〔 PRODUCT DATA 〕 EJECTOR PIN / EJECTOR SLEEVE STRENGTH CALCULATION
■ Ejector Pin / Ejector Sleeve Strength Calculation Ejector pins and sleeves are subjected to compressive loads when the cavity is filled with molten plastic. When long, thin objects are subjected to such compressive loads, "buckling", bending of the pin, or breakage can occur. In order to prevent "buckling," we recommend that you select an appropriate configuration after performing strength calculations beforehand. (1) Computing buckling load P [kgf]: Euler's formula is normally used to calculate the buckling strength of ejector pins. P = n π 2 AE ( K L ) 2 (2) Computing compression load P 1 [kgf]: Compression load refers to load that is applied to the ejector pin during filling and pressurization with φ d P Cross section A [mm 2 ]
《 Example 2 》 Stepped Ejector Pin We will examine buckling strength when an internal cavity pressure of = 4kgf/mm 2 is applied to a stepped ejector pin with a tip diameter (d) of φ 1.2mm, total length (L) of 100mm, and a tip section length ( L ) of 40mm. (A) Buckling strength calculation for the tip section (1) From Euler's formula P = nπ 2 AE ( K ℓ ) 2 = 2.05 × π 2 × π × 1.2 2 4 × 21000 × ( 1.2/4 40 ) 2 = 27.0 (kgf) (2) Whereas, the compression load P 1 exerted on the ejector pin is P 1 = × A = × πd 2 4 = 4 × π × 1.2 2 4 = 4.5 (kgf) (3) Therefore, the safety factor (S) is S = P P 1 = 27.0 4.5 ≒ 6.0 (B) Buckling strength calculation for the retention diameter section As strength calculations for stepped ejector pins are extremely complex, we will make the calculation here assuming a straight ejector pin with tip diameter of φ d and total length of L. ( 1) From Euler's formula P = nπ 2 AE ( K L ) 2 = 4 × π 2 × π × 1.2 2 4 × 21000 × ( 1.2/4 100 ) 2 = 8.42 (kgf) (2) Whereas, the compression load P 1 exerted on the ejector pin is P 1 = × A = × πd 2 4 = 4 × π × 1.2 2 4 = 4.52 (kgf) (3) Therefore, the safety factor (S) is S = P P 1 = 8.42 4.52 ≒ 1.9 ■ Ejector Sleeve Strength Calculation Examples 《 Example 3 》 Straight Ejector Sleeve We will examine buckling strength when an internal cavity pressure of = 4kgf/mm 2 is applied to the straight ejector sleeve shown in the figure on the right.
P
＝ 4kgf/mm 2
φ d ＝ 1.2
φ d 1 ＝ 2
molten plastic. P 1 = × A
Core
Movable side die plate
n: terminal condition constant For straight
n = 4
< Stepped Ejector Pin >
n = 2.05
For stepped
Ejector pin
A: cross–sectional area [mm 2 ]
P
＝ 4kgf/mm 2
π 4 π 4
d 2
For round
φ d ＝ 8 φ d 1
(d 2 –d 1 2 )
For cylinder
Upper ejector plate
＝ 4
E: Young's modulus
21000 [kgf/mm 2 ]
Lower ejector plate
K: radius of gyration of cross section K =
/A [mm]
K = d/4
For round
K = d 2 + d 1 2 /16
For cylinder
: geometrical moment of inertia [mm 4 ]
πd 4 64
=
For round
π 64
（ kgf ）
=
P 1
P 1
(d 4 –d 1 4 )
For cylinder
Internal cavity pressure
: internal cavity pressure [kgf/mm 2 ]
（ kgf/mm 2
）
< Straight Ejector Sleeve >
Cross section A [mm 2 ]
(3) Computing safety factor:
P
＝ 4kgf/mm 2
S = P P 1 《 Considerations regarding safety factor values 》 : S afety factor (S) is affected by a wide variety of elements, including those listed below. • Inaccuracy of load estimates /inconsistent strength of materials /effect of heat treatment • Notch effect/finished surface roughness /abrasion and corrosion during use • Expansion and contraction due to heat /fatigue/shock/mold separation resistance during ejection of the molded object; etc. In specific terms, we recommend that you decide in advance on an in–company design standard taking into consideration the empirical values of the various companies, and then use this to gauge the appropriateness of the computed results.
d 2 + d 1 2 16
P = nπ 2 AE (
K L )
2
＊ K =
(1) From Euler's formula
× 21000 × (
)
＝ 3 φ d ＝ 6 φ d 1
2
π (8 2 –4 2 ) 4
(8 2 + 4 2 )/16 100
= 4 × π 2 ×
φ d 1 φ d
φ d
= 15600 (kgf) (2) Whereas, the compression load P 1 exerted on the stepped ejector sleeve is P 1 = × A = × π (d 2 –d 1 2 ) 4 = 4 × π × (8 2 –4 2 ) 4 = 151 (kgf) (3) Therefore, the safety factor (S) is S = P P 1 = 15600 151 ≒ 103 《 Example 4 》 Stepped Ejector Sleeve
φ D ＝ 8
φ D 1
＝ 3.5
■ Ejector Pin Strength Calculation Examples 《 Example 1 》 Straight Ejector Pin
P
＝ 4kgf/mm 2
We will examine buckling strength when an internal cavity pressure of = 4kgf/mm 2 is applied to a straight ejector pin with a diameter (d) of φ 2mm and total length (L) of 100mm. (1) From Euler's formula P = nπ 2 AE ( K L ) 2 = 4 × π 2 × π × 2 2 4 × 21000 × ( 2/4 100 ) 2 ≒ 65 (kgf) (2) The compression load P 1 exerted on the ejector pin is P 1 = × A = × πd 2 4
< Stepped Ejector Sleeve >
We will examine buckling strength when an internal cavity pressure of ＝ 4kgf/mm 2 is applied to a stepped ejector sleeve with a tip diameter (d) of φ 6mm, hole diameter (d 1 ) of φ 3mm, total length (L) of 100mm, tip section length ( L ) of 40mm, retention diameter (D) of φ 3mm, and recessed hole diameter (D 1 ) φ 3.5mm. (A) Buckling strength calculation for the tip section (1) From Euler's formula P = nπ 2 AE ( K R ) 2 ＊ K = (d 2 + d 1 2 ) 16 = 2.05 × π 2 × π (6 2 –3 2 ) 4 × 21000 × ( (6 2 + 3 2 )/16 40 ) 2 = 15810 (kgf) (2) Whereas, the compression load P 1 exerted on the stepped ejector sleeve is P 1 = × A = × π (d 2 –d 1 2 ) 4 = 4 × Q × (6 2 –3 2 ) 4 = 84.8 (kgf) (3) Therefore, the safety factor (S) is S = P P 1 = 15810 84.8 ≒ 186
φ d ＝φ 2mm
(B) Buckling strength calculation for the retention diameter section As strength calculations for stepped ejector sleeves are extremely complex, we will make the calculation here assuming a straight ejector sleeve with tip diameters of d and d 1 . (1) From Euler's formula P = nπ 2 AE ( K L ) 2 = 4 × π 2 × π (6 2 –3 2 ) 4 × 21000 × ( (6 2 + 3 2 )/16 100 ) 2 = 4940 (kgf) (2) Whereas, the compression load P 1 exerted on the ejector pin is P 1 = × A = × π (d 2 –d 1 2 ) 4 = 4 × π × (6 2 –3 2 ) 4 = 84.7 (kgf) (3) Therefore, the safety factor (S) is S = P P 1 = 4940 84.7 ≒ 58.3.
= 4 × π × 2 2 4 ≒ 12.6 (kgf)
< Straight Ejector Pin >
(3) Therefore, the safety factor (S) is S = P P 1
= 65
≒ 5.2
12.6
1191
1192
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