Is there any relationship between shear strength and tensile strength of a metal?

yes friend that is the reason why tensile test results are very important.

mostly if we do not have the experimental data on shear strength we relate using shear stress failure theory, that is " yield strength /2"

Generally for all practical purpose for ductile materials vonmises stresses are recomended. if we examine the graph of max shear stress theory and maximum normal stress theory and vonmises theory, vonmises theory is conservative.

you can refer book by timoshenko vol2 strenght of materials for further details

Many aeronautics steel and AA have a Fsu (shear strenght) quite close 60% of Ftu (tensile strenght). So in first approx use this ratio. Also for titanium alloy it is valide in 1st approx.

In a multiaxial stress state (and if the metal is homogeneus and isotropic) use VM stress as applied equivalent stress.

See also my pubblication on material strength for better details.

regards.

The relationship varies from 0.5 to 0.577, whether you use Tresca or von Mises criterion. Also, a value of 0.6 (which is an approximation of 0.577) is recomended in Shigley´s textbook.

No, you shouldn't apply Mises approach for the ultimate stress.

Mises theory is related to the onset of yield; therefore, it is related to the yield stress in the material.

I typically use 1/2 of tensile to estimate the shear strength. However, We can use the 45 degree shear plane theory. the relationship to tensile to shear is 1/sqrt(2) which is .707.

Some sources still use .577 factor

Tensile strength represents how hard you can pull on something without it breaking. Shear strength represents how hard you can try to cut it without it breaking.

for different kind of materials refer this link you will get what u needs ..

Your proposal (von mises equation) is very good. You must remember which that shear strength is not real parameter but also is a formal parameter. I will hope that your attemption on your idea will be upbeat.

As you said there is a relationship but i need to know that for which one we can use Von mises or Tresca relationship: Yield Stress, Ultimate Stress, Failure Stress?

For example can we say the ultimate shear stress is 0.577 of ultimate tensile stress or we can use this relationship between failure shear stress and failure tensile stress?

All PLASTIC deformation (at moderate temperatures, anyway) may be regarded as happening in shear. The shear strength of a material is determined by its resistance to dislocation motion (which may come from the lattice itself or from any strengthening mechanisms).

Schmid's law relates the shear stress on a material to the applied tensile stress. A useful introduction is here:

For a single crystal, the Schmid factor S = cos φ cos λ depends on the orientation of the slip planes within the crystal, and can take any value between 0 and 0.5. For a polycrystalline material, it can be assumed that the most favourably oriented grains will slip first - shear deformation therefore occurs on planes at 45° to the tensile axis, where S =0.5.

This is the origin of the factor of two (or 1/2) others have described above: To reach a given shear stress, the applied tensile stress must be greater by a factor 1/S. The tensile strength is therefore approximately twice the shear strength in a polycrystalline material.

The relationships exist for YIELD shear and tensile stress. von Mises plasticity evolution depends on only 1 parameter (equivalent uniaxial plastic strain). Thus, von Mises cylindrical yield surface expands proportionally, and the consecutive yield surfaces are self-similar and the proportion between the shear and the yield remains the same, according to this fundamental theory. However, this theory is a simplification, and the actual hardening under shear or tension may not be proportional. However, since von Mises approach has worked for the industry for many years, it is probably OK for most applications. Thus, I would assume that the ratio remains the same throughout hardening.

The relationship between shear strength and tensile strength is characterized by the criteria of plasticity. There is a lot of criterias, but the most known and applied in metals are VON MISES and TRESCA. The Von Mises criterion is more conservative. We could say to the relationship es 1/sqrt(3) if we consider the Von Mises criterion.

You can consult this question in Federico Paris's book "Elasticidad".

Thanks for all answers, is there any publication that focus on this relationship and have some experiments too?

Farhan, you got a lot of good answers. If a metal fails in ductile manner, it fails by shear deformation. The von Mises yield criterion related the applled stress state to the shear stress. If you are interested in the behavior of a single crystal, look at the Schmidt's law, the taylor factor is the extension of the Schmidt concept to polycrystalline materials. Any textbook on mechanics of materials will tell you more.

Tensile strength represents materials' max pulling strength before fail

Shear strength represents materials' max failing (cutting) load acting parallel to the direction of the load of the material plan.

Ductile material fails in shear force whereas brittle fails in shear force.

Engineering calculation, normally 40% of tensile strength is used for shear force calculation(i.e. SYS=0.4 TYS)

Typo error...brittle material fails in tensile force

For the first question, i think there is no different for one dimensional tension. But for 3D problem there are: 1.73*Tau_max<=Sigma_v<=2*Tau_max. with the assumption that material has no volumen change. you can see some more from the plasticity theory.

For steel stamping process application, i used to calculate the cutting force. We will consider 80% of UTS is Shear Strength. This is common in stamping industry.

As far as Vonmises stress, it is related to Yield strength.

Vonmises criteria is principal stress ≤ Yield strength, otherwise you structure will fail.

the stated laws are for yield stresses. if you want the relations for Ultimate and Failure stresses you need to have Hardening laws included. it mans you should pay attention to the constitutive laws and hardening laws simultaneously.

for ore information, go for plasticity books.

According to von Mises the yielding will occur if the octaedric shear stress of a point in the structure reaches the octaedric shear stress of the tensile specimen at the moment of yielding process during the test.

According to Tresca the yielding will occur if the maximum shear stress of a point in the structure reaches the maximum shear stress of the tensile specimen at the moment of yielding process during the test.

Just now

Hi!!! The strength of a material determines the amount of force or load it can withstand before it fails. The failure criterion used in design can be different for different materials and therefore , there are multiple criteria of strength. For metals , the design criterion is usually based on the yield strength. The Tensile strength (TS) or Ultimate tensile strength (UTS) is the maximum load divided by the original cross section of the specimen (UTS=Pmax/Ao), And, the shear strength determines the amount of shear force (is a force that tends to produce a sliding failure on a material along a plane that is parallel to the direction of the force) on a material it can withstand before it fails. So the von Mises yield criterion can be also formulated in terms of the von Mises stress or equivalent tensile stress to obtain a relationship between shear strength and tensile strength. See also the book "Principles of Materials Selection for Engineering Design" by Mangonon

Regards….

Poisson's coefficient also provides an indication (not necessarily linear) of that relationship specially in anisotropic materials.

Hello there, have a good read of Shigley's Mechanical Engineering Design. this book with answer not only to yur current question but also related question around what are doing sisnce you have this specific question.

Hi, i think answer is given but i just want to add this,

relation depent on type of material also.

if it is brittle or ductile metal, ductile metal fail because of shear and brittle because of tension stress.

regards

It depends on the crystal system i.e. FCC, BCC, HCP (Schmid factor)

Normally shear strength is 0.6 to 0.7 times tensile strength for the metals.

yes. I agree with Ch. Rodopulos

Coming back to the original question

"Can we use the von Mises theorem for ultimate stress or not? "

In principle we are relating two things which are truly incompatible. Misses is based on statement that there is a critical value of the distortional energy stored in an isotropic material. In its know format Misses relates the three components of stress to either the positive dilatation strength or the shear strength. The question therefore is whether the term positive dilatation strength can be related to tensile strength?

Perhaps by reading the following article you can get some basic idea.

The second point in this incompatibility is the question

since fracture is a cumulative effect (involving CHANGES OF SOMETHING OVER TIME) and therefore it points towards SOMETHING THAT RUNS OUT could be for example ductility exhaustion or other post yield mechanisms how fracture, i.e. tensile strength can be related to something like Misses where the nature of time is missing?

The third point

Is the positive dilatation and distortional energy under a linear relationship?

The fourth point

Misses played with brittle steel (202 years ago). Is it possible that brittle steel might be different compared to a more ductile material?

Reading

Budiansky, B. and O’Connell, R. J., 1976, “Elastic Moduli of a Cracked Solid”, Int. J. Solids Structures, 12, 81-97.

Cottrell, A. H. (1981), The Mechanical Properties of Matter, Krieger, Malabar, FL

The relationship between shear strength and tensile strength varies from material to material and also depend on crystal structure. For example in case of steel, shear strength is 0.75 times of tensile strength, but in case of cast iron the shear strength is approximately 1.3 times of tensile strength.

If your material is ductile and isotropous, the ratios of Von Mises (3^1/2) or Tresca (1/2)will provide you with a good approximation. A better general approximation is to use the average orientation factor (i.e., the Taylor factor) corresponding to the texture and crystal plasticity of your material, as pointed out by Prof. Raabe below.

Sorry, in my previous answer I made a mistake, the correct Von Mises ratio between the shear and tensile equivalent stresses is 3^(-1/2) = 0.577.

If you use linear elasticity, shear stress is not coupled with tensile stress... so the relation strongly depends on the material you are using.

The question posed by Mr. Khodaee is about the tensile strength, and tensile strength refers to the plastic flow in tension (often, the ultimate tensile strength, UTS, in a tension test), not to the elastic stresses or strains.

By the way, the Von Mises criterion mentioned in the question is not any theorem but a good compact and close-form analytical approximation to the experimental behaviour of texture-free polycrystalline materials deforming by dislocation-mediated plasticity.

The relationship between the two properties is purely material dependent and not generalised. Failure theories such as Von Mises' and Tresca criterion, attempt to predict the onset of plastic deformation in ductile metals. However, these models cannot accurately estimate the critical upper range of deformation.

For isotropic materials, as metals, the most popular relationships between tensile strenght s and shear strenght t are the following, given by Tresca and von Mises, respectively:

Tresca : s=2 t

Von Mises s= (3^(1/2)) t .

In attachment please you will find same pages of my book edited by McGraw-Hill in italian language, explaining the theoretical reason for the given ralationships.

If you have any question do not hesitate to contact me once more.

Best regards,

L.Nunziante.

Prof.Luciano Nunziante: Full Professor of Structural Mechanics

University of Napoli Federico II. Via Claudio 21, 80125 Napoli . Italy.

e-mail: nunsci@unina.it ; Phone: +39 81 768 3727 .

Home phone & fax: +39 81 575 6147 ; mobile phone: +39 338 8211 918.

The use of von mises and tresca criteria is based on ideal modelling of material's yielding. Usually these materials are assumed as linear elastic / perfect plastic. so strain hardening will make it very complicated and you can not use a straight formula for it. The best way is to have experimental test for your specific material.

you can consult with prof. Eslami in your university.

See the following link that gives you some practical information for machine design application:

Dear Mr. Khudaee,

Practical experiments have revealed the shear strength to be one third of

the tensile strength practically for all steels on an average. For quick design

calculations this relationship is taken into account.

A.V.Sethuraman

The ratio between the shear and tensile equivalent stresses is 0.577

for ductile material, ratio of shear strength to tensile strength is roughly around 0.25, while for brittle material, the ratio is close to unit one

The Shear strength of a material under pure shear is usually 1/√3 (0.577) times its tensile yield strength  in case of Von mises criterion and 0.5 times its tensile yield strength in case of Tresca criterion.

It basically depends on the failure theory  you use.  

I have already answered this question. The point here is that the material must be thoroughly homogenous.

A.V.Sethuraman

Dear Farhan Khodaee,

in my opinion the answers you received on Research Gate to your ask are not correct. As a matter of fact, you did not asked: what is the ratio between shear strength and tensile strength for metals . You asked somewhat about the mechanical reasons of the relationship.

From this point of view, the mean answer is: the relationship depends on the proper yielding criteria assumed for the material. For further explanation see the attached PDF.

Yes there are relationships between shear strength and tensile strength in metals. 

One fairly easy and conservative theory to use is the Maximum shear stress theory, however it is only used to predict yielding and hence only applies to ductile materials.

where the yield strength in shear is half the yield strength in tension

S_sy = 0.50 X S_y

Another failure theory more accurate for designing to close limits is called the Distortion energy theory, again for ductile materials only and defining the beginning of yield. Other names for this theory are the Shear-energy theory and he von Mises-Hencky theory.

where the yield shear strength is 0.577 the yield strength in tension

Ssy = 0.577 X Sy

We ran tests on Q&T 4140 steel with measured uts=133,000 psi. Bottom line, avg of 5 tests, ultimate torsional shear strength = .72 x ultimate tensile strength.  (.67-.75)

Which stress criterion could I use in the case of rubber failure? In that case, the stress state is σ1 ≠ σ2 ≠ σ3 ≠ 0.

I have answered this question already. The ratio will be 1/3 of Tensile

strength

I read many answers, I want contribute too:

Is there any relationship between shear strength and tensile strength of a metal? YES, buts depends of materials properties, don't assume always are the same.

More info about:

Tensile Strength. In tension testing, the ratio of maximum load to original cross-sectional area. Also known as ultimate strength (UTS).

Shear strength is the load along a plane that is parallel to the direction of the force. The ratio of shear stress to the corresponding shear strain for shear stresses below the proportional limit of the material. Values of shear modulus are usually determined by torsion testing. Also known as modulus of rigidity.

I would like to know if there is any relationship between shear and tensile strength of metals, and can we relate them with von Mises theorem or not? Depends, Using adequate models (like von Mises, Tresca, Nadai-Tresca, and Nadai-von Mises), one can convert mechanical properties from tension and compression testing to shear stress and shear strain; however, this conversion seldom extends into a high strain range.

Can we use the von Mises theorem for ultimate stress or not? YES, but von Mises equation is often used in engineering design in a different form where, instead of principal stresses, the engineering stresses are used, including torsion stress.

This similarity has been used in

the limit load analysis where plastic finite-element (FE) stress analysis is used to determine representative failure loads for various components. Second, both von Mises and Tresca distributions are constant through the

wall thickness. Third, it should be noted that the Tresca value is equal to the maximum hoop stress value on the outer surface.

Please see this article from Google, Relationship between tensile and shear strengths of the mushy zone in solidifying aluminum alloys, Metallurgical and Materials Transactions, January 2003, Volume 34, Issue 1, pp. 105–113.

The ultimate shear stress is 1/√3 (0.577) times of ultimate tensile stress in case of Von mises criterion and 0.5 times in case of Tresca criterion.

Yes, I agree, ad already said several months ago.

I agree with 0.577

It depends if the material in question is ductile or fragile. For ductile materials 0.577 is used.

For fragile materials, the theory that best adapts is Mohr theory , and in that case the value is 0.5

0.577 for steel

Dear Akeel,

In structural and mechanical engineering, the shear strength of a component is important for designing the dimensions and materials to be used for the manufacture or construction of the component (e.g. beams, plates, or bolts). In a reinforced concrete beam, the main purpose of reinforcing bar (rebar) stirrups is to increase the shear strength.

The relationship between shear strength and tensile strength (yield) is characterized by the criteria of plasticity. There is a lot of criterias, but the most known and applied in metals are VON MISES and TRESCA. We can say to the relationship 1/sqrt(3) if we consider the Von Mises criterion.

There are no published standard values for shear strength like with tensile and yield strength. Instead, it is common for it to be estimated as 60% of the ultimate tensile strength. Shear strength can be measures by a torsion test where it is equal to their torsional strength.

For Steel,

Shear strength = appr. 0.578* yield strength.

Hope it is helpful for you.

Ashish

Although there is no direct approach between shear and ultimate you could combine the relationship between shear and elasticity limit as mentioned about 10x above and the ration between elasticity limit and ultimate strength which depends on the material you consider.

For steels -based on catalogue / data sheet- with the equation is

Rm </= 700 M Pa

Re =-8e-4*Rm^2+2.166*Rm-438.28 in MPa

for Rm>700 MPa

Re=1.0144*Rm-33.457.

The errors are less the usual dispersion of those values.

This allows you to establish, for your case, the value Tau/Rm but valid only in the elastic domain up to Re if you have only the Rm value available.

If you want to "estimate" the ultimate shear value it is possible to assume an increase proportional to (1+ Rm/Re) for the value at Re BUT it is not any thing more than a " rough estimation".

Please, read the attached files, best regards Luciano Nunziante.

files

Dear Luciano,

As I stated in my statement above, your content gives in detail about the von misses and tresca.

Ashish

Yes, I agree, best wishes Luciano

60% of tensile may be for shear strength

Good rule of thumb relationships between ultimate tensile strength, Su and Shear strength, Ssu for different materials are:

Wrought Steel: Shear Strength, Ssu = 0.82 Su

Malleable Iron: Shear Strength, Ssu =0.90 Su

Cast Iron : Shear Strength, Ssu =1.30 Su

Copper and Copper alloys: Shear Strength, Ssu = 0.90 Su

Aluminium and Aluminium Alloys: Shear Strength, Ssu = 0.65 Su

Please see (A. D. Deutschman, W. J. Michels and C. E. Wilson, in Machine Design: Theory and Practice, 1975, pg. 89)

Thank you

I read your question one day and finally decided to try and put an end to this eternal argument. Because you, Farhan, are (or was) not the only one wondering. This is one of those ghosts that have been haunting mechanical engineers since the dawn of time. Tresca or von Mises criteria are nice but theory is one thing, experiment is another. Check this out and you might be surprised: