Structural System Design

Structural System Design

12 Structural System Design Injection molds are subjected to high levels of pressure from the heated polymer melt. When this pressure is integrated ...

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Structural System Design

Injection molds are subjected to high levels of pressure from the heated polymer melt. When this pressure is integrated across the surfaces of the mold cavities, forces result that typically range from tens to thousands of tons. The structural design of the mold must be robust enough to not only withstand these forces, but also to do so while producing high quality molded products. To develop a robust structural design, the mold designer should understand the relationships between the pressures, forces, and stresses in an injection mold. Figure 12.1 shows the typical flow of stresses through the mold, platens, and tie-bars. During molding, the melt pressure is exerted against all surfaces of the mold cavities. This pressure results in both compressive and shear stresses in the cavity inserts, core inserts, and support plates. This mixed state of stress is indicated by the diagonal arrows in Figure 12.1. The molding machine’s platens are also under a significant state of bending to transfer the forces to the tie bars, which are in tension.

Figure 12.1: Flow of stresses during molding

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12 Structural System Design

12.1 Objectives in Structural System Design In general, the mold designer wishes to provide a structural design for the mold that will not break due to fatigue across many molding cycles, excessively deflect under load, and will not be overly bulky or expensive. These objectives can be explicitly stated as: • • •

Minimize the stress, Minimize deflection, and Minimize cost.

Each of these objectives will be briefly discussed after which the detailed analysis and design of molds are presented.

12.1.1 Minimize Stress The state of stress varies significantly between the moving and the stationary sides of the mold. For most molds, the cavity inserts are directly supported by the top clamp plate and the stationary platen. As such, the cavity inserts are generally in a state of pure compression so very little out of plane bending occurs. On the moving side, however, the pocket required to house the ejector assembly provides no support for the core inserts. As a result, the core inserts and support plates must transmit the load via both compressive and shear stresses, which will tend to result in significant plate bending. Figure 12.2 plots the predicted von Mises stress in a hot runner mold for the laptop bezel when the surfaces of the mold cavity are subjected to 150 MPa of melt pressure. The von Mises stress, σMises, is a commonly used criterion to predict failure, which is defined as: σ Mises =

σ12 − σ1 σ 2 + σ 22 < σ limit

(12.1)

where σ1 and σ2 are the first and second principle stresses. To avoid failure, the von Mises stress should be less than some specified stress limit, σlimit. The limit stress is usually determined by two primary concerns. The first concern is that the stress should not be so high so as to plastically deform the mold. When a material is subjected to stress, it will deform, or strain. For most materials, the amount of deformation is proportional to the applied stress. The ratio of stress to strain is related to the elastic modulus, E, as: ε=

σ E

(12.2)

where ε is the strain that results from an applied stress, σ. A material with a higher elastic modulus will tend to deform less given an applied stress. The stress-strain behavior of two common mold metals, P20 and aluminum QC7, are plotted in Figure 12.3. P20 has a much higher modulus than QC7, meaning that it will

12.1 Objectives in Structural System Design

Figure 12.2: Von Mises stresses during molding

1000 900 QC7 P20

800

Stress (MPa)

700

 Yield = 830 MPa E P20 = 205 GPa

600 500 400

 Yield = 420 MPa

300

E QC7 = 71 GPa

200 100 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Strain (%)

Figure 12.3: Stress-strain behavior of P20 and Al 7076

0.8

0.9

1

1.1

1.2

1.3

301

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exhibit less strain and deformation given an applied load or stress. The yield stress is the point at which the material departs from its linear behavior. The yield stress is also the stress at which the material plastically deforms, meaning that the components with a higher stress will not return to its original shape after the load is removed. Once the material exceeds the yield stress, it can continue to carry load up to the ultimate stress, after which it fails completely. All mold designs should be engineered to operate at stresses lower than the yield stress. With respect to relating the yield stress to the limit stress, there are two common approaches. One approach is to simply set the limit stress equal to the yield stress, but then assume a worst case scenario with respect to the load condition. For example, a mold designer working with P20 could assume a limit stress equal to the yield stress of 830 MPa. To ensure the mold does not yield under load, the mold designer should then perform analysis assuming the highest melt pressures that the mold would ever be expected to see, perhaps 200 MPa. Another approach is to set the limit stress equal to the yield stress divided by a factor of safety: σ limit =

σ yield f

(12.3)

where f is the factor of safety, whose value is related to the level of uncertainty and the cost of a potential failure. Typical values range from 1.5 for non-critical mold components to 6.0 for hoist rings. When a factor of safety is used, the mold designer should apply the expected melt pressure for the mold, perhaps 100 MPa. To avoid over-designing the mold, the mold designer should not jointly apply a factor of safety with the worst case scenario. While the mold designer might expect that a design based on the yield stress with a conservative factor of safety would be robust, this approach may result in molds that fail after a significant number of molding cycles. The reason is that the continued cycling of clamping loads and melt pressure in the mold cavity causes cyclic stresses as shown in Figure 12.4. Each stress cycle allows small cracks in the mold to open and close, causing a few more crystals at the crack tip to fail. Over the course of thousands of molding cycles, these cracks will grow and propagate through the mold like a wedge driven by a hammer. Once the crack reaches a critical size, the stress concentrations around the crack will cause the mold to fail even when the mold was properly designed with a limit stress specified well below the yield stress. This failure mode is generally referred to as fatigue. Fatigue is a well understood mechanism, and the behavior of various materials has been characterized through cyclic stress testing to a million cycles or more. In general, the number of cycles that a mold can withstand will decrease with the applied stress. Figure 12.5 plots the expected number of cycles before failure due to fatigue as a function of the imposed stress; this data is generally referred to as “s-n” curves where the “s” implies stress and the “n” implies number of cycles. The “endurance stress” is defined as the stress at which a theoretically infinite number of stress cycles can be applied without failure. For most steels, the endurance stress is approximately one-half the yield stress. For P20, the endurance stress is approximately 450 MPa.

12.1 Objectives in Structural System Design

303

500 450 400

Stress (MPa)

350 300 250 200 150 100 50 0 0

20

40

60

80

100

120

Time (s)

Figure 12.4: Cyclic stresses in molds

The data in Figure 12.5 indicate that QC7 has a much lower endurance stress than P20. There are two very important differences in the behaviors of QC7 and P20. First, the s-n curve for QC7 has a greater slope than that for P20. Second, QC7 (and all known aluminums) do not exhibit an endurance stress limit. In other words, the continued cycling of any stress on aluminum will eventually cause failure due to fatigue. For this reason, the mold designer working with aluminum should carefully select the limit stress according to the desired number of molding cycles. If the mold is to be used for less than 1000 molding cycles, then the mold designer may select a limit stress equal to the yield stress of 545 MPa. If approximately 10,000 molding cycles are expected, then the allowable limit stress drops to 370 MPa. If the mold is to be operated for up to a million molding cycles, then the limit stress should be set to 170 MPa. To summarize, the limit stress is specified according to whether issues related to yielding or fatigue will dominate during the mold’s operation: ⎛ σ yield ⎞ , σ endurance ⎟ σ limit = min ⎜ f ⎝ ⎠

(12.4)

If the mold is to be designed for a low number of molding cycles, then the limit stress can be set to the yield stress and designed using a safety factor or a worst case scenario. If the mold is to be operated for a large number of molding cycles, then the endurance stress should be used as the limit stress. These data are provided for some common mold materials in Appendix B.

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1000

Stress (MPa)

 Yield = 830 MPa

QC7 P20

 Endurance = 456 MPa  Yield = 545 MPa

100 1.0E+02

1.0E+03

1.0E+04

1.0E+05

1.0E+06

1.0E+07

Number of Cycles

Figure 12.5: Stress-failure curves for QC7 and P20

12.1.2 Minimize Mold Deflection While excessive stresses in the mold components can cause damage to the mold, excessive mold deflection is an even greater concern in many molding applications. The primary reason is that excessive mold deflection can cause flashing at the parting lines between the core and cavity inserts. In tight tolerance applications, mold deflection can also cause part dimensions to be out of specification. As such, the mold design may be driven more by the need to minimize deflection rather than minimizing stress. Figure 12.6 plots the deflection of the mold and platens for the stress distribution shown in Figure 12.2. It is observed that the maximum mold deflection occurs at the center of the mold cavity, with the core surface deflecting 0.24 mm to the left and the cavity surface deflecting 0.12 mm to the right. As a result, the melt pressure in the mold cavity causes the surfaces to separate by a total of 0.36 mm (0.014 in). This deflection will cause any nearby parting lines to open a similar amount. Since this amount is much greater than the vent thickness (typically on the order of 0.02 mm), a significant amount of flashing is expected. The mold design must be improved to reduce this deflection.1 1

The contours of Figure 12.6 indicate that there is platen deflection as well. Mold designers and molders usually assume platens to be flat and infinitely rigid. While outside the scope of this book, platen deflection can be a significant issue. In Figure 12.6, the deflection of the stationary platen is approximately 0.04 mm, roughly twice that of the moving platen. The reason is that the mold’s ejector housing tends to transfer forces closer to the sides of the moving platen so there is less applied load and deflection in the center of this platen compared to the deflection of the stationary platen.

12.1 Objectives in Structural System Design

305

Figure 12.6: Deflection during molding

12.1.3 Minimize Mold Size The mold designer can investigate different mold materials and sizes to reduce the mold stress and deflection. A review of the material properties in Appendix B will reveal that harder materials (such as H13) can withstand much higher stresses than softer materials (such as 1045 or QC7). However, the mold designer should be aware that all steels have nearly the same elastic modulus, around 200 GPa. As a result, the mold designer can not change the deflection by steel selection, but rather must resort to changing the geometry of the mold. The simplest method to reduce deflection is to increase the thickness of plates. As later analysis will show, this approach is effective since the stiffness of the mold plates is related to the cube of the plate thickness. Even so, the repeated use of very large and thick plates can result in an overly heavy and expensive mold with a stack height that limits the availability of molding machines. For this reason, the mold designer should seek to minimize the size of the mold through appropriate analysis and careful specification of plate thicknesses and support structures such as pillars and interlocks.

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12.2 Analysis and Design of Plates In mold design, the term “plate” refers to a prismatic or rectangular structural member with a length and width typically greater than the thickness. The “face” of the plate generally refers to the top and bottom surfaces of the plate that span the width and length directions. The “sides” of the plate refer to the four sides of the plate that traverse the thickness direction. Mold plates are widely available from a number of suppliers in a variety of sizes and materials. Plates can be provided oversized with a slight (1 mm) stock allowance, or finish ground to tolerances on the order of ±0.02 mm. The majority of the mold consists of plates, including the top clamp plate, A plate, cavity inserts, core inserts, B plate, support plate, ejector plate, ejector retainer plate, and the rear clamp plate.2 Each of these plates is typically subjected to a load on one face of the plate. While the sides of the plate may be constrained by surrounding plates, the majority of the applied load is carried by compressive and shear stresses and thus transmitted through the thickness and across the plate. Plate compression and bending are next separately analyzed.

12.2.1 Plate Compression If the plate is fully supported by subsequent mold plates and the mold platen (as typical on the stationary side of the mold), then all plates are in compression and there is negligible plate bending. It should be noted that compressive forces due to mold clamping will tend to cause uniform compressive stresses through the mold plates. The compressive stress, σ, is defined as the force, F, per unit of compressed area, Acompression: σ =

F Acompression

(12.5)

The strain, ε, that develops is the stress divided by the elastic modulus, E: ε=

σ E

(12.6)

The amount of deflection, δcompression, is equal to the strain multiplied by the length across which the strain exists: δcompression = ε L

(12.7)

Deflection due to compression is not usually an issue since 1) it is relatively small and 2) it is uniform across the mold. As such, it does not cause flashing or significant dimensional 2

One notable exception is the design of molds with deep cores, in which the core insert is not constructed from a plate but rather from a rod. This type of mold design has a separate set of issues that are subsequently discussed in Section 12.3.

12.2 Analysis and Design of Plates

307

change in the molded part. As the following example will show, however, the mold designer should slightly increase the depth of the mold cavity to compensate for plate compression if a tight tolerance is specified on the thickness of a part with a deep cavity. Example: Estimate the change in the stack height of the bezel mold when clamped with 200 metric tons of force. To provide an accurate estimate of the mold deflection, the compressive stress and strain in each plate and rail could be calculated. Each component’s deflection in the height direction would then be computed and summed to provide the total mold compression. However, a faster but approximate estimate can be readily provided by assuming the mold is a monolithic block in a uniform state of compression. Figure 12.7 provides the outside mold dimensions along with a few other critical dimensions later discussed. The compressive stress in the mold is approximately: σ =

F Acompression

=

200 tons ⋅ 9807 N/ton = 17 MPa 0.381 m ⋅ 0.302 m

This stress level corresponds to a strain of: ε=

σ 17 MPa = = 8.3 ⋅ 10−5 E 205 GPa

The deflection across the entire mold during clamping is then: δmold = ε L = 8.3 ⋅ 10−5 ⋅ 403 mm = 0.03 mm In actual molding, the total mold deflection may be twice this amount since the rails on the sides of the ejector housing will exhibit a significantly higher state of stress. However, the above analysis provides a quick estimate on the correct order of magnitude.

Figure 12.7: Bezel mold dimensions for compressive stress analysis

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Example: Estimate the change in the height of the mold cavity due to the clamping of the surrounding A plate with 200 tons of force. The change in the height of the mold cavity is due to the compression of the plate surrounding the mold cavity. To provide an accurate analysis, the compressive stress in the plate surrounding the mold cavity is based on the area of the supporting material shown in Figure 12.8. This area does not include the mold cavity, leader pins, and guide bushings since these components do not transmit any of the clamping force from the stationary to the moving side of the mold. As a result, the compressive stress in the surrounding B plate will be somewhat higher than the 17 MPa previously calculated. The projected area of the cavity retainer plate is approximately: Acompression = 0.381 m ⋅ 0.302 m − 0.248 m ⋅ 0.168 m ⎡ π (0.032 m)2 π (0.020 m)2 ⎤ 2 −4⎢ + ⎥ = 0.069 m 4 4 ⎣ ⎦

Given 200 metric tons of clamping force, this projected area leads to a compressive stress in the plates surrounding the cavity of: σ plate =

F 200 tons ⋅ 9807 N/ton = = 28.5 MPa A 0.069 m 2

This stress level corresponds to a strain of: ε=

σ 28.5 MPa = = 1.4 ⋅ 10−4 E 205 GPa

Figure 12.8: Support area around cavity

12.2 Analysis and Design of Plates

309

Given a height of 12 mm, the deflection across the height of the mold cavity during clamping is then: δmold = ε L = 1.4 ⋅ 10−4 ⋅ 12 mm = 0.002 mm The change in the cavity height due to the mold clamping is very small. Furthermore, the melt pressure will be exerted on the surfaces of the mold cavity during molding, which will tend to counteract the mold clamping force. As such, the change in the cavity height during molding does not need further consideration.

12.2.2 Plate Bending If the back face of the plate is not fully supported, then shear stresses will develop and cause the plate to bend. Plate bending is a typical issue for the plates located between the ejector housing and the mold cavity on the moving side of the mold. The shear stress, τ, is defined as the force, F, per unit of area in shear, Ashear: τ=

F Ashear

(12.8)

Figure 12.9 provides an example of a static force analysis of a portion of the bezel mold that is in shear. While the actual shear stresses will vary with the distribution of the melt pressure across the mold cavity, a reasonable estimate can be achieved by assuming a uniform distribution around the perimeter of the mold cavity. As such, the area in shear is:

Figure 12.9: Shear stresses around perimeter

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Ashear = perimeter ⋅ height = (2 Wcavity + 2 Lcavity ) (H b_plate + H support_plate )

(12.9)

Example: Calculate the shear stresses in the core insert and support plates if the melt pressure exerts a total force of 200 metric tons across the mold cavity. The dimensions from the mold design are provided in Figure 12.10.

Figure 12.10: Bezel mold dimensions for shear stress analysis

The area in shear is: Ashear = (2 ⋅ 0.248 m + 2 ⋅ 0.168 m) (0.12 m − 0.012 m) = 0.090 m 2 The shear stress is: τ=

F Ashear

=

200 tons ⋅ 9807 N/ton = 21.8 MPa 0.090 m 2

This shear stress is very low relative to the limit stress of either aluminum or steel. The fundamental issue with plate bending in mold design is not the existence of shear stresses in the plates, but rather the development of large deflections across any long unsupported spans of the mold plates. Most molds utilize a moving ejector assembly, and so do not fully support the support plate between rails of the ejector housing. Accordingly, the mold plates behave like a beam in bending. The idealized case is represented in Figure 12.11 in which the entire load, F, is assumed to be applied to the center of the mold section. This assumption is made to provide a conservative estimate of the maximum deflection. The deflection of the plate is conservatively estimated assuming the beam bending equation with a central load as: δbending =

F L3 48 E I

(12.10)

12.2 Analysis and Design of Plates

Fully supported

311

F

Figure 12.11: Plate bending modeled as a beam

where L is the length of the span and I is the moment of inertia. For a rectangular section, the moment of inertia is: I =

1 W H3 12

(12.11)

where W is the width of the mold section in bending (in the direction normal to the section of Figure 12.11) and H is the combined thickness of the core insert and the support plate. Example: Compute the deflection due to plate bending in the bezel mold assuming a loading of 200 metric tons from the melt pressure. The width of the mold that is in bending is conservatively assumed to be equal to the width of the mold cavity, which is shown as 248 mm in Figure 12.10. The combined height of the core insert and support plate is 120 mm. Then, the moment of inertia is: I =

1 0.248 m (0.120 m)3 = 3.6 ⋅ 10−5 m 4 12

The free span in bending is taken as the distance between the inside surfaces of the ejector housing, shown as 215.9 mm in Figure 12.11. The deflection due to bending can then be estimated as: δbending =

200 tons ⋅ 9807 N/ton ⋅ (0.2159 m)3 = 0.056 mm 48 ⋅ 205 GPa ⋅ 3.6 ⋅ 10−5 m 4

This deflection is roughly twice the 0.024 mm deflection presented in Figure 12.6 from the finite element analysis. The reason is that the finite element analysis assumed a uniformly distributed load, which reduces the predicted deflection by 60% compared to the single, centrally located load used above. The presented analysis will tend to over predict the mold deflection due to bending, but is on the correct order of magnitude and should lead to robust mold designs.

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H W

L

F

Figure 12.12: Decomposition of separate bending areas

With respect to multiple cavity molds, the analysis should be applied to separate portions of the mold cavity as appropriate. Figure 12.12 provides a top and side view of a layout design for a six cavity mold. One analysis approach is to lump the melt pressure across three cavities together to compute the applied force, F, which acts primarily on the effective width, W. It should also be noted that the effective plate thickness, H, should not include the thickness of the cores when the cores do not contribute significantly to the stiffness of the mold assembly.

12.2.3 Support Pillars Plate deflection can be reduced significantly through the use of support pillars located between the rear clamp plate and the support plate. In general, support pillars are best placed directly under the portions of the mold cavity that generate significant force. By providing direct support of the mold plates, shear stresses and bending close to the support pillar are significantly reduced. A typical design is provided in Figure 12.13. In this design, a clearance is provided through the ejector plate and the ejector retainer plate. The support pillar is then located using a dowel that mates the center of the support pillar to a hole is in the rear clamp plate. Since the support plate is secured to the rear clamp plate with socket head cap screws, the support pillar is fully secured upon mold assembly. Unfortunately, the location of support pillars can conflict with other components including the ejector pins and the ejector knock-out rod(s). For this reason, different layouts and sizes of support pillars should be analyzed. If mold deflection is a critical issue, then the ejector layout can be adjusted to provide space for several large support pillars at ideal locations. Three potential support pillar locations are provided in Figure 12.14. At left, two smaller support pillars are located outside the ejector blades; the support pillars are fairly evenly spaced with regard to the span of the bezel. However, the support pillars can not be placed directly under the bezel face without rearranging the ejector layout.

12.2 Analysis and Design of Plates

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Figure 12.13: Typical mold design with support pillar

Another design may call for one very large support pillar at the center of the mold so as to avoid interference with the ejector pin layout. However, this support pillar will not greatly reduce the deflection of the mold plates since the majority of the plate bending will occur due to the loading on the left and right sides of the molding. Furthermore, this design could conflict with the use of a centrally located ejector rod from the molding machine, which is quite common. As another alternative, the layout at right uses a single support pillar of intermediate size. This design requires fewer support pillars than the first design, but has a larger span between the support pillar and the ejector rail and so will likely provide more deflection. The number, location, and size of the support pillars should be analyzed. One of the complexities of the analysis is that the support pillars are structural members of finite diameter and stiffness. This means that the support pillars will deflect under the compressive load. The core insert and support plate will deflect with the support pillar. Furthermore, the core insert and the support plate will exhibit bending between the support pillar and the ejector rail. To estimate the total plate deflection, superposition must be used to add the deflection due to compression and bending. This concept is shown in Figure 12.15.

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Figure 12.14: Different support pillar placements

Compression

Bending

Total

compression bending

total

Figure 12.15: Superposition of compression and bending

To perform the analysis, the forces across the mold must be converted to a set of load cases that is suitable for manual analysis.3 Figure 12.16 provides the conversion from the melt pressure imposed on the surface of the mold cavity to compression and bending load cases. The total force, F, is the integral of the melt pressure across the length and width of the cavity. To estimate the bending, this force is broken into two equal parts applied at the center of the span between the support pillar and the ejector rail. A force balance can then be applied to determine the forces that must be carried by the support pillar(s) and the ejector rails. 3

Alternatively, a finite element analysis model can be constructed using the detailed mold geometry. While such structural analysis techniques are very capable, modeling and computational burdens remain significant barriers to routine implementation.

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F/4 P P P P P P P P P

F/2 F/2 F/2 F/4

Figure 12.16: Compression and bending load cases

Once the load cases are developed for superposition, the previously presented compression and bending analyses may be performed to estimate the stress in the support pillar, σ, the deflection due to compression, δcompression, and the deflection due to bending, δbending. The deflection across the surface of the mold cavity can be estimated as a function of the distance, x, from the centerline of the support pillar as: ⎛ 3 L2 x − 4 x 3 ⎞ x⎞ ⎛ δtotal (x) = δcompression ⎜1 − ⎟ + δbending ⎜ ⎟ ⎝ L⎠ L3 ⎝ ⎠

(12.12)

where L is the length of the span from the support pillar to the ejector rail and the range of x is restricted to one-half the length of the span. The maximum deflection of the mold will occur either at the center of the support pillar or half-way between the support pillar and the ejector rail, depending on the relative magnitudes of the deflections. This leads to the following formula for the maximum deflection under superimposed compression and bending: δcompression ⎛ ⎞ δmax = max ⎜ δcompression , + δbending ⎟ 2 ⎝ ⎠

(12.13)

Example: Design a support pillar for the bezel mold so that the total deflection is less than 0.1 mm. Analyze the stress in the pillar and resulting deflections. Since the support pillar will support the core insert and support plate underneath one side of the bezel, only this local area of the mold cavity is analyzed as shown in Figure 12.17. The top and bottom sides of the bezel are close to the ejector rails and thus will not cause significant plate bending.

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Figure 12.17:

Area of mold cavity local to support pillar

Assuming a melt pressure of 150 MPa, the total force, F, exerted by the plastic on this portion of the mold is: F = P ⋅ A = 150 MPa ⋅ (168 mm ⋅ 13 mm) = 327,600 N = 33 tons which is still a significant amount of force. A support pillar diameter of 37.5 mm is initially analyzed. Given that the support pillar must convey one-half of the force, the stress in the pillar is: σ =

F /2 327,600 N = = 297 MPa A π (0.0375 m)2 /4

This stress is very close to the endurance stress for the support pillars if they are produced from SAE1040 steel. The resulting strain is: ε=

σ 297 MPa = = 0.014% E 205 GPa

The length of the support pillar is 89 mm (3.5 in). The resulting deflection is: δcompression = ε L = 0.014% ⋅ 88.9 mm = 0.13 mm which is greater than the specified maximum deflection of 0.1 mm. To reduce the deflection, the support pillar diameter is increased to 50 mm. The stress is thereby reduced to 167 MPa while the deflection due to compression is reduced to 0.07 mm. The analysis of bending is very similar to the previous example. However, the effective width of the mold plate is conservatively estimated as the width of the mold cavity (13 mm). The moment of inertia is then: I =

1 0.013 m (0.120 m)3 = 1.9 ⋅ 10−6 m 4 12

With the use of the support pillar, the span has been reduced from 216 mm as shown in Figure 12.11 to 108 mm as shown in Figure 12.17. Applying the bending equation for a force of 33 tons in this area of the mold, the deflection due to bending is then estimated as:

12.2 Analysis and Design of Plates

δbending =

317

33 tons ⋅ 9807 N/ton ⋅ (0.108 m)3 = 0.02 mm 48 ⋅ 205 GPa ⋅ 1.9 ⋅ 10−6 m 4

Since the deflection due to bending is quite small, the maximum deflection will occur at the center of the mold due to compression of the support pillar. It should be noted that the thickness of the B plate and/or support plate could be slightly reduced while still meeting the deflection requirement. Furthermore, the deflection due to compression could be greatly reduced by pre-loading the support pillars. Specifically, support pillars of 88.97 mm (88.9 mm plus the 0.07 mm) can be used such that the pillars compress to their nominal 88.9 mm (3.5 in) length so that cavity becomes flat during molding.

12.2.4 Shear Stress in Side Walls The foregoing analysis focused on plate deflection across the parting plane. However, the shear stresses in the side walls of the mold plates can also result in excess deflection and even failure. This concern becomes especially significant for molds with deep cavities. The cup mold represents a typical scenario with the load case shown in Figure 12.18. Specifically, the deep cavity provides a tall side wall along which the melt pressure, P, is exerted. If the cavity is very deep, then significant shear stresses and bending deflections can develop. The width of the side wall, from the surface of the mold cavity to the side of the mold, is generally referred to as the “cheek”. A common guideline in the mold design is that the width of the cheek, Wcheek, should be equal to the height of the mold cavity, Hcavity. The maximum shear stress in the side wall can be estimated as a function of the height of the mold cavity, Hcavity, the width of the cheek, Wcheek, and the molding pressure, P: τ=

P ⋅ H cavity ⋅ Wcavity σ F = < limit A Wcheek ⋅ Wcavity 2

(12.14)



Bending

Figure 12.18: Shearing and bending of side walls

P P P P P P

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To avoid failure, the shear stress should be less than one-half of the limit stress for the material. Applying this constraint to Eq. (12.14) and solving for the width of the cheek provides: Wcheek > 2 H cavity ⋅

P σ limit

(12.15)

If the mold is made of SAE4140, then the endurance stress is 412 MPa. For a typical maximum injection pressure of 150 MPa, the width of the cheek should be specified as: Wcheek > 2 H cavity ⋅

150 MPa = 0.73 ⋅ H cavity 412 MPa

(12.16)

Accordingly, the rule of thumb that the width of the cheek should equal the thickness of the cavity provides a slight factor of safety under typical assumptions. Even though the shear stress may not exceed the specified limit, the mold designer should also verify the deflection of the side wall under load. Assuming that the side wall acts as a simply supported beam with a uniform load, then the deflection due to bending of the side wall can be estimated as: δbending =

4 3 P H cavity 3 2 E Wcheek

(12.17)

Example: Estimate the shear stress and deflection of the side wall in the mold for the cup. Given the 3 mm wall thickness of the cup, a maximum melt pressure of 80 MPa is assumed. The height of the cavity from the parting plane to the top of the cup is 50 mm, and the width of the cheek is 45 mm. The shear stress in the side wall is approximately: τ = 80 MPa ⋅

50 mm = 89 MPa 45 mm

The maximum deflection of the side wall occurs at the parting plane, and will be approximately: δbending =

3 ⋅ 80 MPa ⋅ (0.05 m)4 = 4 ⋅ 10−5 m = 0.04 mm 2 ⋅ 205 GPa ⋅ (0.045 m)3

This stress and deflection should not be an issue so no changes are required to the mold design.

12.2 Analysis and Design of Plates

319

12.2.5 Interlocks In the previous example, the deflection of the side wall was not an issue. However, this issue would likely be significant if the melt pressures were higher, the mold cavity was deeper, or the molding tolerances were tighter. The mold designer could increase the width of the cheek to reduce the side wall deflection. However, this approach adds significant size and expense to the mold. Another alternative is to use interlocks on the parting plane near the edges of the mold to transfer part of the bending load from the stationary half of the mold to the moving half of the mold. Round and rectangular mold interlocks are shown in Figure 12.19. Both types of interlocks should be placed on the parting plane and as close to the mold cavities as possible. In general, the rectangular interlock will provide greater resistance to deflection due to its larger size and cross sectional area across the interlock. However, round interlocks are available in smaller sizes and are easier to install in a mold. A detail view of a mold design incorporating a round interlock is shown in Figure 12.20. In this design, the male interlock is fit into a through hole in the B plate of the mold. The female interlock is fit into a blind pocket in the deeper A plate of the mold. Both interlocks tightly fit into the surrounding plates, and are retained in the height direction with socket head cap screws. It is important that the mold designer does not jeopardize the structural integrity of the side wall by removing excess mold material when incorporating the interlocks. When the melt pressure is exerted on the side wall, the interlock will transfer part of the load from the A half of the mold to the B half of the mold. The use of the interlock effectively doubles the stiffness of the side wall, resulting in a halving of the amount of the side wall deflection.

Figure 12.19: Round and rectangular interlocks

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12 Structural System Design

Figure 12.20: Mold design with round interlock

Since larger interlocks can carry higher loads, the largest interlock should be used that can be readily incorporated into the mold design. If the interlock is exposed to a lateral force, Flateral, exerted by the side wall, then the shear stress in the interlock, τinterlock, can be estimated as: τ interlock =

Flateral Ainterlock

(12.18)

where Ainterlock is the cross-section area of the interlock at the parting plane. If the interlock is made of S7 tool steel, then the design should provide a shear stress less than 300 MPa to avoid failure. Example: Estimate the shear stress in the 19 mm diameter interlock used to support the side wall of the cup cavity. The primary uncertainty in this analysis is the estimation of the lateral force applied to the interlock. This estimation of the lateral force is complicated by the round shape of the cup that provides a non-uniform cheek width between the guide pins. However, an estimate can be provided by assuming that the interlock is exposed to the lateral force from the nearby surface of the mold cavity. As shown by the hatched section of Figure 12.21 that represents the nearby plastic, the effective area can be estimated as the product of the interlock width and the cavity height.

12.2 Analysis and Design of Plates

Figure 12.21:

321

Projected view of interlock and cavity

Of course, the interlock will not be exposed to all of the lateral force from the melt pressure exerted on the side wall of the mold cavity. A conservative estimate is that half of the force will be carried by the interlock, so: 1 Pmelt ⋅ φinterlock ⋅ H cavity 2 1 = 40 MPa ⋅ 19.05 mm ⋅ 50 mm = 19,050 N 2

Flateral =

The shear stress in the interlock can than be estimated as: τ interlock =

Flateral 19,050 N = = 67 MPa Ainterlock π (0.019 m)2 /4

Since this shear stress is less than the 300 MPa limit stress, the interlock is structurally sufficient to transfer half the loading from the side of the mold cavity to the moving half of the mold.

12.2.6 Stress Concentrations In mold plates, stress concentrations will occur wherever material has been removed between the mold cavity and the supporting plates. Stress concentrations are especially common in injection molds due to the installation of water lines and ejector holes. The resulting stress distribution about the hole will be similar to that shown in Figure 12.22. In this example, a hole has been provided in a mold plate at a distance of 1.5 times the hole’s diameter. A pressure of 100 MPa has been applied to the top surface. The resulting maximum von Mises stress is 340 MPa, which corresponds to a stress concentration factor of 3.4.

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12 Structural System Design

Figure 12.22: Stress concentration about hole

As the hole is moved further away from the mold cavity, the stress concentration is reduced. To evaluate the stress concentration factor, a series of finite element analyses were performed with varying mold geometries. Figure 12.23 plots the stress concentration as a function of the number of hole diameters from the cavity surface to the centerline of the hole. A model of the stress concentration factor, K, was fit to the data, providing: ⎛φ ⎞ K = 3.1 + 0.75 ⎜ hole ⎟ ⎝ H hole ⎠

2.29

(12.19)

where φhole is the diameter of the hole and Hhole is the distance from the cavity surface to the center of the hole. This model is plotted as the dashed line in Figure 12.23. Holes located close to the cavity surface obviously cause significant stress concentrations. However, it is observed that a stress concentration of 3 results even when a hole is located far from the cavity surface.

12.2 Analysis and Design of Plates

323

5

Finite Element Analysis Model Fit

Stress Concentration Factor

4

3

2

1 0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

Number of Diameters from Cavity to Hole Centerline

Figure 12.23: Stress concentration as a function of distance

This explains why many molds develop cracks emanating from the waterlines in molding applications with high melt pressures, even when the cooling lines are located far from the cavity surface. For this reason, molding applications with high melt pressures should be constructed of materials with high endurance stresses such as A6, D2, or H13. Example: A thin wall molding application will utilize a filling pressure of 200 MPa with a core insert constructed of H13. Specify the closest allowable distance for a cooling line with a diameter of 9.5 mm. H13 has an endurance stress of 760 MPa. Since the melt pressure will provide a nominal compressive stress of 200 MPa, the allowable stress concentration factor is: K =

σ endurance 760 MPa = = 3.8 σ nominal 200 MPa

The distance may be evaluated using Figure 12.23 or calculated using Eq. (12.19). Solving Eq. (12.19) for the distance, HHole, provides: ⎛ 0.75 ⎞ H hole = φhole ⎜ ⎝ K − 3.1 ⎟⎠

2.29

⎛ 0.75 ⎞ = 9.5 mm ⋅ ⎜ ⎝ 3.8 − 3.1 ⎟⎠

2.29

= 11.1 mm

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12 Structural System Design

Cooling lines seem to cause more significant problems than ejector holes in practice. Cracks emanating from cooling lines will eventually leak and cause quality issues with the moldings. By comparison, cracks emanating from ejector holes may not ever cause a catastrophic failure. The reason is that the deformation of the ejector hole under load can cause the plate around hole to be supported by the ejector, thereby reducing the stress around the hole. As such, cracks propagating from ejector holes will reach a critical length at which point the elastic deformation of the core insert onto the ejector prevents further crack propagation. Example: A prototype molding application will utilize a melt pressure of 100 MPa with QC7. A mold designer is contemplating placing a 4 mm diameter ejector with 0.5 mm of aluminum between the edge of the cavity side wall and the edge of the ejector pin. Calculate the stress level and estimate the deflection of the ejector hole. The distance from the cavity side wall to the center of the ejector, Hhole, is 2.5 mm. The stress concentration factor, K, is: ⎛ 4 mm ⎞ K = 3.1 + 0.75 ⎜ ⎝ 2.5 mm ⎟⎠

2.29

= 5.3

Given a melt pressure of 100 MPa, the von Mises stress level at the pin will be approximately 530 MPa. This stress level is just below the yield stress of 545 MPa provided in Appendix B so the material should not immediately yield. However, the von Mises stress is far above the endurance stress of 166 MPa required to prevent failure across one million molding cycles. A review of the fatigue behavior of Figure 12.5 indicates that the design will likely fail around 1,000 cycles. To estimate the deflection of the ejector hole, the compressive stress in the mold plate adjacent to the ejector hole must be estimated. One approach would be to assume that the entire von Mises stress is compressive in nature. This approach will over estimate the hole deformation since the von Mises stress also includes a component of the shear stress, and is thus larger than the compressive stress. Continuing with this assumption, the strain in the adjacent plate material is estimated as: ε=

σ 530 MPa = = 0.73% E 72.4 GPa

The material that is deforming about the ejector hole has a length equal to the hole diameter. The deflection of the hole can then be estimated as: δhole = ε φhole = 0.73% ⋅ 4 mm = 0.03 mm This amount of deflection is on the order of the clearance provided around the ejector pin for venting. Over many molding cycles, the hole will plastically deform and cause binding of the ejector pin. For validation, a finite element analysis of this load case was conducted and indicated that the deflection was actually 0.10 mm as shown in Figure 12.24.

12.3 Analysis and Design of Cores

Figure 12.24:

325

Deformation around ejector hole near the cavity

The reason for the large variance between the analysis and the simulation was that the close proximity of hole to the cavity surface caused local bending at the top of the hole as shown in Figure 12.24, which was not considered in the analysis. By counting the 0.01 mm displacement lines, the results do indicate that the vertical deflection of the mold plate to the left and right of the holes is very close to the 0.03 mm predicted by the previous analysis.

12.3 Analysis and Design of Cores For the purpose of structural design, a core can be considered shallow when the height of the core is less than both the width and length of the core. Shallow cores, such as for the bezel mold, will not be subjected to excessive stress or deflection caused by the application of the melt pressure to the side walls of the core. As such, shallow cores can be designed according to the previously described analysis for mold plates.

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12 Structural System Design

12.3.1 Axial Compression The vertical deflection of cores due to compression by the melt can be modeled as previously discussed in Section 12.2.1, though with the compressive forces and cross-sectional areas appropriate for the core. Example: Estimate the vertical deflection of the core shown in Figure 12.25 assuming a melt pressure of 80 MPa. It may appear that the cooling insert in the design of Figure 12.25 will fully support the core insert. While this may be fine in theory, a more robust design may be provided by assuming that the cooling insert provides no support. There are two reasons for making this assumption. First, the outer surfaces of the cooling insert may not tightly fit the inner surfaces of the core insert. Any gap greater than the deflection of the core will completely prevent the cooling insert from supporting the core. Second, the cooling insert may be made from a different material than the core insert. As such, the cooling insert may not be able to withstand the stresses imposed while supporting the core insert. The total deflection of the top surface of the core can be estimated by superimposing the compression of the side walls with the bending of the top surface. While the compressive stress distribution in the side walls of the core is not entirely uniform, the average stress is approximately: σ side_wall =

Figure 12.25:

Fvertical 80 MPa ⋅ π (63 mm)2 /4 = = 216 MPa Aside_wall π [(63 mm)2 − (50 mm)2 ]/4

Axial compression of hollow core

12.3 Analysis and Design of Cores

327

This is a fairly high stress level indicating that a mild steel or aluminum should not be used in this application when considering the cyclic loading and possible fatigue. Assuming a steel core insert, the vertical compressive strain in the side walls is: εvertical =

σ side_wall 216 MPa = = 0.11% E 205 GPa

The total height of the core insert is 58 mm. With a strain of 0.11%, the total vertical deflection at the top of the side walls equal to: δvertical = εvertical ⋅ H core = 0.11% ⋅ 58 mm = 0.06 mm Bending from the edge of the top surface to the center of the core can be calculated using beam or plate bending equations of Section 12.2.2 and added to the displacement due to compression to estimate the total vertical deflection. The structural design of deep cores is further complicated since deep cores can bend due to the application of lateral forces from the melt pressure against their sides. Core bending may be a significant problem when cores are slender and have a low stiffness associated with their cross-section area, especially when the cores are hollowed out to provide for mold coolant. Analysis and design of the core must ensure that the potential core deflection is not excessive and that the compressive stresses around the perimeter of the core are acceptable. These two concerns are next addressed.

12.3.2 Compressive Hoop Stresses When a core insert includes a hollow section for a cooling line or other purpose, the side walls must withstand the compressive forces imposed by the melt pressure. The load case is shown in Figure 12.26.

Figure 12.26: Core insert loaded by melt pressure

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12 Structural System Design

The compressive stress, σhoop, caused by the melt pressure is: σ hoop =

P φcore 2 hcore

(12.20)

where P is the melt pressure, φcore is the outer diameter of the core insert, and hcore is the thickness of the core’s side wall. To avoid compressive failure in the side walls, the hoop stress must be less than the specified limit stress for the material. This leads to the following constraint on the thickness of the side wall: hcore >

P φcore 2 σ limit

(12.21)

The constraint can also be used to find the maximum inner core diameter, φinner: ⎛ P ⎞ φinner < φcore ⎜1 − σ limit ⎟⎠ ⎝

(12.22)

A general guideline can be developed for inserts produced from P20 steel with an assumed melt pressure of 150 MPa. Since P20 has an endurance stress around 450 MPa, the thickness of the side wall should be greater than: hcore >

φ ⋅ 150 MPa φ ≈ 2 ⋅ 456 MPa 6

(12.23)

and the maximum internal diameter of the core is: ⎛ 150 MPa ⎞ 2 φinner < φcore ⎜1 − ≈ φcore 456 MPa ⎟⎠ 3 ⎝

(12.24)

In practice, the mold designer should customize the above analysis by utilizing the maximum melt pressure and endurance stress that are specific to the molding application. Example: Compute the compressive hoop stress in the core insert for the cup mold assuming that the outer diameter, φcore, is 60 mm, the wall thickness, hcore, is 10 mm, and the melt pressure is 80 MPa. Also recommend a maximum inner diameter if the core insert is made of aluminum QC7. Given the above assumptions, the compressive hoop stress is: σ hoop =

80 MPa ⋅ 60 mm = 240 MPa 2 ⋅ 10 mm

This is a safe but significant amount of stress. If the core insert is to be made of aluminum QC7, then two different loadings might determine the allowable inner diameter of the core. First, the mold designer should consider the cyclic loading at the 80 MPa melt

12.3 Analysis and Design of Cores

329

pressure with an endurance stress of 166 MPa. This fatigue analysis indicates that the inner diameter should be: ⎛ 80 MPa ⎞ φinner < φcore ⎜1 − = 0.51 φcore = 31 mm 166 MPa ⎟⎠ ⎝

Second, the mold designer should consider an overpressure situation wherein the molder accidentally injects the melt at the maximum pressure of the molding machine. A single cycle at too high a pressure could cause the core insert to fail. To check this, a melt pressure of 200 MPa can be used with QC7’s yield stress of 545 MPa. This yield stress analysis indicates that the inner diameter should be: ⎛ 200 MPa ⎞ φinner < φcore ⎜1 − = 0.63 φcore = 38 mm 545 MPa ⎟⎠ ⎝

Comparing the above two results indicates that cyclic fatigue is a more critical issue than yield in an overpressure situation. The maximum inner diameter when using QC7 is 31 mm if a high number of molding cycles is desired.

12.3.3 Core Deflection Another common issue with deep cores is excessive deflection or “core bending” due to variations in the melt pressure around the periphery of the core. The variation in melt pressure is often due to the side gating as shown in Figure 12.27. However, slight variations in melt flow can cause significant bending in center gated designs when the cores are very slender (e.g., a core length on the order of ten times the core diameter). The problem is compounded by the fact that the core bending effect is self-reinforcing, which means that a slight bending of the core facilitates more melt flow and pressure to the thicker portion of the cavity and further bending of the core.

Figure 12.27: Lateral loading of core insert

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12 Structural System Design

Core bending can be analyzed through appropriate use of bending equations. Typically, the core is held to the moving side of the mold, and the top of the core is free to bend. The deflection due to the pressure difference, ΔP, across the core is: δbending =

4 ΔP φcore H core 8EI

(12.25)

where I is the moment of inertia. For a hollow core with an outer diameter, φcore, and an inner diameter, φinner, the moment of inertia is: I =

π 4 4 (φcore − φinner ) 64

(12.26)

The magnitude of the pressure difference around the core will vary with the geometry of the molding application. A shorter core, like that shown in Figure 12.27, will have a pressure difference that is a significant fraction of the pressure required to fill the mold (perhaps 50% of the filling pressure). As the core becomes longer relative to its diameter, the pressure difference around the core will become less compared to the pressure difference along its height (perhaps 10% of the filling pressure). However, the core deflection is proportional to the fourth power of the core height, so a small asymmetry of the melt pressure can cause a large deflection. Example: Estimate the magnitude of the deflection for the cup’s core assuming that the outer diameter is 60 mm, the inner diameter is 40 mm, the height is 58 mm, and the pressure difference around the core is 40 MPa. The moment of inertia for the core is: I =

π [(0.060 m)4 − (0.040 m)4 ] = 5.1 ⋅ 10−7 m 4 64

Assuming a steel core with a modulus of 205 GPa, the deflection is: δbending =

40 ⋅ 106 Pa ⋅ 0.06 m ⋅ (0.058 m)4 = 0.03 mm 8 ⋅ 205 ⋅ 109 Pa ⋅ 5.1 ⋅ 10−7 m 4

Since the deflection is small, core bending will likely not be an issue in this application even if the pressure difference around the core was significantly greater. Naturally, core bending becomes much more significant as the core becomes more slender. To minimize core bending, the mold designer should utilize solid cores with a minimal length to diameter ratio. When possible, slender core pins should be interlocked with the stationary side of the mold as shown in Figure 12.28. Such interlocking of the core pin reduces the lateral deflection of the pin to approximately 10% of the deflection for a pin that is supported on only one end. When interlocking or increasing the size of the core is constrained by the geometry requirements of the molding application, the mold designer should strongly recommend using a center gate at the top of the corner or two opposing gates at the bottom of the core to minimize the pressure gradient exerted on the core.

12.3 Analysis and Design of Cores

331

Figure 12.29 shows the use of flow leaders as another approach to reduce the core deflection. In this design, the flow leaders will assist the melt to travel down the cavity with lower filling pressures. At the same time, the melt will propagate into the thinner adjacent sections of the cavity and partially freeze, thereby preventing the core from deflecting a significant amount even if significant pressure differences arise later in the filling stage. The flow leaders shown on the core in Figure 12.29 may be undesirable as protrusions on the inner surface of the molded part if in contact with fluids. As such, the flow leaders may be integrated on the outside surface of the molding according to a variety of design configurations set into the cavity insert.

Figure 12.28: Interlocking of slender core into cavity

Figure 12.29: Use of flow leaders to minimize core deflection

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12 Structural System Design

12.4 Fasteners The mold design must also include fasteners to rigidly fasten the multiple components of the mold. There are three types of fasteners commonly used in molds. First, fits are used to tightly locate one component within another, such as a cavity or core insert being located within a retainer plate. Second, locating pins or dowels are used to locate one components above another, such as the ejector housing to the support plate. These first two fastening methods only provide fastening across the length and width directions of the mold. To fasten the mold components together in the height direction, socket head cap screws are used wherein the screw’s head is retained in a mold plate and the screw’s threads engage the component to be fastened. Each of these fastening methods is next analyzed.

12.4.1 Fits A “fit” refers to the mating of two components. A clearance fit refers to a mating in which a nominal clearance between the surfaces of the two components. While a clearance fit provides for easy assembly with no insertion forces, the clearance between the two components permits the precise location of components to remain unknown. Since tight tolerances are required in molds, interference fits are commonly used to locate the mold components. Interference fits occur when the male component has a nominal dimension that is larger than the nominal dimension of the female component, as shown in Figure 12.30 for a core insert and a retainer plate. Since metals have a high elastic modulus, a rigid interference fit can result

Figure 12.30: Location-interference fit for inserts

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12.4 Fasteners

when the difference between the nominal dimensions is very small, on the order of 0.01% of the nominal dimension. The tightness and rigidity of the interference fit increases with the amount of interference between the two components. Unfortunately, the implementation of interference fits is impeded by the dimensional variations imposed in the components’ machining processes. For this reason, standard systems of fits have been developed to provide limits on the dimensions of the components. The fits analyzed here are based on a lateral hole basis and have been converted from U.S. customary units to metric units.4 In this method, rectangular members with width, W, and length, L, are modeled as a circular member with apparent diameter, D, computed as: D=

W ⋅L

(12.27)

The tolerance limit, λ, on a given dimension is then calculated according to a formula: λ = 0.001 ⋅ C ⋅ D

1 3

(12.28)

where C is a coefficient corresponding to the lower and upper limit for the male or female component provided by international standards. Table 12.1 provides coefficients for locational-interference fits (LN1 to LN3) and drive-interference fits (FN1 to FN3). Locationalinterference fits are used when the accuracy of location is critical and the components require lateral rigidity. However, locational-interference fits do not provide significant retention force in the height direction, so the components must be secured in the height direction to another component via screws or other means. FN1 to FN3 correspond to drive fits with increasing interference and requiring increasing insertion forces. While drive fits provide semi-permanent assemblies, mold designs usually provide screws or other means for positively retaining the components in the height direction. Table 12.1: Location tolerance interference coefficients [mm]

Fit

4

Cinterference

Female (hole in plate)

Male (insert)

Lower limit

Upper limit

Lower limit

Upper limit

5.67

9.05

LN1

4.89

0.00

4.93

LN2

7.14

0.00

7.84

8.59

13.52

LN3

12.22

0.00

7.84

13.67

18.60

FN1

13.57

0.00

4.93

14.34

17.73

FN2

22.02

0.00

7.84

23.47

28.41

FN3

30.85

0.00

7.84

32.30

37.24

Two of the most common standards for fitting include “Preferred Limits and Fits for Cylindrical Parts”, ANSI B4.1-1967 (R1999), and “Preferred Metric Limits and Fits” ANSI B4.2-1978 (R1999). ANSI B4.1 is analyzed here due to its relative simplicity and broad applicability, though the mold designer may conform to whatever standard is most appropriate.

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12 Structural System Design

Example: The base of the core insert for the cup mold is 88.90 mm on each side. Specify the tolerances for a light drive (FN1) fit. The apparent diameter of the core insert is: D=

88.90 mm ⋅ 88.90 mm = 88.90 mm

The lower tolerance limit for the insert dimension is computed with C equal to 14.34: lower λinsert

1 3

= 0.001 ⋅ 14.34 ⋅ 88.9 = 0.064 mm

The upper tolerance limit for the insert dimension is computed with C equal to 17.73: 1 3

upper λinsert

= 0.001 ⋅ 17.73 ⋅ 88.9 = 0.079 mm

The lower tolerance limit on the mating hole in the retainer plate is 0: λlower plate

1 3

= 0.001 ⋅ 0.0 ⋅ 88.9 = 0.000 mm

The upper tolerance limit on the mating hole in the retainer plate is computed with C equal to 4.93: 1 3

λupper plate = 0.001 ⋅ 4.93 ⋅ 88.9 = 0.022 mm The minimum and maximum dimensions on the insert are specified as 88.96 and 88.98 mm, respectively. The minimum and maximum dimensions on the hole in the plate are specified as 88.90 and 88.92 mm. These dimensional limits are shown in Figure 12.31.

Figure 12.31:

Insert and plate dimensions for an FN1 fit

12.4 Fasteners

335

It may be of interest to estimate the insertion force required to achieve various interference fits, so that excessive insertion forces may be avoided. The insertion force may be estimated by the compressive stress required to strain the components during assembly. The expected amount of interference can be computed as the average male dimension minus the average female dimensions. Alternatively, the expected amount of interference, λinterference, can be computed using the formula: λinterference = 0.001 ⋅ Cinterference ⋅ D

1 3

(12.29)

where Cinterference is a coefficient derived from the limit coefficients provided in Table 12.1. Assuming that the plate is much larger than the insert, the compressive stress, σ, in the insert is estimated as: σ =

λinterference ⋅ E 2D

(12.30)

where E is the modulus of the material. The factor of 2 in the above equation stems from the fact that the compressive stress in the insert will also drive a tensile stress in the plate. Accordingly, the interference causes equal strain in both the insert and the plate. The insertion force can then be estimated as the compressive stress multiplied by the contact area and the friction coefficient: Finsertion = f σ (π D H )

(12.31)

where f is the friction coefficient and H is the height of the contact zone between the two components. Example: Estimate the insertion force for the core insert for the cup mold. Assume an FN1 fit with a contact height between the plate and the insert of 42 mm. The expected dimension for the core insert is 88.97 mm while the expected dimension for the hole in the retainer plate is 88.91 mm. The expected amount of interference, λinterference, is 0.06 mm. The resulting stress in the steel components is: σ =

0.06 mm ⋅ 205 ⋅ 109 Pa = 69 MPa 2 ⋅ 88.9 mm

Assuming a coefficient of friction of 1.0, the resulting insertion force is: Finsertion = 1.0 ⋅ 69 MPa (π ⋅ 88.9 mm ⋅ 42 mm) = 808 kN An insertion force of approximately 808 kN or 180,000 lbs is required to drive the core insert into the retainer plate. If a press is not available with this capacity, the mold designer can utilize a location-interference fit. Also, it is desirable to provide a slight taper along the leading edge of the core insert to assist in alignment during assembly.

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12 Structural System Design

12.4.2 Socket Head Cap Screws A ½″-13 socket head cap screw is shown in Figure 12.32. Socket head cap screws like this screw are the most common fastener used in molds. The primary reason is that socket head cap screws have been carefully designed such that the strength of the head, threads, and bolt are matched. As a result, the socket head cap screw provides a standard and efficient method for retaining multiple components along the screw’s axis.

Figure 12.32: Typical socket head cap screw

The sizes and load carrying capability of socket head cap screws are related to their size, material, and treatments. Analysis of standard socket head screw designs indicates that the head height is equal to the thread diameter, and that the head diameter is approximately 150% of the thread diameter. While the strength of the fastener varies somewhat with the coarseness of the thread, the tensile strength of standard DIN/ISO screws can be fairly well estimated by assuming an ultimate stress, σultimate, of 800 MPa multiplied by the cross-section area of the outer thread diameter: Ftensile = σ ultimate

2 π Dthread 4

(12.32)

Example: Specify the size of the socket head cap screws used to fasten the stationary and moving halves of the laptop bezel mold shown in Figure 12.7. Since this socket head cap screw is used in a critical application where failure may result in loss of equipment or life, a worst case scenario is assumed. First, the maximum mass of the mold is estimated assuming a solid block of steel according to the dimensions provided in Figure 12.7. The maximum mass of the mold is: M mold = ρmold H mold Lmold Wmold = 7800

kg ⋅ 0.403 m ⋅ 0.381 m ⋅ 0.302 m = 362 kg m3

12.4 Fasteners

337

Next, the worst case scenario is assumed. The worst case scenario occurs when the mold is clamped to only one side of the molding machine without the support of the moving platen, which may occur when the mold is being installed in the molding machine. Furthermore, the worst case scenario will assume that the entire mass of the mold must be supported by only one tightened screw, which may occur if the other cap screws are not tightened or tightened to lesser amounts. The resulting load case is shown in Figure 12.33. The exerted force on the screw by the mold can be estimated by summing the moments about the locating ring to find: Fscrew = M mold ⋅ ng ⋅ g ⋅

LCOG Lscrew

where g is the acceleration due to gravity (9.8 m/s2), LCOG is the distance between the platen and the mold’s center of gravity, and Lscrew is the distance from the locating ring to the screw. The coefficient ng relates to the number of gravities that may be exerted on the mold, and is usually set quite high for safety purposes. Due to the shock of a crane, ng is set equal to 10. Substituting the approximate values from Figure 12.33 provides: Fscrew = 362 kg ⋅ 10 ⋅ 9.8

Figure 12.33:

m 0.2 m ⋅ = 47,000 N s2 0.15 m

Worst case analysis for screw loading

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12 Structural System Design

Solving Eq. (12.32) for the diameter yields: Dscrew =

4 Fscrew = π σ ultimate

4 ⋅ 47,000 N 3 ′′ = 8.65 mm → 10 mm or 6 8 π ⋅ 800 ⋅ 10 Pa

The analysis indicates that a 3/8″ or M10 socket head cap screw should be sufficient. For reference, the mold base selected for this application was provided with ½” socket head cap screws. Failure of cap screws in this mold base is not expected.

12.4.3 Dowels Cap screws should not be relied upon to locate mold components given their relatively large radial clearances. As previously discussed, an interference fit should be used to locate one component within another. For parallel plates or components, however, dowels or other locating pins should be used as shown in Figure 12.34. In this design, concentric holes are provided into the coplanar surfaces of the two plates. A dowel then mates with the two holes to locate the two components along the axis of the dowel. Manufacturing variances in the holes’ location, diameter, and roundness limit the ability to precisely locate the two components relative to each other. Equation (12.28) can be used for various types of fits by varying the limit coefficients, C, for the dowel and holes according to international standards. Table 12.2 provides coefficients for a locational-clearance fit (LC1), locational-transitional fits (LT1 and LT3), as well as the loosest locational-interference fit (LN1). Locational clearance fits are intended for parts that are typically stationary but can be readily disassembled and reassembled. This fit provides the same order of tolerance as threaded fasteners, so is not recommended for injection molds since the large clearance can allow accelerated wear of sliding surfaces. Locational-transition fits provide for tighter control of location, but with the possibility of interference between the dowel and the hole which hinders the mold assembly.

Figure 12.34: Typical locating dowel design

12.4 Fasteners

339

Table 12.2: Location clearance and transitional coefficients [mm]

Fit

CInterference

Female (hole in plate)

Male (dowel)

Lower limit

Upper limit

Lower limit

Upper limit

LC1

–4.16

0.00

4.93

–3.39

0.00

LT1

–6.38

0.00

7.84

–2.43

–2.51

LT3

–0.73

0.00

7.84

0.72

5.65

LN1

4.89

0.00

4.93

5.67

9.05

Example: A 12 mm dowel is to be used to mate the ejector housing to the support plate. Specify the dimensions for an LT3 fit. Estimate the expected clearance between the dowel and the hole, as well as the insertion force in the event of the worst case interference. The apparent diameter of the core insert is: D=

88.90 mm ⋅ 88.90 mm = 88.90 mm

The lower tolerance limit for the dowel diameter is computed with C equal to 0.72: 1 3

lower λdowel = 0.001 ⋅ 0.72 ⋅ 12 = 0.002 mm

The upper tolerance limit for the dowel diameter is computed with C equal to 5.65: 1 3

upper λdowel = 0.001 ⋅ 5.65 ⋅ 12 = 0.013 mm

The lower tolerance limit on the mating hole in the retainer plate is 0. The upper tolerance limit on the mating hole in the retainer plate is computed with C equal to 7.84: 1 3

λupper plate = 0.001 ⋅ 7.84 ⋅ 12 = 0.018 mm

.

The minimum and maximum dimensions on the dowel are specified as 12.002 and 12.013 mm, respectively. The minimum and maximum dimensions on the hole in the plate are specified as 12.000 and 12.018 mm. This design is shown above in Figure 12.34. The average clearance between the two components is 0.0015 mm (or 1.5 μm, equal to the hole’s average diameter of 12.009 mm minus the dowel’s average diameter of 12.075 mm). Given that manufacturing variation exists, it is important to check on the magnitude of the dowel’s insertion force when the hole and the dowel are at their specified limits. The worst case interference will occur when the hole’s diameter is 12.000 mm and the dowel’s diameter is 12.013 mm. The maximum amount of interference, λinterference, is 0.013 mm. The resulting stress in the steel components is:

340

12 Structural System Design

σ =

0.013 mm ⋅ 205 ⋅ 109 Pa = 111 MPa 2 ⋅ 12 mm

Assuming an insertion length of 12 mm and a coefficient of friction of 1.0, the application of Eq. (12.31) results in an insertion force of: Finsertion = 1.0 ⋅ 111 MPa (π ⋅ 12 mm ⋅ 12 mm) = 50 kN This magnitude of insertion force for a dowel is clearly undesirable since separation of the mold plates can not be accomplished manually. The mold assembler would require grinding to reduce the pin diameter to avoid such excessive insertion forces.

12.5 Review Molds are mechanical assemblies that must withstand high levels of stresses imposed by the pressure of the polymer melt. Several constraints drive the structural design of the mold. First, the mold must be designed to avoid yielding given a single over pressure situation. Second, the mold must be designed to avoid failure due to fatigue associated with the cyclic loads associated with molding thousands or millions of cycles. Third, the mold must be designed to avoid excessive deflection while molding, which would lead to flashing of the molded parts and accelerated wear on the mold’s parting line. Of these issues, fatigue and deflection tend to dominate though the relative importance is a function of the number of mold cavities, the molding pressures, the mold geometry, and the production quantity. Analyses were provided to model the compression of mold plates, cores, and support pillars as well as the bending of plates, side walls, and cores. Superposition of compression and bending can be used to estimate the total deflection of the cavity surfaces. Analyses were also developed for stress concentrations in mold plates. In general, all analyses indicate that increasing the amount of steel between the load and support points provides for lower levels of stress and deflection. As such, the mold designer must perform analysis to develop robust designs that are not uneconomical. The uses of support pillars, interlocks, and other designs were demonstrated to reduce deflection. Common fastening means were also analyzed including interference fits, socket head cap screws, and dowels with clearance and interference fits. The mold designer must remember to provide means for fastening the cavity and core inserts to the rest of the mold while providing tight control of location relative to other mold components. In practice, the provision of fasteners may interfere with other subsystems of the mold including part ejection and mold cooling. In such cases, iterative redesign of the mold may be required to efficiently locate all the mold’s subsystems without increasing the size and cost of the mold.

12.5 Review

341

After reading this chapter, you should be able to: • Describe the flow of forces from the mold cavity to the machine tie bars; • State the relationship between modulus, stress, and strain; • State the relationship between ultimate stress, yield stress, and endurance stress; • Specify the limit stress and maximum deflection based an application’s requirements; • Estimate the compressive, shear, and hoop stresses in various mold components; • Estimate the deflection of a plate, core, or support pillar due to compression; • Estimate the deflection of a plate, core, or side wall due to bending; • Specify the plate thickness and use of support pillars to avoid excessive mold deflection; • Specify the mold cheek and use of mold interlocks to avoid excessive stress or mold deflection; • Specify the design of mold cores to avoid excessive hoop stress and core bending; • Specify the distance between the mold cavity and stress concentrations (such as ejector holes and cooling lines) as a function of the material properties and application requirements; • Specify the dimensional limits on male and female components to achieve clearance, transition, interference, and drive fits; • Estimate the expected clearance or insertion force for a specified fit; and • Specify the use of socket head cap screws to securely fasten mold components. The analysis and design of each of the mold’s subsystems has been completed. The next chapter is intended to increase the mold designer’s awareness by providing a critical examination of available mold technologies.