
.
From Table 9.2 the material density is 7.9 g/cm3 and the cost is $0.8/kg. Therefore the
material cost/part is:
The total manufacturing cost is thus:
2. The part illustrated in Fig. 9.39 is the tool part of a barbeque spatula which is to be
assembled to a wooden handle. The head is 15 cm long, and has a width of 9 cm across
the front and 12 cm across the back. The shaft is 20 cm long by 3 cm wide, and increases
to 4 cm wide over the angled portion. The two bends are parallel and separated by a
distance of 6 cm. To reduce material scrap the parts are to be blanked in pairs from strip
as shown in Fig. 9.40. The distance between the centers of each spatula blank along the
strip is 7.0 cm, and the strip width is 36.0 cm.
(a) Estimate the total manufacturing hours for the die elements for a blanking die for one
of the spatula blanks.
(b) Assuming an 85% learning curve, estimate the total manufacturing hours for the die
elements for a blanking die for a pair of the spatula blanks.
(c) Including a dieset with useable area large enough for the nested pair of blanked
shapes, estimate the likely cost of the blanking die.
(d) If the spatula is to be made from 16 gage T304 stainless steel, estimate the material
cost per part using the cost and scrap value given in Table 9.2. How much greater would
be the material cost per part if the parts where blanked out individually with a spacing of
12.75 cm?
(e) Using the material values estimated in part (d), how many spatulas would need to be
produced to justify the additional investment in a blanking tool for blanking of pairs?
Figure 9.39
Figure 9.40
Solution:
a) The approximate perimeter of the spatula blank is approximately equal to the rectangle
which surrounds it, i.e:
The plan area is LW = (15+20) 9 = 315 cm2.
The perimeter complexity is
From Eq. 9.4 the basic die manufacturing hours are
From Fig. 9.9 the area correction factor for LW = 315 cm3 and Xp = 5 is approximately
fLW = 3.2 and using fp = 1.0 for an unknown production quantity give the total points for
the blanking die are:
b) Using an 85% learning curve the multi-die index (multi-cavity index described in
Chapter 8) is m = 0.766. Thus the hours to make the two aperture blanking die can be
estimated as:
c) Allowing 5 cm space around the pair of blanks, which have a width across the strip of
35 cm and a distance along the strip of 14 cm (equal to twice the pitch), gives a required
useable die area:
cm2.
The equivalent tool making hours for the die set from Eq. 9.2 is:
Thus the estimated die cost for pairs is:
Pair blanking die cost = (12.7 + 180.6) 40 = $7,700
d) Each blank uses an area of strip equal to As = 35 x 7 = 245 cm2. Since the thickness of
16g steel is 1.52 mm and the density and cost (Table 9.2) are 7.9 g/cc and $6.60/kg, the
cost of strip material per part is:
The surface area of the part is approximately:
cm2
Thus the approximate area of scrap is 217.5 cm2, and using the scrap value of $0.40/ kg
gives the return on scrap as 27.5 x 0.152 x 7.9 x 0.4/1000 = $0.01. (an insignificant
amount at the individual part level.
If the parts were blanked out individually the material cost increases by factor 12.75/ 7
which gives:
Cm = 1.94(12.75/ 7) = $3.53
The area of scrap is is now (12.75 x 35) – 217.5 = 228.8 cm2 and the return on scrap
would be 228.8 x 0.152 x 0.4 x 7.9/1000 = $0.11. Taking this into account the increase in
material in material cost/part would be (3.53 – 0.11) – 1.93 = $1.49, an increase of 76
percent.
e) From part (a) the basic manufacturing hours for a single-part blanking die is 106.2.
With 5 cm space around a single part the required useable die area is:
cm2
The equivalent toolmaking hours for the die set are thus:
Thus the cost of the die would be (12.2 + 106.2) 40 = $4,700. Finally the number of
spatulas needed to justify the added die cost is (7,700 – 4,700)/1.49 = 2000 parts.
3. The machine support bracket illustrated in Fig. 9.41 Is to be manufactured from 6
gage low-carbon, commercial quality steel using turret press with plasma profile cutting
capability, followed by press brake operations. The dimensions of the blank prior to
bending are 20 x 15 cm. After bending the horizontal base is nominally 5 cm wide and
the vertical wall is 10 cm high, tapering down over a horizontal length of 8 cm to a height
of 4 cm. The vertical slots are 6 cm long and the horizontal ones 7.5 cm long. All four
slots have a width of 1.25 cm wide. Punches are available on the turret press for both slot
sizes.
Estimate the following elements of the manufacturing process steps:
(a) The plasma profile cutting time per bracket.
(b) The turret press punching time per bracket.
c) The processing cost per part prior to the bending operation, assuming sets of 20 are
made from custom sized sheets measuring 800 x 840 mm.
(d) The press force required for the bending operation.
(e) The cycle time and operation cost per bracket for the press brake operation.
Figure 9.41
Solution:
a) The approximate outer perimeter length is :
P = 20 + 15 + 12 + 10 + 9 = 66 cm.
From Table 9.6 the plasma cutting speed, sp = 60 mm/s. From Eq. 9.32 the speed for
cutting 6 gage carbon steel is:
.
The plasma cutting time per part is thus:
b) All of the long slots will be punched over the sheet area first, followed by a single
turret index and the punching of the short slots. The average table move per punch is of
the order of 10 cm, which from Table 9.5 gives the time per hole punch as:
or t = 2.8 s for the four slots per part.
c) From Table 9.5, assume the loading plus unloading per sheet is 30 s. The processing
time per part is thus:
14.3 + 2.8 + 30/20 = 18.6 s
and with a machine rate of $72/hr this gives a processing cost/part, with an assumed 15%
downtime, of:
d) The bend length, lb = 20 cm, the gage thickness, h, =5.08mm and, from Table 9.2, the
UTS of the material is U = 330 x 103 kN/m2. From Eq. 9.24 the required press force is:
e) Part length + width = 200 +150 =350 mm. From Eq 9.36, the time for brake bending of
one bend (Nb =1) is:
with a corresponding operation cost of (7.8/3600)28/0.85 = $0.07.
4. If the machine support bracket illustrated in Fig. 9.41 is to be made from annealed
low-carbon commercial quality steel with properties as in Table 9.2, what is the smallest
inside bend radius you would recommend? If for the structural requirements of the
design, you wish to use half-hard commercial quality steel (not annealed after the finish
rolling passes of the strip) for which the maximum allowable strain is reduced to 0.10, to
what value would you need to increase the inside bend radius?
Solution: From Table 9.2, the maximum tensile strain to which annealed low carbon
steel should be subjected is e = 0.22. From Eq. 9.43, e = 0.22 gives r/h = 1.77, where h is
the gage thickness and r is the inside radius. Therefore:
For half-hard sheet: e = 0.10 and r/h = 4.5 from Eq. 9.43 and therefore:
5. The body of a rotor component shown in Fig. 9.42 Is to be made by deep drawing from
2.03 mm thick Al 3003 aluminum alloy sheet. The part is 13 cm high. The large diameter
is 20 cm which steps down to 17.5 cm 6.5 cm from the base.
(a) Estimate the required blank diameter to produce this part and determine that it is
possible to produce the initial 20 cm diameter cup in a single drawing operation, so that
only one redraw operation will be needed.
(b) The set of operations required to make the rotor includes: blank, draw, redraw, and
trim. Estimate the costs of the dies for each of these operations.
(c) Estimate the press forces required for each of the operations.
(d) Estimate the cycle times for each of the operations. Assume that hand loading and
unloading are required for each operation except blanking.
(e) Using rates for the appropriate machines from the Table 9.4, estimate the processing
cost for the set of four primary operations.
Figure 9.42
Solution:
From Tables 9.1, 9.2 and 9.3
Material: Al 3003
Gage thickness, h: 2.03mm
Cost: $3.00 per kg
Scrap value: $0.80 per kg
Yield strength, Y1: 53 MN/m2
UTS, U: 221 MN/m2
Density, ρ: 2.7 g/cc
a) For deep drawn shape: β0 = 1.8; β1 = 1.3
Surface area:
cm2.
Therefore
To produce an initial deep-drawn cup, 20 cm in diameter, the drawing ratio would be β =
1.80. This is equal to the initial drawing ratio β0 = 1.8 for Al 3003, so only one redraw to
step down to 17.5 cm diameter will be needed.
b) All of the dies are circular and so from Eq. 9.3, Xp = π and the basic die manufacturing
points are
. The other die cost parameters are:
Plan area
D2, cm2
1282
400
306
400
D, cm
1. Blank
2. Draw
3. Redraw
4. Trim
35.8
20
17.5
20
flw
6
3.2
3.0
3.2
Die set area
AU = (10 + D)2
2098
900
756
900
Using fd = 1 for the die types 1 and $ (Eq. 9.4), fp = 1 for unknown quantity (Eq. 9.7),
fd = 1.3 for die types 2 and 3 we get:
1. Blank
2. Draw
3. Redraw
4. Trim
Mds
Eq. 9.2
21.9
11.1
9.8
11.1
c) With U=221 MN/m2, h = 2.03 mm,
1. Blanking (Eq 9.2):
2. Drawing with Y1 = 53 MN/m2 (Eq. 9.27):
Mp
Eq. 9.8
209.1
141.9
131.5
110.9
Total
Die cost
Mp ($40 / hr)
$8,364
$5,676
$5,260
$4,436
$24,000
3. Redrawing operation substituting U = 221 and D = 17.5 gives:
4. Trim operation (Eq. 9.20) with D = 20 cm: Calculations same as for blanking with D
reduced from 35,8 cm, i.e. f = 252(10/35.8) = 140 kN.
d) Using loading, pres operation and loading times from Eq. 9.31, with L = W = D gives:
Blanking: t = 3.8 + 0.11 (75.7) = 12.1 s
Drawing: t = 3.8 + 0.11 (40.0) = 8.2 s
Redraw: t = 3.8 + 0.11 (35) = 7.7 s
Trim:
t = 8.2 s
e)
Press, kN
Rate $/hr
Process Cost*, $
Blank
500
76
0.3
Draw
200
55
0.15
Redraw
200
55
0.14
Trim
500
76
0.20
Total
* with downtime of 15% for setup and miscellaneous stoppages.
0.79
6. The completed rotor component involves the production of the body, as described in
problem 5, with the addition of a centered 12.5 cm diameter hole, surrounded by twelve
1.5 cm diameter holes as shown in Fig. 9.43.
(a) Estimate the press force required to punch these 13 holes simultaneously, and select
an appropriate press from Table 9.4.
(b) Estimate the cost of the required die.
(c) Estimate the cycle time and process cost for this piercing operation.
Figure 9.43
Solution:
a) Gage thickness, h = 2.03 mm; UTS, U = 221 MN/m2.
Total perimeter to be sheared is: ls = π(12.5 + 12 +1.5) x 10-2 = 0.96 m. From Eq 9.20
press force is:
Required press size is 500 kN (Table 9.4).
b) LW = 202 = 400 cm2. Number of holes, Np = 13. Perimeter of custom punch, Pp = π20
= 63 cm. This from Eq. 9.12, the approximate cost of the piercing die is:
c) From Eq. 9.31, the estimated cycle time is: t = 3.8 + 0.11(20 + 20) = 8.2 s.
Using $76/hr from Table 9.4 and assuming and assuming 85% plant efficiency gives
processing cost = 8.2(76/3600)/0.85 = $0.20.
Chapter 10 Design for Die Casting
In all of the questions below use the data from Tables 10.1 to 10.7 for material costs,
material properties, cycle time calculations and machine hourly rates. Where needed
assume a die making rate of $40/ hour.
1. The rotor assembly housing shown in Fig. 10.12 is to be die cast using Aluminum
Alloy A360. The part is 13 cm high. The large diameter is 20 cm which steps down to
17.5 cm at the midpoint. The bottom of the housing is open and the top has large hole
12.5 cm diameter in the center surrounded by twelve 1.5 cm diameter holes. The part has
a uniform wall thickness value of 2.50 mm. It has an estimated weight of 584g. Taking
account of the through holes, it has a projected, Ap, of 170.3 cm2.
(a) Determine the appropriate cold chamber machine, from Table 10.4, on which to cast
the part using a single cavity die.
(b) Estimate the die casting cycle time and use it to obtain the estimated die-casting
process cost.
(c) Estimate the time and process cost for trimming.
(d) Estimate the material cost.
Figure 10.12
Solution:
a) Projected area, Ap = 170.3 cm2. From Eq. 10.17 the estimated projected area of the
cavity, overflow wells and runners is:
cm2.
Using a cavity pressure of 48 MN/m2 for Al. alloy from Table 10.6 gives the separating
force as:
From Table 10.4 we would like to select the 1800 kN cold-chamber machine.
Check: i) the required minimum clamp stroke Ls = 2D + 12.5 = 16 + 12.5 = 38.5 cm.
ii) the shot size (Eq. 10.22) is: Vs = Vp(1 +2/h), where Vp = part volume = 584/2.74 = 213
cm3 from the given weight and material density. Therefore Vs = 213(1 + 2/2.5) = 384
cm3.
Note that Vs is OK for the 1800 kN machine, but we need to step up to the 6,000 kN
machine for large enough die opening. The 3,500 kN machine is marginally short but
may not accommodate the part extraction device or the lubricating mechanism.
b) Components of the cycle time are:
Eq. 10.25
Ladle time = 0.0048Vs = 1.84 s.
Eq. 10.26 and Table 10.7 Fill time =
, always
negligibly small
For cooling time, we first need to calculate the cavity projected area, Af. A good enough
approximation to this value can be obtained by dividing the part volume by the wall
thickness (provided there are no large thickness variations). This gives:
cm2.
Thus from Eq. 10.32, cooling time,
s.
Eq. 10.33 Extraction time,
s.
Eq. 10.35 Lubrication time,
s.
Eq. 10.34 Reset time,
s.
Thus the cycle time is t = 1.84 + 2.63 + 4.2 + 4.5 + 10.15 = 23.3 s.
Using the machine rate of $94/hr for the 6000 kN machine and 15% downtime gives the
processing cost as:
(23.3/3600)94/0.85 = $0.72
c) From Eq.10.37, the process time for trimming is
and using a trim press rate of $40/hr gives: trim cost = (12.6/3600)40/0.85 = $0.16
d) Using shot and part volumes from above, density of 2.74 g/cc and material cost of
$2.74/kg gives (Eq. 10.23) material cost per part:
Piece-part cost = 0.72 + 0.16 + 1.13 = $2.01
2. For the rotor assembly housing described in Problem 1, estimate the cost of:
(a) A single-cavity die; assuming tolerance level 2 (Table 8.8 Chapter 8) and medium
surface finish (Table 10.9)).
(b) A corresponding trim die. Note that trimming will be required on two levels – the
bottom edge of the part, and the holes at the top. Make appropriate cost allowance for this
step in the trim die construction.
Solution: a) Die set (following section 8.8.2 with 25% increase for die casting)
i) Required plate area: LW = 400 cm2
Clearance = 7.5 + 0.5(400 – 100)/100 = 9.0 cm
Width, including overflow wells = 20 x 1.2 = 12.5 cm
Space for runner system = 12.5 cm
Therefore plate area, Ac = (20 + 18)(24 + 18 + 12.5) = 2071 cm2 and the required total
plate thickness, hp = 13 + 18 = 31 cm. Eq. 10.38 gives equivalent hours for die set as:
hrs.
ii) Part: Complexity: Both the inner and outer surfaces comprise 10 surface patches. In
addition there is one large hole and 12 small ones each counting one cyclindrical surface
patch. The total complexity, X, is thus:
Eq. 8.14 with 25 % increase gives:
hrs.
With Ap = 170.3, Eqs 8.13 and 8.17 give:
(1)
(2)
hrs
(3)
Adding 10% to (1) for increased tolerance level gives + 1.5 hrs, and adding 18% to (1),
(2) and (3) for medium surface finish gives +20.3 hrs and total hours for cavity and core
inserts is:
hrs.
Using $40/hr for die making gives a die cost as C1 = (298 + 136.9)40 = $17,500
Trim die: Profile complexity is (10.39)
Basic manufacturing points are (10.40):
Correction factor for plan area is
Allow 2 hours for each of 13 holes gives (Eq. 10.42) manufacturing hours:
Finally allow 25.5 hrs for stepped configuration of trim die (see addition factors in
section 10.11.3) gives total hours = 83.0 + 25.5 = 108.5 hrs. Therefore ar $40/hr the
estimated trim die cost 108.5 (40) = $4,340
Figure 10.13
3. Figure 10.13 shows a support platform for a precision electrical instrument. The
platform is 100 mm high, and the platform base and top have outer dimensions 75 x 60
mm. The cutout in the top plate has dimensions 45 x 30 mm. The platform is to be
diecast with Zamak 5 Zinc alloy with eight through holes as shown.
Section thicknesses are 5 mm everywhere, with the exception of the stiffening ribs along
the length and both sides of the legs. These are 3 mm thick.
From the solid model of the part, the part volume, Vp, is 42.0 cm3. The projected area for
machine size selection, Ap, equals the top plate area of 31 cm2.
(a) Determine the appropriate hot chamber machine, from Table 10.3, on which to cast
the part using a single cavity die.
(b) Estimate the die casting cycle time and use it to obtain the estimated die-casting
process cost.
(c) Estimate the time and process cost for trimming.
(d) Estimate the material cost.
Solution:
Following identical steps as in Problem 1, but with a hot-chamber machine for Zamak 5
gives:
a)
cm2,
cm2,
kN.
The required clamp stroke, L = 20 + 12.5 = 32.5 cm. The required shot
size,
, but Vp =
42 cm3 as given, and therefore
cm3.
The part requires a 4000 kN machine, which is the smallest one with sufficient opening.
b) Cavity surface area is
cm2. Therefore using β = 0.4 for zinc,
cooling:
s.
Extraction time,
s
Lubrication time with two side pulls is:
s. (divide by 3 because once every three cycles)
Reset time, tr = 1.75(4.6) = 8.1 s.
Therefore cycle time is t = 3.27 + 2.08 + 1.50 + 8.10 = 15.0 s.
Using $70/hr for the 4000 kN machine gives process cost = (15/3600)(70)/0.85 = $0.34.
c)
Trim time, tp = 5.4 + 0.18(6 + 7.5) = 7.8 s.
Therefore trim process cost = (7.8/3600)(40)/0.85 = $0.10.
d)
Using density and material cost for Zamak 5 gives:
Cp = (42 + 0.2(58.8 – 42))(6.6)(1.74)/3600 = $0.52.
4. For the support platform shown in Fig. 10.13, and described in Problem 3, estimate the
likely cost of a single-cavity casting die and a trim die. Die opening will be in the vertical
direction so that the top and bottom surfaces can be flat and parallel as required. The die
will have two opposing side-pull mechanisms to form the T-shaped legs and the opening
between them. Assume a single trim die is to be used, which has mechanisms to allow
edges of the feet to be shaved from the front and rear at the same time as the vertical
trimming action. A total length, P, of 82 cm is to be trimmed as well as the eight standard
through holes.
For geometry such as this one, there is no simple breakdown into inner and outer
surfaces. In this case it is easier to use a breakdown of the main features. The total
complexity, X, has been calculated in this way in the table below.
Name of feature
Surface
Patches
(Ns)
6
12
5
1
Top plate
Leg
Feet
Holes
Repeats
(r)
1
4
4
8
Complexity
X
(0.1Ns*r0.766)
0.60
3.47
1.45
0.49
X =6.0
Solution:
a) Following identical steps to problem 2, but including two side-pulls gives:
Plate area, Ac = (7.5 + 15)(6.0 + 27.5) = 754 cm2
Total plate thickness, h = (2D +15) =35 cm
Total die set hours is,
Hours associated with the projected area,
hrs.
hrs.
Hours for ejector system,
hrs
Additional hours for complexity (X = 6.0) are:
Added hours for two identical opposing side-pull mechanisms:
hrs.
For medium surface finish add
hrs.
For tolerance level 2 add 0.1Mx = 7.1 hrs.
hrs.
Thus the estimated total die cost is:
(153 + 10.3 + 14.0 + 70.9 + 136 + 17.1 + 7.1)40 = $16,000
b) Using P = 82 cm, the complexity of the total profile to be trimmed is:
Therefore basic points for trim die are:
hrs
With 8 holes to be trimmed
Adding two opposing side-action mechanisms adds
hrs
hrs.
Total die making hours are 47.9 + 68.0 = 115.9 hrs and estimated trim die cost is:
115.9(40) = $4,600.
5. Assume production of the Rotor housing (Fig. 10.12) has started using a single cavity
mold and trim tool. The actual cycle times have been measured as 22 seconds for die
casting and 14 seconds for trimming. In addition the actual casting die cost was $16,000
and trim die cost was $5,000. Your company has now been asked to quote for 450,000
housings.
(a) Estimate the likely optimum numbers of cavities to use for this order, using the 85%
learning curve.
(b) Select the appropriate machine from Table 10.4, for use with your selected number of
cavities.
(c) Make a quick estimate of the likely cost of the multi-cavity mold and the multiaperture trim tool.
(d) Is your multi-cavity die capable of making the entire order?
(Use coefficients k1 = 62.0 and m1 = 0.0036 for cold-chamber machine hourly rates (see
Section 10.8) and assume a trim press rate of $40/ hour)
Solution: a) From equation 10.3 the optimum number of cavities will be:
Therefore use 4 cavities in a 2 x 2 arrangement.
b) The separating force on 4 cavities will be (from Problem 1 solution):
f = 4(1431) = 5,724 kN
Therefore can still use the 6,000 kN machine as for a one cavity die.
c) Using an 85% learning curve the 4 cavity dies would cost:
Casting die cost:
Trim die cost:
d) From Table 10.2 the typical cavity life for aluminum casting is 100,000, so we would
anticipate 400, 000 from a 4-cavity casting die. The trim die life is greater,
Chapter 11 Design for Powder Metal Processing
1. The two level part shown in Figure 11.32 is made from iron powder with the
compaction characteristics given. The part is compacted to a density of 7.0 gm/cc.
a) Determine the dimensions of the loose powder fill in the compaction tooling.
b) If the loose powder costs $2.2 per kg. Estimate the material cost for the part, assuming
a powder loss of 1.5%.
Solution: a) The compression ratio of for the loose powder is 7.0/2.5 = 2.8
Therefore the total height of the loose powder in die (fill height) is 20 x 2.8 = 56 mm and
the fill height corresponding to the flange thickness is 10 x 2.8 = 28 mm.
b) The volume of loose powder is:
.
Therefore powder mass is 2.5 x 25.77 = 64.43 gm and the cost, with 1.5% powder loss,
is:
64.43(1.015)2.2/1000 = $0.144.
2. a) Based on the information given in Figure 11.32 select the most appropriate press
from Table 11.3 for the part.
b) Determine the compaction cycle time, setup time and the compaction costs per part
assuming a batch size of 5000 parts. Assume a press operation cost rate and set up rate of
$90/hr.
Solution: a) Assuming the compaction characteristics of the powder follow the form
, then:
600 = A(7.2)b and 220 = A(6.0)b, which give
or
, from which b = 5.50. Therefore,
.
The compaction pressure for a density of 7.0 gm/cm3 is therefore:
N/mm2.
The L/D ratio of the part is 20/30, which is less than 1.0 and therefore no correction of
this pressure for part thickness is necessary. The projected area of the part is:
mm2.
The required compaction force is:
kN. Allowing a 15%
safety margin the minimum press capacity is 388 kN. The required die insert diameter is
3(30 + 20) = 150 mm and the required fill height is 56 mm (from Problem 1). Therefore
from Table 11.3 the 400 kN press will be suitable.
b) The 400 kN press has a maximum stroke rate of 40 / min and a minimum stroke rate of
7 / min. Using the rule for estimating the stroke given in Section 11.10.2, the stroke rate
for the part will be 40(1 – 0.1 – 0.05 x 2 – 0.5 x 337/400) = 15 / min. The cycle time is
60/15 = 3.96 s. The corresponding cost is 90 x 3.96/3600 = $0.099.
From the rule for setup time in Section 11.10.2, the setup time is 2hr. For a batch
size of 5000 parts the setup cost is 90 x 2/5000 = $0.036. Therefore the total compaction
cost per part is 0.099 + 0.036 = $ 0.135 or 13.5 cents per part.
Figure 11.32
3. For the part shown in Figure 11.32 estimate the initial cost of the compaction tools
(die, punches and core rods). Assume that tool steel costs $20 per kg and carbide for the
die insert costs $5/cc.
Solution: This is a two level part and therefore requires one upper punch and two lower
punches. A single core rod passes through the other punches. The following data is
needed for the estimation of the costs of the tools for compaction.
Fill height, hf = 56 mm.
Enclosing diameter of whole part, D0 = 30mm.
Enclosing diameter of lower level 1, Dl1 = 30 mm.
Enclosing diameter of lower level 2, Dl2 = 20 mm
a) Tool Material Costs
From these data the following determine the die material costs:
Die thickness, T = hf +17.8 = 73.8 mm.
Carbide insert diameter, Dc = D0 + 20 = 50 mm.
Die case diameter for selected press (problem 2) = 20.32 cm.
Cost of carbide insert material =
Cost of die case material =
.
The following determine the punch material costs:
Length of upper punch,
Length of lower punch 1,
+88.9 = 56 + 88.9 = 144.9 mm
Length of lower punch 2, L2 = hf + 119.4 = 56 + 119.4 = 175.4 mm
Stock diameter for upper punch, 30 + 38.1 = 68.1 mm
Stock diameter for lower punch 1, 30 + 38.1 = 68.1 mm
Stock diameter for lower punch 2, 20 + 38.1 = 58.1 mm
mm
Material costs for these punches are as follows:
Upper punch cost, π68.12 x 96.6 x 7.86 x 20/(4 x 106) = $55.32
Lower punch 1 cost, π68.12 x 144.9 x 7.86 x 20/(4 x 106) = $82.98
Lower punch 2 cost, π58.12 x 175.4 x 7.86 x 20/(4 x 106) = $73.11
The total material costs for the die and punches are:
724.62 + 353.5 + 55.32 + 82.98 + 73.11 = $1289.53
b) Die Manufacturing Costs
The die profile in this case is circular with a diameter of 30 mm. This means that the die
profile complexity factor is unity. The die machining time is given by:
hr
The die finishing time is given by:
hr
Therefore, assuming a die shop manufacturing rate of $45/hr the die manufacturing cost
is:
45(6.91 + 11.78) = $841.05
c) Punch profiles
The machining and finishing times for punch profiles are given by Equations 11.18 and
11.19.
Machining time (h) = 1.25 + 0.1∑Pr/25.4 + 0.89(0.6Li + 9.53) × (π /4 –
At)/25.43 + 0.18Li∑Pr/25.42.
Finishing time (h) = 0.4Li∑(FcPr)/25.42
Upper punch:
Pr = π30, Li = 96.6 mm, Dpr = 30 + 38.1 = 68.1, At = π302/4 = 706.95, Fc = 1
Substituting these values gives: Machining time = 14.92 h and finishing time = 5.65 h.
Lower punch 1:
Pr = π30, Li = 144.9 mm, Dpr = 30 + 38.1 = 68.1, At = π302/4 = 706.95, Fc = 1
Substituting these values gives: Machining time = 20.81 h and finishing time = 8.47 h.
Lower punch 2:
Pr = π20, Li = 175.4 mm, Dpr = 20 + 38.1 = 58.1, At = π202/4 = 314.2, Fc = 1
Substituting these values gives: Machining time = 19.14 h and finishing time = 6.83 h.
Holes through punches:
The machining and finishing times for circular holes through punches are given by
Equations 11.19 and 11.20.
Machining time
Finishing time (h) = PtLi/25.42.
, plus 15 min setup time per hole
Upper punch:
Dh= 8, Li =96.6 and Pr = π8
Substituting these values gives: Machining time = 0.31 h and finishing time = 3.76 h.
Lower punch 1:
Dh= 20, Li =144.9 and Pr = π20
Substituting these values gives: Machining time = 0.79 h and finishing time = 14.11 h.
Lower punch 2:
Dh= 8, Li =175.4 and Pr = π8
Substituting these values gives: Machining time = 0.35 h and finishing time = 6.83 h.
Punch manufacturing costs:
Upper punch:
Total hours = 14.92 + 5.65 + 0.31 + 3.76 = 24.64, for a cost of 45 x 24.64 = $1108.4.
Lower punch 1:
Total hours = 20.81 + 8.47 + 0.79 + 14.11 = 44.18, for a cost of 45 x 44.18 = $1988.1.
Lower punch 2:
Total hours = 19.14 + 6.83 + 0.35 + 6.83 = 33.15, for a cost of 45 x 33.15 = $1491.75.
Core rod: A single core rod 8 mm diameter is need and the equivalent hours for this core
rod is given by Equation 11.23:
which gives an equivalent hours of 4.08, for a cost of 45 x 4.08 = $183.6.
Total Costs: The total cost of each tool element is the sum of the material and
manufacturing costs.
Die:
Upper punch:
Lower punch 1:
Lower punch 2:
Core rod:
Total tool cost:
$726.62 + $353.5 + $841.05 = $1921.17
$55.32 + $1108.4
= $1163.72
$82.98 + $1988.1
= $2071.08
$73.11 + $1491.75
= $1564.86
= $183.6
$6904.43
4. The part shown in Figure 11.32 is sintered using a mesh belt continuous flow furnace
with the following characteristics:
Furnace length (m)
High heat zone length (m)
Belt width (cm)
Throat height (cm)
Load capacity (kg/m2)
Operating cost ($/hr)
9.14
1.83
30.48
10.16
73.24
120
Assuming that the parts must be separated by at least 2mm, estimate the sintering cost per
part for a batch size of 5000.
Solution: The part material has a sintering time from Table 11.11 of 25 minutes at a
temperature of 1121 0C. The speed of the furnace belt, vf = 1.83/25 = 0.0732 m/min. The
number of parts across the width of the belt is 30.48/(3.0 +0.2) = 9.525 or 9 parts.
Therefore the length of the batch of parts on the belt is BL = 5000 x 3.2/9 = 1777.8 cm.
(Note: this assumes a slightly conservative pattern of rows of 9 parts with 2 mm between
rows) The time for the batch to pass through the furnace is (1777.8/100 + 9.14)/0.0732 =
367.73 min., giving a time per part of 367.73/5000 = 0.074 min. This gives a cost per part
of 0.074 x 120/60 = $0.148.
Note: Normally the load on the belt should be checked because the parts may need to
spread out if the loading is too high. The weight of each part is 64.43 gm from Problem 1.
The area taken up by the batch is 30.48 * 1777.8 cm2. The loading on the belt is 5000 x
64.43 x 10/(30.48 x 1777.8) = 59.45 kg/m2. This is less than the maximum loading of
73.24 kg/m2 for the furnace. Thus the parts to not need to spaced out further than the
2mm given above.
5. Feedstock for powder injection molded parts is made up from 8% Nickel-Iron alloy
powder and a Polyethylene/Carnauba binder. Using the data in Tables 11.16 and 11.19
determine the following:
a) The volumetric powder loading of the feedstock.
b) The feedstock density.
c) The mass based powder loading of the feedstock.
d) The thermal conductivity of the feedstock.
e) The specific heat of the feedstock.
f) The thermal diffusivity of the feedstock.
Solution: a) From Table 11.16 the critical solids loading for the material is 0.63 and
therefore the volumetric powder loading is 0.96 x 0.63 = 0.6048.
b) The powder material density (Table 11.16) is 7.899 g/cc and the binder density (Table
11.19) is 0.94 g/cc. The feedstock density (Equation 11.45) is 0.6048 x 7.899 + (1 –
0.6048)0.94 = 5.15 g/cc.
c) From equation 11.48, the mass based powder loading is 0.6048(7.899/5.15) = 0.928
d) The specific heat of the powder material (Table 11.16) is 450 J/kg/0K and that of the
binder 1400 J/kg/0K. From equation 11.48 the feedstock specific heat is 0.928 x 450 + (1
– 0.928)1400 = 518.5 J/kg/0K.
e) The thermal conductivity of the powder material (Table 11.16) is 76 W/m/0K and that
of the binder (Table 11.19) 0.19 W/m/0K. The binder thermal conductivity is given by
Equation 11.50:
, where
, kp is the thermal conductivity of the powder material
and kb is the thermal conductivity of the binder. The volumetric loading
from above is
0.6048. Substituting these values gives the thermal conductivity of the feedstock as 0.82
W/m/0K.
f) The feedstock thermal diffusivity is given by Equation 11.47 and is (0.82 x 103)/( 518.5
x 5.15) = 0.307 mm2/s.
6. Repeat Problem 5 for a feedstock made from Tungsten-7% Nickel-3% Iron alloy
powder and a mixed wax binder.
Solution: a) From Table 11.16 the critical solids loading for the material is 0.63 and
therefore the volumetric powder loading is 0.96 x 0.62 = 0.5952.
b) The powder material density (Table 11.16) is 16.884 g/cc and the binder density
(Table 11.19) is 0.927 g/cc. The feedstock density (Equation 11.45) is 0.5952 x 16.884 +
(1 – 0.5952)0.927 = 10.42 g/cc.
c) From equation 11.48, the mass based powder loading is 0.5952(16.884/10.42) = 0.964
d) The specific heat of the powder material (Table 11.16) is 480 J/kg/0K and that of the
binder 300 J/kg/0K. From equation 11.48 the feedstock specific heat is 0.964 x 480 + (1 –
0.964)300 = 473.52 J/kg/0K.
e) The thermal conductivity of the powder material (Table 11.16) is 150 W/m/0K and that
of the binder (Table 11.19) 0.12 W/m/0K. The binder thermal conductivity is given by
Equation 11.50:
, where
, kp is the thermal conductivity of the powder material
and kb is the thermal conductivity of the binder. The volumetric loading
from above is
0.5952. Substituting these values gives the thermal conductivity of the feedstock as 0.508
W/m/0K.
f) The feedstock thermal diffusivity is given by Equation 11.47 and is (0.508 x 103)/(
473.52 x 10.42) = 0.103 mm2/s.
7. A part with a finished volume of 10 cm3 is produced by powder injection molding in a
single cavity mold, from the feedstock material in Problem 5. The final sintered part has a
density which is 0.99 of the true density of the metal alloy. Using the data in Tables 11.16
and 11.19 determine the following:
a) The cost per kg of the feedstock.
b) The material cost per part, assuming that the reject rates for molding, debinding and
sintering are 1% and the yield from regrinding the waste feedstock material after molding
is 95%.
Solution: a) The feedstock cost is given by equation 11.51. The unit cost of the powder
(Table 11.16) is $14/kg and that of the binder (Table 11.19) is $2.2/kg. From problem 5
the mass based solid loading is 0.928. The feedstock unit cost is:
0.928 x 14 + (1 - .928)2.2 = $13.15/kg.
b) The cavity volume is given by Equation 11.53 and the volumetric loading from
problem 5 is 0.6048. This gives 10 (0.99/0.6048) = 16.37 cm3. The feedstock density
from problem 5 is 5.15 g/cc and the mass of material in the cavity is 16.37 x 5.15 = 84.28
g. The volume of material in the runners is given by Equation 11.55 and gives 2.1 x
16.37.52 = 8.98 cm3, with a corresponding mass of 8.98 x 5.15 = 46.26 g.
The material cost per part is given by Equation 11.57:
,
which gives:
13.15(84.28/(0.99 x 0.99) + (84.28(1 – 0.99) + 46.26)(1 – 0.95))/1000 = $1.16.
8. Determine the molding cycle time for the part in Problem 7. The part has a maximum
wall thickness of 5mm and depth of 4 cm. The complexity factor for the part is 2.0 and a
packing pressure of 7MPa is used in the molding process. The molding machine used has
a dry cycle time of 2 s and the ejection stroke is 15 cm.
Solution:
i) Fill time. For the part the shot volume, Q, is the sum of the cavity volume, Vc, and the
runner volume, Vr, and is therefore equal from Problem 7 to 16.37+8.98 = 25.35 cm3.
From Equation (11.59) the injection time, ti, is given by Q/15 which equals 25.35/15 or
1.69 s.
ii) Packing time. The complexity factor, Xc, is 2.0 and the packing pressure is 7MPa.
Equation (11.60) gives the packing time, tp.
iii) Cooling time. The maximum wall thickness of the part is 5 mm and in the cavity this
becomes larger due to the shrinkage during sintering. The volumetric expansion factor,
Kv, is given by Equation (11.57):
and therefore the maximum
cavity thickness, hcm, is
= 5.89 mm. From Table 11.19, the molding injection
temperature, Ti, is 140 0C; the mold temperature, Tm, is 35 0C; and the ejection
temperature, Tx, is 80 0C. The thermal diffusivity of the feedstock was determined from
Problem 5 to be 0.307 mm2/s. The cooling time for the molding is given by Equation
(11.61):
.
iv) Reset time. The part depth is 4 cm and from this the cavity depth, Dc, will be
cm. Assuming that the dry cycle time of the molding machine is 2.0 s
and the ejection stroke is 15 cm, then the mold resetting time, tr, is given by Equation (11.
62):
s.
v) Cycle time. The molding cycle time is therefore given by 1.69 + 22.23 + 15.45 + 6.43
= 45.8 s. The molding cost is then determined by multiplying the cycle time by the
operation cost rate of the molding machine.
Chapter 12 Design for Sand Casting
For the problems below use casting metal data from Tables 12.1 and 12.2, and the
production data provided with the equation variable definitions in Section 12.7.
1. Figure 12.8 shows a large Idler Arm, which is part of the mechanism on a construction
machine, and is to be sand cast using malleable iron. A total production volume of 2000
castings are anticipated. The three large holes will be cast into the part; the axial bore
requiring a cylindrical core. Secondary machining operations will be carried out to finish
machine the three holes including the key slot, mill the faces of all the bosses and drill the
small cross bore. Pattern and core dimensions will be adjusted to leave material for the
finish machining but this will not affect the cost estimating for the casting.
The basic geometry of the part and the core are given below.
Part:
Core (for axial bore):
Finished part volume Vfc = 16,100 cm3
Part projected area Ap = 1,188 cm2
Part length L = 80 cm
Part width W = 30 cm
Part depth D = 22 cm
No. of surface patches = 42*
(*allowing for repeated features)
Core volume Vc = 1,861 cm3
Core projected area Apc = 258 cm 2
Core length Lc = 28 cm
Core width Wc = 9.2 cm
Core depth Dc = 9.2 cm
No. of surface patches = 3
Obtain estimates for the following:
(a) The cost of the malleable iron in the casting.
(b) The mold and sand core costs per part.
(c) The cores and casting manufacturing costs
(d) The casting cleaning cost
(e) The piece part cost
Figure 12.8
Solution: a) Part weight = 16,100(7280)/106 = 117.2 kg. From Eq. 12.1:
Poured weight =
kg
Using malleable iron cost at the spout of $0.44/ kg (Table 12.2) and a scrap value of
$0.07/ kg from Table 12.1, gives, with an assumed 2% scrap rate,:
Iron cost per casting = (117.2 x 0.44 – 0.07(189 – 117.2))/(1 – 0.02) = $79.83
b) Equation 12.6: Mold sand cost,
Equation 12.7: Core sand cost, assuming 8% scrap rate, is,
c) Core manufacturing cost
From Equation 12.12 with production data provided below the equation:
Mold and casting manufacturing cost
From Equation 12.14 the burdened automated line worker rate, to provide a production
rate of 13.6 molds / worker/ hour is:
hour
From Equation 12.13 the cost per molded part is:
d) Cleaning cost from Equations 12.15 and 12.16 is:
e) The sum of the costs estimated in (a) through (d) gives the piece part cost:
(79.83 + 3.41 + 0.24 + 3.11 + 17.12 + 9.25) = $113
2. For the Idler Arm described in Problem 1, determine,
(a) The likely cost of the pattern
(b) The likely cost of the core box required to make the axial core
Solution:
a) Pattern cost:
Plate area, Apl = (80 + 10)(30 + 10) = 3600 cm2
Plate thickness, hp = 7.5 cm
Cost of mounting plates (Equation 12.8) is:
Using geometry information in Problem 1, complexity X = 4.2 and using $40/hr for
pattern making gives (from Equation 12.9) the cost for stainless steel impression inserts:
Thus the estimated cost of patterns is (Equation 12.10):
Cpt = 1.25(2,088 + 18,287) = $25,000
b) Core box cost: Using now the core geometry data in Problem 12.1,
Plate area, Apl = (28 + 10)(9.2 + 10) = 730 cm2
With a core depth of 9.3 cm, the required thickness of plates in the core box is:
hp = (9.3 + 7.5) = 16.8 cm
From Equation 12.9, the cost of finished plates for the cavity inserts is:
Cpm = (0.1 + 0.064(16.8))730 = $857
With geometrical complexity of 0.3, the estimated cost of the cavity inserts is (Equation
12.10):
Thus the estimated cost of the core box is (Equation 12.11):
3. Figure 12.9 illustrates a Machine Bracket which is to be sand cast in stainless steel.
The part is to be manufacture at a mid-size automated foundry and the part can be
manufactured in pairs in their standard mold size. The layout to be used is shown in Fig.
12.10. The two cores required to make the transverse bores in each casting are 1.8 cm
diameter, with lengths 7.4 cm and 10 cm respectively. Note that the cores project beyond
the casting to sit in core prints in the mold impression.
The basic geometry of the part is given below.
Finished part volume Vfc = 55.6 cm3
Part projected area Ap = 54.8 cm2
Part length L = 19.0 cm
Part width W = 6.0 cm
Part depth D = 3.0 cm
No. of surface patches = 48*
(*allowing for repeated features)
(a) Estimate the likely cost of the pattern for the two cavity mold (use an 85% learning
curve for the repeated pattern impression, and use your estimated length and width of the
envelope surrounding the pair of impressions (Fig. 12.10) to estimate the cost of the
plates).
(b) The likely cost of the two core boxes.
Figure 12.9
Figure 12.10
Solution: a) Pattern cost
Plate area, Apl = (19 + 10)(2 x 6 + 15) = 783 cm2
Plate thickness, hp = 7.5 cm
Cost of mounting plates (Equation 12.8) is:
Cpm = (0.1 + 0.064(7.5))783 = $454
Using geometric information in Problem 1, complexity X = 4.8 and using $40/ hour for
pattern making gives (from Equation 12.9), the cost for stainless steel impression inserts:
Thus the estimated cost of the patterns is (Equation 12.10):
b) Core box cost # 1
L = 10, D = 1.8, Ap = (10 x 1.8) = 18, nsp = 3
Plate area, Apl = (10 + 10)(1.8 + 10) = 236 cm2
With a core depth of 1.8 cm, the required thickness of plates in the core box is:
hp = (1.8 + 7.5) = 9.3 cm
From Equation 12.9 the cost of finished plates for the cavity inserts is:
With geometric complexity of 0.3, the estimated cost of the cavity inserts is (Equation
12.10):
Thus the estimated cost of the core box is (Equation 12.11):
Core box cost #2 Using exactly the same equations with L changed to 7.4 cm gives:
cm2
Cbox2 = 1.25(143 + 329) =$589
Therefore the total cost of core boxes = 655 + 589 = $1250
4. For the Machine Bracket described in Problem 3, determine,
(a) The cost of the stainless steel in the casting.
(b) The mold and sand core costs per part (allow for the two pattern impressions per plate
in appropriate equations).
(c) The cores and casting manufacturing costs
(d) The casting cleaning cost
(e) The piece part cost
Solution: a) Part weight
kg and, from Equation 12.1, :
Poured weight =
kg.
Using stainless steel cost at the spout of $2.62 /kg (Table 12.2) and scrap value of
$0.40/kg from Table 12.1 gives, with an assumed 2% scrap rate, Iron cost/casting
= (2.62 x 0.62 – 0.40(0.62 – 0.43))/(1-0.02) = $1.52
b) Equation 12.6, mold sand cost,
Equation 12.7, core sand cost, assuming 8% scrap rate, is:
c) Core manufacturing cost: From Equation 12.12 with production data provided below
the equation,
Mold and casting manufacturing cost: From Equation 12.14 the burdened automated line
worker rate, to provide a production rate of 13.6 molds/worker/hour is, using the length
and width around the twp cavities:
From Equation 12.13 the cost per molded part is, allowing 2 castings/mold,
d) Cleaning cost from Equations 12.15 and 12.16 is:
e) Piece part cost is (1.57 + 0.02 + 0.07 + 3.94 + 0.74) = $6.35
Chapter 13 Design for Investment Casting
1. Using the example in this Chapter, estimate the pattern cost (including tooling cost per
part) if the part were of the same material and geometric shape but were twice the size in
terms of the linear dimensions (i.e. eight times the volume). Assume use of the same
injection molding machine for the wax pattern and one cavity per mold.
Solution:
Cost of wax
From Table 13.1, the No. 2 wax material has a density of 0.97 gm/cm3 and a cost per unit
weight of 2.87 $/kg. The part volume is 26 608 cm3 and, from Table 13.5 the volume
shrinkage allowance for phosphor bronze is 4%.
From Eqn.13.1, the cost of wax for one pattern is:
Cpm = 0.97 x 2.87 x 26.608 (1 + 0.04)/1000 = $0.077
Process cost per pattern
The number of cavities per mold is 1, and the projected area of the part is 35.52 cm2. The
proportion of runner volume can be estimated at 31% by interpolation from Table 8.2.
From Eqn. 13.2, the total projected shot area is:
As = 1 x 35.52 (1 + 0.31) = 46.53 cm2
For the required shot size we will assume 1.5 times the part volume, i.e. approximately
40 cm3. From Table 13.6, the maximum wax injection flow rate is 82 cm3/s.
From Eqn. 13.3, the fill time is: tf = 40/82 = 0.488 s
The maximum wall thickness of the part is 3mm. From Table 13.1, the thermal
diffusivity of the wax is 0.092 mm2/s, the injection temperature is 67oC, the
recommended mold temperature is 25oC and the ejection temperature is 50oC.
From Eqn. 13.4, the cooling time is:
s
We assume 2 for the pattern clearance factor. The pattern depth is 4.4 cm, the hand
clearance is 10 cm and Table 13.6 gives the press closing and opening speeds as 2.54
cm/s.
From Eqn. 13.7, the machine open and close time is:
toc = (2 x 4.4 + 10)(1/2.54 + 1/2.54) +1 = 20.59 s
The number of cores per pattern is zero and the number of patterns per mold is 1.
From Eqn. 13.8, the total reset time is:
tr = 20.59 + 0.4 + (3 x 0 + 2) x 1 = 22.99 s
Now, from Eqn. 13.11, the total cycle time is:
tt = 0.488 + 7.534 + 22.99 = 31.01 s
For a machine and operator rate of 30 $/hr, Eqn. 13.12 gives:
Cip = 30 x 31.01/(3600 x 1) = $0.258
Pattern cost
The minimum clearance around the cavity in semi-automatic injection is 7.5 cm and the
number of clearances is normally 2.
From Eqn. 13.16, the combined thickness of core and cavity plates is:
hp = 4.4 + 2 x 7.5 = 19.4 cm
The projected area of the mold base is:
Ac = (7.0 + 15.0)(5.4 + 15.0) = 448.8 cm2
Eqn. 13.15 gives the cost of ejector, riser and stripper plates:
Cfp = 0.66 Ac + 366 = $662.2
From Eqn. 13.14, the cost of the plates containing the cavity is:
Cvp = 0.0125 x 19.4 x 448.8 + 0.428 x 448.8 x 1 + 14.27 x 19.4 + 32.18 x 1 = $688.3
The mold base cost is now given by Eqn. 13.13:
Cb = 688.3 + 662.2 = $1,350.5
The cost with custom work is given by Eqn.13.18:
Cab = 1.1 x 1,350.5 = $1,485.6
The manufacturing hours to generate the mold shape which has 30 surface patches is
given by Eqn. 13.19:
Mx = 0.1 x 30 = 3 hr
The total projected area of the part in the mold is 35.52 cm2. From Eqn. 13.20, the time
required to remove the material to form the mold depends on this area:
hr
One side-pull takes about 16 hr to manufacture. For this part two side-pulls would be
required for the holes in the sides of the part, requiring 32 hr.
The time to make ejection pins and fit them into the mold is given by Eqn. 13.21:
hr
Using the same tolerance and surface finish as the example in the text, we get:
For tolerance: 0.1(3 + 6.02) = 0.902 hr
For surface finish: 0.15(3 + 6.02) = 1.35 hr
The total manufacturing time is:
Mtot = 3 + 6.02 + 32 + 2.98 + 0.902 + 1.35 = 46.25 hr
For a mold-making rate of 40 $/hr, Eqn. 13.23 gives a mold-making cost of:
Cdm = 40 x 46.25 = $1,850
Adding the cost of two side-pulls at $30 each and the cost of the mold base gives, from
Eqn. 13.24, the total mold cost:
Cd = 1,850 + 60 + 1,485.6 = $3,396
Finally, for a scrap rate of 5%, a machine set up time of 15 min., a batch size of 1,000,
and a product volume of 10,000, Eqn. 13.32 gives a total cost per pattern piece of:
Ctp = (0.258 + 0.077)/(1 – 0.05) + 30 x 0.25/1,000 + 3396/10,000 = $0.7
2. Using the example in this Chapter, estimate the cost of metal per part if the part were
of the same material and geometric shape, but were twice the size in terms of the linear
dimensions (i.e. eight times the volume).
Solution:
From the example in the text, the raw material cost of phosphor bronze is 1.764 $/kg and
the cost of ready-to-pour liquid metal increases this cost to 1.879 $/kg. The cost of
pouring the metal is 0.072 $/kg.
From Eqn. 13.29, the casting yield is:
The final metal cost per kg is given by Eqn. 13.42. If the value of scrap metal is 0.7 $/kg
and the metal loss is 0.02, we get:
Cmat = [1.879 + 0.072 – 0.7(1 – 0.02 – 0.3856)]/0.3856 = 3.98 $/kg
This gives a metal cost per part of 3.98 x 0.2376 = $0.946
3. Using the example in this Chapter, estimate the cost of the shell mold per part (not
including the pattern cost) if the part were of the same material and geometric shape but
were twice the size in terms of the linear dimensions (i.e. eight times the volume).
Solution:
To determine the number of parts per cluster, we need to know the casting yield Yd (Eqn.
13.29) and the shell mold yield Ysm(Eqn. 13.30):
From Eqn. 13.31, the number of parts per cluster is:
npc = 16(0.2376/0.3856 + 0.2376/0.6303 = 16
From Eqn. 13.28, the time to assemble the cluster is:
tca = 10(16 + 3) = 190 s
Using 23$/hr for the assembly worker rate and 7 min for set up, the total cluster assembly
cost from Eqn. 13.27 is:
Ctca = 23 x 190/3600 + 1.5 + 23 x 7/60 = $5.397
From Eqn.13.33, the dry weight of the shell mold is:
Wsm = 16 x 0.2376/0.6303 = 6.031 kg
And for a cost of $1 per kg the cost of the dry mold material = $6.031
Eqn. 13.34 gives the cost to apply primer coats to the cluster. With a rate for the machine
and operator of 31.2 $/hr, a time for each primer coat of 20 s, a time for subsequent
primer coats of 15 s, the total cost for 3 primer coats is:
Cpr = 31.2 x (20 + (3 – 1) x 15)/3600 = $0.433
Eqn. 13.35 gives the cost to apply back-up coats robotically to the cluster. With a rate for
the machine and operator of 35.15 $/hr, a time for each back-up coat of 10 s, The total
cost for 5 back-up coats is:
Cbu = 5 x 35.15 x 10/3600 = $0.49
The totals for the shell mold are:
Assemble cluster
5.397
Clean and etch
0.06
Apply 3 primer coats
0.433
Apply 5 back-up coats
0.49
Pattern melt-out
0.15
Burnout, sinter, preheat 0.61
Shell mold material
6.031
Giving a total of $13.17 and resulting in a cost per part for the shell mold of:
13.17/16 = $0.823
4. Using the example in this Chapter, estimate the cost of breakout, cleaning and cut-off
if the part were of the same material and geometric shape, but were twice the size in
terms of the linear dimensions (i.e. eight times the volume).
Solution:
Breakout
To determine the number of parts per cluster, we need to know the casting yield Yd (Eqn.
13.29) and the shell mold yield Ysm(Eqn. 13.30):
From Eqn. 13.31, the number of parts per cluster is:
npc = 16(0.2376/0.3856 + 0.2376/0.6303 = 16
From Eqn. 13.47, the cost of breakout per cluster is:
Cbo = 0.09 + 2.08 x 16/1000 = $0.1233
Cleaning
From the text the cost of cleaning per cluster is Ccl = $0.07
Cutoff
For a part weight of 0.2376 kg, Eqn. 13.54 gives the number of supplemental cuts:
and Eqn. 13.53 gives the time per cluster for cutoff, assuming a thickness for the one gate
of 1 cm, tco = 4 + (5 x 12 + 2)(1 x 16 + 3 x 6) = 242 s, with an operator rate of 25 $/hr,
the operator cost per cluster is: Ccf = 242 x 25/3600 = $1.681
And, with a setup time of 5 min. the setup cost per cluster is: Cfs = (5/60) x 25 = $2.08
The total cutoff cost per cluster is now: 1.618 + 2.08 x 16/1000 = 1.7133
Finally, adding the costs for breakout, cleaning and cutoff and dividing by the number of
parts on the cluster gives a cost per part of: (0.1233 + 0.07 + 1.7133)/16 = $0.1204
5. For the example in this Chapter, what would be the savings if the two side holes were
eliminated? Would you consider that it would be less expensive to produce the holes by
machining after casting?
Solution:
The mold shape will now have 28 surface patches saving 0.2 hr in mold manufacturing.
Eliminating the installation of two side pulls will save 32 hr in mold manufacturing time.
Thus, the total mold manufacturing time will be reduced by 32.2 hr and with a mold
making rate of 40 $/hr this will mean a savings of $562.
In addition, a further $60 will be saved due to the cost of the two side pulls resulting in a
total savings of $622 or 622/10000 = $0.0622 per part.
The cost of drilling the holes can be estimated using the data in Chap.7.
The diameter of the holes is 12 mm and their depth is 0.25 mm.
From Table 7.3 and with a part weight of 0.2-4.5 kg, the loading and unloading time for a
horizontal fixture would be 33.1 s per hole.
From Table 7.4, the time to engage the tool in a drilling machine is 9 s and the basic
setup time for the batch of 1000 is 1.0 hr giving a cost per hole of $3.6.
The drilling of the holes would take a minimum of about 2 s.
Totaling these results and multiplying by 2 gives 90 s.
For a machine and operator rate of 30 $/hr we get a cost for the two holes of:
95 x 30/3600 = $0.79 which is much greater than the savings in tooling cost for the
investment casting if the-side pulls were eliminated.
Chapter 14 Design for Hot Forging
The sample forgings (A, B and C) shown in Figures 14.30 -32 are used for the following
problems.
Part Data
Material
Medium carbon steel
Material cost ($/kg)
1.00
Material density, ρ (g/cm3)
7.83
Length, L (mm)
254
Width, W (mm)
51
Thickness, T (mm)
23
Part volume, V (cm3)
43
Projected area, Ap (cm2)
48
Outer perimeter, Pr (mm)
594
Number of surface patches
30
Figure 14.30 Sample Forging A
Part Data
Material
304 Austenitic stainless stl.
Material cost ($/kg)
4.0
Material density, ρ (g/cm3)
7.97
Length, L (mm)
83
Width, W (mm)
83
Thickness, T (mm)
20
Part volume, V (cm3)
28
Projected area, Ap (cm2)
47
Outer perimeter, Pr (mm)
320
Area of through hole, AH, (cm3)
16
Perimeter of through hole, Pw (mm)
145
Number of surface patches
30
Figure 14.31 Sample Forging B
Part Data
Material
Medium carbon steel
Material cost ($/kg)
1.00
Material density, ρ (g/cm3)
7.83
Length, L (mm)
76
Width, W (mm)
50
Thickness, T (mm)
50
Part volume, V (cm3)
65
Projected area, Ap (cm2)
34
Outer perimeter, Pr (mm)
240
Area of through hole, AH, (cm3)
5
Perimeter of through hole, Pw (mm)
80
Number of surface patches
50
Figure 14.32 Sample Forging C
1. For sample forging A shown in Figure 14.30 determine the following:
a) The flash thickness
b) The flash land width
c) The projected area of the flash land
d) The volume of the flash for this forging
e) The gross weight of the forging
f) The material cost for this forging
Solution: a) The flash thickness is given by Equation 14.1:
and V = 43 cm3. Thus the flash
thickness for this forging is 1.13 + 0.0789 x 430.5 – 0.000134 x 43 = 1.64 mm.
b) The ratio of the flash land width to the flash thickness is given by Equation 14.2:
and this ratio is 3 + 1.2e-0.00857 x 43 = 3.83.
The flash land width is therefore 3.83 x 1.64 = 6.29 mm
c) The projected area of the flash land is given by the land width multiplied by the length
of the flash line (outer perimeter of the part) or 6.29 x 594/100 = 37.35 cm2.
d) The volume of the flash produced per unit length of flash line is given by Equation
14.3:
or 0.1234 x 430.5 = 0.81 cm3/cm. The total flash volume
is therefore 0.81*594/10 = 48.07 cm3.
e) From Equation 14.5 the gross weight of the forging is given by:
gross weight =
In this case AH = 0. Assuming that the scale loss is 5% then the gross weight becomes:
7.83[(43 + 48.07)1.05]/1000 = 0.75 kg.
f) The material cost for this forging is the gross weight multiplied by the unit material
cost and this is 0.75 x 1.00 = $0.75
2. For sample forging B shown in Figure 14.31 determine the following:
a) The flash thickness
b) The flash land width
c) The projected area of the flash land
d) The volume of the flash for this forging
e) The web thickness needed for the through hole in this part
f) The gross weight of the forging
g) The material cost for this forging
Solution: a) The flash thickness is given by Equation 14.1:
and V = 28 cm3. Thus the flash
thickness for this forging is 1.13 + 0.0789 x 280.5 – 0.000134 x 28 = 1.54 mm.
b) The ratio of the flash land width to the flash thickness is given by Equation 14.2:
and this ratio is 3 + 1.2e-0.00857 x 28 = 3.94.
The flash land width is therefore 3.94 x 1.54 = 6.09 mm
c) The projected area of the flash land is given by the land width multiplied by the length
of the flash line (outer perimeter of the part) or 6.09 x 320/100 = 19.48 cm2.
d) The volume of the flash produced per unit length of flash line is given by Equation
14.3:
or 0.1234 x 280.5 = 0.65 cm3/cm. The total flash volume
is therefore 0.65 x 320/10 = 20.9 cm3.
e) The web thickness for the hole is given by Equation 14.4:
and this gives 3.54 x 160.227 = 6.64 mm
f) From Equation 14.5 the gross weight of the forging is given by:
gross weight =
Assuming that the scale loss is 5%, then the gross weight becomes:
7.97[(28 + 20.9 + 16 x 6.64/10)1.05]/1000 = 0.49 kg.
g) The material cost for this forging is the gross weight multiplied by the unit material
cost and this is 0.49 x 4.00 = $1.95.
3. For sample forging C shown in Figure 14.32 determine the following:
a) The flash thickness
b) The flash land width
c) The projected area of the flash land
d) The volume of the flash for this forging
e) The web thickness needed for the through hole in this part
f) The gross weight of the forging
g) The material cost for this forging
Solution: a) The flash thickness is given by Equation 14.1:
and V = 65 cm3. Thus the flash
thickness for this forging is 1.13 + 0.0789 x 650.5 – 0.000134 x 65 = 1.76 mm.
b) The ratio of the flash land width to the flash thickness is given by Equation 14.2:
and this ratio is 3 + 1.2e-0.00857 x 65 = 3.69.
The flash land width is therefore 3.69 x 1.76 = 6.48 mm
c) The projected area of the flash land is given by the land width multiplied by the length
of the flash line (outer perimeter of the part) or 6.48 x 240/100 = 15.55 cm2.
d) The volume of the flash produced per unit length of flash line is given by Equation
14.3:
or 0.1234 x 650.5 = 0.99 cm3/cm. The total flash volume
is therefore 0.99 x 240/10 = 23.88 cm3.
e) The web thickness for the hole is given by Equation 14.4:
and this gives 3.54 x 50.227 = 5.1 mm
f) From Equation 14.5 the gross weight of the forging is given by:
gross weight =
Assuming that the scale loss is 5%, then the gross weight becomes:
7.83[(65 + 23.88 + 5 x 5.1/10)1.05]/1000 = 0.0.75 kg.
g) The material cost for this forging is the gross weight multiplied by the unit material
cost and this is 0.75 x 1.00 = $0.75.
4. For sample forging A shown in Figure 14.30 determine the following:
a) The classification of the forging
b) The complexity factor, Ffc, of the forging
c) The sequence of forging operations for this part
d) The equivalent energy capacity, Ef, of the forging equipment needed for this forging
e) An appropriate size of forging hammer to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
f) An appropriate size of mechanical press to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
Solution:
a) For this forging L/W = 254/51 = 4.98 and L/T = 254/23 = 11.04. The first digit of the
classification is 2 and the second digit is 0.
b) The complexity is given by LWT/V = 254 x 51 x 23/ (1000 x 43) = 6.93
c) The number of operations required is 5 – fuller 1, fuller 2, edger, blocker and finisher.
d) From Table 14.7, the material load factor is 0.065 kg-m/mm2. From the classification
(Figure 14.25), the shape load factor is 1.2. The total projected area from Problem 1 is 48
+ 37.35 = 85.35 cm2. The equivalent energy capacity of the forging equipment required is
given by Equation 14.8 equals 85.35 x 100 x 0.065 x 1.2 = 665.71 kg-m.
e) The equivalent energy capacity corresponds to a power hammer of about 500 lb rating
or a 1200 lb drop hammer. From Figure 14.21, the Relative Cost per Operation is 0.9.
The forging cost is given by Equations 14.9 and 14.10. Cost = 0.9 x 5 x 0.15 = $0.675.
f) The equivalent energy capacity corresponds to a mechanical press of about 300 ton
capacity. From Figure 14.21, the Relative Cost per Operation is 0.45. The forging cost is
given by Equations 14.9 and 14.10. Cost = 0.45 x 5 x 0.15 = $0.338.
5. For sample forging B shown in Figure 14.31 determine the following:
a) The classification of the forging
b) The complexity factor, Ffc, of the forging
c) The sequence of forging operations for this part
d) The equivalent energy capacity, Ef, of the forging equipment needed for this forging
e) An appropriate size of forging hammer to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
f) An appropriate size of mechanical press to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
Solution: a} For this forging L/W = 83/83 = 1 and L/T = 83/20 = 4.15. The first digit of
the classification is 1 and the second digit is 0.
b) The complexity is given by LWT/V = 83x 83 x 20/ (1000 x 28) = 4.92
c) The number of operations required is 3 – scale break, blocker and finisher.
d) From Table 14.7, the material load factor is 0.13 kg-m/mm2. From the classification
(Figure 14.25), the shape load factor is 1.4. The total projected area from Problem 1 is 47
+ 19.48 = 66.48 cm2. The equivalent energy capacity of the forging equipment required is
given by Equation 14.8 equals 66.48 x 100 x 0.13 x 1.4 = 1210 kg-m.
e) The equivalent energy capacity corresponds to a power hammer of about 1000 lb
rating. From Figure 14.21, the Relative Cost per Operation is 1.0. The forging cost is
given by Equations 14.9 and 14.10. Cost = 1.0 x 3 x 0.15 = $0.45.
f) The equivalent energy capacity corresponds to a mechanical press of about 500 ton
capacity. From Figure 14.21, the Relative Cost per Operation is 0.5. The forging cost is
given by Equations 14.9 and 14.10. Cost = 0.5 x 3 x 0.15 = $0.225.
6. For sample forging C shown in Figure 14.32 determine the following:
a) The classification of the forging
b) The complexity factor, Ffc, of the forging
c) The sequence of forging operations for this part
d) The equivalent energy capacity, Ef, of the forging equipment needed for this forging
e) An appropriate size of forging hammer to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
f) An appropriate size of mechanical press to produce this forging and the estimated
forging cost per part, assuming that the operating cost per operation of a 1000 lb power
hammer is $0.15.
Solution: a} For this forging L/W = 76/50 = 1.52 and L/T = 76/50 = 1.52. The first digit
of the classification is 0 and the second digit is 0.
b) The complexity is given by LWT/V = 76 x 50 x 50/ (1000 x 65) = 2.92
c) The number of operations required is 3 – scale break, blocker and finisher.
d) From Table 14.7, the material load factor is 0.065 kg-m/mm2. From the classification
(Figure 14.25), the shape load factor is 1.9. The total projected area from Problem 1 is 34
+ 15.55 = 49.55 cm2. The equivalent energy capacity of the forging equipment required is
given by Equation 14.8 equals 49.55 x 100 x 0.065 x 1.9 = 611.98 kg-m.
e) The equivalent energy capacity corresponds to a power hammer of about 500 lb rating
or a 1200 lb drop hammer. From Figure 14.21, the Relative Cost per Operation is 0.9.
The forging cost is given by Equations 14.9 and 14.10. Cost = 0.9 x 3 x 0.15 = $0.405.
f) The equivalent energy capacity corresponds to a mechanical press of about 300 ton
capacity. From Figure 14.21, the Relative Cost per Operation is 0.45. The forging cost is
given by Equations 14.9 and 14.10. Cost = 0.45 x 3 x 0.15 = $0.203.
7. For the forging A shown in Figure 14.30, assuming that the part is produced using
multi-impression dies on a hammer, determine the following:
a) The estimated initial cost of the forging dies assuming a die manufacturing rate of $50
per hour.
b) The die cost per part assuming a life volume of 200,000.
Solution: Die Material Costs
For this part, nbd = 0, nbl = 1, nsf = 0, nfin = 1, nedg = 1, nsb = 0, and nfl1 = n fl2 = 1 (Fig.
14.25). Therefore, Nimp = 2 and Nfl = 0. The width of the platter is equal to the width of
the part = 51 mm, if only one part per cycle is produced. The average cavity depth dave is
V/Ap = 10 x 43/48 = 8.96 mm. Equivalent bar diameter Dbar = (4daveW/π)0.5 = 16.2 mm.
From Section 14.5.1, the cavity spacing,
mm.
The cavity edge distance,
mm. The depth of
the die block is 5 x 0.5T = 57.5 mm.
The width of the die block (Equation 14.16),
,
0
which gives 265.83 mm, assuming the angle of the fullers is 15 .
The length of the die block, (Equation 14.17),
mm.
The die material cost is given by Equation 14.18:
, which assuming the unit die material cost is $20/kg
and the density of tool steel is 7.9 g/cc, gives $1338.
Die Manufacturing Costs
i. Block Preparation Time. The time for initial preparation of the die block is given by
(14.19)
where Tbt is the base time and is 6 hours because Ffc is greater than 6.0. Therefore,
h
ii. Layout Time. The time for laying out the die block is given by
(14.20)
where Nc is the number of forgings per cycle, Sc is the cavity standard, Slk is the lock
standard, and m is the multicavity index—usually taken as 0.7. The cavity standard, Sc, is
(14.21)
In this Nc = Slk = 1 and Sc = 0.6 + 3 x 0.4 = 2.2.
Therefore, Tlay = 0.008 x 48 x 6.93 x 2.2 = 5.85 h.
iii. Milling Time. The time for milling the die cavities is obtained from
(14.22)
where Sml is the milling standard given by
(14.23)
where Ms is the number of surface patches per unit projected area, Nsp/Ap and in this case
= 0.625. Also, Κ = 0.9(1 – exp(–0.0098dave)) and b = 0.4 + 0.7 exp(–0.0039dave), from
which K = 0.076 and b = 0.40. These give Sml as 0.131 and so 0.2 must be used.
Substituting into Equation 14.24 gives a milling time of 3.27 hours.
iv. Bench Work Time. The bench work time, tbw, on the dies is given by
(14.24)
where Sbn is the bench standard, which depends on the forging complexity and average
cavity depth. The benchwork factor, Fins = (Ap/6.54 + 0.5Ns) and Sbn = B0 + 0.26(Fins –
15). The constant B0 depends on the average depth, dave, as follows:
dave ≤ 12.7 mm
dave > 22.86 mm
dave > 12.7 and ≤ 22.86
B0 = 0.056dave
B0 = 4.5 + (0.04dave – 0.9)2.19
B0 = 0.5 + (0.04dave – 0.35)7.27
In this case dave = 8.96 and B0 = 0.056 x 8.96 = 0.5, from which Fins = 22.34 and Sbn =
2.41. These give a bench work time of 5.3 hours.
v. Planing Time. The block planning time
(14.25)
where the cavity time Tcav = Tlay + Tmill + Tbw = 14.43 hours and Tpl = 0.44 hours
vi. Dowel Time. The dowel time, Τdl, is 3h if the die material volume is less than 4260
cm2, or else is 4h. The die volume is 8468 cm3 and then the dowel time is 4 hours.
vii. Flash Gutter Time. The time to machine flash gutters on the die cavities,
(14.26)
where Pr is the outside perimeter in millimeters, which is 594 and then Tfl = 0.94 h.
viii. Edger Time. The time for manufacture is given by Equation 14.27:
= 254(16.2/25.4 +1)0.005 = 2.08 h.
ix. Finish-Polish Time. The time to finish-polish the dies cavities, Tpol, is given by
Equation 14.28:
h
The total die manufacturing time is the sum of the above times and consequently the die
manufacturing cost is:
or = 50(11.74 + 5.85 + 3.27 + 5.3 + 0.44 + 4 + 0.94 + 2.08 + 4.56) = $1909.18..
The total initial die manufacturing costs are thus
(14.30)
Thus the total initial die cost is CDIE = 1338 + 1909 = $3247
b) The die resink quantity, Qrs, is determined by
and in this case βm = 1 and
βs = 0.75. Qrb is the basic resink quantity (say 40,000) and Qrs = 30,000. Assuming 5
resinks the total die life, LD will be 180,000
The cost of each resink, Crs, is assumed to be equal to:
= $656
The forging die cost per part, CD, is given by:
CD = (3247 + 5 x 656)/180,000 = $0.036 per part.
8. Repeat question 14.4 assuming two identical forgings are made at the same time. You
will need to nest the parts together to form a suitable forging platter. Some of the data
determined in question 14.1 may be needed to complete the calculations.
Solution: The two parts can be nested together to produce a platter with a length of about
275 mm and a width of about 60 mm. The volume is doubled and it is reasonable to
assume that the area of the flash land will be about double that of the single forging,
because of the flash between the two cavities.
a) For this forging platter L/W = 275/60 = 4.58 and L/T = 275/23 = 11.96. The first digit
of the classification is 2 and the second digit is 0.
b) The complexity is given by LWT/V = 275x 60 x 23/ (1000 x 86) = 4.41
c) The number of operations required is 4 – fuller, edger, blocker and finisher.
d) From Table 14.7, the material load factor is 0.065 kg-m/mm2. From the classification
(Figure 14.25), the shape load factor is 1.1. The total projected area is 170 cm2. The
equivalent energy capacity of the forging equipment required is given by Equation 14.8
equals 170 x 100 x 0.065 x 1.1 = 1215.5 kg-m.
e) The equivalent energy capacity corresponds to a power hammer of about 1000 lb
rating. From Figure 14.21, the Relative Cost per Operation is 1. The forging cost is given
by Equations 14.9 and 14.10. Cost = 1 x 4 x 0.15/2 = $0.3.
f) The equivalent energy capacity corresponds to a mechanical press of about 500 ton
capacity. From Figure 14.21, the Relative Cost per Operation is 0.5. The forging cost is
given by Equations 14.9 and 14.10. Cost = 0.5 x 4 x 0.15/2 = $0.15.
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