Understanding 3D Solid Deformation: A Cube with No X-Direction Force

[DrVirz]

Hi all,
Having some trouble getting the final answer on the question below, the fact that the cube doesn't have a force in the x direction is throwing me off. Once I fine the force(??) in the x-direction, I can just use the generalized Hooke's law to obtain strain? Any help is appreciated.

Homework Statement

2. Equations in upload of solution.

See my solution for relevant equations.
http://[ATTACH=full]199721[/ATTACH]

 

[Chestermiller]

In the block is constrained in the x-direction, what is the strain in the x-direction? What does your first equation give you in this case?

Chet

 

[DrVirz]

In the block is constrained in the x-direction, what is the strain in the x-direction? What does your first equation give you in this case?

Chet

So I am assuming Sigma(x) = 0, therefore, Strain(x) = -1.00005 x10^-4

However the question says to find the stress in the x-direction, which I have taken as 0?

Or, do I sub Epsilon(x) as 0 and therefore would be able to find Sigma(x) from the first equation? After second thought I think this is correct due to 0 deformation in the x direction hence strain is 0?

 

[Chestermiller]

So I am assuming Sigma(x) = 0, therefore, Strain(x) = -1.00005 x10^-4

However the question says to find the stress in the x-direction, which I have taken as 0?

How can you say that the strain in the x direction is not zero, when the problem statement says that strain in the x direction is zero? What does the word "constrained" mean to you?

Chet

 

[DrVirz]

I realized this soon after I posted my previous response and edited it soon afterwards, it make sense now.

Due to the tensile force in the y direction and the compressive force in the z direction the block obviously expands in the z direction and contracts in the y direction? Therefore, shouldn't Epsilon(y) be negative and Epsilon(z) be positive. Is this done by changing the sign (+ or -) for the Sigma values in the generalised Hooke's law? I.E. Sigma(y) should in fact be -50MPa?

You can see the two different answers on either side of my page.

 

[Chestermiller]

You have a sign error in the calculation of the stress in the y direction. It should be negative.

Chet