Trajectories for Principal Stresses 1

[releky]

This is the trajectories for principal stress for uniform load

http://imagizer.imageshack.us/v2/800x600q90/20/jyq8.jpg

I'm looking for illustration of trajectories of principal stress for concentrated load at midspan. Has anyone encountered this?

 

[BAretired]

I have encountered examples on the internet. It doesn't look much different from the uniform load example which you have shown. See the link below. The left hand picture is for a cantilever with a point load at the end. Imagine it mirrored about the fixed end and you have the same as a point load at center.

http://classes.mst.edu/civeng110/concepts/12/principal/maximum/trajectories.gif

BA

 

[releky]

I didn't just imagine it.. but actually do it.. see below.. they are not equal (I read over 5 structural books and nowhere is the concentrated midspan load stress trajectories illustrated, hope someone who encounters one can show it based on actual tests, plots and theoretical justifications)

http://imagizer.imageshack.us/v2/800x600q90/713/ayit.jpg

 

[Ingenuity]

releky

The cantilever model should be mirrored about the fixed-end to replicate a single point load over a simple span. Then the similarities to a uniformly loaded single span beam will be evident.

Why are you requiring such trajectories, anyway?

 

[releky]

Look at the shear and moment diagram of a concentrated load at midspan here:

http://en.wikipedia.org/wiki/Load_diagram

It is constant shear from left support to midspan. But in the trajectories for principal stress there is no shear near midspan. So how do you reconcile with the fact that the shear and moment diagram shows shear at midspan that isn't there in the trajectories above?

 

[BAretired]

As stated by Ingenuity, you have mirrored it incorrectly. Try instead to mirror it about the fixed end as I indicated in my previous post and you will end up with a downward force at each end and an upward force in the middle. There is no difference between a simple span with load at midspan and a cantilever with load at end. The fixed end of the cantilever corresponds to the midspan of the simple span.

I know I have seen principal stress trajectories for a simple span beam loaded with concentrated load in midspan. I will try to find it.

BA

 

[BAretired]

releky,
Your case may be found at this link:

https://www.google.ca/search?q=prin...uYCcj8oASrr4D4CQ&ved=0CEAQsAQ&biw=853&bih=504

See the attached file. The first is half of the beam with point load at midspan. The second is a beam with different loads and locations.

BA

 

[releky]

Thanks for the images. But in realistic beam. The neutral axis is inside the flange of a T-Beam just below the top. This would change the trajectories pattern greatly with more vertical lines. So if you have seen the stress trajectories for a T-beam with the neutral axis at the flange, kindly attach one.

I think someone in the world should have made java program to see changes in the trajectories with changes in shear-to-load span (a/d) and neutral axis.

 

[releky]

I mean shear-to-depth span ratio (a/d).. typo

 

[rb1957]

note the difference of the end conditions ... SS for the distributed load, fixed for the point load.

if both SS, it makes sense to me that the trajectories would be very simple for point load and distributed load as the moment diagrams are quite similar, max moment mid-span. Fixed end condition makes peak moment at the ends and opposite sense to the mid-span, so you'd see tension bending stresses at the ends on say the top surface and compression at the mid-span.

like Ingenuity asked, why do you want this ?

the initial pic was for a plate beam, now you mention T-beam ? of course the flange will affect the bending stress distribution (and so the principal stresses).

Quando Omni Flunkus Moritati

 

[BAretired]

Thanks for the images. But in realistic beam. The neutral axis is inside the flange of a T-Beam just below the top. This would change the trajectories pattern greatly with more vertical lines. So if you have seen the stress trajectories for a T-beam with the neutral axis at the flange, kindly attach one.

I think someone in the world should have made java program to see changes in the trajectories with changes in shear-to-load span (a/d) and neutral axis.

If you want to look for stress trajectories for a tee beam, be my guest but you still haven't told us why you want them. Principal stress trajectories are valid for an elastic beam. I don't know how valid they are after extensive cracks have formed.

For an elastic beam, the upward shift of the neutral axis due to the flange would have some effect, but not a dramatic effect on the stress trajectories. The slope at the neutral axis would be 45[sup]o[/sup] and the curves would be quite similar to the ones we have seen.

As the tee beam approaches its ultimate capacity, the neutral axis shifts upward even further. At that point, stress trajectories would seem to be unreliable and may not have much meaning.

BA

 

[releky]

I'm analyzing stress trajectories because of this important experiment done in which the concrete cracks at much lower load specified by ACI and other codes. See image below from "Design for Shear Based on Loading Conditions":

http://imageshack.com/a/img132/8823/1q1w.jpg

The beams and bars details of the above tests are the following:

http://imageshack.com/a/img534/6797/l0rj.jpg

"SUMMARY AND CONCLUSIONS
In this study, the results of 1200 beams tests were examined. By identifying the effects of loading type and the distance from the applied load to the support, some simple changes in
code provisions were developed. These proposed provisions are applicable only to structural members subjected to a narrowly-defined-type of loading. The shear design of many structural components is left unchanged.

1. The shear strengths of members subjected to uniform, or near uniform, loads are higher than those of member subjected to concentrated loads. Current code provisions provide safe estimates of strength for beams subjected to uniform loads;
2. Test specimens that exhibit shear strengths less than that permitted by ACI 318-05 are by and large limited to specimens subjected to concentrated loads that are applied between 2d and 6d from the face of the support;
3. The primary impact of the proposed provisions will be to increase the size of transfer girders or other elements under concentrated loads and hence increase the shear strength of such critical structural elements;
4. Most beams in a reinforced concrete building are loaded via a slab or a series of joists. Such loads are much closer to a uniform loading, and the shear design provisions for these members will remain unchanged; and
5. The current upper limit on shear strength, 10 sqrt (Fc') bw d (U.S. units) (5/6 sqrt (Fc') bw d [SI units]), should remain in place if the proposed provisions are adopted, that is, if Eq. (4) is
used for concentrated loads acting at distances between 2d and 6d from the face of the support.
"(end quote)

Bottomline of the above article is. The formula for concrete shear Vc = 2 sqrt(Fc) b d is only valid for distributed load or shear to depth span ratio below 2. For concentrated load above a/d of 2, the suggested formula in the paper is Vc = 1 sqrt (Fc) b d. In other words, the Vc for concentrated load applied between 2d and 6d from face of support should be 1/2 that of Vc loaded from uniform load for a/d beyond 2. This means the stirrups may to be much increased to balance decrease of Vc.

What is your experience on this? How often do you make concentrated load at midspan for typical a/d (shear to depth span) beyond 2?

 

[releky]

I saw the above information at

http://freepdfhosting.com/0f6460bf5d.pdf

 

[BAretired]

I don't recall running into that situation before and unless our code has changed recently, high shear span to depth ratios are not treated as a special case so far as I know. The framing diagram which you show in thread507-357341 is not typical in my experience. I would have preferred a different arrangement of members to avoid the large midspan reaction.

Another concern is that the carried beams and the carrying beam have the same depth, flush top and bottom, so the reaction is not bearing on top of the carrying beam as you illustrate in the specimens above. It is transferred through the depth of the beam which would suggest perhaps using a trapeze type of hanger under the carried beam and extending each side at 45[sup]o[/sup] into the carrying beam and anchoring with horizontal extensions each end. The hanger would be designed to carry the full reaction without reliance on V[sub]c[/sub].

In the future, it would seem prudent to avoid that type of framing.

BA

 

[releky]

If you will be given say a quiz and ask where shear cracks will first occur in the following sketch.

http://imageshack.com/a/img34/739/ijgf.jpg

Isn't it you would choose the shear span on the right as it encounters the greater shear (a/d of 1.7 versus 5.8 in the article)?

But in actual experiments (done by the authors shared earlier):

http://imageshack.com/a/img13/9015/1av4.jpg

Shear cracks form on the left (one with longer shear span) and not only that.. but at 1/2 of Vc! Later the right crack forms at much higher load.

Critical idea of the experiment is that in concentrated load at higher a/d, Vc is only 1/2 as suggested in the code. Therefore you must increase stirrups to balance lost of Vc capacity.

Or is there something wrong in the experiments to produce the odd results? Are there other asymmetrical load experiments?

 

[AELLC]

All of the above proves my point, we should go back to pre-1971 ACI, working stress design.

 

[releky]

I don't recall running into that situation before and unless our code has changed recently, high shear span to depth ratios are not treated as a special case so far as I know. The framing diagram which you show in thread507-357341: Horizontal cracks in beam below slabs is not typical in my experience. I would have preferred a different arrangement of members to avoid the large midspan reaction.

Another concern is that the carried beams and the carrying beam have the same depth, flush top and bottom, so the reaction is not bearing on top of the carrying beam as you illustrate in the specimens above. It is transferred through the depth of the beam which would suggest perhaps using a trapeze type of hanger under the carried beam and extending each side at 45o into the carrying beam and anchoring with horizontal extensions each end. The hanger would be designed to carry the full reaction without reliance on Vc.

In the future, it would seem prudent to avoid that type of framing.

BA. The following is file photo of the framing of the girder/secondary rebars taken underneath it without any formworks. The center is the girder/secondary joint with 250kN (56 kips) factored concentrated load at midspan. The a/d of the girder is 5.9 (the same as in the article shared earlier).

http://imageshack.com/a/img845/1701/odqf.jpg

You will see 4 bars below the secondary bars with stirrups on sides so the joint bottom reactions would be transferred to the supporting beam, so no problem about it. Note all beams have same sizes, 300mm width, 500mm depth.

My and the designer problem is the article. If it has basis, then the Vc calculated manually and via ETABs is not enough. It means ETABs has a flaw in the VC calculation in concentrated load for high a/d such as 5.9. If this is the case, we have to remove the entire 2" of floor topping mortars and tiles to decrease the superimposed dead load. Our country codes follow the ACI codes so the only way I can change our country code is to change the USA code by appeal to the ACI committee.

Also in my country, most of my structural engineers friends only work with 4 meter beam span because that's where we are comfortable with its much lower moment and shear. With the framing of this project, it is long span (the girder is 6 meters length with 6 meters secondary at front and back). You mean you never design such long beam span too and only design 4 meter span? Isn't there anyone here who has ever designed 6 meter beam span with 250kN (56 kips) factored concentrated load at midspan?

 

[BAretired]

Your photo is very interesting. I have never seen steel tied in the air prior to placing formwork. That is not a technique common in my country.

I don't think you should start removing floor topping to decrease dead load just yet as this would have an effect on your fire rating. Further investigation is needed to determine the best course of action.

A 6m span is not a problem for a system of one way concrete joists. That avoids the girder beam with the large concentrated load at midspan, a feature which needs to be carefully reviewed.

BA

 

[hokie66]

There have been many changes to ACI and other concrete code provisions over the years. Most of these changes have been due to research designed to confirm (or otherwise) code provisions. Some changes have been in the less conservative direction, some more conservative. These changes have almost never led to imposed modifications of existing structures.

Concrete structures in service rarely mimic our design assumptions exactly. Rather, they often find alternate load paths. Whether of not you take any comfort from it, I suspect that the structure you are concerned about will work in a somewhat different manner than assumed, in that the central point load will be reduced/redistributed because of two way action and deflection compatibility.

 

[CELinOttawa]

This issue is not new, and has a long history. It is similar to the effect of reducing the size of the aggregate in the concrete, and is fundamentally an instance of St. Venant's principle in action. The truth of the matter is that the longitudinal steel keeps the cracked sections together and causes the "failed" beam to continue carrying load, and thus serviceable under ULS even if failed in SLS and requiring repair. So while this paper is entirely correct and the beam has cracked due to the imposed load, the fact of the matter is that the beam should still be correctly under reinforced, not have failed in flexure, and will continue to service the loads.

Should it be of interest, the Canadian A23.3 code, the New Zeaaland Code (since 2006), British Standard, Australian Standard and Eurocode all specifically address this issue and can accurately predict the true failure stress. The US (ACI 318), NZ 3101:1995 and IS codes all predict unreasonably high values, but do not result in unsafe beams thanks to aggregate interlock.

An interesting summary of the effect (with a focus on increase in beam size and decreasing aggregate size) can be found at:

http://http-server.carleton.ca/~tedsherw/Sherwood5.pdf.

Keep one simply thing in mind: It is not the type of load that matters most, but the relative effect of the maginitude of load to the flexural and shear behaviour of the beam. Thus your large point load in effect causes the beam to behave as a deeper beam than it really is, because the load is more propotionate to one which would be carried by a larger, longer beam. Here the profile of depth to length governs the behaviour.

It is important to note that more and more code committees are formally including specific terms to deal with the "true" case in order to eliminate rare situations where the longitudinal reinforment may not be able to hold the aggregate interlock and thus fail to provide a sufficient level of safety.

If you want to be careful, I would advocate taking a look at the NZS 3101:2006 code as it is based on ACI, but has the required specific clauses under 9.3.9.3.4, and a good commentary to go with it.