Standard Components For Plastic Mold 2021

〔 PRODUCT DATA 〕 STRENGTH OF INLAY SECTION OF POSITIONING LOCKING BLOCKS WEDGE AND CUTTING AMOUNT

〔 PRODUCT DATA 〕 METHODS FOR COMPUTING ONE–STEP CORE PIN DIMENSIONS

■ One Step Core Pins: Computing the Gradient θ of the Shaped Section

■ Strength of Positioning Locking Block Inlay Sections

ℓ

Considering the force acting on the inlay section to be cantilever as shown in the figure on the left Bending Moment Section Modulus Mmax. = F∙H Z = A∙ L 2 6 Allowable Stress σ b = Mmax. Z = F∙H Z F = σ b ∙Z H = σ b ∙A∙ L 2 6 × H {maximum stress}

For A = A 1

F

F

H

ℓ

ℓ

F

L

F

A

Step 1A

Step 1B/1E

Step 1C

Step 1D

A–V 2 L –D + A

D–V 2 L

A–V 2 L

A–V 2 ( L –C)

Gradient θ computation

θ= tan –1

θ= tan –1

θ= tan –1

θ= tan –1

① Inlay height H The below table shows that the longer the H, the lower the maximum stress a locking block can endure.

σ b = 1200–1800kgf/cm

A = 25mm L = 10mm H = 4mm, then F =

V = D–2 L tan θ

V = A–2 L tan θ

V = A–(2 L –D + A) tan θ

V = A–2 ( L –C) tan θ

V dimension computation

2 {11760–17640N/cm 2 }

(When the material is hard steel) Assuming that σ b = 1200kgf/cm 2 {11760N/cm 2 }

F { Maximum stress }

H

Strength coefficient

E For the shaft diameter designation 0.01mm increments type, calculate using P for D. Calculation of tan –1 (arc tangent) is simple using a function calculator. • How to derive the tan –1 ( arc tangent ) value from “ Conversion table of trigonometric functions ” ■ To find tan –1 ( x ) , please refer to the Conversion table of trigonometric function on D P.1215 . When the x of tan –1 (x) is less than or equal to 1 When the x of tan –1 (x) is greater than or equal to 1

kgf { N }

4 5 6 7 8 9

1250 {12258} 1000 {9800} 833 {8163} 714 {6997} 625 {6125} 556 {5449} 500 {4900}

100

1200 × 2.5 × 1 2 6 × 0.4

= 3000 2.4

= 1250kgf {12250N}

80 67 57 50 44 40

Assuming stress concentrating factor α = 2.5 { α = 2.5 when inlay corner area R is close to 0} F = 1250 2.5 = 500kgf {4900N}

① Locate the tan θ column from among the trigonometric functions listed on the top section of the Conversion table, then proceed down the column until you find the relevant value. ② The angle for θ on the left–hand column for that relevant value will be almost equal to the calculated value for tan –1 .

① Locate the tan θ column from among the trigonometric functions listed on the bottom section of the Conversion table, then proceed down the column until you find the relevant value. ② The angle for θ on the right–hand column for that relevant value will be almost equal to the calculated value for tan –1 .

10

② Inlay length L In the above case, maximum stress F 1 when L is lengthened from 10mm to 12mm is: F 1 = 1200 × 2.5 × 1.2 2 6 × 0.4 = 4320 2.4 = 1800kgf {17640N} L = 10 c 12 F 1 F = 1800 1250 = 1.44 The calculation indicates that the strength of the inlay section is 1.44 times greater.

③ Inlay corner area R The larger the corner area R, the smaller the α (stress concentrating factor). Therefore, the maximum stress which the locking block can endure increases.

(Example) tan –1 (1.4281) ≒ 55°00’

(Example) tan –1 (0.0875) ≒ 5°00’

R

tan –1 (0.0850) = 4°50’–5°00’

tan –1 (1.4315) = 55°00’–55°10’

θ (theta)

When deg (angle) = 0°00’–11°50’

.5640 .5664 .5688 .5712 .5736 .5760 .5783 .5807 .5831 .5854

.8250 .8241 .8225 .8208 .8192 .8175 .8158 .8141 .8124 .8107 sin θ

.6380 .6873 .6916 .6959 .7002 .7046 .7089 .7133 .7177 .7221 cot θ

1.4641 1.4550 1.4460 1.4370 1.4281 1.4193 1.4106 1.4019 1.3934 1.3848 tan θ

20 30 40 50 10 20 30 40 50

40 30 20 10 50 40 30 20

sin θ .0000

cos θ 1.0000

tan θ .0000

cot θ

deg (angle°)

Large R ←→ Small R 1 ≦ α ≦ 3

∞ 90°00’

0°00’

・ ・ ・ ・ ・ ・ ・ ・ ・ ・ ・ ・ ・ ・ ・

・ ・ ・ 30 20 10 50 40 30

55°00’

35°00’

30 40 50 10 20 30

.0785 .0814 .0843 .0872 .0901 .0929 .0958

.9969 .9967 .9964 .9962 .9959 .9957 .9954

.0787 .0816 .0846 .0875 .0904 .0934 .0963

12.706 12.251 11.826 11.430 11.059 10.712 10.885

■ Relationship Be tween Wedge and Cutting Amount

Sinking quantity of the wedge when cut 0.1mm to inclined plane with an angle of A°

Sinking quantity of the wedge when cut 0.1mm in perpendicular direction with an angle of A°

85°00’

5°00’

54°10’

cos θ

deg (angle°) θ (theta)

When deg (angle) = 54° 10’–66° 00’

(Reference Data) Method for Computing Dimensions During Tip Shape Selection ( ※ V is the dimension prior to tip shape processing. ) C (Chamfering) G (Cone cutting) T (Tapering) R (Rounding) B (Spherical processing)

0.1

A

B

A

B

A

B

A

B

A

C

A

C

A

C

A

C

0° 30´

11.460 5.730

7° 8° 9°

0.820 0.720 0.640 0.580 0.520 0.480 0.440 0.410 0.390

16° 17° 18° 19° 20° 21° 22° 23° 24°

0.360 0.340 0.320 0.310 0.290 0.280 0.270 0.260 0.250

25° 26° 27° 28° 29° 30° 35° 40° 45°

0.240 0.230 0.220 0.210 0.210 0.200 0.170 0.160 0.140

0° 30´

11.460 5.730

7° 8° 9°

0.810 0.710 0.630 0.570 0.510 0.470 0.430 0.400 0.370

16° 17° 18° 19° 20° 21° 22° 23° 24°

0.350 0.330 0.310 0.290 0.270 0.260 0.250 0.240 0.220

25° 26° 27° 28° 29° 30° 35° 40° 45°

0.210 0.200 0.200 0.190 0.180 0.170 0.140 0.120 0.100

x 1

x 1

x 1

S

G

1°

1°

1° 30´

3.820 2.870 2.290 1.910 1.430 1.150 0.960

1° 30´

3.820 2.860 2.290 1.910 1.430 1.140 0.950

2°

10° 11° 12° 13° 14° 15°

2°

10° 11° 12° 13° 14° 15°

2° 30´

2° 30´

3° 4° 5° 6°

3° 4° 5° 6°

Cutting amount to inclined plane when the sinking quantity of the wedge is 1.0mm with an angle of A°

Cutting amount in perpendicular direction when the sinking quantity of the wedge is 1.0mm with an angle of A°

G = Standard:

SR = automatically determined.

S = Standard:

0.1mm increments

0.1mm increments

SR ± 0.1 E The spherical shape of the tip is not a perfect sphere.

K = 1° increments

Q = 0.1mm increments

Precision/Extra precision: 0.05mm increments

Precision/Extra precision: 0.05mm increments

20 < K ≦ 60 and θ= K

L ∙tan θ – A 2 (1–sin θ )∙tan θ –cos θ

0.2 ≦ Q ≦ V/2 x 1 = Q(1–sin θ ) x 2 = Q {1–(1–sin θ )tan θ }

V 2

0.5 ≦ G < θ< 45°

SR =

V 2(tanK–tan θ ) Processing limit value α for L : α = V 2tanK

0.1 ≦ S < V 2tanK K = 1° increments 10 ≦ K ≦ 45 and θ< K

x 1 =

E

x 1 = SR(1–sin θ )

x 2 = G(1–tan θ ) Processing limit value α for L : α = G

Processing limit value α for L : α = Q

Processing limit value α for L : α = V 2

A

D

A

D

A

D

A

D

A

E

A

E

A

E

A

E

x 2 = S(tanK–tan θ )

0° 30´

0.009 0.017 0.026 0.035 0.044 0.052 0.080 0.087 0.105

7° 8° 9°

0.122 0.139 0.156 0.174 0.191 0.208 0.225 0.242 0.259

16° 17° 18° 19° 20° 21° 22° 23° 24°

0.276 0.292 0.309 0.326 0.341 0.358 0.375 0.391 0.407

25° 26° 27° 28° 29° 30° 35° 40° 45°

0.423 0.438 0.454 0.469 0.485 0.500 0.574 0.643 0.707

0° 30´

0.009 0.017 0.026 0.035 0.044 0.052 0.070 0.087 0.105

7° 8° 9°

0.123 0.140 0.158 0.176 0.194 0.212 0.231 0.249 0.268

16° 17° 18° 19° 20° 21° 22° 23° 24°

0.287 0.306 0.325 0.344 0.364 0.384 0.404 0.424 0.445

25° 26° 27° 28° 29° 30° 35° 40° 45°

0.466 0.488 0.510 0.532 0.554 0.577 0.700 0.839 1.000

1°

1°

θ= 0° c SR = x 1 θ> 0° c SR > x 1

θ= 0° c Q = x 1 = x 2 θ> 0° c Q > x 1 > x 2

1° 30´

1° 30´

θ= 0° c G = x 2 θ> 0° c G > x 2

Processing limit value α for L : α = S

2°

10° 11° 12° 13° 14° 15°

2°

10° 11° 12° 13° 14° 15°

2° 30´

2° 30´

3° 4° 5° 6°

3° 4° 5° 6°

1193

1194

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