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TL;DR: I make a cylindrical disc out of material X. Fill it with fluid and apply pressure on top of the disc which pushes the curved part of the cylinder outwards till fluid pressure breaks it. The tensile strength can then be determined by doing $5 X$ (Maximum Fluid Pressure). This should give the same answer for tensile strength as obtained in uniaxial loading.

So I've been looking at a few material tests and they all start with a rectangular sample of the material, loaded into a machine which extends them by increasing load at a constant rate and measures the strain/stress till the point of material fracture. The yield stress is measured in usually MPa which has the same units of pressure.

I was wondering if I could measure the tensile strength of a material in an alternative way. Say I have made a cylinder out of the material whose tensile strength is to be measured. I then fill up the cylinder with some fluid and apply pressure to it and measure the YIELD STRENGTH at the point when my disc ruptures.

I'm aware that compared to the previous uniaxial loading, now I'm loading it in possibly 2 or 3 directions or maybe 5 directions: The material at any point is being pushed in $\pm x$ directions and also in $+z$ direction, maybe even in $\pm y$ directions. This means that I might have to include a multiplication factor of 3 or 5 to the fluid pressure in order to obtain the total stress on the material.

Now my question is, is this quantity the same as tensile strength? I know that the fluid is converting the vertical compression into tangential stress on the material so technically we should also be calling it pressure but intuition tells me that this is the same thing as tensile strength.

If the material has a tested tensile strength of $100 MPa$ then I should be able to rupture the disc at $20 MPa$.

If I'm wrong please correct me. Thank you!

EDIT: Link to additional reading

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  • $\begingroup$ Is the highlighted statement yours, or something given to you? $\endgroup$ – Bob D Jul 8 '19 at 14:50
  • $\begingroup$ @BobD It's mine. $\endgroup$ – Weezy Jul 8 '19 at 15:26
  • $\begingroup$ OK. I am drafting an answer. $\endgroup$ – Bob D Jul 8 '19 at 15:43
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Refer to the section of the cylinder shown in diagram below.

When considering a cylinder there are three types of tensile stress to consider:

  1. Axial stress, $σ_a$, which acts along the axis of the cylinder.

  2. Radial stress, $σ_r$, which acts perpendicular to the surface of the cylinder.

  3. Tangential stress, $σ_t$, which acts tangential to the surface of the cylinder. This is sometimes referred to as "hoop" stress, named after the stress that a strap of metal (the hoop) wood experience when holding barrel pieces together against to internal pressure.

In the diagram, $t$ is the thickness of the cylinder, the difference between its inside and outside radius.

In many cases when analyzing these stresses, we use what is called the thin wall approximation, meaning that the wall thickness $t$ is less or equal to 0.1 times the average radius $r$ where $r=\frac{r_{i}+r_{o}}{2}$. Then these three stresses are, where $P_i$ is the internal liquid pressure equal to the sum of the the pressure of the liquid (which includes the pressure applied to the disc) and atmospheric pressure, are:

$$σ_{t}=\frac{P_{i}r}{t}$$

$$σ_{a}=\frac{P_{i}r}{2t}$$

$$σ_{r}=-P_i$$

You can see from the three equations that the maximum stress will be the tangential, or hoop stress $σ_t$. If the cylinder is made of a single homogenous, isotropic material (properties the same in all directions), then the critical stress would be the tangential stress and not the longitudinal stress as in the case of axial loading. So if you want a 5 X safety factor for rupture, it would be 5 X the tangential stress, not 5 X the uniaxial (longitudinal) stress.

Hope this helps.

enter image description here

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  • $\begingroup$ This is a great answer. Thank you very much! Where can I find more on this topic, like what do I search for? Also you can correct the suffix for hoop stress on your last paragraph. Again, thank you! $\endgroup$ – Weezy Jul 8 '19 at 16:40
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    $\begingroup$ Not sure where you can get more on the topic. I learned about it in my undergraduate strength of materials class (and taught it in a PE FE exam review class). Suggest you google up "hoop stress" as a start. Made the correction. Thanks. $\endgroup$ – Bob D Jul 8 '19 at 16:45
  • $\begingroup$ I'm asking for more resources because I'm not exactly doing a thin wall approximation as stated in the answer. Also I might looking into altering the cylinder into elliptical shapes. $\endgroup$ – Weezy Jul 8 '19 at 16:46
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    $\begingroup$ @Weezy Without the thin wall approximation the equations become more complex. The NCEE reference handbook for the FE exam has them. You might be able to download the handbook on line. $\endgroup$ – Bob D Jul 8 '19 at 16:53

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