I read in a documentation to some adhezive: the tensile strength is 13 N/mm2. How many kilograms can the 10x10mm bonding hold?

I tried to calculate it by the following:

$$\begin{align} F &= m/g \\ \therefore m &= Fg \end{align}$$

the strength is in N/mm2 e.g. that is

$$\begin{align} p = F/A \\ \therefore F = pA \end{align}$$

substituting the $F$ in $m = Fg$, I got

$$m = pAg$$

Using real numbers, - tensile strength = 13 N/mm2 = 13*10e6 N/m2 - A = 10x10mm (100mm2 = 0.0001m2)

$$m = 13*10e6 * 0.0001 * 9.81 = 12753 \text{ kg} $$

The result doesn't look right - the 10x10mm bonding imho can't hold 13.000kg weight.

Where did I make a mistake? So the question is: how to convert the 13 N/mm2 tensile strength to kilograms (e.g. how many kilos I can hang to the bonded 10x10mm plates?)

  • $\begingroup$ @Wasabi thanx for the editing. I definitely need learn the math-markup. $\endgroup$ – kobame Apr 24 '16 at 21:07

Pay attention to your unit analysis. You multiplied by the acceleration due to gravity, rather than dividing by it. Unit analysis would have shown this to be nonsense.

$$13 \frac{N}{mm^2} \cdot 100 mm^2 = 1300 N = 1300 \frac{kg \cdot m}{s^2}$$

Therefore, to get kg, you need to divide by an acceleration:

$$\frac{1300 \frac{kg \cdot m}{s^2}}{9.81 \frac{m}{s^2}} = 132.5 kg$$

EDIT: In terms of the symbolic analysis, your very first statement was wrong: Force equals mass times acceleration, so you should have written it as

$$F = m\cdot g \text{ and } m = \frac{F}{g}$$

Then, when you combine that with the equation for pressure, you get

$$m = \frac{pA}{g}$$

  • $\begingroup$ OMG! That's my stupid simple-math error. :( Thank you very much. $\endgroup$ – kobame Apr 24 '16 at 18:41

Performing the same calculations with the same values I obtain approximately 1275 kg. Hence there must be at least an arithmetic error in your calculations.

As Dave Tweed pointed out in his answer, this is not the correct calculation to obtain the tensile force required to cause failure. The correct calculation results in about 130 kg.

That isn't unreasonable if the adhesive is loaded purely in tension. However, if the load is slightly off center, or there is a torque, then the adhesive will be loaded unevenly and begin to fail at one end first. Once the adhesive has failed at a corner or edge, the remaining adhesive will have higher stress and failure will occur more rapidly. Adhesives tend to fail by peeling more readily than in pure tension. Peel strength is generally a more useful measure of adhesive strength, as adhesives tend to fail by peel long before they fail in tension.

  • $\begingroup$ Your calculation is completely wrong, but the rest of your answer is very appropriate. $\endgroup$ – Dave Tweed Apr 24 '16 at 16:11
  • $\begingroup$ Yes, agree with the peeling. The only info what i got from the data-sheet is the sheer and tension strength. It isn't for some rocket-science, just for my own approximation - learning. $\endgroup$ – kobame Apr 24 '16 at 18:46
  • $\begingroup$ @DaveTweed My mistake, good catch. Just goes to show double-checking your approach is always wise if an answer doesn't seem reasonable. $\endgroup$ – wwarriner Apr 24 '16 at 19:26

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