The Complete Part Design Handbook

The Complete Part Design Handbook

E. Alfredo Campo

For Injection Molding of Thermoplastics

ISBN 3-446-40309-4

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115

2 Engineering Product Design

When designing plastic components, success will depend on one prime factor:how well we use the variety of plastic properties and the processing methods for obtaining optimum results.

The designer should select the best resin, realizing that it is essential for the resin’s full potential to be exploited to ensure that the molded part will satisfy both functional and cost requirements.

Plastics are governed by the same physical laws and the same rules for good designas other materials. These principles can be applied if the polymer properties are suitable for the operating environment of the product being considered.

It is necessary to know and understand what the end product must do and under what circumstances it will operate, before a design analysis can be done.

2.1 Understanding the Properties of Materials

There is a big difference between the properties, processing methods, and appli-cations of materials manufactured by various industries. There is not a single mate rial that can be used for all applications. Each new outstanding propertydeveloped in a material opens the door for new applications, technologies, and innovations that will improve the efficiency and quality of life of the end users.

Product designers should compare the properties of various groups of materials(steels, thermoplastics, aluminum alloys, rubber, etc.), because each material has different properties developed for specific applications and markets and uses different manufacturing processes. All materials have benefits and defi ciencies (properties, processes, and quality), making it diffi cult to compare the cost offinished products made of different materials and processes.

The material properties are directly related to the end use applications whether or not one material is better than another. To illustrate this point, a thermoplasticresin cannot replace a structural steel beam used in building construction; the thermoplastic resins do not have the strength, creep resistance, or melt strength to be extruded into thick walled shapes. Thermoplastic beams would alsowarp in all directions. However, structural beams can be made of thermoset composites, although this is expensive. In less critical applications, such as the housing industry, wood composite structural beams are replacing steel beams,because of their performance and light weight; they are easy to work with and offer a competitive price.

A thermoplastic resin cannot replace the steel in automotive disc/drum brake housings, because the product requires dimensional stability, low thermalexpansion, and high strength and rigidity at elevated temperatures. Thermo-plastic resins do not meet the requirements. However, brake pads made ofthermoset polyimide have been successfully used in airplanes.

Metals cannot replace automotive rubber tires, bellows, diaphragms, or com-pression seals, because metals do not have the elasticity, fatigue endurance,wear resistance, and toughness of rubber. Metals are not used for light-weight and compact cellular phone housings, because metals are electrical conductors,heavy, corrosive, and expensive.

116 2 Engineering Product Design

Automotive engine cast iron and aluminum intake manifolds are being replaced by fiber glass reinforced nylon to improve efficiency, lower weight, creating new manufacturing processes, and cost reduction.Automotive steel bumpers, externalside panels, and hoods have been replaced with TPE, thermoset composites, and PC alloys to reduce weight, improve styling, and reduce costs.

Portable electrical tools and small kitchen appliance housings are no longer made of die cast steel or aluminum but have been replaced with nylon and ABS, improving toughness, electrical insulation, and styling, lowering weight and cost reduction.

Water faucet valves made of die cast steel, brass, or copper are being replaced bynew designs, updated styles, and colors, using acetal, which eliminates corrosion,providing cost reduction and opening new markets.

High performance, large size irrigation valves (from 1.50 to 3.0 in dia.) and small valves (0.75 and 1.00 in dia.) made of die cast steel and brass were successfully replaced with GR nylon 6/12 for the large valves and with GR nylon 6/6 or acetal for the small valves. This improved performance and reliability, eliminatedcorrosion, and provided cost reduction. Other low performance commercial valves made of rigid PVC (lower cost) are also produced for the irrigationmarket.

Toilet anti-siphon (ballcock) valves made of several brass and copper componentswere replaced with a multi-functional design in acetal, improving performance,eliminating corrosion, and providing cost reduction. The acetal valves hadexcellent performance over a 30 year period.

The comparison of properties is an effective tool when applied to materials in the same family. To illustrate the point that properties between different material families cannot be compared, Figure 2-1 shows several graphs using different generic property values of the different material families.

The ferrous metal bars include cast iron, cold rolled steels, structural steels,alloy steels, stainless steels, and tool steels. The nonferrous metal bars include magnesium, aluminum, copper, nickel and brass alloys, and titanium. The rubberbars include acrylic, butadiene, butyl, chloroprene, nitrile, silicone, urethane,EPDM, EPM, fluorocarbon, and natural rubbers. The thermoset bars include phenolics, silicones, alkyds, DAP, polyimides, aminos, unsaturated polyesters,epoxies, and urethanes. The thermoplastic bars include ABS, acrylics, acetals,nylons, LCP, PBT, PET, PS, PE, PP, PC, PPO, PEI, PEKK, PSU, PPS, PTFE, PVC,and SAN.

The specific gravity graph shows the unit weight of a material compared towater and reveals that metals are two to eight times heavier than plastics. On a strength-to-weight basis, plastics have a more favorable position, as indicated by the specific gravity graph. In general, the cost of metals is much higher than plastics.

The continuous exposure temperature graph shows that metals have widertemperature ranges than plastics; metals can be used at colder and at elevated temperatures. This property is used for the classification and temperature range of plastics.

The tensile strength (kpsi) graph shows that metals are much stronger than plastics; metals resist higher forces when being pulled apart before breaking. The tensile strength of a plastic varies with temperature; it decreases with increasing temperature over a much smaller temperature range.

Ferrous metals

Nonferrous metals

Thermosets

Thermoplastics0 2 4 6 8 10

Specific gravity

Ferrous metals

Nonferrous metals

Thermosets

Thermoplastics

-460 0 500 1.000 1.500 2.000

Continuous exposure temperature(˚F.)

Ferrous metals

Nonferrous metals

Thermosets


Tensile strength(kpsi)

Ferrous metals

Nonferrous metals

Thermosets

Thermoplastics

0 5 10 15 20 25 30

Modulus of elasticity(Mpsi)

Ferrous metals

Nonferrous metals

Thermosets


Coefficient of linear thermal expansion(in/in/˚F) x 10-6

Figure 2-1 Comparison of generic properties of materials


The modulus of elasticity (Mpsi) graph shows that metals have higher resistance to deflection for short-term, intermittent, or continuous loading than plastics.Metals have better dimensional stability at elevated temperatures than plastics.Since plastics deflect more than metals under the same loading, it is important that metal and plastic parts be loaded using different techniques. Plastics require that the load be distributed in compression mode.

The coefficient of linear thermal expansion graph shows that increasing the temperature causes more dimensional changes for plastics than for metals.Whenplastics and metals are used together and are exposed to the same temperatures,plastic parts become larger than metals; therefore, design compensations should be provided to compensate dimensional change in plastics.

The thermal conductivity graph shows that metals are good conductors ofheat while plastics are excellent insulators. Despite their relatively low effectivetemperature range, plastics may be superior to metals as high temperature heat shields for short exposures. A plastic part exposed to a radiant heat source soon suffers surface degradation. However, this heat is not transmitted to the opposite surface as rapidly as in metals.

The electrical volume resistivity graph compares only the insulation materials used in electrical applications, (metals are conductors).

The dielectric strength graph shows the voltage gradient at which electrical failure or breakdown occurs as a continuous arc; the higher the value the betterthe material. Plastics have excellent electrical resistance properties, while metals are conductors.

2.1.1 Plastics Selection Guidelines

More than 20,000 thermoplastic grades and over 5,000 thermoset grades havebeen developed for the plastics industry. Because of the enormous diversity ofplastic materials, the selection of the best plastic material for a given application is relatively difficult and time consuming, especially for inexperienced plastic designers.

Table 2-1 provides a comparison of plastics and their properties. The tableincludes the most widely used unreinforced, 30% GR thermoplastic, andreinforced thermoset materials; basic mechanical, thermal and electrical pro-perties, and process temperatures, indicating the process characteristics of the resins. Table 2-1 should be used as a preliminary plastic selection guide.

The material properties listed in Table 2-1 were obtained by the resin producers by testing molded bars using ASTM procedures under laboratory conditions.Because most applications are not flat bars, but complex confi gurations, the actual properties will be different from the published ASTM properties. The values given are only approximate guides used to compare the values between resins for material selection and for preliminary product design calculations.To obtain precise properties for the new product design and confi guration,a prototype mold is required, molding the selected materials, and testing the performance under actual service conditions.

This chapter provides detailed information for all important plastics, theirchemistry, characteristics, advantages, limitations, and applications. Severalplastic organizations, such as ASTM, Modern Plastics, D.A.T.A., Inc., EngineeringPlastics, IDES “Prospector” and all the resin suppliers provide data properties sheets.

Ferrous metals

Nonferrous metals

Thermosets

Thermoplastics0 0.5 1.0 1.5 2.0 2.5

Thermal conductivity(BTU/hr/ft2/˚F/in) x 103

RubberMica laminationsGlass laminations

ThermosetsThermoplastics

105 107 109 10111013 10151017 1019

Electrical volume resistivity(Ohm-cm)

0 1.0 2.0 3.0 4.0

Dielectric strength(Volt/0.001 in) x 103

RubberMica laminationsGlass laminations

ThermosetsThermoplastics

Figure 2-1 (continued)


Table 2-1 Property Comparison for Selected Plastics

Types of Polymers

Spec

ifi c

Gra

vity

Ten

sile

Mod

ulu

s@

73

°F (

Mp

si)

Ten

sile

Str

engt

h@

Yie

ld (

Kp

si)

Not

ch I

zod

Im

pac

t@

73

°F (

ft-l

b/i

n)

Con

tin

ue

Exp

ose

Tem

per

atu

r e (

°F)

Pro

cess

ing

Tem

per

atu

r e (

°F)

Flam

mab

ilit

yU

L -94

Die

lect

ric

Stre

ngt

h(V

ol/M

il)

Dis

sip

atio

n F

acto

r@

1.0

× 1

06 Hz

ABS Unreinforced 1.05 0.30 5.002.50

12.00167185

410518

HB350500

0.030.04

Acrylic Unreinforced 1.17 0.38 7.500.030.50

150190

410575

HB450530

0.09

Acetal Unreinforced 1.42 0.400 10.00 1.30195230

375450

HB 560 0.005

HDPE Polyethylene Unreinforced 0.94 0.20 3.50No

Break158176

400535

HBV2

450500

0.0005

PP Polypropylene Homo Unfi lled 0.90 0.17 4.000.50

20.00212

390525

HBV2

450600

0.002

PS Polystyrene Unfi lled 1.05 0.45 6.000.250.60

122158

390480

HBV2

300600

0.0040.0020

PVC Polyvinyl Chloride Rigid 1.38 0.35 5.900.40

20.00150185

365400

HBV1

600800

0.115

PC – 30% Fiber Glass 1.40 1.25 19.001.703.00

220265

430620

V1V2

450 0.001

PPO – 30% Fiber Glass 1.25 1.10 14.501.702.30

200240

520600

HBV0

550630

…

PBT – 30% Fiber Glass 1.53 1.35 17.50 0.90200250

470530

HBV0

750 0.004

PET – 30% Fiber Glass 1.67 1.50 22.0 1.60 392510565

V05V

430 0.002

LCP – 30% Fiber Glass 1.62 2.25 23.00 1.30430465

660680

V05V

6401,000

0.0019

HTN – 30% Fiber Glass @ 73 °F – 50% RH 1.44 1.50 32.00 1.80 315580620

V2V0

500 0.004

Nylon 6/6 – 33% GR @ 73 °F & 50% RH 1.38 0.90 18.00 2.50 265530580

HBV2

400 0.006

PEI – 30% Fiber Glass 1.50 1.30 24.50 1.90356390

640800

V0495630

0.0025

PPS - 30% Fiber Glass 1.38 1.70 22.0 1.10390450

600750

V05V

450 0.0014

PSU – 30% Fiber Glass 1.46 1.35 14.50 1.10350375

600715

V05V

450 0.002

DAP – (TS) Fiber Glass 1.94 1.40 7.50 1.00390430

290350

V1V0

400450

0.0110.017

(EP) Epoxy – (TS) Fiber Glass 1.84 3.00 18.00 0.50350

4450300430

HBV0

380400

0.020.05

(PF) Phenolic – (TS) Fiber Glass1.741.88

1.902.28

6.5010.00

0.750.90

350450

330390

V1V0

300 0.03

(UP) Polyester – (TS) Fiber Glass1.751.90

1.902.00

10.5015.00

0.5018.00

200250

170320

V05V

450530

0.010.04

(PI) Polyimide – (TS) Graphite Fiber 1.65 0.70 7.50 0.70600740

690V05V

500560

0.0100.003

119

Designer Check List

General ConsiderationsPerformance requirements

(structural, loading cycle, aesthetic,

etc.)

Multifunction design

Product design for assembly

Structural load (static, dynamic,

cyclic, impact, etc.)

Product tolerance specifi cations

Life of product

Resin selection based on performance

of similar applications and end use

Product design for assembly process

Quality of product vs. process

Secondary operations

Packaging and shipping

Environmental RequirementsEnd use temperature

Time, weather, strain, and stress

cracks

Others (chemical, lubricants, water,

humidity, pollution, gasoline, etc.)

Design FactorsType, frequency, direction of loads

Working stress selected (tensile,

compression, fl exural, combination)

Strain percentage selected

Load deformation (tensile, shear,

compression, fl exural, etc.)

Tensile, flexural, initial, secant, yield

modulus used (temperature, creep)

Correlating the test results to end use

environment conditions

Safety factor

Design product for effi cient molding

Economic FactorsCost estimate of the new product

Resin cost vs. molding performance

Number of mold cavities vs. size of

machine and automatic fast cycles

Eliminate secondary operations

Redesign part to simplify production

:

:::

:::

::::

::

:

::

::

:

:

::

:::

::

Quality Control Tests RequiredTension

Compression

Flexural

Impact (drop weight, dynatup, etc.)

Torsion, fatigue

Creep (tension, fl ex, temperature)

Chemical resistance

Weather (outdoors or accelerated)

UL electrical classifi cation

UL continuous service temperature

UL temperature index

Final product UL approvals

Resin Processing CharacteristicsViscosity and crystallization

Difficulties in molding the resin

Melt and mold temperature

Sensitivity to thermal degradation

Directional layout of reinforcements

Frozen stresses

Mold shrinkage control

Molding problems (fl ashing, voids,

warpage, short shots, brittleness,

tolerances, surface fi nishing, etc.)

Material handling

Percentage of reground (runners and

rejected molded parts) allowed to

mix with the virgin material

Drying the virgin resin and reground

material.

Prototype molding the product (resin

behavior unknown)

Appearance of ProductAesthetic product application

Dimensional control, warpage, etc.

Color matching, discoloration

Surface fi nishing

Weld lines, sink marks, fl ow lines

Parting line fl ash

Gate type, size, number, location

Decoration

::::::::::::

::::::::

::

:

:

::::::::



If the product information and the quality of data available about a material have not been developed by the resin supplier, the designer should develop a check list by gathering all the facts related to the application. A typical designer’s check list has been included here (Table 2-2). It may be used as a guideline todevelop a specific check list for any application.All aspects of the part are covered,including the product end use requirements, the structural considerations,the operating environment, the economics, and the appearance factors. This information is provided for making a quick analysis of the part requirements,such as temperature, environment, product life expectancy, and cycle and rateof loading.

Designing with plastics requires maximizing the performance and effi ciency ofthe product and the injection molding process. The following basic principles should be adopted in designing plastic products.

Design freedom is achieved using multifunctional design concepts.

When comparing materials that satisfy the requirements, remember that most metals have greater strength than plastics, and that all plastic material properties are time, temperature, and environment dependent.

Metal design principles are very different from the concepts used in plastic parts design.

Polymers are not substitutes for metals; in most designs the product geometrymust be redesigned using plastic principles to be successful.

We need to remember that there are no bad thermoplastic materials, only bad plastic applications.

2.2 Structural Design of Thermoplastic

Components

This section will present principles for structural design of molded plasticparts. The only data provided are what is necessary to illustrate the type ofinformation needed for analysis of plastic design structures. The mechanical properties described are the properties frequently used by designers of plastic components.

Figure 2-2 shows two regions of the stress-strain curve. First, the region of lowstrain (O – L) will be discussed. This region is known as the elastic range; it is pertinent to applications where minimum deformation of the part under load is of prime concern. The second region of low stress (O – P) is known as the stress limit, which is important when the specimen springs back without deformation.The following discussion of creep and relaxation describes the effect of loading time on strength properties within the stress-strain curve. Specific attention is paid to creep under a constant load and relaxation from a fi xed deformation.

The design methods present the recommended methods for using the mechanicalproperties and concepts for designing with plastics. Illustrations are included to show how the equations, originally developed for metal designs, can bemodified. Designing within the viscoelastic modulus utilizes modifi ed elastic design equations. This method is normally used when deformation of the part is of prime concern. Yield design uses design principles that originate from the principles of plasticity. In this section, the yield stress is the controlling material

•

•

•

•

Strain, ε, (%)

Ten

sile

str

ess,

σ, (p

si)

Elastic range

E = Modulus of elasticity

P

O L

Stress limit

Figure 2-2 Stress-strain curve

121

variable. It is emphasized that the major difference between metal and plastic designs is the necessity of allowing for the time dependence of the mechanical properties of polymeric materials over the entire range of temperatures and environmental conditions that the part may encounter in use.

2.2.1 Stress-Strain Behavior

To understand the response of the material, design engineers have been using a set of relationships based on Hooke’s law, which states that for an elastic material,the strain (deformation) is proportional to the stress (the force intensity).

Roark and Young, Timoshenko, and others have developed analyses based on elastic behavior of materials that exhibit a good approximation of simple elastic behavior over a wide range of loads and temperatures. For high stress levels and repeated loading and creep, more sophisticated analyses have been developed todeal with these types of applications.

Unfortunately, Hooke’s law does not reflect accurately enough the stress-strain behavior of plastic parts and it is a poor guide to successful design, because plastics do not exhibit basic elastic behavior. Plastics require that even thesimplest analysis take into account the effects of creep and nonlinear stress-strain relationships. Time is introduced as an important variable and, because polymers are strongly influenced in their physical properties by temperature,that is another important parameter to be considered.

In order to analyze these effects, mathematical models exhibiting the same type of response to applied forces as plastics are used.

The elements that are used in such an analysis are a spring, which represents elastic response because the deflection is proportional to the applied force, and the dashpot, which is an enclosed cylinder and piston combination that allows the fl uid filling the cylinder to move from in front of the piston to behind the piston through a controlled orifi ce.

The retarded elastic response which occurs in plastic materials is best representedas a spring and dashpot acting in parallel. The creep or cold flow, which occurs in plastics, is represented by a dashpot. The combination best representing the plastic structure would be a spring and dashpot in parallel combination, in series with a dashpot. The basic elements and the combinations are shown in Figure 2-3.

One of the results of the viscoelastic response of polymers is to vary therelationship between the stress and strain, depending on the rate of stressapplication. The standard test used to determine structural properties for many materials is the analysis of the stress-strain curve. Figure 2-4 shows the slope of the curve, which is the elastic constant called Young’s modulus; the stress at which the slope of the curve deviates from the straight line is referred to as the tensile strength; and the stress at which the material fails by separation is called the ultimate tensile strength. In the case of viscoelastic behavior, the shape of the curve will depend on the rate of loading or on the rate of straining, depending on the way in which the test is performed. The modulus can vary over a range ofthree or four to one within the usual testing range and the material can exhibit ductile yielding at the lower straining rates. The value of the tensile strength and the ultimate strength can frequently vary by a 3 : 1 ratio.

It is apparent that, when tensile tests are done on plastics, the loading rates must be specified to make sure the data have any meaning at all. It also becomes clear

A

1

A

2G2A

G1A

Model "A"

1

B

2

B

G1B G2

B

Model "B"

Figure 2-3 Plastic resin structural models,elastic and plastic range

Strain, ε, (%)

Ten

sile

str

ess,

σ, (p

si)

Elastic range

E = Young´s modulus or modulus of elasticity

Ultimatetensile stress

Failure

Stress limit

Figure 2-4 Young’s modulus

2.2 Structural Design of Thermoplastic Components


that such conventional data is essentially useless for the design of plastic parts,unless the end use loading rates happen to be the same as those of the test. Inorder to be useful, the tensile test would have to be run over a wide range of rates and the form in which the data is best presented is a three dimensional plot ofstress-strain and time as illustrated in Figure 2-5.

2.2.2 Tensile Testing of Viscoelastic Materials

In this section we will address the internal effects of forces acting on a structure.The thermoplastic components will no longer be considered to be perfectly rigid, such as in the static analysis cases. Structural design is concerned with the analysis of material strength, such as the deformations of various structures under a variety of loads.

The simple tensile test is probably the most popular method for characterizing metals and so it is not surprising that it is also widely used for plastics. However,for plastics, the tensile test needs to be very carefully performed, because plastics,being viscoelastic, exhibit deformations that are very sensitive to such things as cross head speed rate in tension testing, moisture, stress level, temperature, and creep time.

The stress-strain curves as shown in Figures 2-10 and 2-11 illustrate an interestingphenomenon observed in some flexible plastics, such as thermoplastic elastomers.This behavior is known as the plastic range, cold drawing, or continuouselongation of the specimen beyond the yield point without breaking. It occurs because, at low cross head speed rates, the molecular chains in the plastic havetime to align themselves under the influence of the applied stress. Therefore, the plastic specimen’s molecular chains are able to align at the same rate at which the material it is being strained.

The simplest case to consider is the application of a straight tensile load on a test specimen of constant cross section. The specimen is loaded at both ends with an equal force applied in opposite directions along the longitudinal axis and through the centroid cross section of the tensile test specimen. Underthe action of the applied tensile forces, internal resisting forces are set upwithin the tensile test specimen. The tensile test assumes that the forces are applied through an imaginary plane passing along the middle of its length and oriented perpendicular to the longitudinal axis of the tensile test specimen. The magnitude of these forces must be equal and directed away from the test specimen(tension loading) to maintain an equilibrium of these forces. Typical tensile test equipment, including an extensometer, is shown in Figure 2-6.

Some assumptions are made regarding the variation of these distributed inter-nal resisting forces within the specimen. Because the applied tensile forces act through the centroid, it is assumed that they are uniform across the specimen’scross section. The load distribution depends on the tensile test specimengeometry, dimensions, and manufacturing process. It also depends on thecrystalline molecular structure of the polymer, the coupling agent used toreinforce the compound, and the flow orientation of the material reinforcement.However, to determine the mechanical properties of a polymer by performing either test in compression or tension, the cross head speed rate, at which loading is applied, has a signifi cant influence on the physical properties obtained when running the tests at different loading rates. Ductile materials exhibit the greatest sensitivity of physical property variations at different cross head speed loading rates, whereas these effects are reduced and sometimes negligible for brittle materials.

Stress

Log time

Isochronousstress-straincurves

Surface at specifictemperature

Creep curve

Strain

Figure 2-5 Three-dimensional graph,stress-strain-log time

123

2.2.2.1 Stress or Tensile Strength (σ)

Instead of referring to the internal force acting on some small element of the area, it is easier to use the ratio between the force acting over a unit area of the cross section. The force per unit area is termed as the stress (σ) and is expressed in units of force per unit area, e.g., lb/in2 (psi). If the forces applied to the ends ofthe tensile test specimen are such that the bar is in tension, then the term stress or tensile strength (σ) condition can be applied to the specimen. It is essential that the forces are applied through an imaginary plane passing through the centroid cross section area of the tensile test specimen.

2.2.2.2 Tensile Test Specimen

The tensile test specimen is held in the grips of either an electrically driven gear or hydraulic testing equipment. The electrically driven gear testing equipment is commonly used in testing laboratories for applying axial tension or compression loads.

To standardize material testing procedures, the American Society for Testing Materials (ASTM) has issued standard specifications and procedures fortesting various metallic, non-metallic, and thermoplastic resins in tension and compression tests. The ASTM test procedures for thermoplastic materials can be found in Chapter 11. Figure 2-7 shows a tensile test specimen specifi ed for plastic materials. The dimensions shown are those specified by ASTM for tensile test specimens to fit the grips of the tensile test equipment.

The elongations of the tensile test specimen are measured by a mechanical extensometer (see Figure 2-6), an internal gauge (micro-processor tester),or by cementing an electric resistance type strain gauge to the surface of the tensile test specimen. This resistance strain gauge consists of a number of veryfine wires oriented in the axial direction of the tensile test specimen. As the test specimen elongates, the electrical resistance of the wire changes and this change of resistance is detected on a Wheatstone bridge and interpreted as elongation.

Extensometer

Figure 2-6 Tensile test equipment and temperature chamber

8.50 inch

0.50 inch0.75 inch 0.125inch

Figure 2-7 Thermoplastic tensile test specimen



YB

Strain, ε, (%)

Stre

ss,σ

, (p

si)

PL

0

Med. carbon steel

Figure 2-8 Stress/strain curvefor medium carbon steel

Strain, ε, (%)

B

Stre

ss,σ

, (p

si)

PL

0

Alloy steel

Figure 2-9 Stress/strain curvefor alloy steel

PL

0

B

Strain, ε, (%)

Stre

ss,σ

, (p

si)

High carbon steel

Figure 2-10 Stress/strain curvefor high carbon steel

Y

00 01

B

Strain, ε, (%)

Stre

ss,σ

, (p

si)

Non ferrous alloys

ε0

Figure 2-11 Stress/strain curvefor nonferrous alloys and cast iron materials

Strain, ε, (%)

Stre

ss,σ

, (p

si)

0

Rubber / elastomers

Figure 2-12 Stress/strain curvefor rubber or elastomeric materials

Y B

Strain, ε, (%)

Stre

ss,σ

, (p

si)

PL

0

Unreinforced resins

Figure 2-13 Stress/strain curvefor unreinforced resins

Stress-Strain Curves for Various Materials

Y

B

Strain, ε, (%)

Stre

ss,σ

, (p

si)

PL

0

Reinforced resins

Figure 2-14 Stress/strain curvefor reinforced resins

B

Strain, ε, (%)St

ress

,σ, (

psi

)

PL

0

Brittle resins

Figure 2-15 Stress/strain curvefor brittle resins

125

2.2.2.3 Strain (ε)

The elongation over the tensile test specimen gauge length is measured for any predetermined increment caused by the tensile load. From these values the elongation per unit length, called strain and denoted by ε, may be found bydividing the total elongation ΔL by the original gauge length L, i.e., ε = ΔL / L. The strain is usually expressed in units of inch per inch and consequently is dimensionless.

2.2.2.4 Stress-Strain Curve

As the tensile load is gradually increased at a cross head speed rate, the total elongation over the gauge length and the load are measured and recordedcontinuously at each increment of the load until fracture of the specimen takes place. Knowing the original cross sectional area of the tensile specimen, the stress (σ), may be obtained for any value of the tensile load by applying the following formula:

Tensile Stress = σ = W / A

where W denotes the tensile load in pounds, and A the original cross sectional area in square inches. Having obtained the numerous values of stress (σ) and strain (ε), the test results are plotted with these quantities considered as ordinate and abscissa, respectively. This is the tensile stress-strain curve or diagramof the material in tension. The stress-strain curve represents the mechanical characteristics or behavior for each type of material, therefore the stress-strain curves assume widely differing geometries for various materials. Figure 2-8represents the stress-strain curve for a medium carbon steel, Figure 2-9 the curvefor an alloy steel, Figure 2-10 the curve for a high carbon steel, Figure 2-11 the curve for nonferrous alloys and cast iron materials, and Figure 2-12 the curvefor rubber or elastomeric materials.

Tests conducted at room temperature using ASTM recommended proportional limits showed that polyethylene resin, PP copolymer resin, TPE resins, acetal resin, and unreinforced nylon resin (at 50% relative humidity) are materials that yield gradually until break as shown in Figure 2-13. Reinforced nylon resin (at 50% relative humidity), PC glass reinforced resin, and other compounded polymers that have limited elongation characteristics yield a curve as shownin Figure 2-14. Acrylic resin, PET glass reinforced resin, PBT glass reinforced resin, LCP, PF, PAI, PEI, PEAK, dry as molded nylon glass reinforced resins and most brittle compounded resins usually break before yielding occurs, as shownin Figure 2-15.

2.2.2.5 Hooke’s Law

For any material having a stress-strain curve of the form shown in Figure 2-16,the relation between stress and strain is linear for comparatively small values of the strain. This linear relation between elongation and tensile stress was fi rst noticed by Sir Robert Hooke in 1678 and is called Hooke’s law. This initial linear range of action of the material is described by the following formula:

Stress (σ) = Modulus of Elasticity (E) × Strain (ε)

or

Strain (ε) = σ / E

where E (Modulus of Elasticity) denotes the slope or the straight line 0-PL (originto the proportional limit) as shown in the stress-strain curve Figure 2-16.



E. Alfredo Campo


ISBN 3-446-40309-4




201

Table 2-5 Flat Circular Plate Equations, Part I

W = Concentrated load (lb); w = Unit load (psi); M = Moment (in-lb/in); δ = Defl ection (in); θ = Change in slope (radians);E = Modulus of elasticity (psi); H = Deflection factor (in.); υ = Poisson’s ratio; σ = Stress (psi); t = Wall thickness (in);a = Outer radius (in); b = Inner radius (in); d = Shaft radius; r0 = Radius of load (in); K = Plate constant

Case Type Stress and Deflection Equations (Constant Thickness)

Concentrate Center LoadEdge Simply Supported

W

2 a

y⎛ ⎞−+

= + −⎜ ⎟+⎝ ⎠

20

2 20

13 (1 ) 1log

12 4

rW a

rt a

υυσ

υπ υ

− +=

2

Max. 3

3 (1 )(3 )

4

W a

E t

υ υδ

π

Uniform Distribute LoadEdge Simply Supported

w

2 a

yedge simply supported ⎛ ⎞+

= ⎜ ⎟⎝ ⎠

2

Max. 2

3 (3 )

8

w a

t

υσ

− +=

4

Max. 3

3 (1 )(5 )

16

w a

E t

υ υδ

Concentrated Center LoadOuter Edge Fixed

2 a

W y

For a > r0

⎛ ⎞+= +⎜ ⎟⎝ ⎠

20

Max. 2 20

3 (1 )log

2 4

rW a

rt a

υσ

π

−=

2 2

Max. 3

3 (1 )

4

W a

E t

υδ

π

Uniformly Distributed LoadOuter Edge Fixed 2 a

w y

=2

Max. 2

3

4

w a

tσ

−=

4 2

Max. 3

3 (1 )

16

w a

E t

υδ

−=

4 2

Max. 3

3 (1 )

16

w aH

E t

υδ

For thicker flat circular plates having (t / a = 0.1), multiply the defl ection equation by the constant (H), where *H = 1 + 5.72 (t / a)2.

Central CoupleOuter Edge Simply Supported

2 dM

2 a =+

2

2

0.49

( 0.7 )

aK

d a

⎡ ⎤−= + +⎢ ⎥

⎣ ⎦Max. 2

3 2 ( )1 ( 1) log

4

M a d

K ad tσ υ

π

Central CoupleOuter Edge Fixed

M

2 a

2 d

=+

2

2

0.10

( 0.28 )

aK

d a

⎡ ⎤−= + +⎢ ⎥

⎣ ⎦Max. 2

3 2 (0.45 )1 ( 1) log

0.454

M a d

K ad tσ υ

π

Radial Center LoadEdge Simply Supported

W

2 b

2 a

y

⎡ ⎤⎛ ⎞+⎜ ⎟⎢ ⎥⎝ ⎠ ⎛ ⎞⎢ ⎥= + −⎜ ⎟⎝ ⎠+⎢ ⎥⎢ ⎥⎣ ⎦

2

Max. 2 2 2

12 1

3 1log 1

2 ( )

aW a

bt a b

υ υσ

μπ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞⎢ ⎥= + ⎜ ⎟⎝ ⎠⎛ ⎞ ⎛ ⎞⎢ ⎥+ −⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠−⎣ ⎦

2 2 2 2 222

Max. 32 2

1 3 13 1 ( ) 1 4 1

log1 14 1 ( )

1

W a b a ba

bE t a b

υυ υυ

δπ

υ υ

2.16 Flat Circular Plates


Table 2-5 Flat Circular Plate Equations, Part II

W = Concentrated load (lb); w = Unit load (psi); M = Moment (in-lb); δ = Defl ection (in); θ = Angular change (rad.);Q = shear (lb/in); E = Modulus (psi); υ = Poisson’s ratio; σ = 6M/t2 (psi); t = Wall thickness (in); a = Outer radius (in);r0 = Radius of load (in); D = E t 3 / 12 (1 – ν2); N = Equivalent radius (in); K, C, L, G = Constants (ratio-dependent)

Case Type Boundary Values Special Cases

Outer and Inner EdgeSimply Supported;Central Radial Load

rOW

2 a

b rb a ra0, 0, 0, 0M Mδ δ= = = =

b

2

b

K W a

D

θθ

−=

b QbQ K W=22

a b 4 b 6 6W aa

C Q C LD D

θ θ= + −

oa b

W rbQ Q

a a= −

b / a 0.10 0.50 0.70

ro / a 0.50 0.70 0.70 0.90 0.90

K δMax. –0.0102 –0.0113 –0.0023 –0.0017 –0.0005

K θa 0.0278 0.0388 0.0120 0.0122 0.0055

K θb –0.0444 –0.0420 –0.0165 –0.0098 –0.0048

KMrb –0.4043 –0.3819 –0.0301 –0.0178 –0.0063

KMro 0.1629 0.1689 0.1161 0.0788 0.0662

KQb 2.9405 2.4779 0.8114 0.3376 0.4145

Outer & Inner Edges Fixed;Change in Slope

rO

O2 a

b b a a0, 0, 0, 0δ θ δ θ= = = =

Mrb orb a b;

K D bM Q Q

a a

θ= =

Qb ob Max. o2

;K D

Q K aa

θδ δ θ= =

ra rb 8 b 9 o 7D

M M C Q a C La

θ= + +

b / a 0.10 0.50 0.70

ro / a 0.50 0.70 0.70 0.90 0.90

K δo –0.1071 –0.0795 –0.0586 –0.0240 –0.0290

KMrb –2.0540 1.1868 –3.5685 2.4702 0.3122

KMra –0.6751 –1.7429 –0.8988 –5.0320 –6.3013

KQb –0.0915 –17.0670 4.8176 –23.8910 –29.6041

Uniform Distributed Load;Edge Simply Supported

g yrO

w

2 a

2a ra C 170, 0;M M w a Gδ = = =

417

C 1122 1

GW aG

Dδ

υ

− ⎛ ⎞= −⎜ ⎟⎝ ⎠+

2 2 2a o( )

8 (1 )

wa r

D aθ

υ= −

+

ro / a 0.00 0.20 0.40 0.60 0.80

K δC –0.0637 –0.0576 –0.0422 –0.0230 –0.0067

K θa 0.0961 0.0886 0.0678 0.0393 0.0124

KMC 0.2062 0.1754 0.1197 0.0621 0.0177

2 2a o o 11 14 17

(3 )( ); If 0, 0.015, 0.062,

2 16

wQ a r r G G G

a

υ− += − = = = =

4 3 4 2 3

11 14 C C a(5 )

; ; ; ;64 (1 ) 16 8 (1 )

(3 )w a w a w a w a w aL T G L T G M

D D D Dδ θ

υδ θ

υ υ

υ− − − += = = = =

+ ++

Linear Increase Load;Edge Simply Supported

rOw

2 a

2a ra C 180, 0;M M w a Gδ = = =

418

C 1222 1

Gw aG

Dδ

υ

− ⎛ ⎞= −⎜ ⎟⎝ ⎠+

2 2a o o(2 )

6

wQ a r a r

a

−= − −

ro / a 0.00 0.20 0.40 0.60 0.80

K δC –0.0323 –0.0249 –0.0164 –0.0083 –0.0023

K θa 0.0512 0.0407 0.0278 0.0148 0.0043

KMC 0.0955 0.0708 0.0449 0.0222 0.0061

318

a 15 o 12 15 18(4 )

2 ; If 0, 0.004, 0.022,1 45

Gw aG r G G G

D

υθ

υ

+⎛ ⎞= − = = = =⎜ ⎟⎝ ⎠+5 4 2 3

o12 C C a

o

(6 ) (4 ); ; ;

15 (1 ) 45 15 (1 )

w a r w a w a w aL T G M

D a r D Dδ

υ υδ θ

υ υ

− − − + += = = =

− + +

Central Circular Load;Edge Simply Supported

Wr

rO

2 a

2 2 2o

(3 ) 1For ; ( ) 2 ln ; ln

16 (1 ) 4 (1 )

W rW a ar r a r r

D r D r

υδ θ

π υ π υ

+⎡ ⎤ ⎡ ⎤> = − − = +⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦2 2 2

2 2r o o2 2

( )4 (1 ) ln (1 ) ; 1.6 0.67 ; If 0.5

16

W a a r NM N r t t r t

r a rυ υ

π

⎡ ⎤−= + + − = + − <⎢ ⎥⎣ ⎦

2

o o t 2or , If 0.5 ; 4 (1 ) ln (1 ) 4

16

W a NN r r t M

r rυ υ

π

⎡ ⎤⎛ ⎞= > = + + − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2

Max. Max. Max.(3 )

at ; ; ; (1 ) ln 116 (1 ) 4 (1 ) 4

W a W a W ar a M

D D N

υδ θ υ

π υ π υ π

− + ⎡ ⎤= = = = + +⎢ ⎥+ + ⎣ ⎦

203

Besides the usual loadings, Table 2.5 Part II also includes several loading cases that may be described best as externally applied conditions that force a lack offlatness into the flat circular plates.

The first time we look at Table 2.5, Part II it appears to be a formidable task tocalculate the strength of these structures. However, when we consider the numberof cases it is possible to present in a limited space, the reason for this method ofpresentation becomes clear. With careful inspection, we find that the constants and functions with subscripts are the same except for the change in variables.In Table 2.5, Part II, the tabulated values in the Special Cases are listed for the preceding functions for the most frequently used denominator values of the variable ratios, such as b / a and r0 / a.

Example 2-41

A flat circular plate is made of nylon 6/6 with 33% fiber glass reinforcement at 73 °F and 50% relative humidity. The radius is 3.00 in with a wall thickness of 0.25 in. The plate is simply supported around its edge and it is loaded with 500.00 lb at the center. The load is distributed through a round area of0.125 in radius. Determine the maximum bending stress at the surface of the plate and the maximum deflection at the center of the plate.

Solution

This flat circular plate and loading are covered in Table 2.5, Part I, case load at center with the outer edge simply supported. The diagram and equations in Figure 2-96 were obtained from this table:

t = 0.250 in, w = 500 lb, a = 3.00 in, r0 = 0.125 in,

E = 900,00 psi, υ = 0.39, σ = 18,000 psi

20

Max. 2 20

2

2 2

13 (1 ) 1log

12 1 4

3 500 (1 0.39) 1 3.00 1 0.39 0.125log

0.39 1 0.1252 3.1416 0.25 1 0.39 4 3.0

10,794 psi

rW a

rt a

υυσ

υπ υ

⎛ ⎞−+= + −⎜ ⎟+ +⎝ ⎠

⎛ ⎞× + − ×= + −⎜ ⎟+× × + × ×⎝ ⎠=

2 2

Max. 3 3

3 (1 ) (3 ) 3 500 3.0 (1 0.39) (3 0.39)

4 4 3.1416 900,00 0.25

0.158 in

W a

E t

υ υδ

π

− + × × − += =× × ×

=

Example 2-42

A thick flat circular plate is made of nylon 6/6 with 33% fiber glass reinforce-ment at 73 °F and 50% relative humidity with a radius of 4.00 in and a uniform wall thickness of 0.50 in. The plate’s outer edge is fixed and it is uniformly loaded along the round area of plate with 200.00 lb/in.

Determine the maximum bending stress at the surface of the plate and the maximum deflection at the center of the plate.

W

2a

δ

Figure 2-96 Flat circular plate,concentrated center load, and simply supported edge



Solution

This thick flat circular plate and type of loading is presented in Table 2.5, PartI, case Uniformly Distributed Load with the Outer Edge Fixed. The diagram and equations in Figure 2-97 were obtained from the table.

Because this example case deals with a thick plate, we need to investigate if the thickness / radius ratio is greater than 0.1 to modify the maximum deflection by multiplying the value by the constant (H).

t = 0.500 in, w = 200 psi, a = 4.00 in, E = 900,00 psi,

υ = 0.39, σ = 18,000 psi

For thicker flat circular plates with a ratio t / a > 0.1, multiply the defl ection equation by the constant (H), where H = 1 + 5.72 (t / a)2.

0.500.125 0.1

4.00

t

a= = >

2 21 5.72 ( / ) 1 5.72 (0.50/4.00) 1.089H t a= + = + =

2 2

Max. 2 2

3 3 200 4.012,223.00 psi

3.1416 0.50

w a

tσ

π

× ×= = =×

4 2 4 2

3 3

3 (1 ) 3 200 4.0 (1 0.39 )0.072 in

16 16 900,00 0.50

w a

E t

υδ

− × × −= = =× ×

Max.1.089 0.072 0.079 inH x yδ = = × =

Example 2-43

A flat circular plate, made of acetal homopolymer, has a wall thickness of0.187 in and a 5.00 in outside diameter, and is simply supported with a uni-formly distributed load of 6.0 psi. Calculate the maximum deflection in the center, the maximum stress, and the deflection equation for Figure 2-98.

This flat circular plate and type of loading is presented in Table 2.5, Part II,case Uniformly Distributed Load Edge Simply Supported. First, we need todetermine the maximum moment, the bending stress, the plate constant,and the deflection caused by the load. Second, we need to calculate the total deflection of the plate caused by the load, the moment, and the loading constant. Finally, we need to check the deflection at the outer edge.

t = 0.187 in, w = 6.0 psi, a = 2.50 in, r0 = 0,

E = 410,000 psi, υ = 0.35, σ = 10,000 psi

2 2

Max. Center(3 ) 6.0 2.50 (3 0.35)

7.85 lb-in.16 16

w aM M

υ+ × += = = =

Max. 2 2

6 6 7.851,339.97 psi

0.187

M

tσ

×= = =

w

2a

δ

Figure 2-97 Flat circular plate,uniformly distributed load, and fi xed edge

w

2a

rO

Figure 2-98 Flat circular plate,uniformly distributed load with simply supported edge

205

3 3

2 2

410,000 0.187256.66

12 (1 ) 12 (1 0.35 )

E tD

υ

×= = =− −

4 4

C(5 ) 6.0 2.50 (5 0.35)

0.0565 in64 (1 ) 64 256.66 (1 0.35)

w a

D

υδ

υ

− + − × += = = −+ × +

The total deflection equation for the flat circular plate is:

2C

C 2 (1 )a yM y

L TD

δ δυ

= + ++

, where for this case4

11yw a

L T GD

−=

Where the constant G11 = 0.015, when r0 = 0.

2 4

2 4

7.85 6.0 0.0150.0565

2 256.66 1.35 256.66

0.0565 0.01132 0.000365

aa a

a a

δ× × ×= − + −

× ×= − + × − ×

Checking the deflection at the outer edge, when a = 2.50 in

2 40.0565 0.01132 2.50 0.000365 2.50

0.0565 0.07075 0.01425 0.0aδ = − + ⋅ − ⋅

= − + − =

Example 2-44

A flat circular plate is made of acetal homopolymer with a wall thickness of 0.125 in and 4.00 in outside diameter. It is mounted in a fi xture toproduce a sudden change in slope in the radial direction of 0.05 radiant at a radius of 0.75 in. It is then clamped between two fl at fixtures as shown in Figure 2-99.

Calculate the maximum bending stress.

This is an example of forcing a known change in slope into a fl at circular plate, clamped (fixed) at both inner and outer edges. This flat circular plate and type of loading is presented in Table 2.5, Part II, case Outer and Inner Edge Fixed and Change in Slope, where: θ0 = 0.05, b / a = 0.10, r0 / a = 0.50 and Poisson’s ratio of υ = 0.35.

t = 0.125 in, a = 1.50 in, b = 0.15 in, r0 = 0.75 in,

θ0 = 0.05 rad., θb = 0.0 rad., δb = 0.0 in, E = 410,000 psi,

υ = 0.35, σ = 10,000 psi

3 3

2 2

410,000 0.12576.04

12 (1 ) 12 (1 0.35 )

E tD

υ

×= = =− −

2.054 0.05 76.045.20 lb-in.

1.50Mrb

rbK D

Ma

θ× × − × ×= = = −

2 2

0.0915 0.05 76.040.154 lb/in.

1.50

Qbb

K DQ

a

θ× × − × ×= = = −

2a

rO

θO

4.00 dia.

0.125

0.15 r.

1.50 r.

0.05 rad.

0.75 r.

Figure 2-99 Flat circular plate having a change in slope with both outer and inner edges fi xed



8 9 0 7

8 9 70.05 76.04

5.20 ( 0.154)

ra rb bD

M M C Q a C La

C a C La

θ= + +

×= − × + − × × +

Max. 2 2

6 6 5.201,996.80 psi

0.125rbM

tσ

× ×= = =

Max. 0 0 0.1071 0.05 1.50 0.008 inK y rδ θ= = − × × =

Example 2-45

A flat circular plate, made of acetal homopolymer, has a wall thickness of0.250 in and 5.00 in outside diameter, it is simply supported at the outeredge and subjected to two types of loads. One center load provides a uniform pressure over a diameter of 0.0625 in. The other is axis-symmetrically loaded with a distributed load that increases linearly from the center to the outside radius rO = 1.00 in;, this load has a value of 10.00 psi at the outer edge.

Calculate the maximum bending stress.

This example requires analyzing two different cases and to superposition the results. The first case is the linear increase of the distributed load with simply outer edge supported (Figure 2-100), the second case is the central circular uniform load with simply supported outer edge (Figure 2-101). Both cases are presented in Table 2.5, Part II.

t = 0.250 in, a = 2.50 in, r01 = 1.00 in, r02 = 0.031 in,

E = 410,000 psi, υ = 0.35, σ = 10,000 psi

From the special case data, the following variable ratios are obtained:

r01 / a = 1 / 2.5 = 0.40, K yC = –0.0164, K θa = 0.0278, KMC = 0.0449

3 3

2 2

410,000 0.250608.38

12 (1 ) 12 (1 0.35 )

E tD

υ

×= = =− −

4 40.0164 10 2.50.0105 in

608.38Cw a

K yD

δ− × ×= = =

2 20.0449 10 2.5 2.80 lb-in.MCM K w a= = × × =

2 2

Max.(3 ) 2.50 (3 0.35)

0.0105 in16 (1 ) 16 608.38 (1 0.35)

P a P

D

υδ

π υ π

− + − × += = =+ × × +

20.76 lb.P = −

The second moment component is calculated by using the equations providedin Table 2.5, Part II, case Central Circular Loading and Simply Outer Edge Supported.

w

2 a

rO1

Figure 2-100 First case:Linear decreasing distributed load and edge simply supported

P

2 a

r

rO2

Figure 2-101 Second case:Center uniformly circular load and edge simply supported

207

2 2 2 2021.6 0.675 1.6 0.03 0.25 0.675 0.25

0.085 in

N r t t= + − = × + − ×=

Max.20.76 2.50

(1 ) ln 1 (1 0.35) ln 14 4 0.085

9.19 lb-in.

P aM

Nυ

π π

−⎡ ⎤ ⎡ ⎤= + + = + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦= −

Max. 2 2

6 6 ( 9.19 2.80)613.44 psi

0.250

M

tσ

− += = = −

2.17 Torsion Structural Analysis

A bar is rigidly clamped at one end and twisted at the other end by a torque T = F × d, applied in a plane perpendicular to the axis. Plane sections remain plane and radii remain straight. There is at any point a shear stress (τ) on the planeof the section; the magnitude of this stress is proportional to the distance from the center of the section and its direction is perpendicular to the radius drawn through the point. The deformation and stresses are shown in Figure 2-102.

In addition to these deformations and shear stresses, there are the longitudinal strain and stress. The longitudinal strain is reduced while the stress is in tension on the outside and a balancing compression stress is exerted on the inside.

Assumptions

The torsion equations are based on the following assumptions:

The bar is straight, of uniform circular cross section (solid or tubing), and of homogeneous isotropic material.

The bar is loaded only by equal and opposite twisting couples, which are applied at its ends in a normal direction to its axis.

The bar is not stressed beyond the elastic limit of the material.

Angle of Twist (θ)

If a shaft of length (L) is subjected to a constant twisting moment (T) along its length, then θ is the angle through which only one end of the bar will be twisted.

Twisting Moment (T)

The twisting moment T for any section along the bar is defined to be the algebraicsum of the moments of the applied couples that lie to one side of the section in question.

Shearing Strain

If a bar is marked on the surface (unloaded), then after the twisting moment (T) has been applied, this line moves as shown in Figure 2-102. The angle (θ) is

•

•

•

LF

d

T = F x d

Figure 2-102 Deformation and stress under torque



measured in radians; the final and original position of the generator is defi ned as the shearing strain at the surface of the bar.

Shearing Stress (τ)

For a solid circular cross section bar, let T = Twisting moment; L = Length ofthe bar; r0 = Radius; J = Polar moment of inertia; τ = Shear stress; θ = Angle oftwist (radians); G = Modulus of rigidity. Then:

( )/( )T L G Jθ = ; Max. 0( )/T r Jτ =

By substituting for 40( )/2J rπ= in the equation above for a solid circular cross

section with radius r0, the following equations are obtained:

40(2 )/( )T L r Gθ π= ;

3Max. 0(2 )/( )T rτ π=

For a circular tube cross section with outer radius r0 and inner radius ri:

4 40 i(2 )/ ( ) ]T L r r Gθ π= − ;

4 4Max. 0 0 i(2 )/[ ( )]T r r rτ π= −

The torsional stiffness of the bar can be expressed by the general equation:θ = (T L) / (G K), where K is a factor dependent on the bar cross section. Forcross section bars, the factor K is equivalent to the polar moment of inertia J.In Table 2-6, the equations for the factor K and for the maximum shear stress (τMax.) for a variety of cross section bars are given.

Example 2-46

Compare the strength and stiffness of a circular injection molded tube made of a plastic material, 1.00 in outside diameter and 0.187 in wall thickness,versus an extruded solid circular bar of the same material with the same diameter.

The strengths of both cross sections will be compared by using the twisting moments (T) required to produce the same shear stress. The stiffness will be compared by using the values of factor (K) for both cross sections.

For the circular tube bar:

4 4 4 4 40 i( )/2 3.1416 (0.50 0.313 )/2 0.083 inK r rπ= − = − =

4 4 4 40 i 0( )/(2 ) 3.1416 (0.50 0.313 )/(2 0.50)

0.166 lb-in.

T r r rτ π τ

τ

= − = × − ×= ×

For the solid circular bar:

4 4 4/2 3.1416 0.50 /2 0.098 inK rπ= = × =

3 3( )/2 3.1416 0.50 /2 0.196 lb-in.T rτ π τ τ= = × × = ×

The solid circular cross section bar is therefore 1.182 times as stiff as the circular tube cross section bar and 1.18 times as strong.

209

Table 2-6 Torsion Equations

Cross section Constant K in T L

K Gθ = Shear stress max.

Solid circle

2rO

4O

2

rK

π= Max. 3

O

2 T

rτ

π=

Circular tube

rOri

4 4O i( )

2

r rK

π −=

OMax. 4 4

O i

2

( )

T r

r rτ

π=

−

Solid ellipse

2rS

2rL

3 3L S

2 2L S

r rK

r r

π=

+Max. 2

L S

2 T

r rτ

π=

Solid square

a

a 40.1406K a= Max. 30.208

T

aτ =

Solid rectangle

b

a3 4

45.33 3.36 1

16 12

b a a aK

b b

⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

Max. 2 2

(3 1.8 )T b a

b aτ

+=

θ = Angle of twist (radians); T = Twisting moment (lb-in); τ = Shear stress (psi);G = Modulus of rigidity (psi); J = Polar moment of inertia (in4); K = Constant equivalent to J (in4); ro = Outer radius (in); ri = Inner radius (in); rS = Elliptical shortradius (in); rL = Elliptical large radius (in); a = Height (in); b = Width (in).


211

3 Structural Designs for Thermoplastics

3.1 Uniform and Symmetrical Wall Thickness

The ultimate design rule for injection molding thermoplastic products is toensure that the wall thickness is uniform and symmetrical.

Non-uniform and/or heavy wall thicknesses can cause serious warpage and dimensional control problems in the injection molded products. Heavy wall sections cause not only internal shrinkage, voids, and surface sink marks, but also nonuniform shrinkage resulting in poor dimensional control and warpage problems.

Figure 3-1 shows a poor cross section design of perpendicular corner walls that causes molding problems, such as differential shrinkage, warpage (concave) of both walls, and internal voids in the corner of the thicker wall. The last two designs are recommended to avoid these molding problems.

Figure 3-2 shows a heavy wall cross section design that could cause molding problems and the recommended design using a thin wall and proportional ribs.

Figure 3-3 shows a nonuniform wall section that should be replaced with a thin uniform wall having the same strength of the original heavy wall section.

Figure 3-4 shows another poor and the recommended uniform wall design.

Figures 3-5 and 3-6 show cross sections of two nonuniform wall designs and the recommended designs with a uniform wall thickness to avoid warpage, internal voids, long molding cycles, and surface sink marks.

Poor design

Molding problems

VoidsWarpage

Sharpcorner

Good design

r.

R.

Good design

Seat

r.

Figure 3-1 Perpendicular walls,end corner designs

Poor design Good design

Figure 3-2 Heavy wall vs. thin uniform ribbed wall designs


Figure 3-3 Nonuniform wall vs. thin uniform wall designs








E. Alfredo Campo


ISBN 3-446-40309-4




VII

Preface

This handbook was written for the injection molding product designer who has alimited knowledge of engineering polymers. It is a guide for the designer to decidewhich resin and design geometries to use for the design of plastic parts. It can also offer knowledgeable advice for resin and machine selection and processing parameters. Manufacturer and end user satisfaction is the ultimate goal.

This book is an indispensable, all inclusive, reference guide that can be used byany plastic product designer. There is no need to search through many books and catalogs for needed information. New illustrations, graphs and equations have been included to provide additional clarity for complex ideas. The equationshave been verified to ensure correctness and not just copied from anothersource. Thousands of hours of research and cross referencing have gone into the completion of this work. In addition, more than 35 years of the “hands-on”experience of a plastics expert have been incorporated in this handbook.

The following topics are covered:

Chapter 1 Plastic Materials Selection Guide: Includes an introductionto plastic materials, the beginning of plastics, classifi cation ofpolymer families. Each resin is discussed by its basic chemistry,properties, processing characteristics, advantages, disadvant-ages and limitations, typical applications and several productillu strations. Thermoplastic materials (35 generic families),thermoplastic elastomer materials (8 generic families), liquidinjection molding of silicone, thermoset materials (16 generic families).

Chapter 2 Engineering Product Design: Starts with the introduction tostructural product design principles, mechanical strength pro-perties of thermoplastics. Centroid, section area, moment ofinertia equations and tables. Beam defl ection analysis methods.Structure analysis of beams, columns, flat circular plates, andtorsion.

Chapter 3 Structural Design for Thermoplastics: Discusses the product wallthickness, structural rib design, sharp corners, bosses, threads,undercuts, integral life hinges, pin hinges. Encapsulation of inserts,types of metal inserts and anchorage, and electrical lead inserts.

Chapter 4 Thermoplastic Gearing Design: An introduction to and classi-fication of gears. Standard spur, helical, bevel, and worm gears;properties required for thermoplastic molded spur gears, mount-ing gears on metal shafts, tolerances and mold shrinkage of gears.Plastic spur and helical gearing technology design, strength,horsepower rating, equations, tables, analysis examples and gear specifi cation illustrations.

Chapter 5 Plastic Journal Bearing Design: An introduction to types ofmaterials for journal bearings. Theory and design for lubrica-tion. Design principles, performances, dimensions, clearances,molding effects, PV limits and surface fi nishing. Self-lubricated thermoplastic bearings. Equations, tables, and analysis exam-ples.

VIII Preface

Chapter 6 Thermoplastic Spring Design: Introduces cantilever beam springdesign, applications, and analysis examples. Locating, fi xing clip,flexible hinges, and torsional spring applications. Belleville spring washers’ equations, tables, and analysis examples.

Chapter 7 Thermoplastic Pressure Vessel Design: Discusses thin- andthick-walled pressure vessels’ basic principles, equations, tables,analysis examples, design guidelines, applications, and pressure vessel regulations.

Chapter 8 Thermoplastic Assembly Methods: Joining two or more compo-nents together: assembly method is selected based on product design geometry, size, end use requirements, thermo plastic mate-rial characteristics, automatic or manual assembly operation, and manufacturing costs. Each assembly method provides a descrip-tion, process sequence, advantages and limitations, typical applica-tions, equipment, product joint design, and its variations.

Chapter 9 Thermoplastic Effects on Design: Starts discussing the polymer melt behavior, reinforcement, degradation, moisture characteris-tics, mold shrinkage and critical properties. The molding process effects caused by molding cycle, melt/mold temperature, injectionpressure and speed, etc. on product design dimensions, surface finishing, weld line strength and impact resistance and othermolding problems.

Chapter 10 Thermoplastic Injection Mold Design: Provides an introduction of injection molds, classification and effects on product design.Types of steels, chemical composition, effects of alloying, heat treatment, properties and characteristics. Types of steels usedfor mold bases and mold components. Cavity surfaces fi nishprocedures and specifications. Types of injection mold designs.Cold runners (two- and three-plate molds, interchangeable mold inserts and vertical insert encapsulation mold). Hot runner molds (internally and externally heated, insulated). Mold design system and other considerations, such as number of cavities, partingline, ejection, cooling, cold runner, gating, venting, cavity inserts sidewall strength, support pillars, molded parts tolerances, mold designer check list, general specifications for mold construction are covered.

Chapter 11 Performance Testing of Thermoplastics: It introduces various tests to which thermoplastic polymers are subjected, describes their properties (statistical analysis), such as mechanical, thermal,chemical resistance, rheometer melt viscosity, soldering heatresistance, electrical, flammability, smoke generation, weathering and micro-organism resistance. Test description, procedures,apparatus, test specimen and conditioning, and their signifi cance are discussed here.

Chapter 12 Thermoplastic Product Cost Analysis: It discusses moldingprocess variables and capital equipment cost. Three cost analysis methods are used to estimate the molded product user’s price.

Documents

The Complete Part Design Handbook