Trajectories for Principal Stresses
Trajectories for Principal Stresses
(OP)
This is the trajectories for principal stress for uniform load

http://imagizer.imageshack.us/v2/800x600q90/20/jyq...
I'm looking for illustration of trajectories of principal stress for concentrated load at midspan. Has anyone encountered this?

http://imagizer.imageshack.us/v2/800x600q90/20/jyq...
I'm looking for illustration of trajectories of principal stress for concentrated load at midspan. Has anyone encountered this?






RE: Trajectories for Principal Stresses
http://classes.mst.edu/civeng110/concepts/12/princ...
BA
RE: Trajectories for Principal Stresses
http://imagizer.imageshack.us/v2/800x600q90/713/ay...
RE: Trajectories for Principal Stresses
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?
RE: Trajectories for Principal Stresses
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?
RE: Trajectories for Principal Stresses
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
RE: Trajectories for Principal Stresses
Your case may be found at this link:
https://www.google.ca/search?q=principal+stress+tr...
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
RE: Trajectories for Principal Stresses
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.
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
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
RE: Trajectories for Principal Stresses
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 45o 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
RE: Trajectories for Principal Stresses
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?
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
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
RE: Trajectories for Principal Stresses
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?
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
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?
RE: Trajectories for Principal Stresses
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
RE: Trajectories for Principal Stresses
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.
RE: Trajectories for Principal Stresses
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....
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.
RE: Trajectories for Principal Stresses
Hokie, the slabs are one way to secondary. So all the loads are focused on the secondary framing into the girder. So there is no 2 way action.
About deflection compatibility. See picture below:
http://imageshack.com/a/img594/2109/gpfi.jpg
One thing we noticed where the girder is different from a concentrated load at midspan loaded from above the beam is that in the girder/secondary joint, the secondary bars are rested on the girder 4 main bars (see the red lines in the picture above). So it's like the load of the secondary beam is downward into the girder lower bars. Now is there any formula to calculate how much the stress in the bars would propagate (see red in the picture). This would kinda cause the concentrated load to be wider and the a/d become less at 4.5, but it still falls into concentrated load at midspan (according to the article). Is there possibility that the entire girder main lower bars share the load of the secondary. If so, the stirrups in the whole length equilibrate it upwards making it like uniform load, but I doubt it. Better yet. Please show stress trajectories for girder/secondary joint especially the stress in the main lower bars carrying the secondary bars framing into the joint as all examples in the books are only for loads above the beam (not below in the bars).
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
You should read this thread in conjunction with thread507-357341: Horizontal cracks in beam below slabs.
BA
RE: Trajectories for Principal Stresses
First, my designer is http://rbsanchez.net/ who designs 40 storey buildings and major public buildings in the Philippines. This is the illustration of how the stirrups are spread in the girder
http://imageshack.com/a/img41/7246/v3py.jpg
Fc = 4000 psi ~ 28 MPA
stirrup size 10mm.. 40,000 psi = 276 MPA, Area = 0.000000142 (from 71x2)
At midspan, with spacing of 8" (200mm), the Vc is
english Vc = 2 sqrt(Fc') bw d = 2 sqrt (4000psi) 12 width x 18.115 depth = 27.49 kip x 4.448 = 122.2 kN
metric V(stirrup)= Av Fy d /s = 0.000000142 x 276000000 x 0.46 / 0.2 = 90 kN = 20.26 kips
(I use different units because USA unit is english while my country is metric and the books we use is english)
So Vn = Vc + Vs = 122.2 kN + 90 kN = 212 kN
Vu = 250 kips. There is deficient of 250 kips -212 kips = 38 kN
I discussed this with designer. He said midspan is more of flexure and shear doesn't appear at least 2d from it. So 2d from it with spacing of 150mm (6"), Vs = 120.118 so
Vn = Vc + Vs = 122.2 kN + 120.118 = 242.4 Still deficient of 8 kN.
The designer said ultimate load can't be reached in service load because the load combination has huge safety factor.
My problem is, if the article is true, then Vc of concentrated load at a/d of 5.8 is only one half
so Vn = 122.2/2 + 120.118 = 181 kN. Then it is a little far from 250 kN.
This is why I plan to remove the floor topping for reducing superimposed load of about 20 kN.
The designer won't believe in the article saying that he never see any of his structure fails. Being management engineer of the project, I don't want the building to be the first to fail. I told him about putting I-beam underneath the girder. He said it's up to me. I wonder if I should put I-beam under the girder. But the metal plate can't frame into the existing column because there are already many bars inside it. Also I wonder if I should let the I-beam carry the entire factored load of 250 kN or only the deflected load. How do you think of it so I can discuss with the designer above (he is laughing at my idea of I-beam but I'm serious about it if the article is true).
RE: Trajectories for Principal Stresses
Vu = 250 kN
Vn = 159 kN
deficiency 250-159 = 91 kN at midspan
The designer said the service load is lower. But I'm not taking chances so the idea of I-beam underneath the girder become a serious possibility. Since they haven't done this before and they told me I'm the one who decide whether to put it or not. Does it make sense to put I-beam that can support only the deficiency of 91 kN? Shear I-beam under concrete beam! Has anyone heard of this before?
Also I saw the following after googling for stress trajectories where there is blank stirrups at midspan (f). Do you believe that no shear of any kind can form at midspan because there is only flexure? Or won't you ever consider it? The designer forgot about the concept of stress/strain and trajectories since they only use Etabs in designing all the buildings.
http://dc99.4shared.com/doc/yK4BwyBe/preview_html_...
from http://dc99.4shared.com/doc/yK4BwyBe/preview.html
RE: Trajectories for Principal Stresses
so? you still have to design to the required loads.
could you make the beams wider ? i know deeper is more efficient, but more of an impact to the design.
i know we want to help, but this problem sounds pretty complicated and i don't think we clearly see the complete design.
Quando Omni Flunkus Moritati
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
http://imagizer.imageshack.us/v2/800x600q90/835/2h...
My designer designs it this way (the 5.5meter is shorter so they make it girder with one way slab framing into the secondary):
http://imagizer.imageshack.us/v2/800x600q90/707/k4...
BA thought of floor joists across the 5.5meter span. If I knew concentrated load at midspan not good. I should have let the designer make it this way (but this costs more because of more beams but at least uniformly loaded)
http://imagizer.imageshack.us/v2/800x600q90/809/9v...
Anyway. Can't change the past anymore. Just need to decide whether its feasible to put steel I-beam under the girder to serve as replacement for the girder or other extra support to handle the missing 91kN shear capacity at girder/secondary beam joint in the following (as designed).
http://imagizer.imageshack.us/v2/800x600q90/707/k4...
RE: Trajectories for Principal Stresses
I don't understand. Here is what you said earlier:
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).
If the factored load is 250 kN, then the factored shear each side of the load is only 125kN, not 250. Kindly clarify.
BA
RE: Trajectories for Principal Stresses
I'm sorry. In the designer calculation sheet, 1.2DL+1.6LL is 170 kN. 250kN is the Demand Vu written which they said is probable shear strength from probable moment strength that forms during plastic hinging when you subject the girder into seismic cyclic loading. The location of my project doesn't have seismic activity because foundation is rested on rocklike bed... the analysis assumes it's soil. But then you never know what happens to rock in a rare once in a lifetime earthquake.
Anyway. I'd like to see stress trajectories of a beam undergoing cyclic loading. I want to see if the area below midspan would suddenly have non-horizontal tension stress trajectories as the moment concentration swings back and forth in the beam. Can the moment of a girder/secondary beam changes location to other parts of the beam that is not the joint? The designer doesn't know the concept of stress trajectories as they just input everything in ETABs including soil density which they put as soft because I thought it was soft but really hard during excavation and the ETABs just outputted the demand Vu and the steel required but too bad they didn't realize that girders have shear demand at the midspan at concentrated load. They use the stirrups spacings for uniform load and the Etabs operator realized this when I showed him the shear and moment diagram of a concentrated load at midspan which has constant shear from support to midspan.
Therefore to convince him to put I-beam. I have to show him stress trajectories of a beam undergoing cyclic loading that the area under midspan of a concentrated load would have non-horizontal tension component or he won't do anything as he emphasized his job is only to know rebars steel requirement and it is up to the contractor how to construct the building. This is the reason why he didn't forsee the congested top bars of the beam which produce horizontal settlement cracks. He is not even aware of the concept of plastic settlement cracks. Most designer is like this in my country. This is why quality control engineers like me have to review everything with more knowledge than the designers and tell them what to do or they would just ignore everything. Most designers in country don't even visit their projects because their job is to sit at office operating Etabs program and printing the steel output for other clients.
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
By the way, the Vg is 1.2DL + 1.6 LL = 170 kN. Vg means gravity shear, hence the factored load is twice that.. about 340 kN. 250 kN is the probable shear strength.
They didn't use solid slab for the 5.5 m span because they said if the top and bottoms bars of the slabs are not done property, the deflection at middle would be big. This is the reason why the slabs are made smaller, to avoid inconsistencies in construction at the cost of bigger secondary beams.
I will meet the designer later. I don't know what to say and thinking what to say. He doesn't like to discuss concepts as he said in engineering, actual experience is better. I will tell him if that is the case, why need computations.
RE: Trajectories for Principal Stresses
It sounds like your designers like to complicate things unnecessarily. A solid slab would have been simpler and easier to construct, albeit using a bit more concrete.
RE: Trajectories for Principal Stresses
You guys need about three more layers of people who don’t know what the hell they are doing in your building design and construction project management process, all of whom apparently don’t have much experience either. Then you could make really really good, bad designed buildings. Why don’t you make ‘stress trajectories’ your PhD thesis subject, and that is truly “pilled higher and deeper,” and report back to us once you have published the thesis in a respected and peer reviewed structural journal. If the shear is the same from the load to the reaction, you might expect the shear reinf’g. to be the same over that beam length. You seem to have a whole bunch of people over there who plug things/crap into Etabs, and the like, and hope for the best without really knowing what they are doing. And then, a whole bunch of other people managing projects who don’t know much either, trying to tell them that they are wrong. And finally, none of you will listen to what some mighty experienced people, here, are trying to tell you about how concrete design works and is done. God help you if there is another hurricane or EQ in your area, you will not want to be pointing toward your engineering and construction prowess as your salvation at that time. You guys would do well to reorganize your construction process so that one knowledgeable person was in charge and took responsibility for wrangling all you know-it-alls into one productive design/build group.
RE: Trajectories for Principal Stresses
hokie, the Philippines is so poor, we don't develope our own codes. We just copy everything from the USA ACI. And in the ACI (see below illustration and text) it is stated that in seismic design, the shear and moment derived from factored load of 1.2 DL + 1.6 LL have to be multiply further (let's say 1.3) so the effect is like 1.3 (1.2 DL + 1.6 LL) to come up with probable moment and probable shear strength. Etabs used this. Since they all use etabs. They just follow the output.
http://imagizer.imageshack.us/v2/800x600q90/823/0i...
http://www.nehrp.gov/pdf/nistgcr8-917-1.pdf
5.3.1 Beam Design Shear
The beam design shear is determined using the capacity design
approach as outlined in Section 3.2. Figure 5-10 illustrates this
approach applied to a beam. A free body diagram of the beam is
isolated from the frame, and is loaded by factored gravity loads
(using the appropriate load combinations defined by ASCE 7)
as well as the moments and shears acting at the ends of the
beam. Assuming the beam is yielding in flexure, the beam end
moments are set equal to the probable moment strengths Mpr
described in Section 5.1. The design shears are then calculated
as the shears required to maintain moment equilibrium of the
free body (that is, summing moments about one end to obtain
the shear at the opposite end).
This approach is intended to result in a conservatively high
estimate of the design shears. For a typical beam in a special
moment frame, the resulting beam shears do not trend to zero
near mid-span, as they typically would in a gravity-only beam.
Instead, most beams in a special moment frame will have nonreversing
shear demand along their length. If the shear does
reverse along the span, it is likely that non-reversing beam
plastic hinges will occur (see Section 5.1).
http://imagizer.imageshack.us/v2/800x600q90/20/u8j...
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
http://imagizer.imageshack.us/v2/800x600q90/59/lqc...
For a girder with secondary beams framing into it. Is the above marked with red what would happen in cyclic loading where the moment shift left and right? If so, then the concentrate loading kinda shift left and right too? If so, then the midspan bottom can develope shear too. Do you generally agree?
I'd show this to the designer. He doesn't think of this because Etabs program outputs in numbers only (rebars). I'll convince him it can happen then proceed to have him design I-beam underneath the girder.
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
http://imagizer.imageshack.us/v2/800x600q90/855/od...
http://imagizer.imageshack.us/v2/800x600q90/844/7o...
But even carbon fiber is not enough because the strain may not occur together with stirrup so Vn won't be Vc+Vs+Vfrp. Also we needed 3 ply to satisfy the seismic load combination of
1.2 DL + 1.6 LL + 1.4 Earthquake + 1 Wind
And the carbon fiber installed is just one U-wrap.
We will carefully design steel I-beam to support the girder. We need very lightweight I-beam although column still has reserved of 300kN axial load. Again. Should I-beam carry the entire factored load or just the shear deficiency? Designer is not sure. And are there special lightweight I-beam of exotic material (like titanium?)
RE: Trajectories for Principal Stresses
Yes, it is called UNOBTAINIUM. See http://en.wikipedia.org/wiki/Unobtainium in case you are not familiar.
So you installed carbon FRP - who designed the FRP system? Are you walking a 'slippery slope' with liability and responsibility - what does the engineer-of-record say about you taking on this strengthening design/install too?
Seems this has added to the complexity of an already problematic situation.
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
Quando Omni Flunkus Moritati
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
rb1957, the building is already finished.. we can't use shoring and scalfolding forever.
The steel I-beam to be put under the concrete beam has to be very stiff or else it would deflect along with the concrete beam which means it can't stop the shear tension in the main beam. I wonder if there are very stiff 6 meter steel I-beam or a truss has to be specially made, and then the column metal plate has to bear the load. This is why composite action of them needed so the I-beam won't take the entire load and stress. Since only two I-beam is needed. It can be specially fabricated and being expensive is not a problem.
RE: Trajectories for Principal Stresses
In this beam, what is the ratio of actual dead load stress to live load stress (unfactored, working stresses?)
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
Quando Omni Flunkus Moritati
RE: Trajectories for Principal Stresses
2) Has anyone actually checked span-depth ratio for compliance with ACI? Was the ratio exceeded, and an actual deflection calculation ever made, cracked section and all? You seem to be fixated on the mistake of calculation shear due to the concentrated load from the secondary beam, but to me, those girders are too shallow.
RE: Trajectories for Principal Stresses
You mean d/2 stirrup spacing was exceeded where the secondary beam occurs on the girder?
RE: Trajectories for Principal Stresses
But in seismic design, you have to subject the entire structure to cyclic loading (magnitude 7 earthquake) so the Vu of 107 kN become more than twice... 250 kN. Hence I-beam still needed to be put under concrete beam to resist solely earthquakes.
In the Philippines, we never use braced frame or shear walls... all are moment resisting frames but unfortunately.. construction is not of good quality.
RE: Trajectories for Principal Stresses
There is no actual cracking yet of the beams. No deflection yet because load is so low. The building beam shear satisfied only 1.2 DL + 1.6 LL but in seismic cyclic loading you have to double or multiply by 2 the entire 1.2 DL + 1.6LL and it is here where the building is not designed for earthquake as far as beam shear is concerned because the column sizes, bars, beam bars are designed for seismic load.. just the stirrups coming up short.
RE: Trajectories for Principal Stresses
I can't get too worried about the shear at the secondary beams. But in a seismic event, I get worried about the shallowness of the girders and possible understrength of the concrete mix. Also, since you say construction is not good quality, what about proper placement of concrete, vibration while placing, etc? That is what worries me.
RE: Trajectories for Principal Stresses
I will talk with the team leader tomorrow or Monday again. I need to know what to ask. The depth of the girder is 500 mm (half meter). The girder with midspan concentrated load has shear to depth span of about 5.9. It's shallow. In the event of section cracks. ACI said the stirrups will hold it like truss. What can you comment about this depth thing during cracked section?
You asked earlier about the ratio of actual dead load stress to live load stress (unfactored, working stresses), the unfactored dead load is about 65 kN (14.6kips), superimposed dead load about 22 kN (4.94 kips), so total dead load + SD = 87 kN (19.5 kips). I plan to remove the entire superimposed dead load (tiles with 2 inches of mortar bed) to lower the dead load. The unfactored live load (typical office load) is 2.4kpa x 18 = 43.2 kN or if 4.8 kpa used is 86.4 kN but 4.8 kpa will be rare because the second floor will be office only (the building is just 2 storey). Hence, ratio of actual dead load stress to live load stress can be 1 or real life.. unfactored live load 1/2 (one half) less than dead load. What do you ask about the ratio?
About the concrete strength. It's 4000 psi and placement not good because space is tight. This is the reason we must put I-beam. Someone suggested that concrete beam - steel I-beam hybrid will work provided shear links/rods from the steel member to concrete element to be connected to have a monolithic response from both the elements. What can you say about this before I explore more ideas with the designers who haven't done such a thing before and since they are busy designing 7 highrise/midrise.. I have to say sense in the discussions because they don't have much time.
RE: Trajectories for Principal Stresses
http://imageshack.com/a/img23/9432/50os.jpg
Please see above layout of the floor. The designers don't do manual computations because they are designing so many buildings so they need sophisticated program like ETABS. They even forgot to manually compute so we are discussing the manual computations and comparing to their ETABS. Please comment on the following. I know the other formula to get tributary area of the slabs in each beam. But since we are computing the concentrated load in the girder/secondary joint, then the shaded area in the picture is the influence of dead & live load. I know there is contribution in the far side of the slab and beam as load varies along the spans but remember the other beams also contribute in supporting the shaded area load but the computation is similar if the entire shaded area is 100% of the load for estimation purpose. Isn't it this approximation is used?
Calculations of dead load
23.54 kN/m^3 (23.54 kN in a cubic meter of concrete)
Beam sections = 500mm depth by 300mm width
Slab thickness - 100mm (4")
Dead Load of secondary beams framing into joint
Beam framing into girder joint 3meters each so 23.54 x 0.5mx0.3m = 3.531x 6 = 21.186 kN
Dead load of the slabs.
23.54 kN/m^3 x 0.1m (thickness of slabs)=2.354 kN/m^2
2.354 x 18 square meters (the shaded area has 18 sq.m) = 42.372
So total dead load of the beam and slab in the girder is 63.558 kN. I didn't include the girder beam self weight because we are getting the concentrated load framing on the joint. Shear Vdl from the deadload is about 63.558/2 = 31.779 kN.
Live Load
assume 4.8 kpa maximum load 4.8 kPa x 18 sq.m - 86.4 kN live load
shear V = 86.4/2 = 43.2 kN
Total load
wu = 1.2 DL + 1.6 LL = 1.2 (63.558) + 1.6 (86.4) =76.2696 + 138.24 = 214.5 KN
or add SD load of tile floor = 23.54 x 0.05m (2") = 1.117 x 18 = 21.186
214.5kN + 21.186 kN = 235.686 Kn
but I plan to remove the entire 2" flooring mortar bed & tiles to reduce load so let's use 214.5kN (I won't allow any partition too in the shaded lines and ceiling weight below is so light because it is just light wooden board so let's ignore it for calculation sake esp since I round off the horizontal span from 5.5m to 6m)
From total wu of 214.5 kN
Vu = 214.5kN/2 = 107.2548 kN (24 kips) concentrated load framing into the girder midspan.
Vc and Vs of the concrete and stirrup is enough to handle 107.2548 kN.
So far, are the calculations above sound at least for 1.2 DL + 1.6 LL?
Of course if you take in seismic, the effect is like twice of 1.2 DL + 1.6 LL and I won't manually compute it. I'd just like to know for now if at least the building shear can support gravity load of 1.2 DL + 1.6 LL
Thanks.
RE: Trajectories for Principal Stresses
OK, the designer are engineers, but obviously they are very good at inputting into ETABS but lack practical knowledge.
I was curious about the ratio of live to dead because I was hoping the problem was not so bad (if the design live was very high) - I like to get an estimate of what the actual FoS is when something is realistically loaded (the actual live load is frequently much less).
Those secondary beams look much too shallow - if you exceed the ACI span/d ratio for beams, you have to justify by checking deflection of a cracked section.
When you deepen those secondary beams, the girders may still be overstressed since this is a moment frame. They look way too narrow. Girders with short spans relative to the beams they support are frequently wider than their overall depth.
RE: Trajectories for Principal Stresses
When you mentioned span/d ratio, do you mean the whole span over depth or shear span over depth? If the latter,
In the aci shear span/d ratio
a/d = 1 very deep beam
a/d = 1-2.5 deep beam
a/d = 2.5 6 shallow beam
a/d = very shallow beam
In the kani curve, very shallow beam would first fail in flexure before it fails in shear.
For shallow beam, it can fail in shear and flexure. My shear span/depth a/d of the girder is 5.9
If you are talking about clear span/depth ratio.. the secondary is 6 meters long, depth 0.5m so span/depth is 12, it's uniform load of the one way slab. There is 3 pcs of 20mm rebars grade 60 (60,000 psi or 414 MPA) bottom bars. Since the load is very low only... the end of short one way slab (see figure earlier), won't the bottom longitudinal bars be enough to resist deflection? I can put carbon fiber for flexure in the secondary beam if needed. But the moment of the secondary center dead load is only 18 kN. What do you think of this based on your experience so I talk about this to the designer... since if I don't mention anything, he won't say anything, so I have to question him... like he admit there is deficient shear after I question him about it.
About the girder. This is why we will put steel I-beam below it. Are you saying even for gravity load only, the beam depths are insufficient based on your experience or are they only deficient in seismic loading? What is your opinion about putting steel I-beam under the girder? Thanks for the tips.
RE: Trajectories for Principal Stresses
The secondary beams are 600mm deep. On a 6m span, L/d = 10 so I can't understand why you believe they are too shallow.
The girder is the same depth, namely 600mm deep. On a 5.5m span, it has L/d = 9.2 which, again seems okay to me.
I am not convinced that any remedial measures are required but if a steel beam is added under the existing girder, how will the steel and concrete beams be attached so they act in unison? How will the steel beam be moment connected to the existing concrete columns? How will the addition of a beam under the girder affect headroom? How will the removal of topping affect the fire separation?
It seems very strange to be reviewing design philosophy after the structure has been built, particularly when the design engineer doesn't seem to recognize or acknowledge that a design problem exists...or does he?
BA
RE: Trajectories for Principal Stresses
Before the structure was built. I asked the designer why all the beam was only 500m deep. He said beam with 600mm depth is only for beam 7 meter span or more. For 6 meters or less, they use 500mm depth. But other engineers I asked said it should be 600mm deep. So is 500mm sufficient?
Worse case, I could categorize the second floor as residential unit to limit any big load.
RE: Trajectories for Principal Stresses
OK I see, the secondary beams are not so long. They just seemed so in the photograph showing the rebar.
So, BA, you are saying there maybe no problem, everything looks ballpark?
Its just that I am so conditioned to see girders that are wide and heavily reinforced. I really can't tell for sure unless I run actual numbers. Also, I prefer to use very old-fashioned techniques, and even so that I graduated from University in 1974 where they taught both the old working stress and ultimate strength concrete designs, I prefer to use working stress, hence my preference for deeper beams.
Releky, have you considered submitting the design for peer review by an engineer from New Zealand or from California or Nevada USA, someone well-trained in seismic?
RE: Trajectories for Principal Stresses
OK I see, the secondary beams are not so long. They just seemed so in the photograph showing the rebar.
So, BA, you are saying there maybe no problem, everything looks ballpark?
Its just that I am so conditioned to see girders that are wide and heavily reinforced. I really can't tell for sure unless I run actual numbers. Also, I prefer to use very old-fashioned techniques, and even so that I graduated from University in 1974 where they taught both the old working stress and ultimate strength concrete designs, I prefer to use working stress, hence my preference for deeper beams.
Releky, have you considered submitting the design for peer review by an engineer from New Zealand or from California or Nevada USA, someone well-trained in seismic?
RE: Trajectories for Principal Stresses
No, 500mm would not be shallow either, but I was under the impression that we were talking about tee beams. The depth of a tee beam is from the underside of beam to top of slab. Are we talking about the beams in thread507-357341: Horizontal cracks in beam below slabs? If so, the depth is 600mm.
BA
RE: Trajectories for Principal Stresses
BA
RE: Trajectories for Principal Stresses
In moment frame, you have to multiply the U = 1.2 DL + 1.6 LL by two to design the longitudinal bars and shear stirrups. All the longitudinal bars of the buildings are design for 2 x (1.2 DL + 1.6 LL), only the stirrups is designed just for gravity load due to the designer not bothering to check the shear and moment diagram of the ETABs outputs.
Now they said they would try to put I-beam to improve the shear resistance but don't know how to start.
RE: Trajectories for Principal Stresses
Here is the sketch shown in the previous thread. The depth looks like 600mm to me. How can that be interpreted another way?
http://img571.imageshack.us/img571/1747/fb1m.jpg
BA
RE: Trajectories for Principal Stresses
BA
RE: Trajectories for Principal Stresses
In the Philippines.. engineers are free from liability in case building is damaged by any seismic activity so designers don't worry about damages caused by earthquake because they can always blame it on the earthquakes. So in a country where 99% of buildings are special moment frames. They are just ordinary moment frames or even just gravity frames.
BA, seismic design is simply making sure the sections have adequate shear strength before developing plastic hinges. To develope plastic hinges, you must exceed up all flexural reserve at factored load. So the shear must be as strong as that able to develope the so called "probable moment strength".
AELLC. Even without peer review, the shear is only design for gravity load... not seismic loading. When you say deflections from crack section.. are you referring to flexural cracks? I'm not worried about flexural cracks because its flexural design is adequate up to probable moments srength. I'm more worried about shear cracks of shallow beams. Have you ever encountered any such shear cracks in your life?
RE: Trajectories for Principal Stresses
You seem to show a calculation that shows the beam is OK for gravity load but then you say you are worried about gravity load. You say you're worried about stirrups at midspan but then say your concern is due to seismic (plastic hinges near supports). Also you list a load combination of "1.2 DL + 1.6 LL + 1.4 Earthquake + 1 Wind" which does not exist (that I have seen). Then say that for seismic the load combination is 2 x (1.2DL + 1.6LL) which I've also never seen. Then you want to use a steel beam to help carry gravity forces which you've already established that the beam is ok. So....
As best as possible try to organize your concerns with your suggested solutions and lets see if we can resolve this.
EIT
www.HowToEngineer.com
RE: Trajectories for Principal Stresses
http://imagizer.imageshack.us/v2/800x600q90/842/ul...
It's in page 732 of the book Reinforced Concrete (A Fundamental Approach) by Edward Nawy 5th edition. I asked the designer why his Vu is so high. He said he also used Wind in addition to Earthquark. Next week I'll try to make him produce what is the biggest load combinations out of the 23 he used to give such a big shear *almost* equivalent to 2 x (1.2 DL + 1.6 LL).
After doing some rough manual computation with him (he doesn't know how to manually compute). I realized the stirrups in the constructed concrete is only good enough for 1.2 DL + 1.6 LL (the exact computations I shared a few messages above).
Now I realized that it is the seismic shear the building is deficient where it can't take on the probable moment strength. I discussed this with the designer and he seemed to realize it and agree I-Beam is needed.
See the logic in the above. So I'm now organized what to discuss with him. Of course it is ultimately the design of the designer I'll follow because I'm of course not authorized to design it my own... because he is the designer. And just giving the designers ideas to think about. I'd discuss with 3 of them next week to get my arguments across the shear is deficient for probable moment strength. The website of the designer is http://rbsanchez.net/ for reference.
RE: Trajectories for Principal Stresses
You are saying the stirrups are OK for gravity but deficient for seismic.
The part I still don't understand - if you can do manual calculations and who you call the designers cannot - why can't you do a complete design as a "go-by", an example of how a typical building should be designed?
A second thing I still don't understand - if these designers are University graduates, why don't they comprehend shear and moment diagrams? That is very basic, something you don't forget even after years of using "black box" software. I have forgotten many things that were studied in University that I have never used at work, but shear and moment are never forgotten.
Perhaps you mean the designers think that the selection of longitudinal reinforcement is of primary importance, and the stirrups are of much less importance. I have reviewed calculations were the designer seems to think beams are very important, and columns and especially footings are much less important - the calculations for those were sloppy and hurried, and full of deficiencies.
RE: Trajectories for Principal Stresses
"You are saying the stirrups are OK for gravity but deficient for seismic." ...
this is the bit that i (in my ignorance) don't understand ... i read the first sentence to say "seismic loads aren't important" and the 2nd "the structure is failing due to seismic loads" ?
Quando Omni Flunkus Moritati
RE: Trajectories for Principal Stresses
In the second sentence. Some of us are very concerned about structures that fail during seismic load.. hence we need absolute quality control check of the design and the execution. And that's what I'm doing.
RE: Trajectories for Principal Stresses
So we are concerned with shear for seismic...
The 1.4xE is based on using a service level seismic force so be careful as ASCE7 now uses strength level loads so if you apply the 1.4 to the strength level load then you are factoring up this force twice.
Also (and you may already have down this) but the shear capacity at each location along the beam should be compared to the shear demand of the load combination. Meaning the shear diagram with lateral forces applied will be different than with only gravity forces.
Sounds like meeting with the designers should get everyone on the same page... hopefully. Let us know how it goes.
EIT
www.HowToEngineer.com
RE: Trajectories for Principal Stresses
RE: Trajectories for Principal Stresses
BA
RE: Trajectories for Principal Stresses
The problem now can be simply categorized as seismic retrofitting of gravity load structures.
If the steel I-beam underneath the concrete girder is difficult to be moment connected to the column and connected to the beam as composite. Does it make sense if you will retrofit those gravity load structures to use steel I-beam simply to support the concrete beam in the event the shear cracks simply make the beam fall down? So the I-beam would be sectioned just to carry the weight of the beam?
In seismic engineering, the idea of plastic hinging is to act like fuse so after the hinges form, you just repair the beam... and the basic idea is not to let the structures fall but just damaged enough to be repaired. So the I-beam would just to make sure the beam won't fall down and give time for occupants to run safely outside and then repair the damages afterwards (in rare seismic event that may happen once a century or twice or never at all.. seismic engineering is about probability as seismic event can't be predicted).