# How to calculate the adhesion I need to keep a 3D printed bending part straight?

3D printed parts cool down during printing, resulting in thermal contraction which causes a bending moment and separation from the printing bed.

The issue is solved by increasing adhesion to the printing bed, which however results in more stresses inside the part which can cause cracking.

The result is a poor printed part:

Provided I know the thermal expansion coefficient of a material and the Young modulus, the temperature delta causing contraction (which should be glass transition temperature minus ambient temperature) and the part length, how can I calculate the adhesion force required to keep the corners of the part sticking to the bed, or the adhesion force required to keep the layers together, without delamination?

I know that complex shapes would require different calculation. For this question I would consider only parallelepipeds ("boxes", "beams").

It may be dependent on layer height, but I think it can be regarded as second order parameter and omitted from a rough estimate.

An approximation is enough, I guess that accurate results would require a finite element analysis, which is beyond the scope of this question.

Let's say each layer thickness is t and the shrinkage between the lower hot edge and upper cold edge after calculating the transient thermal equilibrium s/L. the Radius of the curve of the lift for small-angle deformations will be

$$\frac{s}{t}=\frac{L}{2R} \quad \ R=\frac{tL}{2s}$$

and we know $$M/EI =1/R.$$

So plugging in the numbers and we can find the moment and calculated adhesion needed.

• Let's suppose we have a square beam 1 cm thick and wide, 10 cm long, out of ABS with dT 80°C (100-20 °C) and 2e9 Young modulus. Layers are 0.2 mm thick but maybe I can see everything as a single 1 cm layer? In this case s/L=1e-4 m/mK * 80 K * 0.1 m = 8e-4 m. R=1cm /2/8e-4m=6.25. M=EI/R=2e9 N/m^2*1cm*10cm^3/12/6.25=26.7 newtons? but how to get the Mpa of adhesion at the end of the beam? maybe that's the total force, which is spread over the contact surface. Assuming no force in the center and max at the end, it's like twice the average value at the extremes. 26.7 N/(0.1*0.01)*2=53 kN – FarO Jan 18 at 15:11
• The calculated value seems very low. And it is independent from the part thickness: if it were 10 cm thick, R would be 10 times bigger, 62.5, but I would also be 10 times bigger, so the value of M would not change. This is strange. However, if I calculate the force per layer (0.2 mm) and multiply by the number of layers, I would get quickly big numbers: 53 kPa per layer is 250 kPa per cm, and at 10 cm is 2.5 MPa, which is still very low to cause cracking (typically layer adhesion is around 30 MPa), but the calculations are very rough. – FarO Jan 18 at 15:16
• Can you add an example (for examine mine, with corrections if needed) to your answer? – FarO Jan 18 at 15:16
• Multiplying the stress by the number of layers however cannot be correct: printing the same part with decreasing layer thickness increases the strength of the part, see mdpi.com/2073-4360/10/3/313/pdf table 1. Or maybe it is indeed correct, but other factors play a role. – FarO Jan 18 at 15:51
• you calculate a load on same beam which gives the same radius. i am going to edit my answer to explain this later. – kamran Jan 18 at 17:14