Having ultimate stress of steel, the ultimate strain can be determined by Hook's Law. The strength value can be found in many references. e.g.,

Machine design Theory and Practice .A.D.Deutschman, W.A Michels & C.E. Wilson.. MacMillan Publishing 1975.

For your convenience see table in the following link:

http://www.roymech.co.uk/Useful_Tables/Matter/shear_tensile.htm

Dear Amin,

For pure uniaxial tension test, shear stress = 0.5 Normal stress.

I think using Hooke's law in that zone need assumption of linearity.

Regards,

Mahmoud 

Dear Mehdi and dear Mahmoud,

Thank you for your answers.

Since rupture occurs in nonlinear part of ductile material behavior, it seems Hook’s law cannot be applied there. I am not sure if there is any empirical relationship to find ultimate shear strain.

Hey Amin, 

Actually, you could maybe make a link between the two values that you told us about. 

The value of the ultimate strain in both uniaxial and shear case is equal to the elastic strain plus the maximal plastic strain before rupture. In fact, the plastic strain is often define saying that is first invariant is equal to zero. (The volume is unchanged for plastic strain)

In the case of the uniaxial traction, the plastic strain tensor is then (epu     0     0 ) -------------------------------------------------------------------------------------------- ---(0   -1/2epu  0)------------------------------------------------------------------------------------------------(0       0    -1/2epu)------

On the other hand, the plastic shear strain tensor is just  (0    0    0 ) ---------------------------------------------------------------------------------------------- ------(0    0   eps)-----------------------------------------------------------------------------------------------------(0    eps   0)

with epu: uniaxial plastic strain and eps: shear plastic strain

You could then define a limit value of the plastic strain tensor magnitude. 

For example, if you use its usual expression for the uniaxial test: 

||ep|| = √(3/2) |ep| < √(3/2) eplim    

with eplim :  the ultimate uniaxial strain - the elastic strain (given by experience)

Then, using that result for the shear strain case: 

we obtain : ||ep|| = √2 eps= √(3/2) eplim 

=> ep(shear) = √3/2 eplim  (here, the plastic shear strain is just a bit lower than the             -                                            uniaxial one) 

the ultimate shear strain would be the elastic part of the shear strain (defined by the elastic limit criterion) + the plastic part of the shear strain that we just define. 

I hope that it could help you. 

PS: it could have different way to define theoretically the link between your value. In reality, only an experience is able to help you for defining such a criterion. But this kind of reasoning is maybe better than using an arbitrary value. 

Dear Nicolas,

Thank you for your answer.

I almost got my answer by the equation you presented:

“ep(shear) = √3/2 eplim”

But I got confused by the way you derived that. Could you please explain it more?

Dear Jeff,

Thank you for your answer.

What about the relationship between strains? Do you have any idea about that?

The main idea is to define a definition of maximum plastic strain using

Euclidean norm of tensor .

1) define a limit of the euclidian norm of the plastic strain tensor:  ||ep|| < C (this is the part that we assume )

2) apply this definition to the uniaxial case: you got the relation C and the maximal uniaxial plastic strain

3) apply it to the shear case: you got the relation between the maximal shear plastic strain and the maximal uniaxial plastic strain (using realtion got in point 2)

4) add the elastic part of the strain => you get the ultimate value.

But, the reasoning is purely theorical, then you should check the applicability of the assumption that we make (point 1) before using it in a real problem, or find a more empirical law. But if you can't find any empirical laws maybe this king of reasoning is better than using an arbitrary value for the ultimate shear strain.

You will find that the von Mises strain at rupture under shear loading is larger than the von Mises strain at rupture under uniaxial loading.  

Just convert your shear strain to von Mises Strain.  Then you can use the relationship by Rice and Tracey.  Or if you want to predict the shear strain at rupture then use Rice and Tracey to predict the von Mises strain at rupture under shear loading and then calculate the shear strain from the von Mises strain.  

J R Rice and D M Tracey, On the Ductile Enlargement of Voids in Triaxial Stress Fields, J. Mech. Phys. Solids, Vol. 17, pp. 201-217, 1969.

Dear Michael and dear Nicolas,

Thanks a million for your answers.

There is just another question,

I have found different relationships for von mises strain as below:

e=1/(1+nu)    sqrt(0.5*(e1-e2)^2+(e2-e3)^2+(e3-e1)^2)

e=2/3*sqrt(1.5(e11^2+e22^2+e33^2)+0.75(e12^2+e23^2+e31^2))

Which one is true?

Dear Amin

The link below gives you the details of the most suitable equations for von Mises strain and explains why they sometimes appear to be different. 

For pure shear I would suggest simplifying equation (5) by setting all of the values to zero except for one of the shear strains epsilon_i_j and for your case nu=0.5. 

http://rickbradford.co.uk/T73S04TutorialNotes32.pdf