What are the new dimensions of a cube under biaxial forces?

[Confusedbiomedeng]

Homework Statement

Consider the cube with sides a=b=c=10cm . This block is tested under biaxial forces that are applied in the X and y directions. Assume that the forces applied have equal magnitudes of Fx=Fy=2x10^6 and that the modulus and poissons ratio of the block material is E=2x10^11 Pa and poissons ratio is = 0.3
I) find the new dimensions of sides if both are tensile and I) if Fx is tensile and fy is compressive

Homework Equations

σ=F/A
εX=1/E(σx-νσy)

The Attempt at a Solution

σX=2x10^6/0.1=2x10^7
Same for σy as same variables

εX= 1/2x10^11(2x10^7-0.3(2x10^7))

εX = 7x10^-5

εY= same as same variables

=> a=0.100007m
b=0.100007m
C=0.09993m

I know that this is wrong but I can't seem to figure out why and I can't seem to do when one is tensile and one is compressive could really use help

 

[haruspex]

σX=2x10^6/0.1=2x10^7

Check the area term in there.

Your work would be much easier to follow if you were to use parentheses as appropriate.

 

[Confusedbiomedeng]

Check the area term in there.

Your work would be much easier to follow if you were to use parentheses as appropriate.

Sorry yes your right it should be 0.01
I mistyped as I only have my phone it's quite hard to reread probably

 

[haruspex]

Sorry yes your right it should be 0.01
I mistyped as I only have my phone it's quite hard to reread probably

It's not just that you mistyped that line, though. You carried the error through the rest of the calculation, no?

 

[Chestermiller]

Also, C is wrong. What is the equation for epsilonC?

 

[Chestermiller]

$$\epsilon_C=-\frac{\nu(\sigma_x+\sigma_y)}{E}$$

 

[Confusedbiomedeng]

$$\epsilon_C=-\frac{\nu(\sigma_x+\sigma_y)}{E}$$

And what is the difference then between tensile and compressive ? Also the equations εx=1/E(σx-μ(σy+σz))
εy=1/E(σy-μ(σx+σz))
εz=1/E(σz-μ(σy+σx))

Sorry I got thrown all these equations but no explanation on which or why use them

 

[Chestermiller]

And what is the difference then between tensile and compressive ?

No. That is just the result of substituting $\sigma_z=0$ into the equation below.

Also the equations εx=1/E(σx-μ(σy+σz))
εy=1/E(σy-μ(σx+σz))
εz=1/E(σz-μ(σy+σx))

Sorry I got thrown all these equations but no explanation on which or why use them

Yes. These equations are correct. You can choose whichever directions you please for the three stresses. In your problem, two of them are non-zero (and equal), and the third is zero.

 

[haruspex]

I got thrown all these equations but no explanation on which or why use them

Then let me break this down: εz=1/E(σz-μ(σy+σx))
With only a force in the x direction, the consequence for the z direction would be εz=(1/E)(-μσx). Similarly for a force in the y direction only.
For a force in the z direction only, the consequence for that direction is (1/E)(σz)
With forces in all three directions, to a first approximation, you can just add them together.

 

[Confusedbiomedeng]

i did it out again , does this look more correct??

 

[Chestermiller]

i did it out again , does this look more correct??

Both calculations of C are incorrect. In part (a), you made an algebra error and in part (b) you calculated $\epsilon_z$ incorrectly.

 

[Confusedbiomedeng]

Both calculations of C are incorrect. In part (a), you made an algebra error and in part (b) you calculated $\epsilon_z$ incorrectly.

On the final page is it ? A and B dimensions are correct thou?

 

[Chestermiller]

On the final page is it ? A and B dimensions are correct thou?

Yes, A and B are correct. But, there is no calculation on either page for $\epsilon_z$ in case (b)