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The Complete Part Design

Handbook

E. Alfredo Campo

For Injection Molding of Thermoplastics

ISBN 3-446-40309-4

Weitere Informationen oder Bestellungen unterttp://www.hanser.de/3-446-40309-4 sowie im Buchhandel

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201

Table 2-5 Flat Circular Plate Equations, Part I

W = Concentrated load (lb); w = Unit load (psi); M = Moment (in-lb/in); δ = Deflection (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; r 0 = 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

r W a

r t a

υ υ σ

υ π υ

− +=

2

Max. 3

3 (1 )(3 )

4

W a

E t

υ υδ

π

Uniform Distribute LoadEdge Simply Supported

w

2 a

y +

=

2

Max. 2

3 (3 )

8

w a

t

υ σ

− +=

4

Max. 3

3 (1 )(5 )

16

w a

E t

υ υδ

Concentrated Center LoadOuter Edge Fixed

2 a

W y

For a > r 0 +

= +

20

Max. 2 20

3 (1 )log

2 4

r W a

r t a

υ σ

π

−=

2 2

Max. 3

3 (1 )

4

W a

E t

υδ

π

Uniformly DistributedLoad

Outer Edge Fixed 2 a

w y

=2

Max. 2

3

4

w a

t σ

−=

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 deflectionequation by the constant (H ), where *H = 1 + 5.72 (t / a)2.

Central CoupleOuter Edge SimplySupported

2 d M

2 a =+

2

2

0.49

( 0.7 )

aK

d a

−= + +

Max. 2

3 2 ( )1 ( 1) log

4

M a d

K ad t σ υ

π

Central Couple

Outer Edge FixedM

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 t σ 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 υ σ

μ π

− − + + = + + − −

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

υ υ υ υ

δ π

υ υ

2.16 Flat Circular Plates

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02 2 Engineering Product Design

Table 2-5 Flat Circular Plate Equations, Part II

W = Concentrated load (lb); w = Unit load (psi); M = Moment (in-lb); δ = Deflection (in); θ = Angular change (rad.);Q = shear (lb/in); E = Modulus (psi); υ = Poisson’s ratio; σ = 6 M /t 2 (psi); t = Wall thickness (in); a = Outer radius (in);r 0 = 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

r OW

2 a

b rb a ra0, 0, 0, 0 M M δ δ= = = =

b

2

b

K W a

D

θθ

−=

b QbQ K W =22

a b 4 b 6 6

W aaC Q C L

D Dθ θ= + −

oa b

W r bQ Q

a a= −

b / a 0.10 0.50 0.70r o / 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

K Mrb –0.4043 –0.3819 –0.0301 –0.0178 –0.0063

K Mro 0.1629 0.1689 0.1161 0.0788 0.0662

K Qb 2.9405 2.4779 0.8114 0.3376 0.4145

Outer & Inner Edges Fixed;Change in Slope

r O

O

2 a

b b a a0, 0, 0, 0δ θ δ θ= = = =

Mrb orb a b;

K D b M Q Q

a a

θ= =

Qb ob Max. o2

;K D

Q K aa

θδ δ θ= =

ra rb 8 b 9 o 7

D M M C Q a C L

aθ= + +

b / a 0.10 0.50 0.70

r o / a 0.50 0.70 0.70 0.90 0.90

K δo –0.1071 –0.0795 –0.0586 –0.0240 –0.0290K Mrb –2.0540 1.1868 –3.5685 2.4702 0.3122

K Mra –0.6751 –1.7429 –0.8988 –5.0320 –6.3013

K Qb –0.0915 –17.0670 4.8176 –23.8910 –29.6041

Uniform Distributed Load;Edge Simply Supported

r Ow

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 )

w a r

D aθ

υ= −

+

r o / 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

K MC 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

w Q 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

r Ow

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

w Q a r a r

a

−= − −

r o / 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

K MC 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

W

r

r O

2 a

2 2 2o

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

16 (1 ) 4 (1 )

W r W a ar r a r r

D r D r

υδ θ

π υ π υ

+ > = − − = + + +

2 2 22 2

r o o2 2

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

16

W a a r N M N r t t r t

r a r υ υ

π

−= + + − = + − <

2

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

16

W a N N r r t M

r r υ υ

π

= > = + + − −

2

Max. Max. Max.

(3 )at ; ; ; (1 ) ln 1

16 (1 ) 4 (1 ) 4

W a W a W ar a M

D D N

υδ θ υ

π υ π υ π

− + = = = = + + + +

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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 of

flatness into the flat circular plates.

The first time we look at Table 2.5, Part II it appears to be a formidable task to

calculate the strength of these structures. However, when we consider the number

of cases it is possible to present in a limited space, the reason for this method of presentation 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 r 0 / 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 of 0.125 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, r 0 = 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

r W a

r t a

υ υ σ

υ π 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 simplysupported edge


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Solution

This thick flat circular plate and type of loading is presented in Table 2.5, Part

I, 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 investigateif 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 deflection

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

t σ

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 y δ = = × =

Example 2-43

A flat circular plate, made of acetal homopolymer, has a wall thickness of

0.187 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 to

determine the maximum moment, the bending stress, the plate constant,

and the deflection caused by the load. Second, we need to calculate the totaldeflection 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, r 0 = 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 a M M

υ+ × += = = =

Max. 2 2

6 6 7.851,339.97 psi

0.187

M

t σ

×= = =

w

2a

δ

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

w

2a

r O

Figure 2-98 Flat circular plate,

uniformly distributed load with simply

supported edge

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205

3 3

2 2

410,000 0.187256.66

12 (1 ) 12 (1 0.35 )

E t D

υ

×= = =

− −

4 4

C

(5 ) 6.0 2.50 (5 0.35)0.0565 in

64 (1 ) 64 256.66 (1 0.35)

w a

D

υδ

υ

− + − × += = = −

+ × +

The total deflection equation for the flat circular plate is:

2C

C2 (1 )

a y

M y L T

Dδ δ

υ= + +

+, where for this case

4

11 y

w aL T G

D

−=

Where the constant G11 = 0.015, when r 0 = 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

a

a 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 0.0

aδ = − + ⋅ − ⋅= − + − =

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 fixture to

produce 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 flat 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 flat 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, r 0 / a = 0.50

and Poisson’s ratio of υ = 0.35.

t = 0.125 in, a = 1.50 in, b = 0.15 in, r 0 = 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 t D

υ

×= = =

− −

2.054 0.05 76.045.20 lb-in.

1.50 Mrb

rb

K D M

a

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 achange in slope with both outer and inner

edges fixed


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8 9 0 7

8 9 7

0.05 76.045.20 ( 0.154)

ra rb b

D M M C Q a C L

a

C a C La

θ= + +

×= − × + − × × +

Max. 2 2

6 6 5.201,996.80 psi

0.125

rb M

t σ × ×= = =

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

Example 2-45

A flat circular plate, made of acetal homopolymer, has a wall thickness of

0.250 in and 5.00 in outside diameter, it is simply supported at the outer

edge 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 loadedwith a distributed load that increases linearly from the center to the outside

radius r O = 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, r 01 = 1.00 in, r 02 = 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:

r 01 / a = 1 / 2.5 = 0.40, K y C = –0.0164, K θa = 0.0278, K MC = 0.0449

3 3

2 2

410,000 0.250608.38

12 (1 ) 12 (1 0.35 )

E t D

υ

×= = =

− −

4 40.0164 10 2.50.0105 in

608.38C

w aK y

Dδ

− × ×= = =

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

2 2

Max.

(3 ) 2.50 (3 0.35)0.0105 in

16 (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 provided

in Table 2.5, Part II, case Central Circular Loading and Simply Outer Edge

Supported.

w

2 a

r O1

Figure 2-100 First case:Linear decreasing distributed load

and edge simply supported

P

2 a

r

r O2

Figure 2-101 Second case:

Center uniformly circular load

and edge simply supported

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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 a

M N υ π π

− = + + = + + = −

Max. 2 2

6 6 ( 9.19 2.80)613.44 psi

0.250

M

t σ

− += = = −

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 plane

of 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.

AssumptionsThe 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 ) alongits 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 algebraic

sum 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

•

•

•

L F

d

T = F x d

Figure 2-102 Deformation and stress

under torque


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measured in radians; the final and original position of the generator is defined

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 of

the bar; r 0 = Radius; J = Polar moment of inertia; τ = Shear stress; θ = Angle of twist (radians); G = Modulus of rigidity. Then:

( )/( )T L G J θ = ; Max. 0( )/T r J τ =

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

section with radius r 0, the following equations are obtained:

40(2 )/( )T L r Gθ π= ;

3Max. 0(2 )/( )T r τ π=

For a circular tube cross section with outer radius r 0 and inner radius r i:

4 40 i(2 )/ ( ) ]T L r r Gθ π= − ; 4 4

Max. 0 0 i(2 )/[ ( )]T r r r τ 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. For

cross 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 r π= − = − =

4 4 4 4

0 i 0( )/(2 ) 3.1416 (0.50 0.313 )/(2 0.50)0.166 lb-in.

T r r r τ π τ τ

= − = × − ×= ×

For the solid circular bar:

4 4 4/2 3.1416 0.50 /2 0.098 inK r π= = × =

3 3( )/2 3.1416 0.50 /2 0.196 lb-in.T r τ 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.

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209

Table 2-6 Torsion Equations

Cross section Constant K inT L

K Gθ = Shear stress max.

Solid circle

2r O

4O

2

r K

π= Max. 3

O

2T

r τ

π

=

Circular tube

r Or i

4 4O i( )

2

r r K

π −=

OMax. 4 4

O i

2

( )

T r

r r τ

π=

−

Solid ellipse

2r S

2r L

3 3L S

2 2L S

r r K

r r

π=

+Max. 2

L S

2 T

r r τ

π=

Solid square

a

a 40.1406K a= Max. 30.208

T

a τ =

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 a τ +=

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 = Constantequivalent to J (in4); r o = Outer radius (in); r i = Inner radius (in); r S = Elliptical shortradius (in); r L = Elliptical large radius (in); a = Height (in); b = Width (in).


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211

3 Structural Designsfor Thermoplastics

3.1 Uniform and Symmetrical Wall Thickness

The ultimate design rule for injection molding thermoplastic products is to

ensure 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, internalvoids, 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




Figure 3-5 Nonuniform wallvs. thin uniform wall designs



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