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I have two PCBs (printed circuit board), and they are glued by adhesives, as show in the pictures. And the location of the adhesives are indicated on the picture (please notice that NO adhesive is applied between the PCB boards).

50N force is applied on the upper PCB (Z direction). I use adhesive to prevent the PCB from being separated.

Here are the dimensions of the adhesives:

All the wedge shaped adhesives have height 3mm (Z-direction) (the same as the PCBs), and width is 1mm (x-direction),

The long 80mm adhesive is 1mm thick (y-direction) and height is 6mm (z-direction).

I read from the adhesive specification sheet, the adhesive have the following properties:

$\mathrm{tensile\: strength}: 22N/mm^{2}$

$\mathrm{shear\: strength}: 18N/mm^{2}$

How can i judge whether the adhesive is strong enough? What kind of formula should i use?

enter image description here

I attempted to solve the problem like this:

Step 1:

Total area of glue contact on the blue board

$= 2(3(10) + 3(5.5)) + 80(3) = 333mm^{2} $

Force resisting detachment of the blue board (shear force only in this situation)

$= 333(18) = 5994N $

Step 2:

Total area of the 4 glue contacts on the grey (ground) board

$= 2(1(10) + 1(5.5)) = 31mm^{2}$

Area of the glue contact at the edge of the grey board

$= 80(3)=240mm^{2} $

So, Total force resisting the glue from being pull out from the grey board (tensile+shear force in this situation)

$= 31(22) + 240(18) = 5002N $

Therefore, in 5002N is required to pull the blue out?

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    $\begingroup$ Keep in mind that the failure of any joint does not occur as a single event, but starts at points of greatest stress and propagates. How the various components flex relative to each other when placed under stress (and hence where stresses are greatest, and by how much) is a critical part of the "equation". $\endgroup$
    – Hot Licks
    Jul 27, 2015 at 13:57
  • $\begingroup$ The comment above is absolutely correct. The values reported on the datasheet are based on specific test configurations and cannot be at all applied for a calculation as simple as you envisage. The assumption of constant stress across the surface is underestimating local stresses by a ton. This is a fracture problem, a whole different kettle of fish: have a look at concepts like Energy Release Rate or Linear Fracture Mechanics. It still remains a very specialistic calculation. $\endgroup$
    – Smerdjakov
    Dec 11, 2020 at 12:12

2 Answers 2

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I guess you should just multiply the surface of the contact between adhesive and the part by the tensile strength or the shear strength to obtain the maximum force normal or tangential to the surface of contact, respectively, that the adhesive can withstand. It looks like your design is OK with a large margin for 50N load, but you should compare both the forces and the moments of forces - it is not quite clear where exactly the load is applied.

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  • $\begingroup$ And what happened? :-) $\endgroup$
    – akhmeteli
    Oct 27, 2013 at 16:29
  • $\begingroup$ dear sir, i have updated the answer. i attempted to solve the problem, please comment. Is my answer correct?? $\endgroup$ Oct 27, 2013 at 16:48
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    $\begingroup$ What you did seems correct, but this is only a part of what you should do, so the answer to "is 5002N required to pull the blue out?" is "no". You should also check that moments of forces with respect to any axis are OK. Just imagine for a moment that the load is applied at another, distant point (e.g., the application point is moved far away in the negative direction of y). Then the load would have a large arm of lever and tear off the blue PCB. While in your case that does not look like an issue, I am pretty sure the critical force is significantly less than 5000N. $\endgroup$
    – akhmeteli
    Oct 27, 2013 at 17:11
  • $\begingroup$ So, i should take moment about the edges, and assume that the loading (50N) is locating at the distinct point. And try to find out the minimum required to pull the blue out? $\endgroup$ Oct 28, 2013 at 15:41
  • $\begingroup$ @Delay No More: Ideally, yes, you should calculate the moments with respect to the edges. AS for the load (50N), you should use whatever information you have about the actual point (or area) where this load is expected to be applied. However, do you really need to know the exact minimum force required to tear off the blue PCB? It looks like the adhesive as described is enough to withstand the load with a large margin. $\endgroup$
    – akhmeteli
    Oct 30, 2013 at 6:20
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The force being appled is a force of 50N in a static direction along only one of the 3 dimensional planes (z).

This means the force applied will test the tensile strength of the adhesive tape which is 22N for every square area of tape measuring 1 millimetre one each side. Assuming the 50N force being appled evenly each square millimetre area will be subject to 50N but will break at 22N

The only equation you need is IF tensile strength > force applied THEN adhesive tape is strong enough. IF tensile strength < force applied THEN tape will break

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