Mold Advance Course Book

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Mold Design (Advance) Page 2

1- Temperature Control Basics

Temperature control for a mold refers to a control of receiving and releasing heat on the mold. In this connection knowledge of heat conductivity is important for consideration of heat reception and heat dissipation. Thus the basics of thermal conductivity will be reviewed as follows.

1-1 Heat transfer

When there is a certain temperature difference in an object or between objects, heat will transfer to keep thermal equilibrium in a system. Heat will be transferred from high side to low side and the transfer modes are classified as follows:

Heat conduction

Convection heat transfer (Heat delivery) Heat transfer

Radiation heat transfer

Above three occur in a complex manner, but one normally dominates others.

1-1-1 Heat conduction

Characteristic of heat conduction is that the conductor does not move. Thus heat transfer in a solid object is considered to be the result of genuine heat conduction. To a certain extent heat conduction occurs in gas and liquid but the conductivity there is prohibitively small in comparison with that of solid body. Transfer of heat is made from high temperature area to low temperature area and the transfer rate is proportional to the temperature gradient and the cross section area of heat passage. This is called Fourier’s Heat Conduction Law and the formula is shown below (Fig. 1-1-1.1).

Q = 1 _A₁

ST • ∆ •

λ ... Formula (1.1.1.1) Where Q: Heat transfer rate: Heat flow (kcal/h)

ΔT1 : Temperature difference between 2 points (℃)

S: Distance between 2 points (m)

A1: Cross-section area perpendicular to heat flow (m2)

λ : Heat conductivity (kcal/m•h•℃)

In the case of heat transfer from resin to mold in the molding process, both heat conduction and heat convection occur simultaneously during the injection process but heat conduction dominates during cooling process under holding pressure after the injection process. Heat transfer from cavity surface to wall surface of cooling water pipe is made under genuine heat conduction because it is a heat transfer in a solid body.

Incidentally, heat conductivity of S50C steel, which is often used as mold material, is about 46 kcal/m•h•℃, while heat conductivity of HDPE, which has rather high heat conductivity among resins, is 0.4 kcal/m•h•℃ and that of GPPS, lower heat conductivity among resins, is about 0.1 kcal/m•h•℃. The ratio to S50C is 1:115 and 1:460 respectively (Fig. 1-1-1.2).

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Fig. 1-1-1.1 Image of Heat Transfer

Fig. 1-1-1.2 Heat Conductivity of Various Materials (Ambient Temp. 20℃)

Section A-A Section Area S = Distance Heat Flow "Solid" Temperature Gradient Temperature Difference Air Water Pure Copper

* Notice log scaling.

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1-1-2 Convection heat transfer (Heat delivery)

Looking into heat transfer between liquid and solid, effect of heat transfer along with movement of liquid is much greater than heat conduction. It is called convection heat transfer or heat delivery. The heat transfer rate is proportional to temperature difference between solid and liquid and transfer area between the same. The formula is shown below:

Q = α•ΔT2•A2 (kcal/h) ... Formula (1.1.2.1)

Where A2: Heat transfer area between solid and liquid (m2)

ΔT2 : Temperature difference between solid and liquid (℃)

α : Heat transfer coefficient (kcal/m2•h•℃)

Difference between heat delivery and heat conduction is that in the heat delivery heat transfers along with moving liquid media and heat transfer coefficient α is not a specific constant for material like λ (formula 1.3.8.1) and varies depending upon flow condition. It is considered that there exists a stable film of liquid (or gas), named boundary film, between solid and flow media. This film is not subject to convection heat transfer but conduction heat transfer only. As heat conductivity of flow medium is small in comparison to solid, this boundary film can be treated as a kind of insulation layer made of flow medium (Fig. 1-1-2.1).

Accordingly if a flow makes the film thinner, the heat transfer coefficient α becomes greater and the heat transfer rate becomes faster. Generally in the case of slow flow velocity, the flow forms so called laminate flow in which liquid is not mixed. In this case the boundary film is thicker. On the other hand the film is thinner if the flow is under turbulent flow with high velocity where liquid is well mixed.

As explained, heat transfer coefficient α is the one having a boundary film in between and influenced substantially by the film thickness. Thus it may be called as boundary film heat transfer coefficient. It is important how to determine α in the convection heat transfer. One way is to determine α on the basis of Nusselt Number (Nu) which represents magnitude of heat transfer between solid and flow medium.

α = Nu λ F／D (kcal/m2•h•℃) ... Formula (1.1.2.2)

Where λ F: Heat conductivity of fluid (kcal/m•h•℃)

D: Internal diameter of pipe (m)

Convection heat transfer can be classified by two. One is natural convection heat transfer and another is enforced convection heat transfer. Heat discharge from mold to atmosphere is mainly influenced by natural convection heat transfer together with radiation heat transfer to be explained later. While, heat transfer from mold (internal wall of cooling water tube) to cooling medium (water or oil) is mainly affected by enforced convection heat transfer.

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Fig. 1-1-2.1 Heat Transfer from Solid to Liquid

1-1-3 Radiation heat transfer

Thermal energy from the sun is brought to the earth through a space without any transfer media. This is because heat transfers as electro magnetic wave as same as light and electric wave. This sort of heat transfer is called radiation heat transfer or simply radiation.

Any material radiates heat unless its temperature is 0°K (-273℃) in absolute temperature. The radiation is mutually absorbed, reflected or passed trough. The heat transfer rate in radiation is proportional to difference of the 4th power of absolute temperature (Kelvin’s temperature). It is shown below (Fig. 1-1-3.1).

Q = K (TA4 ―TB4) (kcal/h) ... Formula (1.1.3.1)

Where TA: Absolute temperature of object A (°K)

TB: Absolute temperature of object B (°K)

K: Proportion constant

Proportion constant k includes various elements. This k is not given based on physical property like heat conductivity (λ) but calculation like heat transfer coefficient (α). In the case of radiation from a mold, a formula is given below considering object a (mold) is surrounded by object b (air) and radiation area ratio (AA/AB) is negligibly small.

Solid Liquid

Temperature Difference

Boundary Film A2: Conduction Area

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Mold Design (Advance) Page 6 Q = ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ 100 T 100 T_A4 _B4 (kcal/h) ... Formula (1.1.3.2)

Where AA: Surface area of object a (m2)

σ : Black Radiation constant = 4.88 kcal/m2_•h•k4

ε : Radiation rate of object A

In the formula (1.1.3.2), let us see the influence to heat flow due to radiation of mold temperature by varying the temperature TA like 40℃, 80℃, and 120℃. Room

temperature TB is assumed to be 25℃. Result shows when TA changes to 2 times and 3

times, resultant Q changes 4.5 times and 9.3 times. This tells you that radiation transfer cannot be ignored if temperature difference between room temperature and heated object temperature is big when the object is exposed to atmosphere.

Fig. 1-1-3.1 Image of Heat Radiation Transfer

Absorption

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1-2 Received heat of a mold

In terms of received heat (QI) of a mold, the biggest source must be from resin (QA). Other

sources may be of heat from nozzle (QB) of nozzle touch area of injection machine or

received heat from hot runner manifold and hot tip area (QC) in the case of hot runner mold

(Fig. 1-2.1).

QI = QA + QB + QC (kcal/h)... Formula (1.2.1)

Let’s take up received heat from resin (QA). When W (kg/h) is resin weight injected per

hour, received heat (QA) can be calculated by applying following formula.

QA = W • {CP (TP - TR) + L • C} (kcal/kg•℃) ... Formula (1.2.2)

Where CP: Specific heat of resin (kcal/kg•℃)

TP: Resin Temperature (℃)

TR: Temperature at mold separation (℃)

L: Latent heat of crystalline resin (kcal/kg) C: Crystallinity of crystalline resin (0.1~0.8)

In the formula (1.2.2), temperature at mold separation (TR) can be replaced by thermal

deformation temperature to assure the temperature in the center of the thickest portion of the product to be lower than the heat distortion temperature. In this case, try to set the temperature 10~30℃lower than the heat distortion temperature to entertain safety consideration. A part of the formula {CP(TP-TR) +L•C} can be roughly estimated by resin

material, when the value is represented by total heat amount (Q), formula (1.2.2) can be shown as below.

QA = W•Q (kcal/kg) ... Formula (1.2.3)

Table 1-2.1 shows estimated values of Q by resin material in the safe direction (estimating q

in the bigger side).

3) Released heat from a mold

If there is no temperature control device on a mold (natural radiation only), released heat from a mold (QO) should consist of transferred heat to platen of injection machine (QD) and

radiated heat to atmosphere (QE) (Fig. 1-2.2).

QO = QD + QE (kcal/h)... Formula (1.2.4)

QD is calculated as heat passing through composite wall surfaces, but estimation of heat

resistance between mold clamping plate and platen of injection machine is very difficult. QE

is considered as a mixture of convection and radiation heat transfer. It is influenced by molding conditions such as mold temperature, airflow, mold open time, etc.

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Fig. 1-2.1 Various Heat Received by Mold (QI)

Table 1-2.1 Heat Specifics of Resin Material

Fig. 1-2.2 Released Heat from Mold (Q0) In case of cold runner : QI = Qa + Qb

In case of hot runner : QI = Qa + Qc

Heat from melted resin Heat from

nozzle touch area Heat from hot runner area

Platen of Injection Machine

Mold

a. Released Heat at

Mold Binding b. Released Heat at Mold Opening

Resin Material Specific Heat Cp

Crystallin e Crystalline Rate C Total Heat q Non-crystall ine Latent heat L

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1-3 Heat to be removed from a mold

A mold reaches to thermally balanced condition through heat receiving and heat releasing process. Thus theoretically speaking, molding can be made without temperature control device as long as the balanced temperature is suitable for the plastic molding. However it is advised not to precede molding without temperature control device because a long time will be needed before reaching to a balanced condition and moreover mold temperature cannot be stable being influenced by environmental disturbances.

If receiving heat is more than releasing heat and thermal balanced point is higher than required temperature range, cooling device is needed. On the other hand, if receiving heat is less than releasing heat and balanced temperature point is lower than required temperature range, heating device should be arranged (Fig. 1-3.1).

Here in this section, condition QO< QI, in other words, condition required to cool off a mold,

will be taken and heat to be removed from a mold will be discussed. Removed heat QR can

be expressed in a molding cycle where receiving and releasing heat are to be balanced. QI = QO + QR (kcal/h)... Formula (1.3.1)

Then

QR = QI – QO (kcal/h) ... Formula (1.3.2)

If you understand basics behind the formula (1.3.3), you may simplify the calculation as follows. In the cold runner mold, QB can be traded off by (QD + QE) because (QD + QE) is

usually bigger than QB. By trading them off, cooling calculation will come to safe side

(increased requirement for cooling). In this way, you may treat heat to be removed (QR) is

equivalent to received heat (QA) from resin.

QR ≒QA ≒ W • Q (kcal/h) ... Formula (1.3.4)

In the heat transfer calculation, you may apply formula (1.3.4) for approximate result because q in formula (1.2.3) and table (1-2.1) are given in the safe side. However if you intend to apply formula (1.2.2), it is advised to incorporate 1.5 times safety factor taking account of possible requirement of cycle shortening and expected deterioration in the heat exchanger performance.

QR = 1.5 QA

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Assuming QR can be applicable to all kinds of cooling medium, following formula can be

derived.

QR = WL • CPL (TW －TL) (kcal/h) ... Formula (1.3.6)

Where WL: Required weight of cooling agent (kg/h)

CPL: Specific heat of cooling agent (kcal/kg•℃)

TW: Internal wall temperature of cooling tube (℃)

TL: Average temperature of cooling agent (℃)

TL in formula (1.3.6) is average temperature of the cooling medium other than that in the

boundary film. In the case of water as cooling medium, (TW–TL) can be regarded as about

2~3℃. WL and VL, required volume of cooling agent can be expressed as follows:

WL = VL • ρL (kg/h)... Formula (1.3.7)

VL = V • πD2/4 (m3/h) ... Formula (1.3.8)

Where PL: Density of cooling medium (kg/m3)

V: Flow velocity of cooling medium (m/h) D: Internal diameter of cooling tube (m)

Flow velocity V can be derived from formulae (1.3.4) or (1.3.5), (1.3.6), (1.3.7) and (1.3.8), as follows: ) h / m ( ) T T ( C D Q W 4 V L W PL L 2 ρ − π = • • • • ... Formula (1.3.9)

More accurate formula must be:

{

}

₍_m_/_h₎ ) T T ( C D C L ) T T ( C W 5 . 1 4 V L W PL L 2 R P P − ρ π + − × = • • • • • ... Formula (1.3.10)

Internal diameter of cooling tube (D) in above formulae should be picked up from Table

1-3.1 temporally, and confirm them if it falls in the range of 10,000~30,000 of Reynolds

number (RE) and then finalize the diameter. Be aware that unit of d is in m.

RE = D • ρL • V／µ... Formula (1.3.11)

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Next, Nusselt number (NU), important parameter in the convection heat transfer calculation, will be calculated. Prantle number (PR) in Nusslet number is defined as follows.

PR = ν／A = µ • CPL／λF... Formula (1.3.12)

Where ν : Dinamic viscosity of cooling medium = µ / ρL (m2/h)

A : Heat dissipation rate = λF/ρL • CPL (m2/h)

Nusselt number is given as follows. Be aware NU formula varies slightly depending upon where to get the formula from.

Nu = 0.023 • (RE) 0.8 • (PR) 1/3 ... Formula (1.3.13)

Formula (1.3.13) is effective only for turbulent flow. In the case of laminated flow or transition flow, in which Raynold’s number is less than 10,000, re-evaluation of mold temperature and cooling tube diameter must be carried out.

Once Nusselt number (NU) is decided, heat transfer co-efficient α can be calculated by formula (1.1.2.1). And cooling tube surface area (AL) can be calculated by a converted

formula from (1.2.2).

AL = QR／α • (TW – TL) (m2) ... Formula (1.3.14)

As cooling tube diameter (D) is known, cooling tube length can be calculated as follows: LR = AL／πD (m) ... Formula (1.3.15)

As described, total cooling circuit length of cavity and core can be calculated.

So far, all formulae assumed that heat from injected resin W (kg/h) is transferred perfectly to a mold. Here we should evaluate if such assumption is reasonable or not. Bahlman’s formula should be effective for the evaluation. It is to evaluate if W (kg/h) is possible by estimating molding cycle from theoretical cooling time calculation.

⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − π π = • • • _R _W W P 2 2 C T T T T 4 LN A H T ... Formula (1.3.16) Where Tc: Theoretical cooling time (h)

H: Product thickness (m)

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Mold Design (Advance) Page 12 TP: Resin temperature (℃)

TR: Mold separation temperature (℃)

TW: Mold temperature (℃)

This formula can be used as a guideline because all kinds of condition have to be assumed for calculation. Molding cycle should be evaluated by estimating injection time, mold opening time duration, mold take-out time duration.

If you design temperature control system by applying above explained basics on the heat transfer thermal dynamics, you should be able to provide temperature control system with improved heat exchanging efficiency comparing with traditional system, which was made based on past examples.

In the end of basics on thermal transfer theory, a calculation example is shown below for your better understanding. Try to solve the example before you read the answer to follow. Mind units are to be carefully treated.

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Mold Design (Advance) Page 13 【A calculation example】

Calculate total circuit length of a cooling water system required for a mold for following conditions.

Resin: ABS

Number of product per mold: 2

Product dimension: 100mm×100mm×2mm Molding cycle target: 20 seconds

(Calculation)

① To estimate molding conditions along with physical properties of ABS resin.

• Density: ρ = 1030kg/m3

• Specific heat: CP = 0.35 kcal/kg•℃

• Heat conductivity: λ = 0.2 kcal/m•h•℃

• Resin temperature: TP = 220℃

• Mold separation temperature: TR = 70℃

• Internal wall temperature (water tube): TW = 52℃

• Water temperature: TL = 50℃

② To decide mold specification required for heat transfer calculation. 1) Physical property of water

• Density: PL = 988 kg/m3

• Specific heat: CPL = 1.0 kcal/kg•℃

• Viscosity: µ = 5.58×10-4_P

A•S = 2.009 kg/m•h

• Heat conductivity: λF = 0.552 kcal/m•h•℃

2) To calculate weight of Sprue runner

• Dimension: (φ5×130) mm = (φ0.005×0.13)m • Volume: (0.0052×π/4) ×0.13 = 2.55×10-6_m3

• Weight: (2.55×10-6) ×1030 = 2.63×10-3 kg 3) To calculate weight of product

• Volume: (0.1×0.1×0.002) ×2 = 4.0×10-5 m3 • Weight: (4.0×10-5) ×1030 = 0.0412 kg 4) To assume cooling tube diameter.

• From Table 1-3.1, the diameter φ8 is assumed. D = 8 mm → 0.008 m

③ To check appropriateness of molding cycle 1) To calculate heat dissipation rate (A)

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Mold Design (Advance) Page 14 2) To calculate theoretical cooling time

⎥⎦ ⎤ ⎢⎣ ⎡ − − × π × × π = • − ₇₀ ₅₂ 52 220 4 LN 10 55 . 5 002 . 0 T 4 2 2 C =1.807×10-3h = 6.5 sec

Injection machine (100~150 ton capacity) to be used for this size of product usually required additional 10 seconds maximum for other functions than cooling time. Thus target cycle time of 20 seconds is reasonable. It could be less than 15 seconds.

④ To calculate injection weight per hour

• Weight per one shot: 2.63×10-3 +0.0412 = 0.0438kg • W = 0.0438×3600 / 20 = 7.884 kg/h

⑤ To calculate heat (QR) to be removed

QR = 1.5 QA = 1.5×7.884 {0.35(220－70)} = 620.9 kcal/h

⑥ To calculate flow velocity of cooling water h / m 6251 ) 50 52 ( 1 988 008 . 0 9 . 620 4 V 2× × − = × π × =

⑦ To check Reynolds’s number

RE = 0.008×988×6251 / 2.009 = 24593

RE falls in the range of 10,000~30,000. Thus water tube diameter φ8 is ok. ⑧ To calculate Prantle number (PR).

PR = 2.009×1 / 0.552 = 3.64 ⑨ To calculate Nusselt number

NU = 0.023× (24593) 0.8×3.641/3 = 115.2 ⑩ To calculate heat transfer coefficient (α)

α = 115.2 ×0.552 / 0.008 = 7949 kcal/m2_•h•℃

⑪ To calculate cooling water tube area (AL).

AL = 620.9 / 7949× (52–50) = 0.039m2

⑫ To calculate total cooling water tube length

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Fig. 1-3.1 Temperature Balance vs. Mold Temperature Control Device

Table 1-3.1 Mold Size vs. Channel Diameters

Balanced Mold Temperature without temperature control device.

Balanced Mold Temperature with temperature control device M old T emp er atu re

Time Time Time

a. Without Temperature Control Device (Deviated by outside influences)

b. With Cooling Device c. With Heating Device (natural cooling) Nominal Mold Temperature Cooling Heating Mold Size

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Table 1-3.2 Comparison in Mold Temperature Control Method

Items Coolant Circulation Heater

Objective Cooling Heating

Mold Temperature Lower than naturally balanced temperature (-20～130℃)

Higher than naturally balanced temperature (higher than 100℃)

Cooling Method To circulate heat transfer medium that is cooled by cold water.

Heat transfer to platen or natural radiation.

Heating Method To circulate heat transfer medium heated by heater.

To heat the mold by heater. Temperature Control To detect medium temperature and

control the temperature.

To detect and control mold temperature directly.

Strong Points • Good flexibility in temperature control design.

• Independent to mold heat capacity due to forced cooling.

• Relatively low cost.

• Relatively easy in design and manufacturing.

Characteristics

Weak Points • Relatively high cost.

• Relatively difficult in design and manufacturing.

• Less flexibility in temperature control design.

• Dependent to mold heat capacity due to natural cooling.

Table 1-3.3 Selection Criteria in Temperature Control Method by Resin

Water Circulation Method Resin Mold Temperature _°C

Supplied Pressure Added Pressure

Heater Method PE 30~70 ○ △ × PP 20~80 ○ △ × +++++++P 0S0 40~60 ○ △ × PVC 40~70 ○ △ × ABS 40~70 ○ △ × AS 40~80 ○ △ × PMMA 50~80 ○ △ × MPPE 50~100 △ ○ △ PA 50~110 △ ○ △ PBT 60~110 △ ○ △ POM 70~110 △ ○ △ PC 80~120 △ ○ ○ PPS 120~160 × △ ○

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1-4 Design of Temperature Control Device

Coolant circulation method

Important point for the design of coolant circulation system is to optimize the design in terms of coolant circulation channel, size and location in consideration of product quality and productivity. For the design of pressurized water circulation system, basic design concepts are the same but special attention should be paid for sealing and safety as well. It is suggested to refer technical information supplied by the system suppliers.

Reduced pressure suction type of temperature control device will not be treated in this section because of its specialty in the circuit design. Hereafter, we will discuss about design concept for standard temperature control device utilizing water, an excellent coolant for heat exchanging.

1-4-1 Coolant channel diameter and flow velocity

Cooling efficiency is higher if coolant in the channel is under turbulent flow where boundary film is thin. Thus it is important to decide proper diameter of coolant channel so as to make the flow stable turbulent with Reynolds’s number RE 10.000~30.000. As you may refer to formula (1.3.11), RE seems proportional to diameter d; in other words, a big diameter seems to give a big RE. This may be true if other factors stay the same. But if flow volume is given constant, flow velocity is reversibly proportional to channel cross section area, which is proportional to the 2nd power of channel diameter in case of round channel. Therefore if flow volume is given constant, flow velocity and RE become small with big diameter D referring to formula (1.3.8) and formula (1.3.11). Accordingly if channel diameter D is too big, heat-exchanging performance drops due to smaller flow velocity, Reynolds’s number, Nusselt number and heat transfer coefficient. If the diameter is too small, the heat exchanging performance will also drop due to less flow volume and increased pressure loss in the flow channel. Thus the channel diameter should be appropriately designed referring to Formulae (1.1.2.2), (1.3.8) Table (1.1.2.1). When passage is not of round shape, equivalent diameter should be applicable as explained in the “Gate Runner System”. The equivalent diameter (DE) was as follows:

DE = 4AS/LW... Formula (1.4.1)

Where AS: Cross section area of coolant channel

LW: Circumference length of AS

Fig. 1-4.1 shows equivalent diameter for various cross sections such as half a round or

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Fig. 1-4.1Channel Section vs. Equivalent Diameter

Square Rectangle Half Circle

Channel Section

Equivalent Diameter

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Mold Design (Advance) Page 19 1-5 Heater capacity

Heater capacity can be calculated as follows:

P = WM • CPM (TT－TI) ／TU • 860 • η (kW) ... Formula (1.5.1)

Where P: Capacity (kw) WM: Mold weight (kg)

CPM: Specific heat of mold (kcal/kg•℃)

TI: Atmospheric temperature (℃)

TT: Target temperature (℃)

TU: Heating time (h)

η : Efficiency (0.2~0.5)

η Is a value due to heat transfer loss due to radiation loss or loss due to heater mounting, etc, and normally it is set as 0.5. If you utilize a heater with higher capacity and with adjustable power arrangement, you may ensure stable mold temperature by adjusting heating and radiation conditions in addition to shortened preparation time.

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Fig. 1-5.1stallation Points of Heat Insulation Plate for High Temperature Application

Fig. 1-5.2 Mold with Heated Temperature Control Device

Platen of Injection Machine (Fixed Side)

Heat Insulation Plate Heat Insulation

Plate Platen of Injection

Machine (Movable Side)

Heat Insulation Plate (Movable Mold

Plate, All around Receiving Plate) Heat Insulation Plate (All around Fixed Side Mold Plate)

Heater

Heater Mounting Bolt

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Mold Design (Advance) Page 21 1-6 Clamping force

In order to calculate required clamping force of a mold used for a product, we need to know a force on the mold toward opening direction received from injected resin. This mold opening force (FO) can be expressed as the product of total projected area of a product and a

runner and average molding pressure (cavity inside pressure) as follows

FO = AA • PM • 10-3 (tf) ... Formula (1.6.1)

Where AA: Total projection area (cm2)

PM: Average molding pressure (kgf/ cm2)

Projection area is the area of a product projected in the mold clamping direction (usually perpendicular to PL). Total area is shown below (Fig. 1-6.1)

AA = N • AP + AR (cm2)... Formula (1.6.2)

Where N: No. Of cavity

AP: Product projection area per piece (cm2)

AR: Runner projection area (cm2)

In the case of 3-plate mold, notice that projection of cavity and runner overlaps. If separately calculated, overlapped area will be calculated twice. Assuming mold is of transparent, project parallel light from nozzle side, and then figure projected area on the movable platen surface (Fig. 1-6.2).

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Fig. 1-6.1 Projected Area used for Binding Force Calculation

Fig. 1-6.2 Total Projected Area of 3 Plate Mold

Projection

Projected Area a) Container Type Product b) Disc Type Product *Projected area has no influence from height dimension.

Product _{Sprue, Runner}

Projection

Total Projected Area (Aa)

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Formula (1.6.1) shows that mold-opening force is proportional to projected area if average clamping force is constant. Let us look into molding pressure.

Injected mold by injection machine fills a mold space against pressure loss caused by nozzle, sprue, runner, gate and cavity.

Thus there exists a big pressure difference between sprue area and a part of cavity where resin is filled in the end. Even after resin filling, molding pressure varies from place to place (Fig. 1-6.3). However, mind that you need to know average pressure, not all in different parts.

Although average clamping force varies depending upon product shape, molding condition, mold structure, etc., Table 1-6.1 can be practically used for your guideline. If your calculation reveals clamping force exceeds average mold opening force, the machine should be well justified. In practice, the clamping force should be evaluated as 80% of maximum clamping force against mold opening force compensating estimated average of the opening force. Required clamping force (FC) then is shown as follows.

0.8 FC ≧ FO (tf)... Formula (1.6.3)

FC ≧ FO／0.8 (tf)... Formula (1.6.4)

FC ≧ 1.25 FO (tf)... Formula (1.6.5)

If your selected injection machine has a clamping force more than above described force, you will be able to mold products without burrs on PL surface. But be minded that too big a clamping force may cause you a trouble such as partial concentration of the force in the center or ineffective clamping force due to excessive size of locating hole, etc. (Fig. 1-6.4). Rule of thumb is not to go beyond 20% of above formula.

FO／0.2 ≧ FC ≧ FO／0.8 (tf)... Formula (1.6.3)

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Fig. 1-6.3 Resin Pressure Works toward Mold Opening Incorporated to Projection Area

Table 1-6.1 Various Resin Materials and Average Resin Pressure in Cavity

Fig. 1-6.4 Problems due to too Big Injection Machine

Average Resin Pressure

Binding Force

Resin Material

Average Resin Pressure in Cavity (Kgf/cm2)

Too big a locating ring hole against mold size.

Deformation of platen due to partial binding force on the middle

(particularly for toggle type)

Binding force cannot be applied properly to fixed side mold due to too

big locating ring hole.

Lo

cating

Ring

Ho

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Mold Design (Advance) Page 25 【Calculation example 1】

Calculate required clamping force for a product shown in Fig. 1-6.5. Material is ABS and numbers of cavity are two.

[Calculation]

① Estimate runner layout (Fig. 1-6.6). ② Calculate runner projection area (AR).

AR = 0.5×5 = 2.5 ≒ 3 cm2 (Round up the first decimal.)

③ Calculate product projection area (AP).

AP = (5×8)－(π×12) = 37 cm2

④ Calculate total projection area (AA).

AA = (2×37) + 3 = 77 cm2

⑤ Calculate mold-opening force (FO).

From Table 1-6.1, PM = 300 kgf/cm2,

FO = 77×300×10-3 = 23.1 tf

⑥ Calculate required clamping force (FC).

FC ≧ 1.25×23.1 = 28.9 tf

FC ≤ 5×23.1 = 115.5 tf

Thus, applicable injection machine should be of the clamping force more than 30ton and less than 110 ton (or 100 ton may be).

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Fig. 1-6.5

Fig. 1-6.6

(All Circumference)

Product Projection Area (Ap)

Sprue

Expected Runner Layout _{Runner Projection Area (A}

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1.7 Required injection capacity

A general injection machine provides 3 screw size options per clamping unit. Accordingly you should know which=0.h screw size is selected for evaluation of required injection capacity.

As screw size is given bigger, maximum injection capacity becomes bigger. But injection pressure goes opposite direction. In other words, as screw size is given bigger, maximum injection pressure becomes smaller (Fig. 1-7.1) as long as diameter of hydraulic cylinder is the same.

Thus under the same clamping force, injection machine with small screw is suited for precision thin wall products and injection machine with large screw is suited for large products with thick walls.

Injection capacity is a product of internal cross section area of injection cylinder (screw cross section area) and injection stroke.

VI =AI • SI (cm3)... Formula (1.7.1)

AI = πD2／4 (cm2)... Formula (1.7.2)

Where VI: Injection volume (cm3)

AI: Internal cross section area of injection cylinder (cm2)

SI: Injection stroke (cm)

D: Screw diameter (cm)

Maximum injection capacity shown in the specification of injection machine is calculated by above formula. This means a capacity of an object (air for example) injected under normal temperature and pressure by screw size.

Practically molding is operated by injecting plastic with viscosity and elasticity under high temperature and pressure. Therefore maximum injection capacity for injecting PS-GP (General purpose polystyrene) is usually given together with theoretical value.

Actually, internal pressure of injection cylinder is high, but density of the plastic there is smaller than that under normal temperature and pressure condition because plastic in the cylinder is expanded due to high temperature (Fig. 1-7.2).

Calculation of required injection capacity has two folds. The first step is to calculate expanded capacity of plastic per one shot for product, sprue and runner under high pressure

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and temperature condition, then the second step to compare it with maximum injection capacity assigned for an injection machine. Specifically, total injection capacity can be given as follows:

VA = N • VP + VR (cm3)... Formula (1.7.3)

Where VA: Total injection capacity (cm3)

N: Number of cavity

VP: Mold volume per one piece (cm3)

VR: Volume of sprue and runner (cm3)

To figure volume expansion accurately, we need PVT data per plastic material, but in our purpose it is not necessary to go to those details. Although there is some variations in pressure and temperature conditions, we may approximately estimate 90% of density (1.11 times volume expansion) for amorphous resin, of which specific volume is not much influenced by temperature, and 80% of density (1.25 times volume expansion) for crystalline resin, of which specific volume is much affected by temperature. Furthermore taking account of injection efficiency due to back flow and cushion amount, additional safety factor 80% shall be introduced. To make a formula applicable to all resin, regardless of crystalline and amorphous, density in the injection cylinder is now assumed to be 85% of the value under normal pressure and temperature condition. Then required injection capacity (VS) will

be as follows:

VS≧VA／ (0.8×0.85) = VA／0.68 (cm3) ... Formula (1.7.4)

Or VS = 1.47 VA (cm3)... Formula (1.7.5)

Injection machine with larger injection capacity than above can be utilized, but it should not be too large. Expected problem is that resin starts decomposition in the cylinder if it stays too long in the cylinder. As a minor problem, measuring accuracy drops due to small measuring stroke. Thus similarly to the case of clamping force, calculated injection capacity (VS) should not be less than 20% of theoretical maximum injection capacity of the injection

machine. Accordingly formulae (1.7.4) and (1.7.5) can be expressed as follows. VA／0.16 (VA／0.8×0.2) ≧VS≧VA／0.68 (cm3) ... Formula (1.7.6)

Or 6.25 VA≧VS≧1.47 VA (cm3) ... Formula (1.7.7)

Approximately

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Mold Design (Advance) Page 29 【Calculation example】

Calculate required injection capacity of the product. [Calculation]

① Calculate volume of sprue and runner. Sprue: (π×0.52_{) ／4×7 = 1.4}

(Taper portion is assumed to be cylindrical.) Runner: (π×0.52_{) ／4×5 = 1.0}

Thus VR = 1.4 + 1.0 = 2.4 cm3

② Calculate volume of the product.

VP = (2×5×8)－ (1.8×4.6×7.6) － (π×12×0.2) = 16.5 cm3

③ Calculate total injection amount. VA = (2×16.5) + 2.4 = 35.4cm3

④ Calculate required injection capacity. VS ≧1.5×35.4 = 53.1≒54 cm3

VS ≦6×35.4 = 212.4≒212 cm3

Thus you can apply an injection machine of which theoretical maximum injection capacity is more than 54 cm3 and less than 212 cm3.

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Mold Design (Advance) Page 30 Principle

Hydraulic Pressure (Po)

Pi•D2_{= Po•Do}2

Screw Diameter (D) Big Small

Screw Section Area (Ai) π•D2_/4 _{Big Small}

Injection capacity (Vi) Ai•S Big Small

Injection Pressure (Pi) Po•Do2_/D2_Small _Big

Application _{Thick Wall Product}Big Product. _{Thin Wall Product}Precision Product

Fig. 1-7.1 Internal Injection Cylinder (Screw Diameter) vs. Injection Capacity / Injection Pressure

Fig. 1-7.2 PVT Diagram of Resin

Injection Pressure (Pi)

Material: PMMA Material: PP

Temperature (℃) Specific V olume Specific V olume Temperature (℃)

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1-8 Mold Strength

Mold strength should be evaluated to see if the deformation of the mold due to molding pressure (injection pressure or holding pressure) stays within allowable tolerance limit. Two considerations must be highlighted. One is how to estimate molding pressure. Another is how to decide allowable tolerance limit.

Molding pressure comes from resin. With some exceptions, you need not take up a pressure during injection but a pressure after injection. This is nothing more than the molding pressure inside cavity pressure that was explained and calculated in the previous section for calculating required clamping force. But in this section we choose 500 kgfcm2_{with some margin.}

Tolerable deformation varies depending upon product accuracy, mold structure, locations, etc. One practical reference is if the deformed amount results in generation of burr or not. Clearance to generate burr can be considered as the depth of air ventilation. If burr cannot be a reference, the allowable tolerance should be looked into from the aspect of allowable repeated stress on the mold or product accuracy.

Generally allowable deformation amount is 0.1~0.2mm unless the mold is extremely small in size. 1) Side walls of rectangular cavity

There are two types. One is of split type consisting of sidewalls and bottom plate. Another is made from one block, in other words one-piece cavity.

Split type can be machined easily with high accuracy but weaker in strength. Let us see the difference in strength between split type and one-piece rectangular cavity.

1-8-1) Split type

In the split type, calculation disregards restraint of bottom plate. Actually sidewalls are bolted together with bottom plate or mounting plate. Thus calculation results in the value with safety factor by disregarding binding and friction influence from bolts (Fig. 1-8-1.1).

We apply a model of a beam both side fixed and with equal weight distribution to cavity walls for strength analysis (Fig. 1-8-1.2). Be minded molding pressure uses cm unit, while strength calculation uses mm unit. The maximum deformation (δMAX) on the both side fixed beam appears in the middle

as follows: ) mm ( EI 384 PAL4 MAX = δ ... Formula (1.8.1.1) Where P: Molding pressure (kgf/mm2₎

A: Product height (mm) L: Product length (mm)

E: Vertical elasticity coefficient (kgf/mm2) I: Cross section moment of inertia

Cross section of the beam is of rectangular, thus moment of inertia is as follows: ) mm ( 12 BH I= 3 ... Formula (1.8.1.2) Where B: Cavity thickness (mm)

H: Cavity wall thickness (mm)

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Mold Design (Advance) Page 32 ) mm ( Eb 384 PAL 12 H 3 MAX 4 δ = ... Formula (1.8.1.3)

Fig. 1-8-1.1 Molding Pressure on Cavity with Split Structure

Fig. 1-8-1.2 Model of a Beam, Both Side Fixed, for Cavity Wall

Pm: Molding Pressure (kgf/cm2₎

δMax: Maximum Deformation (mm)

P: Evenly Distributed Weight (Molding Pressure) (kfg/mm2₎

E: Vertical Elasticity Coefficient (kfg/mm2₎

δmax = pal4/384EI I = bh3/12

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Mold Design (Advance) Page 33 【Calculation example】

Calculate wall thickness (H) of split type cavity shown in Fig. 1-8-1.3. Resin material is ABS and vertical elasticity coefficient of the mold is assumed to be E = 2.1×104_kgf/mm2_.

[Calculation]

Molding pressure (P), allowable deformation (δMAX) is assumed as follows. Then formula

(1.8.3) is applied to calculate wall thickness (H). P = 500 kgf／cm2_{= 5 kgf／mm}2

δMAX = 0.025mm (in the case of ABS)

) mm ( 6 . 24 025 . 0 40 10 1 . 2 384 100 20 5 12 H 3 4 4 = × × × × × × × =

Thus, you may decide the wall thickness 25mm if there is no space available, but if possible, decide 30mm for safety consideration.

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1-8-2) One-piece type

One-piece type cannot be simplified as split type. Table 1-8-2.1 shows coefficient (C) corresponding to ratio (L/A) of product length to height. Then allowable deformation (δMAX)

is calculated as below (Fig. 1-8-2.1).

) mm ( EH CPA 3 4 MAX = δ ... Formula (1.8.2.1)

Where P: Molding pressure (kgf/mm2) A: product height (mm)

E: Vertical elasticity coefficient (kgf/mm2) H: Cavity wall thickness (mm)

Cavity wall thickness can be derived from formula (1.8.2.1) as follows.

) mm ( E CPA H 3 MAX 4 δ = ... Formula (1.8.2.2) 【Calculation example】

Assuming last example is of one-piece type; calculate wall thickness (H). Other conditions are the same.

[Calculation]

① Calculate volume of sprue and runner. Sprue: (π×0.52_{)／4×7 = 1.4}

(Taper portion is assumed to be cylindrical.) Runner: (π×0.52_{) ／4×5 = 1.0}

Thus VR = 1.4 + 1.0 = 2.4 cm3

① L/A = 100/20 = 5 kgf/mm2 From Table 1-8-2.1, C = 0.142.

② Calculate cavity wall thickness (H) from formula (1.8.2.2). P = 500 kfg/cm2 = 5 kfg/mm2

δ = 0.025mm (in case of ABS) E = 2.1×104 kfg/mm2 A = 20mm Thus, 6(mm) 025 . 0 10 1 . 2 200000 5 142 . 0 H 3 4× = × × × =

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Wall thickness has come up with 6mm, but it is advised to select 10mm for safety consideration.

Fig. 1-8-2.1 Molding Pressure Operated on Cavity with One-piece Structure

Table 1-8.2.1 Various Resin Materials and Average Resin Pressure in Cavity

Pm: Molding Pressure (kgf/cm2₎

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1-8-3 Wall thickness of cylinder type cavity

Thick wall cylinder is applicable in material strength analysis. Similarly to rectangular type cavity calculation, we will try to make the clearance resulted from deformation smaller than the clearance to cause burr for different resin materials. It will be complicated if we try to calculate wall thickness of cylinder tube. Therefore we evaluate if the deformation is within allowable tolerance under a given wall thickness.

Deformation (δ) in Fig. 1-8-3.1 is given as follows. ) mm ( r R r R E rP 2 2 2 2 ⎥⎦ ⎤ ⎢⎣ ⎡ ₊_λ − + = δ ... Formula (1.8.3.1)

Where P: Molding pressure (kfg／mm2) R: Cavity outside diameter (mm) R: Cavity insider diameter (mm)

E: Vertical elasticity coefficient (kfg／mm2) λ : Poisson’s ratio (λ = 0.3 for steel) 【Calculation example】

Evaluate if the deformation of cylinder type cavity in Fig. 1-8-3.2 can be within tolerance limit or not. Resin material is ABS.

[Calculation]

Calculate δ of formula (1.8.3.2) and compare it with the clearance 0.03mm to cause burr for ABS resin. P = 500 kfg/cm2 = 5 kfg/mm2 E = 2.1×104 kfg/mm2 r = 20mm R = 30mm λ = 0.3 Thus, 0.3 0.014(mm) 20 30 20 30 1 . 2 5 20 2 2 2 2 = ⎥⎦ ⎤ ⎢⎣ ⎡ ₊ − + × = δ

Calculation result reveals that the deformation is less than minimum clearance (0.03mm) to cause burr of ABS resin. Thus the wall thickness is appropriate.

In the case of cylinder type cavity, inserts are used in the plate. As the clearance between inserts and holes on the plate is around 0.01~0.02mm, the plate can share some stress when deformation exceeds the clearance above.

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Fig. 1-8-3.1 Molding Pressure Operated on Cavity with Cylinder Shape

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1-8-4 Mold weight and center of gravity

Calculation for mold weight and center of gravity is required to determine size and position of hook bolts for hoisting a mold. Normally the shape of a mold is of rectangular sections. Thus the volume can be easily figured, so as weight by multiplying specific weight of mold material. In case of steel, apply 7.87.

Regarding center of gravity, it can be determined by estimating the position in the direction of mold thickness. Mold is normally symmetrical with a centerline in the injecting direction. The center of gravity should locate on the centerline.

Calculation proceeds firstly to calculate weight on the center of gravity of each plate and secondly to find a point where each moment can be balanced.

Referring to Fig. 1-8-4.1, calculate each moment as a product of weight on the center of gravity of each plate and distance based on fixed side clamping plate. The total moment should balance with a moment as a product of total mold weight and distance from the reference point to the center of gravity.

w • x = ∑WN • LN... Formula (1.8.4.1)

Where w: Mold weight

x : Distance from reference point to center of gravity.

WN: Weight on each plate.

LN: Distance from reference point to center of each plate.

Distance from reference point to center of gravity can be derived as follows:

w

∑

= WNLN

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Fig. 1-8-4.1 Calculation of Mold Center of Gravity

Fixed Side

M

oun

ting

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Mold Design (Advance) Page 40 【Calculation example】

Calculate center of gravity of the mold shown in Fig. 1-8-4.2. Mold material is S50C. [Calculation]

① Specific weight 7.87 is applied for S50C. ② Calculate weight of each plate.

W1 = 7.87×25×300×300×10-6 = 17.7 kg W2 = 7.87×50×250×300×10-6 = 29.5 kg W3 = 7.87×40×250×300×10-6 = 23.6 kg W4 = 7.87×40×250×300×10-6 = 23.6 kg W5 = 7.87×70×38×2×300×10-6 = 12.6 kg W6 = 7.87×35×170×300×10-6 = 14.1 kg W7 = 7.87×25×300×300×10-6 = 17.7 kg

③ Calculate total mold weight.

W = 17.7 + 29.5 + 23.6 + 23.6 + 12.6 + 14.1 + 17.7 = 138.8 kg

④ Calculate distance from fixed side mounting face to the center of plate. (Fig. 1-8-4.3) ⑤ Calculate moment of each plate.

(No need to convert mm to m. Weight of each plate is assumed to be on the center.) Fixed side clamping plate: W1•L1 = 17.7×12.5 = 221 kg•mm

Fixed side mold plate: W2•L 2 = 29.5×50 = 1475 kg•mm

Movable side mold plate: W3•L 3 = 23.6×95 = 2242 kg•mm

Movable side support plate: W4•L 4 = 23.6×135 = 3186 kg•mm

Spacer: W5•L 5 = 12.6×190 = 2394 kg•mm

Ejector plate (upper ＆ lower): W6•L 6 = 14.1×207.5 = 2926 kg•mm

Movable side clamping plate: W7•L 7 = 17.7×237.5 = 4204 kg•mm

⑥ Calculate gravity position (X) from formula (1.3.9.23).

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Mold Design (Advance) Page 41 Fig. 1-8-4.2 Fig. 1-8-4.3 Fixed Side M oun ting Face

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In the case of calculation example 1.3.9.6, center of gravity locates at 120 mm from fixed side mounting face. If you place a hook bolt at this location, you can hoist the mold without tilting the mold. But notice this location is near parting surface of movable mold plate and support plate. Thus a screw cannot be tapped there. Two solutions can be proposed here. One is to use a lifting bar and another is to move the screw position a bit toward fixed side clamping plate (Fig. 1-8-4.4). The moved amount should be selected for the mold not to tilt more than 10°. If it is moved in the opposite direction, workability to position a locating ring to the nozzle hole of injection machine will be affected (Fig. 1-8-4.5). In our case here, if the screw position is moved toward fixed side mounting face, it comes to much closer to the parting surface. Thus it is recommended to apply a lifting bar.

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Fig. 1-8-4.4 Design Consideration when Gravity Center is Close to Plate Parting Surface

Fig. 1-8-4.5 Mold Inclination and Workability

Hook bolt Lifting Bar

Receiving Plate Movable Side Mold Plate a. By Lifting Bar Calculated Gravity Center

b. By shifting screw position Difficult to position screw hole

for hoisting if gravity center is too close to parting surface of mold plate. Platen of Injection Machine Platen of Injection Machine Gravity Center Receiving Plate Movable Side Mold Plate

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1-8-5 Return force of ejector plate

Ejected ejector plate is normally returned to its original position by spring force of a spring installed on the periphery of return pin.

If the spring force is too weak, the ejector plate cannot fully return to its original position. If it is too strong, operational balance will be affected and galling may be caused on the return pin.

Thus we may define the return force of ejector plate should stand weight of ejector plates (upper and lower) and their friction force.

Friction force is related to friction coefficient of ejector plates (upper and lower). As we know the maximum friction coefficient is 1, it must be enough to estimate the friction force 2 times of the plate weight. Generally, number of springs to be installed on the periphery of return pin is 4. Thus you should calculate the shared friction load per spring is 1/2 of ejector plate weight (2×1/4).

The spring is better to have a smaller spring constant value to assure smoother load transfer to the spring while stroking (Fig. 1-8-5.1). In addition initial deflection of the spring is better not to exceed thread length of stripper bolt (normally 10~15), otherwise you will have a difficulty in installing stripper bolt to female thread hole because of the long spring (Fig.

1-8-5.2). Following checkpoints may be useful for selecting correct spring from available ones

in the market.

① Internal diameter of the spring should be at least 1 mm larger than return pin outside diameter.

② Maximum deflection in usage should be within allowable limit.

③ The spring should have enough returning force at the initial deflection.

④ When spot facing is made for the spring, clearance around return pin can be secured (two times pin diameter) and there should be no interference with coolant channel.

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Fig. 1-8-5.1 Spring Constant vs. Load – Deflection Diagram

Fig. 1-8-5.2 Initial Deflection of Return Pin Spring and Mounting Workability of Ejector Plate

Loa

d

Load – Deflection Diagram for Spring A Load – Deflection Diagram for Spring B Initial Deflection of Spring A

Initial Deflection of Spring B Ejection Plate Stroke Initial Load

Maximum Deflection of Spring A Maximum Deflection of Spring B Deflection

Ejector Plate Fastening Bolt

Screw does not reach to tap hole if initial deflection of return pin spring is too big.

Return Pin Spring

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Mold Design (Advance) Page 46 【Calculation example】

Conditions of an ejector plate are as below. Select appropriate length of springs shown in

Fig. 1-8-5.3.

[Ejector plate conditions]

• Upper ejector plate dimension: 250×110×13mm • Lower ejector plate dimension: 250×110×15mm • Return pin diameter: φ12mm (4pcs)

• Maximum ejector stroke: 17mm • Initial deflection: 8mm

[Calculation]

① Calculate weight of ejector plates.

25×11×(1.3×1.5)×7.87×10-3_≒6.1kg

② Calculate ejector plate return force per spring. 6.1×2/4 = 3.05 kg (or more)

③ Calculate spring constant. 3.05/8 = 0.38 kgf/mm

④ Calculate maximum deflection in use. 17+8 = 25mm

⑤ Select springs in the list to meet required spring constant and deflection.

From Fig. 1-8-5.3, springs having length (L) more than 45mm and less than 100mm are appropriate.

In practice, select one of appropriate springs evaluating influence of spot facing and interference with coolant channel.

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Fig. 1-8-5.3 Partial List of Coil Spring in the Market

(fromミスミフェイス)

1~19 \ Price

N (kgf) Load

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Calculate the appropriate diameter of support pin.

Calculation for the appropriate diameter of support pin and

Deflection of Support pin

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Mold Design (Advance) Page 49 Bending Moment MMAX = Wl Deflection 3EI W = l3 MAX δ

I =Cross Sectional Secondary Moment = 64

d2

π E= Young’s modulus

(Vertical elastic modulus) = ∈ ∂ = Ε = 2.1 x 104 kgf/mm2 ∈= Deflection ∂= Rectangular stress 3EI W 3 l = MAX δ 4 d 3E W 3 ₄ π δMAX = l d = 4 MAX 3 . . E 3 64 . W δ πl Example:-

Support pin length = 250 No. of Support pins = 4

Weight of Cavity Plate = 400 kg E= Young’s modulus = 2.1 x 103 Allowable Deflection = δ = 0.01 MAX

length of SP 400 mm

Weight of Cavity Plate 220 kg Young's Modulus 21000 kgf/mm2 Allowable Deflection 0.01 mm Diameter of SP 102.6792791 mm Diameter of single SP 25.66981977 mm

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Mold Design (Advance) Page 51 `

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Chapter - 2

Mold

Material &

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2- Mold Material

Limiting our scope to a mold for plastic injection molding, steel is the most popular material. Particularly JIS S50C and S55C are mostly applied because these are standard materials of the mold bases in the market.

It is important to select right material to satisfy purpose of the mold and its application on the part of end users. If necessary, heat-treating or surface finish must be carried out to satisfy requirements. Here we will take up materials to be used for main parts of the mold, cavity and core.

2-1 Mold Material

2-1-1 Basics

Normally users’ specification specifies if the material is of heat-treated (quenched) or not treated (raw) for cavity and core material. Note that pre-hardened steel, which is heat-treated when supplied but will not be heat heat-treated after machining, is classified as raw steel.

As-rolled steel → Raw type

Not heat-treated

Materials for Pre-hardened steel → Raw type

Cavity and Core

Heat treated → Heat treatment → Quenched type

Table 2.1.1.1 shows various steel for plastic mold with bland names. Molds made of

as-rolled steel and pre-hardened steel belong to raw type. Pre-hardened steel is heat-treated having 30~40 HRC hardness and yet having a good machineability. Molds made of pre-hardened steel are used without heat treatment. Thus the mold processing is the same as that of as-rolled steel. Cost of a mold is also similar in both cases.

On the other hand, there are two types in quenched type. One is to harden and temper the mold after machining and to finish the mold just by simple polishing. Another is to finish a heat-treated mold with a certain deformation clearance by a grinder or EDM (Electric discharge machine). The former is used for a mold, which does not require high precision but only erosion resistance. Thus the cost is on the same level as raw type. But the latter involves time consuming finishing on the hardened steel surface. Thus the cost is much higher than raw type. Qualified material for mold should satisfy following points.

① Good machineability. ② High abrasion resistance. ③ High corrosion resistance. ④ High toughness.

⑤ High strength.

⑥ Homogeneous property without segregation and pin holes. ⑦ Good heat-treating with less deformation.

⑧ Good heat conductivity. ⑨ Reasonable price. ⑩ Easy procurement.

No single material satisfies all items above. Particularly the extent of abrasion resistance to determine mold life is deeply related to machineability that affects cost of a mold (Fig.

2-1-1.1).

Major factors for determining mold material in the user’s specification are number of injection shot, application and molding material.

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Fig. 2-1-1.1 Wear Resistance and Machineability of Plastic Mold Material

2-1-2 Number of shots and mold material

Total number of shots is a product of monthly production volume and mold life. Total number of shots = monthly production volume×mold life (month)

Naturally, a mold having less number of shots will be made of cost conscious material because depreciation cost per shot needs to be lowered as much as possible, while a mold having high number of shots will be made of life conscious material. Note that in this case cost means machining cost rather than material cost.

Characteristics and application of popular steel materials for mold for plastic molding is shown below in the order of low durability (Table 2.1.1.1).

2-1-2-1 SC steel

S50C and S55C are used for material of mold bases available in the market. They are widely used for cavity and core material of which total shots are less than 100,000. Particularly they are applicable for large molds.

Table 2.1.1.1 shows SC steel under both as- rolled steel and pre-hardened steel. It is

recommended to use pre-hardened one for cavity and core due to better abrasion resistance. 2-1-2-2 SCM steel

Generally machineability of SCM steel is not so good comparing with SC steel. Pre-hardened steel adjusted for better machineability with 28~33HRC hardness is often used for cavity and core material, mold base material, mold plates and holders that require hardness to certain extent.

QT : Quenching and Tempering PH : Prehardened I : Improved Steel W ear Res is tance Machineability Ideal Mold Material 13Cr Class QT 13Cr Class QT I I Winkle Class PH Winkle Class PH Free-Cutting Class PH Free-Cutting Class PH I I I SCM Class SC Class SKD 11 SKD 61 AISI P21

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2-1-2-3 AISI-P21 steel

This is a kind of pre-hardened steel, precipitation hardened with 40 HRC hardness, originated to AISI-P21 of US specification. This should stand for 500,000 shots for usual resins.

There are two types. One is a material with improved machinebility, close to S50C and S55C, by adding lead (Pb) and sulfur (S). Another is a material with improved machinebility for electric discharge machining, texturing ability and polishing. Apply one way or another depending upon mold characteristics.

So far steel materials that are not quenched after machining have been introduced. These materials have benefits of easier machining, costs and delivery comparing with quenched type. Currently 40HRC hardness of pre-hardened steel is the hardest, but it is expected to be 50 HRC hardness in view of recent development in high speed and high precision machining capability for hard metals.

2-1-2-4 SKD-61

SKD-61 steel is normally used for die-cast mold as tool steel for hot processing. But it also is applicable for plastic mold for relatively large production volume.

Table 2.1.1.1 shows this material under pre-hardened steel with 40 HRC hardness. But

normally raw steel is machined and quenched to 50 HRC hardness after machining. A life of quenched mold can stand for at least one million shots for usual resins. If conditions are met, 2~3 times longer life can be expected.

SKD-61 can be nitride to the extent of 0.05mm in depth with 900HV hardness or more. It means that nitride layer still exists after finishing as much as 0.01~0.02mm. Therefore nitride SKD-61 is quite effective for a mold that is subject to galling or seizing.

2-1-2-5 SKD-11

SKD-11 steel is normally used for press mold as tool steel for cold processing. But it is also applicable for mold for plastic with reinforced fiberglass or for mass production.

11 has high resistance to abrasion. When it is quenched at 58~60 HRC hardness, SKD-11 can stand for around 5 million shots without special coating on the surface. Weakness may be poor machineability and toughness. Steel suppliers are developing improved SKD-11 to cover such weakness.

As Table 2.1.1.1 shows, grains are laid out in dense and homogeneity. Thus powder forging is made available. SKD-11 is applicable for molds that requires mirror polishing and abrasion resistance.

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2-1-2-6 Powder metal

This is applicable for a mold for super mass production, super engineering plastics with reinforced fibers, IC, etc.. Similarly to SKD-11, this is made from powder metallurgy process. Powder metal is superior to high-speed tool steel (SKH-51) in terms of hardness and toughness, but the cost is much higher. Therefore this is often used partially in the form of inserts wherever high abrasion resistance is required.

So far we have discussed about typical mold materials in relation with number of shots required for a mold. It is advised to analyze available materials, applications, number of shots, etc. for mold design. a sample of which is shown in Table 2.1.2.6.1

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Table 2.1.2.6.1 Total Shots vs. Mold Material

Total Shots (×1000) Resin, Application 5,000≤ 1,000≤ <5,000 500≤ <1,000 100≤ <500 20≤ <100 <20 Mold Material Heat Treatment Surface Treatment Powder Metal Q/T (62-64 HRC) No/PVD SKD-11 (I) Q/T (HRC 58-60) No SKD-61 (I) Q/T (HRC 48-50) No P/H Steel (M, T) No (HRC 40) No S50C, S55C No No AL alloy (HB 150) No No Resin for General

Application Mold Material Heat Treatment Surface Treatment SKD-11 (I) Q/T (HRC 58-60) No/PVD SKD-61 (I) Q/T (HRC 48-50) No/Nitrided P/H Steel (M, T) No (HRC 40) No/Nitrided SCM (I) No (HRC 33) No Engineering Plastic (Not Reinforced) Mold Material Heat Treatment Surface Treatment Powder Metal Q/T (HRC 62-64) No/PVD SKD-11 (I) Q/T (HRC 58-60) No/PVD SKD-61 (I) Q/T (HRC 48-50) No P/H Steel (M, T) Q/T No (HRC 40) No S50C, S55C No No Mold Material Heat Treatment Surface Treatment Powder Metal Q/T (HRC 64-66) PVD Powder Metal Q/T (HRC 62-64) No/PVD SKD-11 (I) Q/T (HRC 58-60) No/Nitride SKD-61 (I) Q/T (HRC 48-50) No Engineering Plastics (Reinforced) Mold Material Heat Treatment Surface Treatment Cemented Carbide Steel Insert No (HRA85-90) No/PVD SKD-11 (I) Q/T (HRC 58-60) No/PVD SKD-61 (I) Q/T (HRC 48-50) No/Nitride P/H Steel (M, T) Q/T No (HRC 40) No Fire Retarded Grade Mold Material Heat Treatment Surface Treatment

Powder SUS Class Q/T (HRC 56-58) No/PVD 13 CrSUS Class Q/T (HRC 50-52) No/PVD 13 CrSUS Class Q/T (HRC 50-52) No 13 CrSUS Class Q/T (HRC 33) No 13 CrSUS Class Q/T (HRC 33) No Transparent Product, Optical Product Mold Material Heat Treatment Surface Treatment Cemented Carbide Steel Insert No (HRA85-90) No/PVD

Powder SUS Class

Q/T (HRC 56-58) No/PVD 13 CrSUS Class Q/T (HRC 50-52) No/PVD 13 CrSUS Class Q/T (HRC 50-52) No P/H Steel (T) Q/T (HRC 40) No S50C, S55C No No

Note: Q / T: Quenched / Tempered PVD: Physical Vapor Deposition (Ion Plating)

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2-1-3 Plastic materials and mold materials

Depending upon plastic material that may include, reinforced fibers or additives, requirements for abrasion or corrosion resistance vary on the part of cavity and core.

2-1-3-1 Reinforced plastic

Reinforced plastic with filled material such as fiberglass causes high abrasion on the mold. The extent of abrasion is higher if the amount of filled material is greater and the material is harder. For example when glass fiber content is more than 30%, the mold life will become 10~ 20% of the life otherwise.

Mold materials for anti-abrasion were discussed in the previous section. Be minded that hard steel material may cause chipping due to inferior toughness. It may be necessary to lower the hardness and compensate it by surface treatment such as PVD.

2-1-3-2 Flame retardant plastic

Flame retardant plastic that includes halogen (bromine) or fluororesin produces corrosive gas under heat and pressure in the molding. This will shorten the life of a mold. In the case of PVC, chlorine gas is generated. Thus you need to select mold material with high corrosion resistance.

2-1-3-2-1 13Cr stainless steel

This is a stainless steel material to include 13% of chromium. This may be called 13Cr steel or SUS420 in JIS. 13Cr SUS is not quite high in corrosion resistance, but being pre-hardened steel of Martensite structure it can be used as it is due to its reasonable hardness 33HRC or can be quenched to 50HRC if needed.

Thus 13Cr SUS can be used for a mold to be mirror polished or to be used for fire retard resin or fluororesin.

2-1-3-2-2 SUS 630

This is a precipitation-hardened stainless steel having high corrosion resistance. This is supplied as prehardened steel with 35HRC hardness. SUS 630 stainless steel is applicable for a mold for highly corrosive resin such as PVC.

2-1-3-2-3 Transparent resin

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PMMA, PC, etc. are molded. Particularly for photoproducts such as optical discs or lenses, high grade of transparency is required.

Although JIS provides no specific standard for mold steel for plastic injection molding, special steel suppliers made such standard available for our application. Referring to such references select appropriate steel material to be mirror polished for transparent resin molding. Improved materials are often processed by special smelting processing such as vacuum process which brings about homogeneous and dense grain structure with minimum segregation and pin holes so as to assure a mold to satisfy with precise transcription capability. 13Cr stainless steel for such purpose is normally made by vacuum process to suit precision mirror polishing.

2-1-3-2-4 Thin products

Steel material for thin core or fine core is required to be with high rigidity and high toughness, particularly when injection is made from one side only. For such application, Maraging steel to include 18% Ni is recommended.

Maraging steel is supplied as solution treated condition and is to be hardened to 53HRC through age hardening. This material is often used for thin wall core, mirror polished core, and ejector pin with thin wall or small diameter.

2-1-4 Other mold materials

As explained so far, steel material is most balanced in properties as mold material. Thus it is widely used. Other materials than steel are being introduced for particular applications.

2-1-4-1 Aluminum alloy

Mold for extremely small production volume is not necessary is of steel material. Mold made of aluminum alloy can stand for 20,000~30,000 shots. You may extend the life even more by hardening the mold surface with alumite processing. But be aware that aluminum alloy is always subject to damage on its surface because of soft material by nature.

Benefits gained from this material must be low cost, short delivery and improved cycle time due to high thermal conductivity.

2-1-4-2 Copper Alloys

Beryllium copper (BeCu) is a typical copper alloy used for copper alloy mold for plastic molding. This material can be improved in abrasion resistance through age hardening.

Advantage of copper alloy is its high thermal conductivity, while disadvantage must be its high cost. Therefore copper alloy is used for inserts to remove heat from hot spots.