Stress and strain Value for symmetric problems

[Vissu4u]

Hi All,
I have one question on symmetric problems.
For symmetric model i have applied half of the force.
My question is will stress will bocome two times of the symmetric stress.
and same for strain also.
Please explain me with stress and stain relations.

Regards and Thanks
Vinay

[JAlberts]

I think for an accurate response to this question a clearer description of your sample configuration is required.

That being said, assuming the case of a "U" shaped item you must realize that to have a bending stress and strain at the lower center of the "U" an equal and opposite resisting force on the opposite leg is required. As a result, the total applied stress and its corresponding strain calculated for the structure based upon one leg should not be doubled.

[Vissu4u]

I didnot understand please explain with me with some examples.

[JAlberts]

I'm sorry, but that exanple is the simplest and most straightforward one I can offer.

Earlier I posted the above, but I now realize that my example, while correct, only emphasizes original comment regarding the need of a specific example of what type of configuration to which you are referring.

For clarification, although it is more technical, I am going to give a new example that will clarify an issue regarding the application of the value of "strain". In my first example, I was referring to an isolated bending stress at the bottom of the "U" shaped configuration; and, in that case was addessing a calculated value for a single isolated infinitesimal length section of the beam.

So now let's look at the case of a straight piece of bar of uniform cross section and shape for its full length. If you calculate the tensile stress in the bar at any point along that bar that is created by longitudinal load applied at one end of the bar and restrained at the other end of the bar the calculated stress and corresponding "strain" values are uniform throughout the length of the bar for any location along the bar and for any length of bar. However, it is important to remember that the unit of "strain" is "micro meters per meter" of applied stress.
As a result, if you have a bar that is 2 meters long, when you analyze 1/2 of that length, then the "elongation" that you calculate for that 1/2 length of the bar must be doubled to obtain the correct "elongation" for the entire 2 meters length of the bar.

I hope this is of some help in answering your question, but I am concerned that it could do the opposite and confuse you more.

[Vissu4u]

Hi,
Thanks for your explanation. I am able to understand what you have explained for bar. If i take same bar same length and it is half about longitudinal axis. In that case what will happen for stresses and strains.

Regards,
Vinay

[JAlberts]

Just be sure I understand your question, you are asking what happens if, for example, you take a 4 cm square x 1 m length bar and cut it longitudinally to be a 4 cm x 2 cm rectangular x 1 m length bar for analysis?

[Vissu4u]

Hi,

Thanks for your explanation. Yes i am asking same thing. If i do like that stress will same right and strains will be double i guess.

Regards,
Vinay

[JAlberts]

Your conclusion is not correct. If you apply the same logitudinal loading your reduced 2x4 cross section beam as that applied to the original 4x4 cross section beam; then, because the stress in the beam is inversely proportional to the cross sectional area of the beam, the new beam's stress will be doubled and therefore both the longitudinal strain and total beam elongation will be doubled as well. This is because the strain and resulting total beam elongation are directly proportional to the beam stress because strain = stress / young's modulus.