Is Strain a Scalar or Tensor Quantity?

[adabistanesoophia]

Hi,

I know that strain is a unit less quantity (Tensile Strain) but whether Strain is a scalar quantity or not? If it is a scalar quantity then why?

Regards,

Muhammad Rizwan Khalil

 

[berkeman]

What do you think?

 

[adabistanesoophia]

I think its a unitless quantity as for as its types are concerned. It has two types Tensile Strain and Volumetric Starin.

But confused about the term "Strain".

So still confused that it is scaler or not?

MUhammad Rizwan Khalil

 

[berkeman]

Even though strain is unitless (actually unity units), it has a direction, right? What does that mean about whether it is a scalar or not?

 

[adabistanesoophia]

Strain is defined as "Measurement of deformation of a solid when stress is applied 2 it".

So where direction is invlved?

So can v say it is a scalar quantity or not?

Muhammad Rizwan Khalil

 

[berkeman]

Strain is a deformation in a direction, is it not?

 

[adabistanesoophia]

I have coated the defination and in defination there is nothing like this. When we consider its types and solve problem then no unit is there and we have a defination of a scalar quantity is "quantities with unit and magnitude".

But no unit when we solve the problem. SO it can be scalar?

 

[radou]

I have coated the defination and in defination there is nothing like this. When we consider its types and solve problem then no unit is there and we have a defination of a scalar quantity is "quantities with unit and magnitude".

But no unit when we solve the problem. SO it can be scalar?

Every fundamental physical quantity (or one which can be derived out of these) has a unit, regardless of the type of quantity it is mathematically.

 

[adabistanesoophia]

So what unit strain has?

 

[berkeman]

I have coated the defination and in defination there is nothing like this. When we consider its types and solve problem then no unit is there and we have a defination of a scalar quantity is "quantities with unit and magnitude".

But no unit when we solve the problem. SO it can be scalar?

I think you are confusing what a scalar is. Contrast "scalar" with "vector". A scalar has magnitude only (regardless of what its units are). A vector has magnitude and direction.

And as radou says, you should remember that strain actually has units of m/m. You can further say things like m/m = 1, so that's why strain is sometimes referred to as "unitless". But units have nothing to do with the difference between a vector and a scalar.

Now, do you think strain is a scalar or a vector?

 

[adabistanesoophia]

In ur opnion, I think it is a vector coz u have coated that

"Strain is a deformation in a direction, is it not?"

How it has direction?

 

[berkeman]

Here you go:

http://en.wikipedia.org/wiki/Strain_(materials_science)

 

[Gokul43201]

Actually, in its most general definition, strain is a tensor (as is stress).

 

[AlephZero]

Strain is defined as "Measurement of deformation of a solid when stress is applied 2 it".

So where direction is invlved?

So can v say it is a scalar quantity or not?

Muhammad Rizwan Khalil

OK - that is an "beginner level" definition of strain. The whole truth is not so simple.

Strain is neither a scalar nor a vector. In the most general case it's a 2-dimensional symmetric tensor, defined by 9 numbers at any point. (Only 6 of the numbers are independent, because of the symmetry). Stress is also a tensor. The thing that corresponds to "Young's Modulus" is a 4-dimensional tensor with up to 21 independent constants for a general anisotropic material. (For an isotropic material there are only 2 independent constants, not 21. Young's Modulus and Poisson's Ratio are one pair of constants that define the stress-strain behaviour of an isotropic material).

When you are considering a problem like the axial stress in a rod producing an axial strain, the things you are calling "stress" and "strain" are single components of the complete stress and strain tensors. The others are either zero, or you are not interested in them.

You don't need all this heavyweight tensor stuff to handle the simple cases like constant axial stress in a uniform rod. In the equations you have probably seen like

"stress" = force/area
"stress" = Youngs Modulus times "strain"
Extension = "strain" * length

the terms I put in quotes (like "stress") are actually single components of the full stress and strain tensors. They may look like scalars, but they are not.

Obviously until you have studied tensors some of this explanation won't mean much - but as an example, the product of a 2-D tensor times a vector is another vector. The product of the stress tensor with a unit vector gives the force vector on the plane normal to the unit vector. This is completely general and gives you the force on ANY plane section through a body with ANY state of stress in it. For a rod with axial tension, most of the terms in the tensor and the vector are zero, so you can use equations that look like scalar equations, even though they are not really scalars.