youngs modulus


by cmgames
Tags: modulus, youngs
cmgames
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#1
Oct30-07, 07:20 PM
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Hi there, I am currently looking to measure the tensile strength using Young’s Modulus for steel and brass.

My results I have obtained are comparable with published values of E. My question is regarding the formulae, the one I used was: E = F/x X l/a

Where a is the original cross section of the steel and brass. The cross section measured was 8. The length was 50, however it was said that it was equal to 1 because the cylinder using pie r 2 of 50 would be approximating to 8 and therefore would be to one. I don’t really have much understanding of this, could you explain it to me.

Also regarding the application of steel and brass as a structural material. I have found that due to the high Young’s Modulus, steel is more desirable in structures as a high YM means it can span larger distances and is stiffer. Whereas brass is more flexible and so as a appropriate a resource than steel. Is there any other reasons for steel as the main application rather than brass?

I am looking at the difference between material stiffness and component stiffness. It is my understanding that component stiffness measure is by YM only. Whereas material stiffness tests using YM and also the size and shape of the material. Is this correct and is there any other differences between the two?


Lastly, I cant seem to find the YM values for wood – mahogany and parana pine. Where would I be able to find these published values and what would these values tell us when compared to actual results
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PhanthomJay
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Oct30-07, 08:55 PM
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Quote Quote by cmgames View Post
Hi there, I am currently looking to measure the tensile strength using Young’s Modulus for steel and brass.

My results I have obtained are comparable with published values of E. My question is regarding the formulae, the one I used was: E = F/x X l/a

Where a is the original cross section of the steel and brass. The cross section measured was 8. The length was 50, however it was said that it was equal to 1 because the cylinder using pie r 2 of 50 would be approximating to 8 and therefore would be to one. I don’t really have much understanding of this, could you explain it to me.
looks like perhaps you measured the diameter of the cylinder as 8 units in which case its area is about 50 square units. l/a =50/50 = 1.

Also regarding the application of steel and brass as a structural material. I have found that due to the high Young’s Modulus, steel is more desirable in structures as a high YM means it can span larger distances and is stiffer. Whereas brass is more flexible and so as a appropriate a resource than steel. Is there any other reasons for steel as the main application rather than brass?
steel, concrete, wood, and aluminum are the most common structural materials. I don't know much about brass, but it's probably more expensive than steel. Aluminum has a lower modulus than steel, but its strength is for certain types greater than that of certain steels, so it has advantages in cases where deflection is not a major concern, and especially since it is so much lighter in weight than steel.

I am looking at the difference between material stiffness and component stiffness. It is my understanding that component stiffness measure is by YM only. Whereas material stiffness tests using YM and also the size and shape of the material. Is this correct and is there any other differences between the two?
I'm not sure of the terminology, but I think you mean that where YM is a measure of the stiffness of the material, the actual deformation of a given component depends upon its size and shape, i.e., like in your problem, if the area is bigger or the length is shorter, the deformation is lower.


Lastly, I cant seem to find the YM values for wood – mahogany and parana pine. Where would I be able to find these published values and what would these values tell us when compared to actual results
I suppose a web search will get you what you need...southern yellow pine has a YM of about 1600ksi (vs. 30000ksi for steel and 10000ksi for aluminum...I hope you're working in US units). Wood is a funny actor...deflections are oft greater than theory predicts.
cmgames
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Oct31-07, 01:05 PM
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From my results i have found that wood and brass is not inherently stiff. What examples are there of using wood and brass, that by changing it's shape or distribution would make it stiffer?

stewartcs
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Oct31-07, 01:28 PM
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youngs modulus


The modulus will change inversely to the cross-sectional area. So, if you change the shape in such a way that the cross-section decreases, the modulus will increase. That is to say it will become stiffer.
PhanthomJay
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Oct31-07, 08:22 PM
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Quote Quote by cmgames View Post
From my results i have found that wood and brass is not inherently stiff. What examples are there of using wood and brass, that by changing it's shape or distribution would make it stiffer?
Young's modulus (E) of a material is what it is; you can't change its 'inherent stiffness', if that's how you define E. You can change the 'component stiffness' only by changing the size, shape, length, or boundary conditions at supports, of the material, by say increasing its cross section area, shortening its length, increasing its geometric moment of inertia, fixing its supports, etc. For tension/compression components, stiffness is proportional to AE/L. If you double its area, you double its stiffness, for example. In bending, stiffness is proportional to EI/L^3.


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