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I am planning to build a large scale 3D printer that will utilise 2 1000mm linear motion shafts that each have a young modulus of 210 GPA. The ends of the shafts will be secured in a block of plastic and will not be able to move and the distance from one usable end to the other will total 900mm. Due to the large size of the parts being used I have concerns over possible deflection that could occur. Therefore I need to calculate the minimum required diameter necessary for 2 shafts to support a 2kg weight in the centre while deflecting less than 0.05mm (my layer height is going to be 0.1mm). Thanks for any help I am not an engineer so I have not the slightest clue how to do the calculations to determine this. Diagram

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  • $\begingroup$ Hollow shafts or solid? What material of construction for the shafts? The rest is pretty easy after that. $\endgroup$ – Mark Jun 28 '18 at 6:17
  • $\begingroup$ @Mark - The shafts will be solid and be some sort of steel, some of the linear shafts I looked at were using GCr15, High carbon steel. After finding the material properties I discovered that it has a young's modulus of 210 GPA. Thanks $\endgroup$ – Sven Jun 28 '18 at 17:03
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I think that @Donald Gibson answered well. There is no such thing as "will not be able to move" in engineering because everything moves; we just cannot humanly observe those movements in most cases without specialist equipment. Plastic deforms substantially but its elastic modulus returns it to near-original shape whilst marginally deforming permanently under each repetitive stress action.

Cylindrical bars may seem a good idea but your problem is the variable average 19.6133N static force. When that carriage stops dead centre, the inertia will cause a moment of force to develop in excess of the gravitational force of a static object. To counter that, an I beam support would be better and must be in permanent tension to counter deflection in dynamic operation or the supports must be designed as curved rails with some sort of kinetic dampers to absorb inertial acceleration & deceleration forces.

It is not unusual for printer heads to travel up to 30 m.s. if that is combined with gravity, then maybe you need to consider a vertical arrangement instead of horizontal to negate deflection issues better. Use of a third rail would also help somewhat if it acted as a moment energy deflection rail to transfer inertial energy laterally outside the scope of the two cylindrical support rails. Vibration is another issue that may cause harmonic distortions which is why a tubular section under compression may actually be better than a solid cylinder shape. Elliptical bars may actually be the best answer in the application of printer design since they are likely to resist deformation far more readily than cylindrical shapes while being able to withstand greater deflection over the same distance.

This kind of precision engineering is not going to be cheap which is why large printers are so expensive.

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The design you show is not going to work reliably. In any machine predictability is required and your design will produce an unpredictable machine because the load and deflections are unconstrained.

The correct way to look at the problem is to understand the nature and magnitude of the forces and the direction they can be applied. In your tubular design, any side loading will also produce deflections. So the better way to build the machine is to constrain the deflections in the vertical and lateral directions by using a fully supported rail.

Search for linear rails and look for products that have a solid rectangular mounting rail such as this rail from Misumi Corp.enter image description here

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There are equations to calculate the static deflection of your rods and from there design their diameter. But that will give incorrect answers because your print-heads are moving back and forth fast and depending on their movement, can impact the beams in a dynamic resonate mode and create huge vibrations and deflection.

For a start let's look at static deflection. $$ Deflection = Y = \frac { pl^3} {48EI} =< 0.05mm $$ l is 900mm and E = 290 GPA.

$$ I = \pi d^4/64 $$

plugging I in Y we get $$ Y = \frac { pl^3} {48E\times \pi d^4/64} $$

This equation gives us $$ 0.05 \times D^4 = \frac {64 pl^3}{48E\pi} $$ And $$ D = ( 80pl^3/E\pi)^{1/4}$$

However, this D is an underestimate by a large margin. One would need to have data on the range of acceleration and deceleration impact of the print head and then design the shafts accordingly.

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