Shear Stress in an I Beam

[MER3]

I am checking a beam for shear and am trying to decipher the code requirements for shear. .4FyA is the allowable stress, but the issue is with calculating the stress in the beam to compare to allowable.

For an I beam, I ignore the stress in the flanges and focus on the web. Taking the area of the web as DTw, I get the shear area. Divide the shear force by web area and the shear stress is found. Easy enough.

The mechanics of solids method for a rectangular cross section states that maximum shear stress is 1.5 V/A. My method computes average shear stress.

My question is which one is it appropriate to design an I beam using? The mechanics of solids method is more conservative, but usually shear isn't an issue.

 

[jayrod12]

Average shear stress is the way to go.

Check your connection too, if a significant amount of web is removed or modified it's worth checking a net area.

Generally though unless it's short highly loaded beams, shear never governs in steel design.

 

[Hokie93]

Actually, the 0.6 factor in the nominal shear strength equation (Equation (G2-1) from AISC 360-10) arises from shear yield being approximately 60% of the tension yield. A factor of safety equal to 1.50 (or 1.67, depending on the h/t_w ratio) is then applied and the result for the vast majority of rolled steel beams is the familiar allowable shear stress of 0.4F_yA_w.

 

[KootK]

Yeah, shear yield stress comes in somewhere between 0.500 Fy and 0.577 Fy depending on the criterion that you use (Von Mises / Tresca). We deal with this from time to time here as it's a interesting and often misunderstood problem. Check this thread out: Link.

As for addressing your specific, original question:

1) AISC allows you to use the average stress I believe.
2) I personally go with peak stresses (3/2 V_ave).

Using the average stress would seem to imply post-yield stress distribution. A "plastic shear capacity" of sorts. As long as your member doesn't have a constant shear diagram, I can see how that might be rationalized. However, the mechanics of that seem pretty complicated to me. Too complicated to justify any material savings that might accrue.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[PicoStruc]

In canada, for building, we use factored stress as 0.66 Phi * Fy instead of 0.577 Phi Fy because we consider Strain-hardening in post-elastic strain range.

Thus, because we suppose full shear plastification, stress are indeed 'uniform' over the whole shear area...

 

[hokie66]

It is long established practice to use the peak stress in wood, but the average in steel. Steel is just that type of material.

 

[MacGruber22]

Yeah, shear yield stress comes in somewhere between 0.500 Fy and 0.577 Fy depending on the criterion that you use (Von Mises / Tresca)

I don't think about this often, so pardon my ignorance. Why wouldn't pure shear yield be sqrt(2)/2 of pure tension yield?

"It is imperative Cunth doesn't get his hands on those codes."

 

[MacGruber22]

Why wouldn't pure shear yield be sqrt(2)/2 of pure tension yield?

Never mind - I retract that question.

"It is imperative Cunth doesn't get his hands on those codes."

 

[jec67]

Fro a wide-flange beam, calculating the shear stress should be (Shear Force)/(depth x web thickness). The basis for this is the shear stress equation: (Shear Force * Q)/(I * Web Thickness). When a wide-flange beam is analyzed using this formula, one will find the shear stress in the flanges is negligible, and the average shear stress of (Shear Force)/(Depth*Web Thickness) is a close number to the maximum shear stress calculated using (Shear Force * Q)/(I * Web Thickness). For simplicity, the shear stress is calculated using the approximate value obtained with the equation stated above.

Fro a rectangular section (such as dimensional lumber or a plate), the maximum shear stress will be 1.5*(shear force)/(Area). This is a simplification of the same formula of (Shear Force * Q)/(I * Web Thickness).

 

[BAretired]

As stated by jec67, the shear stress in the web of a wide-flange beam is calculated by the usual formula VQ/It. The stress at mid depth is slightly more than it is at the top and bottom but nowhere near 50% more. Average stress is deemed adequate for design purposes.

BA

 

[rb1957]

an I beam is different to a rectangle.

max shear stress in a rectangle is 1.5*(P/A), a result from VQ/It.

in an I beam the web shear stress (VQ/It) is much more constant, 'cause the flanges allow the ends of the webs to develop shear flow; so the average shear stress (P/A) is adequate for design. I'd expect that the "true" answer is that the web shear varies from 0.95 to 1.05*average.

another day in paradise, or is paradise one day closer ?

 

[hokie66]

I agree with others that shear stress in a steel beam can be taken as V/A, but in many cases where shear is maximum, the flanges are coped or blocked. It is still OK to use V/A on the web area which remains, because of the unique and forgiving nature of steel.

 

[rb1957]

thank goodness for "the unique and forgiving nature of steel"

another day in paradise, or is paradise one day closer ?

 

[MacGruber22]

Praise to be upon oh gracious load redistribution GOD! May your enemies be cast away to the gates of Hell!!! ...which is naturally ruled by concrete.

"It is imperative Cunth doesn't get his hands on those codes."

 

[KootK]

Go forth and multiply! Or just redistribute.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[KootK]

I mistakenly assumed that this thread was about weak axis shear rather than strong axis shear. Please disregard my initial post. I'm very much on board with using the average shear stress for wide flange beams loaded about the strong axis orientation.

I agree with others that shear stress in a steel beam can be taken as V/A, but in many cases where shear is maximum, the flanges are coped or blocked. It is still OK to use V/A on the web area which remains, because of the unique and forgiving nature of steel.

In a previous thread here, it came to light that even the area of uncoped flanges can be included in the evaluation of the shear rupture capacity. Neat.

I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.

 

[jec67]

Hokie66: In response to your question, it depends. If you have a beam that is coped top and bottom, you will essentially have a rectangular section. I would use 1.5*V/A in this instance. for a single cope, I don't have a quick answer. I would start with V*Q/I*t and analyze accordingly.

If I recall correctly, the AISC does discuss analysis at copes (in particular long copes). I do not have access to the code at the moment, but it was somewhere in the connections portion of the latest code. Perhaps others in this tread can elaborate if they have ready access to a copy of the code.