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?

 

[releky]

BA, the building already finished so can't do any one-way joints thing.
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).

 

[CELinOttawa]

I don't know about anyone else, but the girder looks to have insufficient shear reinforcement to me... I would have expected to see tight stirrupping similarly to what you need at the supports. The beam loading into the girder is effectively like a support condition where the shear is very high. You've said that the imposed load is 250kN; What is the size of the stirrups and at what spacing? Profile of beam and strength of concrete?

 

[BAretired]

CEL,
You should read this thread in conjunction with thread507-357341.

BA

 

[releky]

I don't know about anyone else, but the girder looks to have insufficient shear reinforcement to me... I would have expected to see tight stirrupping similarly to what you need at the supports. The beam loading into the girder is effectively like a support condition where the shear is very high. You've said that the imposed load is 250kN; What is the size of the stirrups and at what spacing? Profile of beam and strength of concrete?

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).

 

[releky]

I told the designer what if Vc should only be half. Then the deficiency would be about 100 kN especially consider you need to apply safety factor of 0.75 to Vn resulting in Vn of 0.75x212 kN (typo above message of "kips") = 159kN

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_m547c854f.jpg

from

http://dc99.4shared.com/doc/yK4BwyBe/preview.html

 

[rb1957]

"The designer said ultimate load can't be reached in service load because the load combination has huge safety factor." ...
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

 

[releky]

The 2 storey building is already built and finished and there is already a tenant in the ground floor. The lot is very simple.. just 6 meters wide and 12 meters deep. There is no column inside so the 6 meters is the girder beam of size 300mm width and 500mm depth. There is 90kN deficient shear at midspan.. but the designer ignore it saying the midspan has low shear.

 

[releky]

If you are given a blank lot of the following size (the columns in red), how would you design it?

http://imagizer.imageshack.us/v2/800x600q90/835/2hku.jpg

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/k4q9.jpg

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/v2/800x600q90/809/9veq.jpg

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/k4q9.jpg

 

[BAretired]

releky,

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

 

[releky]

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.

 

[hokie66]

I won't try to get between you and your designer, but to answer your question about how I would have designed it, I would have just used the edge beams and a solid slab across the 5.5 metre direction.

 

[releky]

The building is finished already.

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.

 

[hokie66]

If the loads are already factored (1.2DL + 1.6LL), why do you think they need to be doubled? Is this a requirement of your seismic code?

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.

 

[dhengr]

Releky:
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.

 

[releky]

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/v2/800x600q90/823/0ian.jpg

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/u8jl.jpg

 

[slickdeals]

Releky reminds me of Pattontom and his long-winded posts.

 

[releky]

Sorry if this post is long. Situations are desperate in the Philippines. My last question is the following:

http://imagizer.imageshack.us/v2/800x600q90/59/lqcw.jpg

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.

 

[releky]

Hi, I just talked with designer. We decided to put I-beam underneath the concrete girder. But he said the I-beam would just maybe carry the deficient shear. Would this work? He has never attempted putting I-beam underneath girder before and would ask others for more guidance. Won't it work better if the I-beam is to take the entire factored load instead of the deficient shear?

 

[releky]

I installed carbon fiber spaced at 4" (100mm) for emergency seismic precaution (in case seismic activity happens this week) with fiber anchor drilled 3" almost vertical. See illustration and actual picture of the carbon fiber wrap at the following:

http://imagizer.imageshack.us/v2/800x600q90/855/odkl.jpg

http://imagizer.imageshack.us/v2/800x600q90/844/7op4.jpg

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?)

 

[Ingenuity]

And are there special lightweight I-beam of exotic material (like titanium?)

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.