# Calculating Young's Modulus with only deflection

1. Jul 3, 2013

### The3vilburrit0

1. The problem statement, all variables and given/known data

In a project in a summer engineering class I'm taking, I have been tasked with finding Young's Modulus with three different diameters of spaghetti: Thin (1.41 mm), Normal (1.73 mm), and Thick (1.89 mm). Although this is my first time ever attempting to try engineering without taking Calculus 1 or Physics, but I've managed to get as far as finding the stress values. The values are a result from a bending test that has multiple lengths of exposed spaghetti in the middle. In essence, there are three types of support ledges with each support ledge being a different length apart from the pair on the other side of the contraption. The values that I've recorded from testing are below. I tried creating a table using all that I could, but it didn't turn out like I would've liked it to.

THIN (1.41 mm)​
Weight Held @ 75 mm : DIDN'T HAVE TO TEST
Weight Held @ 95 mm : DIDN'T HAVE TO TEST
Weight Held @ 117 mm : 40 g with 16 mm deflection

NORMAL (1.73 mm)​
Weight Held @ 75 mm : 113 g with 7 mm deflection
Weight Held @ 95 mm : 74 g with 10 mm deflection
Weight Held @ 117 mm : 62 g with 14 mm deflection

THIN (1.41 mm)​
Weight Held @ 75 mm : DIDN'T HAVE TO TEST
Weight Held @ 95 mm : DIDN'T HAVE TO TEST
Weight Held @ 117 mm : 85 g with 13 mm deflection

2. Relevant equations

First off, is it even possible to substitute the deformation to find strain with the deflection? Also, once I find Young's Modulus, is that number the slope throughout the elastic deformation time span?

3. The attempt at a solution
Using the area of a circle (A=∏r2), I've been able to find the cross-sectional area of the different thicknesses of the spaghetti. For Thin, the area is 1.561 mm2. For Normal, the area is 2.351 mm2. Lastly for Thick, the area is 2.806 mm2.

I then converted each weight that the types of spaghetti held from grams to kilograms, and then multiplied by 9.81 to convert it into Newtons. Thin: .251 (117 mm); Normal: 1.108 (75 mm), .726 (95 mm), .608 (117 mm); and .883 (117 mm) for Thick. Using the stress formula (σ= force/area), I found the following: Thin: .251 (117 mm); Normal: .471 (75 mm), .309 (95 mm), .259 (117 mm); and .296 (117 mm) for Thick.

To solve the rest of Young's Modulus to find the slope of the Elastic Deformation time span, I need to find what strain is, but to find strain, I also need to know what E (Young's Modulus) is. So pretty much in order to find one thing, I need to know the other and it just keeps going in a continuous loop.

Thank you very much for helping!

-Chris

Last edited: Jul 3, 2013
2. Jul 3, 2013

### Staff: Mentor

Get yourself a copy of Roark's book Formulas for Stress and Strain. You need to study beam bending, and how to quantify the relationship between the beam deflection and the load, in terms of Young's modulus. In beam bending, the strain is not constant across the cross section of the beam. The tensile strain is positive on the outside of the bend, and negative on the inside of the bend. Books on strength of materials lay out the details of how to analyze beam bending. You have some studying to do.

Chet

3. Jul 4, 2013

### cmmcnamara

Roarks is an excellent book for formula look up but for a first timer (as it sounds you are) with beam bending, you may find it hard to follow. Pick up an older Strength of Materials book for cheap to really get the basics. Google Euler-Bernoulli beam theory as well.

4. Jul 7, 2013

### nvn

Yes, probably.

No.

The3vilburrit0: You did not mention whether your supports are fixed supports or simple supports. I.e., is each end of your beam (spaghetti) fixed (clamped), or simply supported (pinned). A simple support means the spaghetti is free to rotate at the supports. A fixed support means the spaghetti is clamped and therefore is not free to rotate at the supports. Which type of supports do you have?

Also, you do not need the cross-sectional area of the spaghetti. You instead need the spaghetti second moment of area (I). Are you familiar with this formula for a solid circular cross section? Look it up, and see if you can find it.

Also, see if you can find the formula for deflection of a simply-supported or fixed-end beam, depending on your answer to my first paragraph, above.

Post these relevant equations, if you want us to see if you are on the right track.

Last edited: Jul 7, 2013
5. Jul 9, 2013

### pongo38

quote "Using the stress formula (σ= force/area)...."
There are shear stresses and there are normal stresses. There are two kinds of normal stress; the first is axial stress, for which you quoted the formula σ= force/area. The second kind of normal stress is bending stress sometimes called flexural stress. This is the one you need.