Any structural or civil engineers here?

[Frixxion]

I think you made some errors
In question A you made an error with surface, you filled in 0.150 but it should be 0.150^2 (to get internal surface area widthxheight)
Question B can be solved with strain = elongation = dL/L0 with dL being the difference in lengt and L0 the initial length.
The modulus of elasticity can be solved with: E= Stress/strain

Also you made some errors with the units.

I included a picture of how I did it , hope you can read my writing

Good luck and just let us know if you have questions

[andy_con]

ignore silly comment i werent paying attention

[tocket]

ignore silly comment i werent paying attention

Obviously, since I already answered this question many posts ago. Oh well, I guess not everyone appreciates my help.

Frixxion, didn't you get the sign of the strain wrong? The material is being compressed, so ΔL is negative.

[Frixxion]

how can i be wrong on A when we got the same anwser???

could you explain B a bit more im unsure where you got 0.4mm from??

thanks

Ok, your answer was 5333333 N/mm^2 and mine was 35555555N/m^2, so I think you misread my notes :P (I know my handwriting sucks, I sometimes mix up my 5's with 3's as wel :P). But you also made a mistake with the unit.

What you did was 800,000/0.150 so you made 2 mistakes, 1st one is to not take the surface of the cube but just it's width. 2nd one is that you said mm^2, that would have been true if you would have taken 800,000/150, but you took 800,000/0.15 which would give you N/m^2.

So, ok, the 0.4 is the amount that the cube has shrunk under the load of the testing machine. So that gives us the dL (Delta L, which is the same as difference in L). So dL/L gives us the strain, which is a unitless number, it's more of a ratio.

I included a quickie paint to help explain this.

[andy_con]

Obviously, since I already answered this question many posts ago. Oh well, I guess not everyone appreciates my help.

Frixxion, didn't you get the sign of the strain wrong? The material is being compressed, so ΔL is negative.

easy......

im greatful for everyones help

[Frixxion]

... Frixxion, didn't you get the sign of the strain wrong? The material is being compressed, so ΔL is negative.

You could take a negative ΔL, but that would result in a negative E, and I never like that :P.

[andy_con]

Ok, your answer was 5333333 N/mm^2 and mine was 35555555N/m^2, so I think you misread my notes :P (I know my handwriting sucks, I sometimes mix up my 5's with 3's as wel :P). But you also made a mistake with the unit.

What you did was 800,000/0.150 so you made 2 mistakes, 1st one is to not take the surface of the cube but just it's width. 2nd one is that you said mm^2, that would have been true if you would have taken 800,000/150, but you took 800,000/0.15 which would give you N/m^2.

So, ok, the 0.4 is the amount that the cube has shrunk under the load of the testing machine. So that gives us the dL (Delta L, which is the same as difference in L). So dL/L gives us the strain, which is a unitless number, it's more of a ratio.

I included a quickie paint to help explain this.

i am on a totally different planet today!!! im getting stressed with this crappy course work and am making silly mistakes.

yes a cube so 150x150 my bad and because we are working in mm i dont need to convert the 150 into anything. just leave it as 150

just gona work though it again make sure i understand everything

[andy_con]

to be honest im really not getting this youngs modulus/elasticity shit and im getting the hump big time!!!

i get stress and strain so i suppose thats something.

im jsut gona have to copy down the youngs modulus stuff you put and just roll with it for now.

ive gota move onto bending moments.........................

oh god!!!!!!!!!!!!!!

[tocket]

You could take a negative ΔL, but that would result in a negative E, and I never like that :P.

Well, ΔL is negative, so it would seem logical that the sign of the stress should also be. A negative elasticity modulus doesn't make sense. So not only do you have to get the units right, but also the signs. I suspect mechanical engineers are a bit sloppy with this though.

But what do I know... my area is chemical engineering.

[danielbriggs]

to be honest im really not getting this youngs modulus/elasticity shit and im getting the hump big time!!!

Andy,
think of it this way; it's basically just a number that tell you how 'stretchy' a material is. Each material will have a different number. Very low values indicate that the material is very very stretchy (or elastic for the proper term)
e.g. rubber has a value of about 0.01 GPa, and diamond has 1200 GPa.
That's basically all it is, but it's very useful in many calculations.

ive gota move onto bending moments.........................

oh god!!!!!!!!!!!!!!

Oh, you'll love them! If you get stuck, post up the q's and I'll have a crack at 'em for you.

Dan