Manually Attempting Empirical Measurement of Flour Dough Compressive Strength

Question

I want to make a DIY dough press, so need to know how much pressure I would need to apply. I don't have mechanical engg background, so based on some googling, I think this is what I need to do, but wanted to validate here.

I took a small dough ball, and placed a very light flat surface on its top, and then put a 16.9 fl oz water bottle (weighing 1.1 lbs) on top of it. It squished the dough ball to a thick circle of 2.09 inch diameter. (3.145 sq inch area).

Then I put 7 more water bottles on top of it (total 8.8 lbs approx), and the dough squished flatter to a circle of 2.91 inch (6.647 sq inch).

Now my questions are,

a) Is this correct way to calculate compressive strength of this dough ball by dividing force / area, where it came to rest. So with 1 bottle on top, when it squished somewhat flatter, the PSI = 1.1/3.145 = 0.349 lbs/sq inch

b) If so, then shouldn't the compressive strength be constant as it is tied to the material. So dividing the final area after adding more weight should yield the same value? The second value, 8.8/6.647 = 1.323 lbs/sq inch.

Answer 1

It sounds like you're assuming the dough follows a perfectly-plastic [2] material model, which might be reasonable but it depends on the material.

a) For a general material, strength is the stress at which failure occurs. So you would find the smallest load which causes any permanent deformation. In practice, that's difficult to measure so you have tricks like what you're doing which depend on assumptions about the material. For a perfectly plastic material, your calculation would be correct but since the stress seems to increase with strain, it looks like the dough is strain-hardening and a different material model may be appropriate, such as linear hardening 1. If it's a rigid/linear-hardening material, you can plot the pressure and strain values on a curve like in the picture below and extrapolate to zero strain to see what the yield stress is.

b) Yes, pressure at equilibrium should be constant for a perfectly-plastic material, but it should be a function of strain for a more general material. However, there may be a mistake in your experiment which is that if the dough grips onto the surfaces then you're measuring a more complicated quantity that also depends on the geometry of the blob and it will become stiffer the flatter it is because it's constrained from spreading outwards by friction with the surfaces. To properly measure the relationship between stress and strain, the dough should be free to slide frictionlessly on the two surfaces. See the picture of different deformation shapes here [3]. Case (e) is the correct one.

[2] https://nptel.ac.in/courses/105108072/mod04/lec4.html

[3] http://www.instron.co.uk/en-gb/our-company/library/test-types/compression-test

Answer 2

Assume you were able to put uniform pressure each time on the dough, and assume the material property is uniform in all directions, then it is correct that the stress readings should be $W_1/A_1$ and $W_2/A_2$. However, the stresses will not be constant, unless $A_2 = 8*A_1$ (for $W_2 = 8*W_1$).

In reality, the stresses will not uniform due to change in strains ($Strain = \frac {\Delta L}{L}$) as shown on the stress-strain curve below.