How to choose an injection molding machine?" alt="How to choose an injection molding machine?" src="http://img.waimaoniu.net/789/789-201808031814533978.jpg"/>
A good injection molding machine must make a consistent plastic part, as will the next rework order. Usually according to injection weight, clamping force, projection area for the selection criteria.
1. Injection Weight
Injection weight is an important parameter of the injection molding machine. It is the most commonly used parameter when choosing an injection molding machine, expressed in ounces or grams. Although this parameter is simple and easy to understand, it is also easily misunderstood. The reason is simple. When the plastic is selected, the injection molder can determine the weight of the plastic parts, so they can easily weight the machine with this injection weight. In fact, the injection weight is defined as the weight of the plastic injection when empty (not injected into the mold) and is measured rather than calculated theoretically. The rubber used is usually the general gravity of 1.05 hard plastic, plastic parts plastic different from the PS, the specifications on the injection weight to be used after the following conversion. For non-PS plastic injection weight = c * b / 1.05, (b = plastic weight and c = injection weight expressed in PS.) For example, Race steel has a specific gravity of 1.42. 8 ounces of steel.
(1) Choose an injection molding machine with a sufficient weight for injection - You should not choose an injection molding machine with an injection weight that is exactly equal to the weight of the molded part plus the plastic weight of the runner. In the less demanding injection molding, such as toy portrait, the total weight should be 85% of the weight of the injection. In demanding injection molding, such as crystal supplies, then use 75%. The injection weights measured on empty injection, respectively, are small at an injection pressure, whereas high injection pressures require high injection pressures.
For example, Injection molded plastic toy with runway plastic with a specific gravity of 1.38 for rigid PVC (UPVC) 4 ounces, injection weight required for an injection molding machine Injection weight expressed as PS = 4 * 1.05 / 0.38 = 3.04 ounces. Using the 85% guidelines, the injection weight required for Injection = 3.04 / 0.85 = 3.58 ounces.
(2) Injection molding machine with too high injection weight is not acceptable - The total weight of plastic and runner plastic should be between 35% and 85% of the injected weight. The lower limit is due to three considerations: the bending of the plate, the residence time of the plastic in the barrel, and the power consumption per kg of injection molded parts.
Using small molds on small injection molded parts can cause excessive bending of the formwork, causing him to flex (affecting product quality) and even rupturing the formwork.
With a large plastic injection molding machine small plastic parts, melt dwell time is too long will lead to plastic decomposition.
The use of a mold cavity or increase the size of the mold can be used to solve large-scale plastic injection molding machine some of the problems. Reducing the temperature of the barrel also relieves the breakdown caused by the long residence time.
2. Clamping Force
Clamping force is the injection molding machine clamping device is a very important parameter, he is the injection molding machine used in injection molding to resist injection pressure maximum clamping force. Not enough clamping force to produce a fist. Most of the injection molding machines are the clamping force (ton) for the model name.
Should be used in the maximum clamping force below the value of a sufficient clamping force and the projection area of the cavity is proportional to the cavity projection area is the mold projection surface area of the mold parting surface. Here, the ton is the metric ton (equal to 1000 kg). Clamping force estimation in the following ways.
The conservative approach is to multiply the projected area by a plastic-dependent constant. For example, hard plastic injection molding thick-walled 1 to 2 tons/inch thin-wall injection is used 3 to 4 tons/inch 2.
For example: To injection a 79mm diameter general hard plastic cup. The thinnest part of the cup is 0.6 mm. Enough clamping force.
The projected area of the cup (and the runner) is 3.1416 * 7.92 / 4 = 49cm². This cup is thin-walled and has a conservative clamping force of 0.62 * 49 = 30.4T.
Calculate the required clamping force, according to our company's machine specifications table, use the previous grade of the machine.
3. The Projected Area
Projection area is an important injection molding machine model selection parameters, here said the projection area is not only the area of the parts but also includes garbage bits and nozzle area.
The area is generally estimated by dividing the clamping force by a constant. Typical hard plastic thick wall injection constants are 1 to 2 t / in 2 and thin wall injection is 3 to 4 t / in 2.
For the existing beer, casting area calculation should be divided into small regular shape calculation.
⑴, d diameter of the circular area of s calculation: s = Лd2 / 4
⑵, side length of a square area s calculation: s = a2
⑶, a long, b-wide rectangular area s calculation: s = ab
⑷, on the bottom of a bottom of the b, high his the calculation of the trapezoidal area s: s = (a + b) h / 2
Note: The above conditions are necessary, but not sufficient conditions, if one of the conditions are not met, you must use the previous grade models.
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