Shear Design for steel beams 3

[NS4U]

The AISC spec, Eq. G2-1 says
Vn= 0.6*Fy*Aw*Cv

What is the 0.6 factor for? Does it have something to do with the fact that maximum shear stress is 1.5 times greater than the average shear stress?

 

[Lion06]

Well, it's really only 1.5 times greater if it's a rectangular section (which WF shapes are not, in actuality the flanges do take some of the shear, though very little). The factor is actually less than that, but it does account for that.

 

[DaveAtkins]

I think it has to do with the fact the the ultimate shear stress in steel is Fy divided by the square root of 3.

DaveAtkins

 

[nutte]

Dave has it.

 

[Lion06]

Dave-

That is interesting. That is what my S&J text says (when shear is acting alone - which is typically pretty close to the case unless there is a moment connection). My other text mentioned what I stated above. I certainly trust S&J over the other text, but that then doesn't account for the increased max shear over the average shear stress. That coupled with the fact that phi and omega for shear is 1.0 is leaving me a little confused. What is used to account for the max shear stress being higher than the average if that 0.6 is really 1/(sqrt3) ?

 

[nutte]

StrEIT, I believe it is just ignored.

For one, the difference between the maximum and average shear stress for a wide flange is much less than the 1.5 for rectangles. My Salmon and Johnson, 4th edition, has an example on page 392 showing the maximum shear stress on the wide flange in the example is 18.1 ksi. The average shear stress, using V/(d tw), is 16.0 ksi, a difference of 11.6%.

We do wind up with less factor of safety against shear than we normally have for other checks, for the three reasons mentioned:
1. Using the average shear stress.
2. Using 0.6 instead of 0.577.
3. Phi is 1 instead of 0.9.

 

[nutte]

The commentary to specification section G2.1 in the 13th edition also discusses this decreased factor of safety, saying it corresponds to test data, and that the consequences of shear yielding of a wide flange shape are minor when compared to tension or compression yielding.

 

[NS4U]

thanks for the responses... i'm curious though why is it 1/sqrt(3).

If Fy is based on a uniaxial tensile test, using mohr's circle would show the shear stress = Fy/2

Unfortunately I don't have S&J right now, could someone please explain the 1/sqrt(3) reasoning.

Thanks

 

[jnadeau]

It has been observed experimentally that the yielding behavior of metals are generally independent of the hydrostatic component of the 3D state of stress--a simple tension test has a hydrostatic component: (sigma_xx + 0 + 0)/3. Using only the deviatoric part of the stress and utilizing the simplest mathematical invariant of this deviatoric part (J_2 = const^2) results in a shear stress of Fy/sqrt(3) ~= 0.6 Fy causing yielding where Fy is the uniaxial yield stress. Thus, 0.6Fy is approximately the shear yield stress (not the ultimate shear stress).

The area of steel used in the Vn calculation is the area of the web, Aw. The code assumes a uniform shear stress of 0.6Fy acting over the area of the web and ignores the flanges due to their relatively small stresses from VQ/It.

 

[hokie66]

And the flanges are often removed at maximum shear by coping for connections.