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- 时间：2016-11-10 21:19 点击：
Plastic rupture in the connector generally has the following two kinds of circumstances：

1、Combined with lack of line strength and rupture. Such as LCP crystallization rate is too fast, the combination of low strength, easy to break, this time can be used to combine the strong strength of the materials, such as PA.

2、Plastic and terminal interference is too large and broken. When the plastic and the terminal interfere too much with the bending strength of the material, the plastic is broken, and the interference quantity can be reduced at this time.

The nature of the plastic rupture is that its actual bending strength exceeds its allowable bending strength. From the point of view of mechanics of materials, the paper tries to analyze it in detail.

According to the mechanics of materials, when L/B>5, the normal work of the beam depends on the normal stress intensity of the beam. When L/B<5, the normal stress strength condition and the shear stress strength condition should be satisfied simultaneously. In the connector, the mechanical model is a fixed beam with rectangular cross section, and it is generally L/B<5, so it should satisfy the condition of normal stress and shear stress at the same time.

No.1 The bending normal stress strength condition is as follows：

σmax=Mmax/W=（0.25*F*L）/(b*h²/6)=(3*F*L)/(2* b*h²)<=[σ] (1)

By f= (F*L³) / (48*E*I)

F=(4*E* f *b* h³)/ L³ (2)

The type (2) into the equation (1), then

σmax=（6* f*E *h）/ L²<=[σ] (3)

Among them, f represents the maximum deflection;

h represents the thickness of the material；

l represents the length of the material；

[σ]Allowable bending strength

No.2 Shear stress strength condition

According to the mechanics of materials, the shear stress strength condition of rectangular section is:

τmax =(Fmax*S)/(I*b)

=(0.5F*b*0.5h*0.25h)/(b*h³/12*b)

=3F/(4*b*h²)<=[τ] (4)

Among them, S below the neutral axis (or more) of the cross-sectional area of the neutral axis.

The type (2) into the equation (4), then

τmax =3* E*h*f/ L³<=[τ] (5)

No.3 problem

1. According to the formula (3), (5), in the case of the same amount of interference, the thickness of H plastic material is thinner, the maximum normal stress or maximum shear stress is smaller, the product is safer, but this is not consistent with the actual situation.

2. The type (3) on both sides at the same time divided by E, is in the same length and thickness, the same amount of interference under the condition of different materials [Sigma]/E if the greater the kind of material is not easy to break. If the above derivation is correct, such as "connector plastic characteristics," as the data, PA6T and PBT, PA66 should be the same as not easy to crack, but the actual PA6T but easy to break, this problem how to explain?

In view of the above two questions, I consulted the Holland chemical industry (DSM) Mr. Dai Jiaqing, his explanation is as follows：

1、 Thin wall thickness and thick wall thickness, the same amount of interference in the terminal, the actual maximum deflection of F is not the same! Because the plastic than metal strength low hardness, the surface generated near the metal plastic deformation, thus the deflection becomes low, thus affecting the sigma Max decreased. The thick plastic wall thickness ratio of thin plastic wall thickness is easy to produce plastic deformation (can be understood as the absorbed energy), and therefore not easy to crack.

2、 On the premise of [Sigma]/E of different materials if the greater the material is not easy to break, is correct. The data referenced by PA6T is PA6T to get the standard test strip under the condition of good forming, and then the data can be measured on the test strip. And because of the fact that in the PA6T forming connector, because of the high processing conditions (such as mold temperature, etc.), it is difficult to get an ideal strength and toughness, as shown in the table, so it is easy to crack.