### Forces impacting on the 3D printer frame

## Frame design priorities:

- Convenient adjustment of XY and Z portals
- Redesign of 3D printer HYPERCUBE.
- Will fit into an extruded profile length of 6 meters. (Stores providing sales and cutting services sell it in multiples of 6 meters, sometimes in multiples of 3 meters.)

I have not seen detailed articles on the Internet about what forces impact on the frame of a 3D printer. So I decided to write about it.

I plan to print infill at 150mm/s

First, let’s determine what forces and where impact on the frame at a given speed.

Let's imagine that the printer is running at a speed of 150 mm/s and the print head suddenly stops.

Let's break it down into several tasks.

- What is the displacement of the frame during heavy braking along the -X and +X axis?
- What is the displacement of the frame during sudden braking along the -Y and +Y axis?
- What is the displacement of the bed console during heavy braking in the X and Y axes?

We determine the forces impacting during sudden braking for the X and Y axes.

**F = A * M**,

**F**- Force

**A**- Acceleration

**M**- Mass

**Fx = A * Mx**

**Fy = A * My**

**Mx**- mass of all moving parts along the X axis

**My**- mass of all moving parts along the Y axis

Everything is simple here, we take it and weigh it.

**A = S1-S0/T**

**A**- acceleration

**S1**- initial speed

**S0**- final speed

**T**- time how long did we stay

**T = L/S**

**L**- braking length

**S**- speed.

Use a calculator and determine the value of the T variable from the diagram.

We don’t need super precise values, I’ll explain why later.

My SI variable values:

**Mx**= 0.105 kg

**My**= 0.555 kg

**T**= 0.00187 seconds (40000 acceleration)

I took these acceleration values specifically in order to load the frame as much as possible at these mass values.

**S1**= 0.15m/s

**S1**= 0m/s

**A**= -80,35714286 m/s^2, minus because we are braking.

**Fx**= 8.4375 newtons

**My**= 44.59821429 Newton

Round up

**Fx = 9 Newton**

**Fy = 45 Newtons**

## Diagrams for force **Fx = 9 Newton ****Fy = 45 Newtons**

Diagram 1. A force of 45 newtons applied in the +Y direction. Isometric |

Diagram 2. A force of 45 newtons applied in the +Y direction. View from above |

Diagram 3. A force of 45 newtons applied in the +Y direction. Side view |

Diagram 4. A force of 45 newtons is applied in the -Y direction. Isometric. |

Diagram 5. A force of 45 newtons applied in the +Y direction. View from above. |

Diagram 6. A force of 45 newtons applied in the +Y direction. Side view. |

Diagram 7. A force of 9 newtons applied in the -X direction. Isometric. |

Diagram 8. A force of 9 newtons applied in the -X direction. Front view. |

Diagram 9. A force of 9 newtons applied in the -X direction. View from above. |

Diagram 10. A force of 9 newtons applied in the +X direction. Isometric. |

### Let's move the carriage to the extreme positions along the Y axis and see how this affects the movement of the frame.

Diagram 11. Force 9 Newton applied in +X direction, carriage in position Y min. Isometric |

Diagram 12. Force 9 newton applied in +X direction, carriage in position Y min. View from above. |

Diagram 13. Force 9 newton applied in +X direction, carriage in position Y min. Front view. |

Diagram 14. Force 9 Newton applied in +X direction, carriage in Y position max. Isometric. |

Diagram 15. Force 9 Newton applied in +X direction, carriage in Y position max. Front view. |

Diagram 16. Force 9 Newton applied in +X direction, carriage in Y position max. View from above. |

## Conclusion.

In real conditions, you will print at accelerations of up to 15,000, and the forces will be 2.6 times less than the displacement, respectively. The cantilevered table also does not experience critical movements relative to the XY portal.

It is also possible to substitute other axle mass values, for example using a carbon tube for the X beam or using a direct extruder.

You can substitute your values in the table. There you will also find T values for other acceleration values.

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