Shear Strength

Can anyone direct me to a reference which actually states the general relationship between shear strength and tensile strength?

I have been told many times (by people that do this type of testing) that shear strength of metals is generally 2/3 to 3/4 the tensile strength. However, no one can seem to give a reference for this relationship.

Thank you in advance.

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I answered this question back in Feb 2003 on sci.engr.mech, so I'll quote myself.

****** The original question was how to estimate shear strength of steel from hardness.

There is one common definition of shear strength. In ASTM E6 it is defined as "the maximum shear stress which a material is capable of sustaining. Shear strength is calculated from the maximum load during a shear or torsion test and is based on the original dimensions of the cross section of the specimen." (The ASM Materials Engineering Dictionary has the exact same definition except for using the word "that" rather than "which").

To answer the original question I just combined two rules of thumb:

A. Convert hardness to tensile strength.

B. To get the shear strength take 50% of the tensile strength.

Rule of thumb B has different percentages of tensile strength depending on where you look. I took the lowest one (50%) to be conservative, and assuming that the question (#1) related to having enough strength in a bolt connection loaded in shear. You might also take the highest one if what you really need to know (question #2) is if your punch press has enough force to shear some cross section of sheet metal.

For question #1 you can see the rules of thumb on page 34 in the notes for the first part of the GMRC Bolted Joint course at:

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where the 50% of tensile strength (UTS) is listed as having come from ASME B1.1 and a value of 60% is listed for carbon steels with hardness of less than 40 HRC as having come from Mil Handbook 5.

Also, in A. Blake's Practical Stress Analysis in Engineering Design,

2nd edition, page 593 the shear strength of a bolt is given as equal to about 60% of the specified minimum tensile strength. Approximately 60% of specified minimum tensile strength is also given for carbon steel fasteners in the IFI Fastener Standards book, 6th ed., 1988, page B-8.

For question #2 you can find 70% of the tensile strength listed on page 658 of G. E. Dieter's book, Mechanical Metallurgy, 3rd edition, McGraw Hill, New York, 1986 (see equation 20-1).

Both Mr. Duerr and Mr. Vojcak took issue with my mentioning the Von Mises relationship as a theoretical basis for doing the conversion in rule of thumb B. They said that the relationship applies only to the ratio between shear yield strength and tensile yield strength (being

57.7% or 1/sqrt(3)).

Actually, experimentally for steels the same ratio applies to much higher strains, so the true stress-true strain curve measured in torsion for a low carbon steel can converted and superimposed with that measured in tension. See Figure 10-7 on page 280 of G. E. Dieter's book, Mechanical Metallurgy, 1st edition, McGraw Hill, New York, 1961. Also see page 344 of the 3rd edition of Dieter.

I said it wasn't rocket science, but there is a reason for the constant in rule B being 0.5 to 0.7.

Pittsburgh Pete



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Can you think of a good reason why there would be a "universal" relationship between tensile strength and shear strength?

I know that people would wish that there were such a relationship, so as to make life easy in terms of analysis and to give them a feeling of a greater knowledge of the fracture phenomena (By simplifying it).

Reply to

Get on Google and enter:

+"shear strength" +metals

You will get a variety of answers depending upon the alloy. Copper is among the more difficult to apply a rule of thumb with the wide range of shear strength values for the various alloys of copper.

Not knowing how you intend to use the shear strength prompt me to state the following.. You must look up, or test, the shear strength of each alloy that you require and not rely on rule of thumb unless you know for a fact the rule is correct for that alloy.. Only then can you make a *safe* decision. If you are shearing the metal, you consider the highest value. If you are designing a device that may cost someone their life or physical harm, use the lowest value and include a *safe* margin for error.

Jim Y

Reply to
Jim Y

No because there is not one.

Reply to
David Deuchar

It varies from alloy to alloy. Check out

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Reply to
Atlas Shrugged

On page 359 of D. R. H. Jones's book "Engineering Materials 3: Materials Failure Analysis", Pergamon Press, Oxford 1993 under the topic of mechanical property correlations he states that the failure stress in shear is approximately 5/8 of the tensile strength. Any good failure analyst would already know this.

The question as stated was naive. Most engineers would have added the details of what materials were being tested and how (single or double shear, etc). However, this is a newsgroup and we entertain naive questions. The phrase "generally" when used by engineers or scientists often just means that "a couple of my drinking buddies and I think this might be right".

The "Guide to Design Criteria for Bolted and Riveted Joints (2nd ed)" notes on page 47 of the hard copy that for ASTM A325 and A490 steel structural bolts in single or double shear the average factor is 0.62 or almost exactly 5/8. Any good fastener testing laboratory already would have a hard copy of this book. It can be downloaded free at:

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Pittsburgh Pete

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Thanks Pete, I had no idea how to argue it, but knew that there had to be some sort of relationship between hardness and tensile strength and shear strength.

Alvin in AZ ps- Hey, DD, even this dumb fly-over know-ed (or at least fiNgured) that. ;)

Reply to

Try a study of Mohr's Circles.

Reply to
Billy H

Many scientists studied the question.

Brinell was one, Mohr another, there is a topic in stuructural engineering called 'complex stresses' the principle stresses set up complementary stresses and they are given in some sources as occuring 45 degrees to each other and other sources state differently. 45 degrees implies 1/2 as a factor.

Tazaghi stated that a set of stresses set up in shear asa result of axial stress always resulted in a stress of a given degree of the applied stress. Testing I did in the labs at uni proved this wrong, it depends upon the degree of consolidation of the material under test, i.e. upon the density of the material. Whether this dependence is relative to some absolute density for each material or not is a matter for investigation but I coined the phrase (as a memory jogger) 'over consolidated iron' to help me remember how steel was working under principle and complementary stresses.

Simply, materials react differently, you cannot rely upon any proportion as a blanket 'rule of thumb' for all materials.

Reply to
Billy H

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