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ME3263E/TME3263

DESIGN FOR MANUFACTURING & ASSEMBLY

Homepage: http://courses.nus.edu.sg/course/mpeleeks/personal/index.htm

A/P Lee Kim Seng Deputy Head (Outreach) Mech Mechanical Engg Dept. National Univ. of S’pore

Room: EA 04-17

Tel: 6516 2574

[email protected]

A TYPICAL INJECTION MOULDING SYSTEM

A/P Lee Kim Seng

Contents

1. Molding machine size selection

2. The molding cycle calculation can be divided into: (i) injection or filling time, (ii) cooling time and (iii) mold resetting time.

3. Mold Manufacturing Cost

(i) Mold base cost

(ii) Core/cavity and parts manufacturing cost

MOLDING MACHINE SIZE

Determination of the appropriate size of an injection molding machine is based primarily on the required clamping force.

This in turn depends upon the projected area of the cavities in the mold and the maximum pressure in the mold during mold filling.

Eg. For a 15 cm diameter plain disk, the projected area is 176.8 cm2. If the disk has a 10 cm dia through hole (78.6 cm2), then the projected area = 98.2 cm2. This is the area over which the polymer pressure will act during filling.

A/P Lee Kim Seng

∅15

∅10

MOLDING MACHINE SIZE (Cont..)

The size of runner system depends upon the size of the part. Typical runner volumes as a % of part volume are shown below:

Runner

Part Volume (cm3)

Shot size (cm3)

Runner % 16

22

37 32

41

28 64

76

19 128

146

14 256

282

10 512

548

7 1024

1075

5

Table 1 Runner Volume (T8.2)

A/P Lee Kim Seng

Thermoplastic

Specific

gravity

Thermal (mm2/s)

Injection Temp

0C)

Mold Temp (0C)

Ejection Temp (0C)

Injection Pressure

(bars)

High-density polyethylene 0.95 0.11 232 27 52 965

High-impact polyethylene 1.59 0.09 218 27 77 965

Acrylonitrile-butadiene -styrene (ABS)

1.05 0.13 260 54 82 1000

Acetal (homopolymer) 1.42 0.09 216 93 129 1172

P{olyamide (6/6 nylon) 1.13 0.10 291 91 129 1103

Polycarbonate 1.20 0.13 302 91 127 1172

Polycarbonate (30% glass) 1.43 0.13 329 102 141 1310

Modified polyphenylene oxide (PPO)

1.06 0.12 232 82 102 1034

Modified PPO (30% glass)

1.27 0.14 232 91 121 1034

Polypropylene (40% glass)

1.22 0.08 218 38 88 965

Polyester teraphthalate (30% glass)

1.56 0.17 293 104 143 1172

Table 2 Processing Data for Selected Polymers (T8.5)

Tx Tm Ti α pj

Clamping force (kN)

Shot size ( cc )

Operating cost ($h)

Dry cycle time ( s )

td

Max. clamp stroke ( cm )

Ls

Driving power (kW )

300

34

28

1.7

20

5.5

500

85

30

1.9

23

7.5

800

201

33

3.3

32

18.5

1100

286

36

3.9

37

22.0

1600

286

41

3.6

42

22.0

5000

2290

74

6.1

70

63.0

8500

3636

108

8.6

85

90.0

Table 3 Injection Molding Machine (T8.4)

A/P Lee Kim Seng


As a general rule, approximately 50% of the pressure generated in the machine injection unit is lost due to the flow resistance in the sprue, runner systems and gates.

This rule will be applied in the costing analyses:

A/P Lee Kim Seng


A batch of 15 cm diameter disks with a thickness of 4 mm are to be molded from ABS in a six-cavity mold. Determine the appropriate machine size?

The projected area of each part = 176.8 cm2. Volume = 176.8 x 0.4 = 70.7 cm3. From Table 1, the % increase in area due to runner system is 18%.

Thus total projected shot area = 6 x 1.18 x 176.8 = 1252 cm2.

Example 1

The recommended injection pressure for ABS = 1000 bars. (Table 2)

Thus max cavity pressure ( 50%x 1000) = 500 bars = 50 MN/m2.

A/P Lee Kim Seng

Note: (Multi-cavities) Total Volume = 176.8 x 0.4 x 6 = 424.32 cm3. From Table 1, the % increase in area due to runner system is 8.03%. Thus total projected shot area = 6 x 1.0803 x 176.8 = 1146 cm2


A batch of 15 cm diameter disks with a thickness of 4 mm are to be molded from ABS in a six-cavity mold. Determine the appropriate machine size?

The maximum seperating force is thus given by: F = ( 1252 x 10-4 ) x 50 x 106 N = 6,260 kN

The available machines as shown in Table 3.

Thus the appropriate machine chosen would be the one with the clamping force of 8,500 kN.

Example 1

A/P Lee Kim Seng

Note:

The maximum seperating force is thus given by: F = ( 1146 x 10-4 ) x 50 x 106 N = 5,730 kN


Note: The selected machine must be checked to ensure that it has a sufficient shot size and a large enough clamp stroke. The required shot size is the volume of the six disks plus the volume of the runner system. Shot size = 6 x 1.18 x ( 176.8 x 0.4 ) = 500.7 cm3.

Checking on machine suitability: For the 8,500 kN machine (Table 3), the max machine shot size = 3,636 cm3, the max clamp stroke, Ls = 85 cm which is sufficient to mold a hollow part up to a depth of 40 cm with a clearance of 5 cm for the part to fall between the end of the core and cavity plates.

Example 1

Or Shot size = 424.32 x 1.0803 = 458.4 cm3


The max stroke ( 40cm) is excessive for molding of 4 mm thick disk.

The stroke is, however, adjustable, and to speed up the machine cycle, it would be reduced to just a few cm.

Example 1

A/P Lee Kim Seng

D D

Cle

aran

ce

Max Clamp stroke

Mold Cycle Time

The mold cycle time estimation is essential in any consideration of the merits of alternative part designs or the choice of alternative polymers.

The mold cycle can be divided into: (i) injection or filling time, (ii) cooling time and (iii) mold resetting time.

A/P Lee Kim Seng

(i) Injection or filling time

A precise estimate of injection time requires an extremely difficult analysis of the polymer flow as it travels through the runners, gates, and cavity passages.

However, modern injection molding machine are equipped with powerful injection units specifically to achieved the required flow rates for effective mold filling.

A/P Lee Kim Seng


It is thus assumed that, at the commencement of filling, the full power of the injection unit is utilized and that the polymer pressure at the nozzle of the injector is that recommended by the polymer supplier.

The flow rate, using elementary mechanics, is given by

Q = Pj / pj m3/s

Where Pj = injection power, W pj = recommended injection pressure, N/m2.

A/P Lee Kim Seng


It is assumed that the flow rate suffers a constant deceleration to reach an insignificantly low value at the point at which the mold is nominally filled.

Thus the average flow rate is given by Qav = 0.5 Pj / pj m3/s

And the filled time would be estimated as tf = ( 2Vs pj / Pj )s

Where Qav = Vs / tf

Vs = required shot size, m3

A/P Lee Kim Seng


For the 15 cm diameter disks molded in a 6-cavity mold, the required shot size is 500.7 cm3.

The recommendation injection pressure for ABS = 1000 bars or 100 MN/m2

The available power at the injection unit of the 8,500 kN machine is 90 kW (Table 3).

Thus the estimated fill time is given by tf = ( 2Vs pj / Pj )s = 2 x ( 500.7 x 10-6 ) (100 x 106) / ( 90 x 103

) s = 1.11 s

A/P Lee Kim Seng

For Vs = 458.4 cm3 , tf = 1.07 s

(ii) cooling time

In the calculation of cooling time, it is assumed that cooling in the mold takes place almost entirely by heat conduction.

Negligible heat is transferred by convection since the melt is highly viscous and it is clear that radiation cannot contribute to the heat loss in a totally enclosed mold.

A/P Lee Kim Seng

(ii) cooling time

Ballman and Shusman suggested an estimate of the cooling time based on truncating Carslaw and Jaeger general series solution to just the first term.

It is assumed that mold opening and ejection are permissible when the injected polymer has cooled to the point where the highest temperature in the mold (at the thickest wall centre plane) equals Tx, the recommended ejection temperature.

A/P Lee Kim Seng

(ii) cooling time

Thus the first-term solution for the cooling time is given by:

Where

hmax = maximum wall thickness, mm

Tx = recommended part ejection temperature, °C varies according to resin used. (eg. Acrylic : 85 °C, ABS: 82 °C )

Tm = recommended mold temperature, °C

Ti = polymer injection temperature, °C

α = thermal diffusivity coefficient, mm2/s

…….. Eq 8.5 tc = h2

max

π2 α loge

4( Ti – Tm) π( Tx – Tm)

s

A/P Lee Kim Seng

(ii) cooling time

tc = h2

max

π2 α loge

4( Ti – Tm) π( Tx – Tm)

s

α = thermal diffusivity coefficient, mm2/s

K

ρ . Cp

α = For Acrylic: α = 0.12 mm2/s = 1.2 x 10-3 cm2/s

Where

K = thermal conductance (eg. Acrylic: 0.18 kcal / m. hr. °C )

ρ = density (eg. Acrylic : 1.19 g/cm3)

Cp = specific heat (eg. Acrylic : 0.35 cal / g °C )

The data needed for making cooling time predictions are given in Table 2 (T8.5) which contains a list of most widely used injection-molded thermoplastics.

…….. Eq 8.5

A/P Lee Kim Seng

(ii) cooling time

It should be noted that Eq 8.5 tends to underestimate the cooling time for very thin wall moldings.

One reason is that for such parts the thickness of the runner system is often greater than the parts themselves and the greater delay is needed to ensure that the runner can be ejected cleanly from the mold.

It is suggested that 3 s be taken as the minimum cooling time even if Eq 8.5 predicts a smaller value.

A/P Lee Kim Seng

(ii) cooling time

tc = h2

max

π2 α loge

4( Ti – Tm) π( Tx – Tm)

s

From the equation 8.5, it is clear that for a given polymer, with given molding temperature, the cooling time varies with the square of the wall thickness of the molded part.

…….. Eq 8.5

This is the principal reason why injection molding is often uneconomical for thick-wall parts.

A/P Lee Kim Seng

h

tc

(iii) mold resetting time.

Mold opening, part ejection and mold closing times depend upon the amount of movement required for part separation from the cavity and core and on the time required for part clearance from the mold plates during free fall.

The summation of these 3 machine operation times is referred to as resetting time. Table 4 (T8.3) shows the approximate machine operation times for flat, box shaped and deep cylindrical parts.

Flat Box Cylindrical

Mold open 2 2.5 3

Part eject 0 1.5 3 Mold close 1 1 1

Table 4 Machine Clamp Operation Times (s)

A/P Lee Kim Seng


Dry cycle time is defined as the time required to operate the injection unit and then to open and close an appropriately sized mold by an amount equal to the maximum clamp stroke.

Values of parameters for a wide range of currently available injection molding machines are shown in Table 3 (T8.4).

Note: The dry cycle time given by the machine supplier bear little relationship to the actual cycle time when molding parts.

This is because the dry cycle is based on an empty injection unit and it takes only milliseconds to inject air through the mold.

Moreover, there is obviously no required delay for cooling and the machine clamp is operated during both opening and closing at maximum stroke and at maximum safe speed.


In practice: The clamp stroke is adjusted to the amount required for the molding of any given part.

If the depth of the part is given by D cm,

Then it is assumed that the clamp stroke is adjusted to ( 2D + 5 ) cm.

Note: Mold opening usually takes place more slowly than mold closing (Table 4). This is because during mold opening, time is needed for the ejection system to eject the part from the core with a significant level of force.

Rapid mold opening may result in warping or fracture of the molded part.

A/P Lee Kim Seng


For present time estimation: Mold opening is assumed to take place at 40% of mold closing speed. (This corresponds to the average of Ostward’s data in Table 4)

If Maximum clamp stroke = Ls, Machine dry cycle time = td,

Then clamp closing time at full stroke = td/2.

Given depth of part to be molded = D, Then the adjusted clamp stroke = ( 2D + 5 ) cm.

Mold Closing time tclose = 0.5 td [(2D + 5 )/ Ls ]1/2 … Eq 8.6

Thus,

A/P Lee Kim Seng


Since: Mold opening is assumed to take place at 40% of mold closing speed. ( topen = 2.5 x tclose ). Given a dwell of 1 s for the molded part to fall between the plates.

Then, Mold opening time

topen = 1 + 2.5x0.5 td [(2D + 5 )/ Ls ]1/2

= 1 + 1.25 td [(2D + 5 )/ Ls ]1/2

tr = 1 + 1.75 td [(2D + 5 )/ Ls ]1/2 ……. Eq 8.7

Mold Resetting time:

A/P Lee Kim Seng


Then the machine resetting time:

tr = 1 + 1.75 td [(2D + 5 )/ Ls ]1/2 …. Eq 8.7

= 1 + 1.75 (8.6) [(2x20 + 5) / 85]1/2

= 11.95 s ( or 12 s)

Plain 15 cm diameter cylindrical containers with a depth of 20 cm are to be manufacture from ABS in a six-cavity mold. Estimate the machine resetting time?

From the previous example, the appropriate machine size is 8,500 kN. From Table 3 (T8.4), Dry cycle time, td = 8.6 s, Maximum clamp stroke = 85 cm.

Given depth D = 20 cm

Example 2

A/P Lee Kim Seng


Then the machine resetting time:

tr = 1 + 1.75 td [(2D + 5 )/ Ls ]1/2 …. Eq 8.7

= 1 + 1.75 (8.6) [(2x10 + 5) / 85]1/2

= 9.2 s

Dry cycle time, td = 8.6 s, Maximum clamp stroke, Ls = 85 cm.

If depth D = 10 cm

Example 2

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Mould Manufacture

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Pocket & Contour milling

Mold Manufacturing Cost

The mold making starts with the purchase of a pre-assembled mold base which includes the main plates, pillars, bushings, etc.

Substantial amount of works have to be performed on the mold base which include:

•Deep hole drilling of the cooling channel.

•Millings of pockets in the plates for core and cavity inserts.

•Custom works on the ejector plate and housing for ejector system.

•Custom works for sliders and lifters etc.

Based on 2-plate mold

As a Rule of Thumb, mold manufacturing cost should be at least double the mold base purchase price to account for the custom work which has to be performed on it.

A/P Lee Kim Seng

A SINGLE CAVITY MOULD

Mould Base

TWO CAVITIES MOULD

A/P Lee Kim Seng

Mold Layout

Mold Cost Estimation The mold cost can be divided into two major categories:

(a) the cost of the fabricated mold base consisting of the required plates, pillars, guide bushings, etc.,

(b) cavity and core fabrication cost

Mold Base Costs

Dewhurst and Kuppurajan have shown that (Fig 8.7) mold base is a function of the surface area of the selected mold base plates and the combined thickness of the cavity and core plates.

Fig 8.7 Principal mold base cost driver

A/P Lee Kim Seng

Mold Base Costs

The mold base costs based on the data in Fig 8.7 can be represented by:

Where

Cb = cost of mold base, $

Ac = area of mold base cavity plate, cm2

hp = combine thickness of cavity and core plates in mold base, cm.

Cb = 1000 + 0.45 Ac hp 0.4 …. Eq 8.8

A/P Lee Kim Seng

Mold Base Costs

The selection of an appropriate mold base is based on the depth of the part, its projected area and the number of cavities required in the mold.

In addition to the cavity size, extra allowance has to be given for molds with mechanical slides and other complicated mechanisms such as unscrewing devices for the molding of screw threads.

A/P Lee Kim Seng

Mold Base Costs

The minimum clearance between adjacent cavities and between cavity surfaces and the edges and rear surfaces of cavity plates should be 7.5 cm.

Extra plate size required to accommodate sliders and unscrewing devices will depend upon the actual mechanisms used.

Note: Two side-pulls (sliders) restricts the mold design to a single row of cavities.

Three or more side-pulls usually implies single-cavity operation.

A/P Lee Kim Seng

Ac = 42.5 x 60 = 2550 cm2

Estimate the mold base cost assuming that 10 cm diameter plain cylindrical cups with a depth, hd, of 15 cm are to be molded in a six-cavity mold.

Example 3

Mold Base Costs

For a 3 x 2 array of cavities:

7.5

∅ 10

7.5 7.5

7.5

60

42.5

A/P Lee Kim Seng

Ac = 42.5 x 60 = 2550 cm2 Example 3

Mold Base Costs


The combined cavity and core plate thickness:

hp = hd + 15 = 30 cm

7.5

7.5

15

Hence

The estimated mold base cost:

Cb = 1000 + 0.45 Ac hp 0.4

= 1000 + 0.45 (2550) (30) 0.4

= $5,473 or $5,500

hd

A/P Lee Kim Seng

Ac = 112.5 x 25 = 2812.5 cm2

Example 3

Mold Base Costs


For a single row of 6 cavities.

7.5

∅ 10

7.5

7.5

112.

5

Plate thickness is 30 cm

Then the estimated mold base cost:

Cb = 1000 + 0.45 Ac hp 0.4

= 1000 + 0.45 (2812.5) (30) 0.4

= $5,933

7.5

7.5

15

Ac = 112.5 x 40 = 4500 cm2

Example 3

Mold Base Costs


If the part requires sliders on both side, then the cavity plate will only hold a single row of 6 cavities.

15

∅ 10

15

7.5

112.

5

If plate thickness is increased to 37.5 cm to support the unscrewing devices.


Cb = 1000 + 0.45 Ac hp 0.4

= 1000 + 0.45 (4500) (37.5) 0.4

= $9,630 15

37.5

Ac = 112.5 x 40 = 4500 cm2

Example 3

Mold Base Costs


If the part requires sliders on both side, then the cavity plate will only hold a single row of 6 cavities.

15

∅ 10

15

7.5

112.

5

If plate thickness remained the same.


Cb = 1000 + 0.45 Ac hp 0.4

= 1000 + 0.45 (4500) (30) 0.4

= $8,894 7.

5 7.

5 15

Core and Cavity Manufacturing Cost Based on 2-plate mold

Ejector Pins

The number of ejector pins required is governed by factors such as the size of the part, the depth of the main core, the depth and closeness of ribs and features contributing to part complexity. The number of ejector pins used was found to be approximately equal to the square root of the cross-sectional area when measured in square centimeters.

Ne = Ap 0.5 ……Eq 8.9

Where: Ne = number of ejector pins required

Ap = projected part area, cm2

A/P Lee Kim Seng

Ejector Pin


Ejector Pins

Sors et. al. suggested a value of 2.5 hours for each ejector pin. Thus, manufacturing hours ( Me) for the ejector system:

Me = ( 2.5 x Ap 0.5 ) hr ……Eq 8.10

Where: Xi = inner complexity of part

Xo = outer complexity of part

The number of manufacturing hours, associated with the geometrical features of the part, for one cavity and matching core is estimated from

Mx = 45 ( Xi + Xo)1.27 hr ……Eq 8.11

A/P Lee Kim Seng


Geometrical complexity counting procedure

The complexity of the inner surface (surface in contact with the main core):

Where: Nsp = number of surface patches

Nhd = number of holes and depressions

Xi = 0.01 Nsp + 0.04 Nhd ……Eq 8.12

Xo = 0.01 Nsp + 0.04 Nhd

A/P Lee Kim Seng


Note: Through hole should not be counted again from the outer surface (counted under inner surface).

When counting multiple identical features on the surface of a part, a power index of 0.7 should be used.

If the surface of a part is covered by 100 spherical dimples, then the equivalent number of surface patches to be counted is 1000.7 = 25

Xi = 0.01 Nsp + 0.04 Nhd ……Eq 8.12

Xo = 0.01 Nsp + 0.04 Nhd

A/P Lee Kim Seng


A plane conical component with recessed base is to be injection-molded. The inner and outer surface complexity levels are established as follows:

Xi = 0.01 x 2 = 0.02

Example 3

The inner surface comprises the following surface segments:

1. Main conical surface 2. Flat base

Thus,

A/P Lee Kim Seng


A plane conical component with recessed base is to be injection-molded. The inner and outer surface complexity levels are established as follows:

Example 3

The outer surface comprises:

1. Main conical surface 2. Flat annular base 3. Cylindrical recess in the base 4. Flat recessed base

The outer surface has an additional depression in the base:

Xo = 0.01 x 4 + 0.04 x 1 = 0.08

A/P Lee Kim Seng

Base on the part area relationship given by Sors et al, for parts with very simple geometry, the manufacturing hours for one cavity and core can be represented by

Mpo = ( 5 + 0.085 x Ap 1.2 ) hr ……Eq 8.13

Where: Mpo = part manufacturing hour

Ap = part projected area, cm2

Core and Cavity Manufacturing Cost

A/P Lee Kim Seng

Core and Cavity Manufacturing Cost To complete a mold cost estimating system, six additional important factors need to be considered.

(a) The need for sliders (50-80 hr) and lifters (100-200 hr)

(b) The need for any unscrewing core to produce molded screw thread (200-300 hr)

(c) The surface finish and appearance needed for the part

(d) The average tolerance level applied to the part dimensions

(e) Surface texturing required? ( approx 5% of cavity cost)

(f) The shape of the surface across which the cavity and core separate – Parting line

A/P Lee Kim Seng


Appearance (Surface finish) % increase

Not critical 10

Opaque, standard (Soc Plastic Engineers #3) 15

Transparent, Standard internal flaws or waviness permissible

20

Opaque, high gloss 25

Transparent, high quality 30

Transparent, optical quality 40

Table 5. % increase in Mold manufacturing hour for different appearance level (T8.6)

A/P Lee Kim Seng

Tolerance level

Description of Tolerance

%

increase

0

All greater than ±0.5 mm

0

1

Most approx. ±0.35 mm

2

2


5

3

Several approx. ±0.25 mm

10

4

Several approx. ±0.05 mm

20

5


30


Table 6. % increase in Mold manufacturing hour for different Tolerances (T8.7)

A/P Lee Kim Seng


Table 7 Parting Surface Classification (T8.8)

Parting plane consideration

Parting surface type Factor ( fp ) 0

1.25 2 - 4 simple steps or a simple curve surface 2

Greater than 4 simple steps 2.5 Complex curve surface 3 Complex curve surface with steps 4

Canted parting surface or one containing a single step Flat parting plane

The additional mold manufacturing hours ( Ms ) for non flat parting surface

Ms = fp Ap ½ hr ……Eq 8.14

Where: fp = parting plane factor given in Table 7 (T8.8)

Ap = projected area of cavity, cm2

A/P Lee Kim Seng

Mold Manufacturing Cost (Mold cost hour system)

The mold manufacturing cost is determined as follow:

(i) Projected area of Part (cm2) Eq 8.10 & 8.13, which include hours for the size effect on manufacturing cost plus hours for an appropriate ejection system

(ii) Geometric Complexity Apply Eg 8.11 to determine the appropriate hours needed.

(iii) Sliders allow 65 hours for each slider

(iv) Lifters allow 150 hours for each lifter

A/P Lee Kim Seng

Mold Manufacturing Cost (Mold cost hour system) Cont..

(v) Unscrewing devices allow 250 hours for each unscrewing device.

(vi) Surface finish / Apperance Apply table 5 (T8.6) to identify the % value for the required apperance. Apply the % value to the sum of hours determined for (I) & (ii) to obtain the appropriate hours needed.

(vii)Tolerance level Apply table 6 (T8.7) to identify the % value for the required tolerance. Apply the % value to the geometrical complexity hours determined for (i)&(ii) to obtain the appropriate hours needed.

(viii)Texture Add 5% of hours needed from (I) & (ii).

A/P Lee Kim Seng


(ix) Parting Plane Determine the category of parting plane from Table 7 (T8.8) and the value of the parting plane factor, fp use fp to obtain the hours needed from Eq 8.14

To determine the cost to manufacture a single cavity and matching core, the total hour needed is multiplied by the appropriate average hourly rate for the mold manufacture.

A/P Lee Kim Seng


It is anticipated that 2,000,000 plain hollow conical components are to be molded in Acetal homopolymer. The component, (Fig 8.8) has a material volume 78 cm3 and a projected area in the direction of molding of 78.5 cm2.

Example 4

The mold manufacturing hours needed are first established: Area = πr2 = π 52 = 78.57 cm2

A/P Lee Kim Seng

Vol = (π42 + πx9x12)t = 77.91 cm3

Assuming base dia = 8 cm; cup mean dia = 9 cm t =2 mm

Hours ( i ) Projected Area of Part

Substitute Ap=78.5 cm2 into Eq 8.10 (ejector pin) & 8.13 (core cavity) Me = ( 2.5 x Ap

0.5 ) = 2.5 x 78.50.5 = 22.15 hrs Mpo = ( 5 + 0.085 x Ap

1.2 ) = 5 + 0.085 x 78.51.2 = 20.97 hrs (Part Mfg hr)

43.12

( ii ) Geometrical Complexity Xi = 0.01 x 2 = 0.02, Xo = 0.01 x 4 + 0.04 x 1 = 0.08 Substitute into Eq 8.11 Mx = 45 ( Xi + Xo)1.27 = 45 ( 0.02 + 0.08)1.27 = 2.42 hrs

2.42

( iii ) No of Sliders (add 65 Hrs per slider) 0

( iv ) No of Lifters (add 150 Hrs per lifter) 0

( v ) No of Unscrewing Devices (add 250 Hrs per device) 0

( vi ) Surface Finish / Apperance Opaque high gross (Table 5) (T8.6) Add 25% of ( i ) & ( ii ), ie 25% x 45.54 = 11.39 hrs

11.39

( vii ) Tolerance Level Category 1 (Table 6) (T8.7), insignificant effect for low complexity

0

( viii ) Texture (add 5% of (i) & (ii)) 0

( ix ) Parting Plane Category 1 (Table 7) (T8.8)

0

Total hours needed = 56.93

(Mold cost hour system) Cont..


Assuming an average rate of $40 per hour for mold manufacturing, the estimated cost for one cavity and core is found to be 56.93 x $40 = $2,277.2

Example 4

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Ac = 42.5 x 60 = 2550 cm2

Estimate the total mold cost if the part given in Example 4 with 10 cm diameter cups and a depth, hd, of 12 cm are to be molded in a six-cavity mold.

Example 5

Total Mold Costs

For a 3 x 2 array of cavities: 7.5

∅ 10

7.5 7.5

7.5

60

42.5 7.5

7.5

12

The combined cavity and core plate thickness: hp = hd + 15 = 27 cm

core plate area

The estimated mold base cost: Cb = 1000 + 0.45 Ac hp

0.4 = 1000 + 0.45 (2550) (27) 0.4 = $5,288

Estimate the total mold cost if the part given in Example 4 with 10 cm diameter cups and a depth, hd, of 12 cm are to be molded in a six-cavity mold.

Example 5

Total Mold Costs

No Item Cost $

1 Mold base cost (6 cavity mold) $5,288

2 Core and Cavity manufacturing cost ( 6 x $2,277.2 )

$13,663.2

Total Mold Cost = $18,951.2

A/P Lee Kim Seng

MOLDING MACHINE SELECTION

The projected area of each part = 78.5 cm2. Volume = 78 cm3. From Table 1, the % increase in area due to runner system is 18%.

Thus total projected shot area = 6 x 1.18 x 78.5 = 555.8 cm2.

Example 5

The recommended injection pressure for Acetal = 1172 bars. (Table 2)

Thus max cavity pressure ( 50%x 1172) = 586 bars = 58.6 MN/m2.

A/P Lee Kim Seng

The maximum seperating force is thus given by: F = ( 555.8 x 10-4 ) x 58.6 x 106 N = 3,277 kN

The available machines as shown in Table 3.

Thus the appropriate machine chosen would be the one with the clamping force of 5,000 kN.


Note: The selected machine must be checked to ensure that it has a sufficient shot size and a large enough clamp stroke.

The required shot size is the volume of the six cups plus the volume of the runner system. Shot size = 6 x 1.18 x 78 = 552 cm3.

Checking on machine suitability: For the 5,000 kN machine (Table 3), the max machine shot size = 2,290 cm3, the max clamp stroke, Ls = 70 cm which is sufficient to mold a hollow part up to a depth of 32.5 cm with a clearance of 5 cm for the part to fall between the end of the core and cavity plates.

Example 5

Thus the estimated mold fill time is given by tf = ( 2Vs pj / Pj )s = 2 x ( 552 x 10-6 ) (117.2 x 106) / ( 63 x 103

) s = 2.05 s

tr = 1 + 1.75 td [(2D + 5 )/ Ls ]1/2 …. Eq 8.7

= 1 + 1.75 (6.1) [(2x12 + 5) / 70]1/2

= 7.87 s

( iii ) Mold Resetting time

From the example, the appropriate machine size is 5,000 kN. From Table 3 (T8.4), Dry cycle time, td = 6.1 s, Maximum clamp stroke = 70 cm.

Given D = 12 cm

Mold Cycle Time

( i ) Mold fill time

pj = recommended injection pressure, N/m2. Pj = injection power, W

(ii) cooling time

(ii) Mold cooling time

From table 2 (Acetal homopolymer) hmax = 2 mm (maximum wall thickness) since vol = 78 cm3

Tx = 129 °C, (recommended part ejection temperature, °C ) Tm = 93 °C, (recommended mold temperature, °C) Ti = 216 °C , (polymer injection temperature, °C) α = 0.09 mm2/s, (thermal diffusivity coefficient, mm2/s)

…….. Eq 8.5 tc = h2

max

π2 α loge

4( Ti – Tm) π( Tx – Tm)

s

A/P Lee Kim Seng

π( 129 – 93) tc =

22

π2 0.09 loge

4( 216 – 93) s

= 6.62 s

Example 5

1 Mold fill time 2.05 s

2 Mold cooling time 6.62 s

3 Mold resetting time 7.87 s

Mold Cycle Time = 16.54 s

A/P Lee Kim Seng

Mold Cycle Time

Thank You

A/P Lee Kim Seng

The size of runner system depends upon the size of the part.

Typical runner volumes as a % of part volume are shown below:

Runner

Part Volume (cm3)

Shot size (cm3)

Runner % 16

22

37 32

41

28 64

76

19 128

146

14 256

282

10 512

548

7 1024

1075

5

Table 1 Runner Volume (T8.2)

A/P Lee Kim Seng

Thermoplastic

Specific

gravity

Thermal (mm2/s)

Injection Temp

0C)

Mold Temp (0C)

Ejection Temp (0C)

Injection Pressure

(bars)

High-density polyethylene 0.95 0.11 232 27 52 965

High-impact polyethylene 1.59 0.09 218 27 77 965

Acrylonitrile-butadiene -styrene (ABS)

1.05 0.13 260 54 82 1000

Acetal (homopolymer) 1.42 0.09 216 93 129 1172

P{olyamide (6/6 nylon) 1.13 0.10 291 91 129 1103

Polycarbonate 1.20 0.13 302 91 127 1172

Polycarbonate (30% glass) 1.43 0.13 329 102 141 1310

Modified polyphenylene oxide (PPO)

1.06 0.12 232 82 102 1034

Modified PPO (30% glass)

1.27 0.14 232 91 121 1034

Polypropylene (40% glass)

1.22 0.08 218 38 88 965

Polyester teraphthalate (30% glass)

1.56 0.17 293 104 143 1172

Table 2 Processing Data for Selected Polymers (T8.5)

pj Tx Tm Ti α

Clamping force (kN)

Shot size ( cc )

Operating cost ($h)

Dry cycle time ( s )

td

Max. clamp stroke ( cm )

Ls

Driving power (kW )

300

34

28

1.7

20

5.5

500

85

30

1.9

23

7.5

800

201

33

3.3

32

18.5

1100

286

36

3.9

37

22.0

1600

286

41

3.6

42

22.0

5000

2290

74

6.1

70

63.0

8500

3636

108

8.6

85

90.0

Table 3 Injection Molding Machine (T8.4)

A/P Lee Kim Seng

me3263e/tme3263 design for manufacturing ...courses.nus.edu.sg/course/mpeleeks/personal/ppt/me...

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