Clamping tonnage calculation injection molding

How do you calculate the clamping tonnage?

To calculate the clamping tonnage calculation injection molding of the machine required to mould the part. We first need to calculate the projected area of the component.
How to determine the projected area:
The Projected area of a component is calculated by multiplying the length and width of the component. This calculation is used for determining the required injection mould clamp force (tonnage of the machine). The projected area is then multiplied by a factor ranging from 2 to 8 depending on the type of material selected for moulding the component.

For example: Let’s say we have a part that is 9″ long by 4″ wide and depth is less than 1 “. The depth is only important if it exceeds more than 1 inch. The projected area for this component is to multiply 9” X 4”, which is 36 square inches.

Clamping tonnage calculation injection molding - Toggle type
Clamping tonnage calculation injection molding - Ram type

Clamping tonnage calculation injection molding: To calculate the tonnage of the machine required to mould the part. We first need to calculate the projected area of the component.
How to determine the projected area:

The Projected area of a component is calculated by To arrive at the Injection mould clamping force required for this mould is to multiply the projected area by a factor ranging from 2 to 8 tons per square inch. This factor depends on the material to be used. Normally, 5 tons per square inch is used for calculating the clamping tonnage for any given mould. In this example, the projected area 36 square inch multiplied by 5 Tons per square inch(36 X 5) results in 180 tons. Always add a safety factor of 10% of the tonnage – (180 Tons +18 Tons = 198 Tons). Hence a 200-ton clamping force machine is the best choice for moulding this component.

The factor 2 tons per square inch is used for components made from material with a high flow rate and 8 tons per square inch is used for low flow materials( stiff ).

When to consider the depth of the component is more than 1“
Whenever the depth( total thickness or height of the part) of the component exceeds 1”, the clamping force must be increased by 10% for every 1” of the component. So if the part is 2” deep then it would be 180 ton +18 ton( 10%)=198 Tons and additional 10% safety factor is 19.8 tons, which means 217.8 tons clamping force is required, ideally, a machine with 225-ton clamping tonnage will be best suited for moulding this component.

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