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2018
DATE
12 - 21
When the slider of the punch moves the stamped part down, the slider will be affected by the slider.
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2018
DATE
12 - 20
Mechanical shearing machine fault diagnosis and elimination.
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2018
DATE
12 - 20
All parts are inspected before installation to verify compliance with design requirements; important parts need to be cleaned before assembly.
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2018
DATE
12 - 20
The blade beam is used to mount the scissors and is swung downwards under the drive of the drive unit. Cutting, the cutting surface of the scissors on the knife beam needs to be machined, and the straightness error of the mounting surface is 1000:0.06mm.
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2018
DATE
12 - 20
How to avoid punch injury accidents Use hand safety tools●Use hand safety tools to avoid accidents caused by unreasonable mold design and sudden equipment failure. Common safety tools are elastic tongs, special tongs, magnetic suction cups, cymbals, pliers, hooks, etc.
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2018
DATE
12 - 20
Grinding tool gaps, materials, heat treatment, stamping wear, guiding structures, and convex and concave molds are the main causes of stamping burrs.Therefore, it is necessary to adjust the mold gap in time,
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2018
DATE
12 - 19
Criteria for bending Regardless of the applicatioand use of available technologies , the production of accurate parts with minimum set -up time must meet and maintain five basic bending criteria.These are:1 . Accurate bend angles2 . Constant bend angle overthe full bend length3 .
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2018
DATE
12 - 19
Shearing machine electrical principleworking principle: Closed power supplies open HK main motor contactor control will be reliable. Pressing the button SB2 main contactor KM1 coil is electrically connected to the main motor. Since the main contactor pull-in shearing machine pulls the electromagnet
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2018
DATE
12 - 19
Pascal's Law is a theory which states that the pressure (P) in a confined fluid, caused by a force (F1), over an area (A1), is transmitted undiminished, causing a force (F2), over the area (A2). This law can be applied to magnify a small force by the ratio of the areas to give a larger force – F2 = F1 (A2/A1).
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