Column Bending angle and Failure Mode
Column Bending angle and Failure Mode
(OP)
Please see figure below of a column pinned on top and bottom. If there is a hole at the side at midspan, the column would deflect more than if there hole is at the base. Is this right? What is the principle called where hole at midspan would make it deflect more than if it occurs at bottom near base? Is it related to K factor or what?






RE: Column Bending angle and Failure Mode
RE: Column Bending angle and Failure Mode
The centroid of the notch move to the right. If you can fix the bottom such that the column below the notch won't move.. What would happen.. is it like converting it into a column of smaller section? or would it still see it as a big column of the original size?
RE: Column Bending angle and Failure Mode
RE: Column Bending angle and Failure Mode
BA
RE: Column Bending angle and Failure Mode
See the following illustration:
What if there is an eccentricity to the right and the hole is under tension. Isn't it that when the other part is in tension, the concrete doesn't have function.. so if the neutral axis started right in the part with the hole. Then it's like having a normal column with the left under tension (and concrete no contribution).. isn't it?
Or do you have to compute for the so called plastic centroid?
RE: Column Bending angle and Failure Mode
Then yes. You could 'back-calculate" the vertical member so the twisting force from the right-eccentric load would balance the yielding forces and strains from the left-eccentric vertical column.
Until something changed the least bit.
RE: Column Bending angle and Failure Mode
BA
RE: Column Bending angle and Failure Mode
Where there is a cavity in the compression side and all the axial load is taken up by rebars (left of neutral axis).. let's say at a unit strain of 0.0021 computing for the axial load capacity from modulus, stress and area of rebars produce bars axial load capacity of 1000 kN. Can you quickly estimate it's corresponding moment capacity (just from it)? Remember no concrete contribution for axial load.
I've been scrutinizing and solving for the interaction diagram but can't yet get a simple procedure for estimating for the moment capacity for the bars when there is no concrete at compressive side and let's assume (for sake of dicussion) a material can prevent the bars from buckling. Thank you.
RE: Column Bending angle and Failure Mode
where As = area of steel on each face
Fy = yield stress
d = c/c distance between reinforcement
BA
RE: Column Bending angle and Failure Mode
I wonder how you count the intermediate bars (?).
Let's say I didn't include the intermediate bars (8 out of 10 only).
As = 8x 0.000314 (20mm area ) = 0.002512
Fy= 414 MPA = 414,000,000 pascal
d = 0.420 meter
M= As.Fy.d = 0.002512 x 414,000,000 x 0.420 = 437kN.m (?)
Without any concrete at compressive side.. the column can still resist 437 kN.m. Is this about right? Isn't it a bit big?
How do you connect it to axial load. let's say the column has axial load of 500 kN. How does this connect to moment capacity of 437 kN.m?
RE: Column Bending angle and Failure Mode
Combining moment with axial load, assume that all bars take an equal portion of the axial load, leaving a reduced effective area of steel to carry moment.
The column would fail when the steel on the compression side reached yield stress (or buckled at a lower stress).
BA
RE: Column Bending angle and Failure Mode
At strain of 0.0021 and Modulus of 29,000,000. Stress is 60900 or 414 MPA.
With bars = 8 x 0.000314 = 0.002512.
Axial capacity is 1054 kN.
Now with the computation of moment capacity in the last message at 437 kN.m
and service load of say 600 kN.
So deducting axial capacity of 1054 kN - 600 kN = 454 kN. capacity left.
How do you connect 454 kN axial capacity left with 437 Kn.m moment capacity. I know eccentricity = M/P. But even if you solve for e=M/P = 437 / 454 = 0.96 what does eccentricity of 0.96 mean and how do you relate axial capacity left to moment capacity left?
RE: Column Bending angle and Failure Mode
19,200,000N or 19,200kN.2,400,000N or 2,400kN.If d=420mm, then M = As.Fy.d = 8(300)*400*420 = 403,000,000 n-mm OR 403Kn-M
That gives you the points on the P and M axes.
When the load is directly over the steel on the compression side, those bars will carry it all by themselves. Their capacity is 8*300*400 = 960kN. That gives you another point on the diagram, i.e. 960kN at an eccentricity of 210mm which means a moment of 960(0.21) = 202 kN-m. At that point, the remainder of the bars, twelve in total, will be doing very little work.
When the eccentricity is less than 210mm, each bar will carry P/20 from axial load alone. If P = 600 kN, then each bar carries 30kN which is 25% capacity. This leaves 75% of the bar capacity to resist moment, so M = 0.75*403 = 302kN-m.
When eccentricity exceeds 210 mm, the force in the outer bars is magnified. So if e=630mm for example, the bar force is 2P or 1200kN which exceeds the capacity of the bars. The point on the interaction diagram is P = 480kN at e = 630mm, which is the same as M = 302kN-m.
BA
RE: Column Bending angle and Failure Mode
But 20*300*400 = 2,400,000 N or 2,400 kN. Not 19,200kN. Would this affect anything below?
Where did you get an eccentricity of 210mm? I can't get it by any M/P above. Is it the eccentricity at balanced point?
Why do you say 25% capacity? If it's full capacity, it's 120kN. 20 pcs x 120kN = 2400 kN. But you get the 25% from axial capacity.. why multiply the remaining 75% to moment capacity??
But 302kN-m is the capacity you said earlier when eccentricity is less than 210mm.. but how can it be the same when eccentricity is 630mm?
Many thanks BARetired!
RE: Column Bending angle and Failure Mode
You are correct. It was past my usual bedtime and my brain was playing tricks on me. I have edited my earlier post.
I arbitrarily assumed the load was centered over the compression steel. Since d=420mm, the steel is 410mm from the center.
The load was stated to be 600 kN which is 25% of the capacity of 20-20M bars. Each bar is taking 30kN of axial load. Axial compression helps the bars on the tension side but reduces the capacity of the bars on the compression side to carry moment. The reduction for those bars is 25%, leaving 75% of their capacity available to resist moment. The bars on the tension side will be loafing.
The column cannot carry 600kN at an eccentricity of 630mm It would fail under that load. It can carry only 480kN at that eccentricity which results in a moment of 302kN-m and provides another point on the interaction diagram.
BA
RE: Column Bending angle and Failure Mode
At failure, this means that P(e+d/2)/d = As*Fy
Where As is the area of steel on the compression side, in the present case, 8-20M bars and P is the failure load at eccentricity e.
BA
RE: Column Bending angle and Failure Mode
Since you mentioned this thing about axial compression helping the bars at tension side? This is the part that perflexed me. Let's go to pure column without holes and concrete both in the compression and tension side. When the moment goes up, the tension side won't have any axial load compressing it anymore. And usually the compression side has the concrete compression block and compression bars not yet yielding. So it's like the point of axial loading pressing on tension side seems to be unimportant.. this only occurs at initial while the compression side has still sufficient capacity. Right?
Anyway. I'm deeply analyzing the derivation of the interaction diagram (also solving for it) for that same column without hole. I noticed something. My interaction diagram for that column is:
The deadload they computed for the column is 357 kN. SD load = 214 kN, Live load is 166 kN. Total load is 737 kN. (I'm just verifying stuff manually)
The nominal axial load capacity of the column (as the above diagram) shows 6927 kN (it was designed for 4-storey but only 2-storey with roofdeck built)
The Mn (nomimal moment capacity) is 726 kN.m
You will notice that the actual load of 737 kN is much below the balanced P and M of the interaction diagram (shown in x in the illustration). (But now I think it's wrong to put it at the point, right? because with low load, moments can be higher and this needs another concept than interaction diagram.
Anyway. My question is this.
I'm analyzing the derivations of the formula. They say axial load can press on the tension side making the moments capacity larger. But below the balanced point.. there is no axial load anymore pressing on the tension side. The formula for Moment in the interaction diagram is
Pn = 0.85fc' ab + As' fs' - As fs
Mn = Pn e = 0.85 fc' ab (h/2 - a/2) + As' fs' (h/2 - d') + As fs (d-h/2)
note 0.85fc' ab is just the role of concrete in the compression side.. but where is the role of concrete in the tension side?
Also for eccentricity larger than balanced point.. I noticed the tensile strength (85 ksi when the bars would break) would be reached for eccentricity of 324mm (compared to balanced e of 284mm). The P computed is 2150 kN and M is 698kN (below the balanced point in the tension failure part). But my total load is only 737kN. Does it mean the bars would break (reach tensile strength) when the moment is 698kN? But it corresponds to axial load of 2150kN. Does it mean I must make the load heavier at 2150kN to avail of the moment capacity? But where is the role of concrete in the tension side at the interaction diagram formula at large eccentricity?
RE: Column Bending angle and Failure Mode
You seem to be misunderstanding the purpose of an interaction diagram. It is a plot of the failure load of a column corresponding to a particular failure moment. Points lying to the left of the line are in bounds. Points lying to the right of the line are out of bounds and represent a failure condition.
737kN is an applied load. You should not be using applied loads or moments with an interaction diagram. You should be using factored loads and factored moments.
The line above the balance point represents compression failures. For a given factored moment, the maximum factored load can be read from the diagram.
The line below the balance point represents tensile failures, i.e. yielding of the tension steel. Without axial load, your diagram shows that failure occurs at a factored moment of 350kN-m. Additional axial load improves the moment capacity until the balance point is reached. If the factored load and moment lie to the right of the line, the column will fail. In the tension branch of the diagram, factored live load should not be included as it will not always be present.
Concrete plays no role on the tension side of a column. In that regard, it is the same as a beam.
BA
RE: Column Bending angle and Failure Mode
Now for a certain P at right eccentric loading.. what is the corresponding P the lower left side (opposite side) would feel? First let's ignore the hole. I'd like want to know if that part can feel the full P of the other side.. or is there a reduction factor or something in this loading case? Many thanks.
RE: Column Bending angle and Failure Mode
The axial load is constant from top to bottom, namely P. The moment varies linearly from top to bottom according to your diagram. A point of zero moment occurs somewhere between the top and the bottom.
Since both end moments are clockwise, equal and opposite horizontal reactions are required top and bottom to maintain equilibrium.
Ht = Hb = (Mt+Mb)/h
where Ht, Hb are the horizontal reactions at top and bottom respectively and h is the height of column.
BA
RE: Column Bending angle and Failure Mode
You mean at the same distance say 1 meter above column base, the compression block 0.85 fc' a b of both pinned and fixed column are same value but on opposite side? This is different concept from buckleness of slender column vs short column isn't it. What topic in structure books does the pinned vs fixed moments concept highlighted?
Etabs show this:
Why is the moments not equal and the zero is not right at midspan?
Again. Very important to know. You mean at the same distance say 1 meter above column base, the compression block 0.85 fc' a b of both pinned and fixed column of same loadings and eccentric at right side are same value but on opposite side? Many thanks.
RE: Column Bending angle and Failure Mode
Edit: At some point above the base, the moment in the pinned base column will be equal and opposite to the moment in the fixed base column.
Completely different.
Any elementary structural analysis book will refer to boundary conditions.
Mt and Mb can be equal under certain load conditions, but a column fixed against rotation at the base and free to rotate but fixed against translation at the top will have Mb = Mt/2. If the top moves horizontally, that will not be true.
BA
RE: Column Bending angle and Failure Mode
See:
How do you compute where in height of the two boundary conditions the compression block 0.85 fc' a b (at ultimate strain 0.003) will be similar for same loading and eccentricity to right side.. at what height for both conditions. This is for the column:
RE: Column Bending angle and Failure Mode
I make a graphics of the above to illustrate the idea. see:
and to zoom graphics see:
http://imagizer.imageshack.us/v2/1280x1024q90/909/...
At Ultimate Strain 0.003 at the balanced point. Moment in the section is PROPORTIONAL to the stress block size 0.85 fc' a b where a is the size of the compressive block, right? This means where there is greater moment, the stress block is smaller (from the strain distribution triangle).
Now since the moment varies in both the pinned and fixed boundary conditions. Then in the fixed, the moment at bottom which you said varies by one half to the above would have twice larger compression stress block than at top, right?
And in the pinned condition. Since there is larger moment at the top, then the compression block at ultimate strain 0.003 is smaller and becoming bigger as it goes down the bottom?
Is the idea generally correct?
Thanks so much.
RE: Column Bending angle and Failure Mode
I think you misunderstand as someone pointed the use of interaction diagrams or their development,
RE: Column Bending angle and Failure Mode
It's eccentric loading. Etabs show:
The building was already built 2 years ago. Just trying to understand it by computing everything manually. The designers used software to model it and they forgot how to manually compute.
RE: Column Bending angle and Failure Mode
what load cases are you running, what are the loads? which one is the column you're analyzing?
Etabs may show many things but if you're not sure what you're inputting and what the program is doing, it might as well be a black box.
RE: Column Bending angle and Failure Mode
The etabs capture is just to show it is eccentric loading. I'm not trying to model anything in etabs. I'm trying to just understand where is the stress block in the column for different boundary conditions. The column concerned is the eccentric column:
Do you think the stress block magnitude is dependent (inversely proportional) on the moment magnitude like the following. I'm still waiting BaRetired reply about this.
RE: Column Bending angle and Failure Mode
the concrete stress blocks needs to balance the stresses in the steel stress block [if you will] to accomodate the moment [by shifting the position of the neutral axis from top to bottom of the section]. A balanced section would give you a balanced stress block between the two, as you shift the neutral axis you get different conditions [meaning a moment and an axial load due to the difference between the resultants of the concrete and steel stress blocks], if you have a point (P,W) in the diagram, you can get a virtual eccentricity e=w/p
so if you're given a problem like you have a load P at e cm from the center of the support you can say W=P*e so you can find it in the diagram.
I don't think you know what you're looking at in etabs, you're showing a column that has moment at the top and moment at the bottom because the conditions where set like that, you basically have a moment frame with fixed supports. It has nothing to do with eccentric loading, the unbalanced moment that the column takes can be turned into a virtual eccentricity [i don't know for which purpose you'd do that] but it makes no sense whatsoever except for the example i stated above.
RE: Column Bending angle and Failure Mode
At Ultimate Strain 0.003 at the balanced point. Moment in the section is PROPORTIONAL to the stress block size 0.85 fc' a b where a is the size of the compressive block, right? The size of stress block is related to both moment and load. It is not directly or inversely proportional to moment. This means where there is greater moment, the stress block is smaller (from the strain distribution triangle). Not necessarily. It depends on the M/P ratio.
Now since the moment varies in both the pinned and fixed boundary conditions. Then in the fixed, the moment at bottom which you said varies by one half to the above would have twice larger compression stress block than at top, right? No, it is not that simple.
And in the pinned condition. Since there is larger moment at the top, then the compression block at ultimate strain 0.003 is smaller and becoming bigger as it goes down the bottom? If moment causes a strain of 0.003 at the top, it will cause a lesser strain further down the column, so the comparison is not valid.
Is the idea generally correct? No!
BA
RE: Column Bending angle and Failure Mode
Oh. I was familiar with beam where the neutral axis can migrate towards the compression edge as the load and steel stress
increase. But reviewing the formulas for column. It seems the compression block is more fixed depending on ultimate strain
of both concrete and bars. For example. If concrete unit strain is 0.003 and steel unit strain is 60/29,000=0.0021, and
neutral axis for the example in the book with 20" column across is cb = 17.5 x 0.003/0.0051 = 10.3". stress-block depth
a=0.85 x 10.3 = 8.76" and concrete compressive resultant is C= 0.85 x 4 x 8.76 x 12 = 357 kips.
In beams. The compressive resultant is not analyze that way but can get smaller as it gets to the compression edge. Any
illuminating idea why? I'd reread the book after getting this critical bird eye view why. Many thanks!
RE: Column Bending angle and Failure Mode
We seem to be drifting a long way from the topic introduced in the original post. You have a problem with a column in an existing building. In particular, you have a void near the bottom of a column where concrete did not fully fill the full area during the pour. The void, if I remember correctly, has been filled with epoxy.
To check the strength in the vicinity of the void, you could neglect the concrete entirely and assume that the steel bars carry all the load. If you do that, you cannot have a tension failure so for the purposes of this discussion, behavior of a beam is not relevant. You are of course, entitled to consider the reduced moment in the vicinity of the void.
Have you considered this approach and if so, have you come to any conclusions regarding the adequacy of the column in question?
BA
RE: Column Bending angle and Failure Mode
I'm not drifting. The reason I'm interested in the stress block size whether it's always one half of the diameter of the column is at least
if it's one half.. the compression block won't become so tiny that it falls right in the void. In the following column, the red is
the part without concrete but filled with epoxy.
But I'm worried that in beams, the compression block can become so tiny near the compressive edge. Can column also do it? I want
to avoid that kind of loading because I want to avoid compressive failure of the bars where it can fracture.
I'm interested in the moments because i'm trying to gauge whether the following really work. By enclosing it with 1 meter concrete
pedestal.. can the column moments be suppressed at that portion and transfer upward (above void):
bars before concrete poured around void with epoxy
concrete poured
RE: Column Bending angle and Failure Mode
position of neutral axis from top to bottom?? Is it not from side to side.. or is your column section upside down
when you state it.
See:
Sponton, what codes are you using because your terms seem foreign. I'm using ACI. Here we don't use steel stress
block. In fact. At balanced point, the steel Asfy at tension and compression side cancel. In your code it doesn't
cancel?
But the column with moment at top and bottom only occur if it's eccentric loading. If it's not, then you earlier
said the 4 sides balance it. See:
from http://www.iitk.ac.in/nicee/EQTips/EQTip17.pdf
Anyway see my previous message to BA above to see picture of the concrete pedestral retrofit. Maybe you can comment on it. Thanks.
RE: Column Bending angle and Failure Mode
correct me if i'm wrong.
1. Concrete stress-strain diagram is for elastic range only at 0.0005.. for near strain limit of 0.0021.. you can no
longer use the stress-strain curve
2. The compressive stress block in the interaction diagram is not the real section of compression when the column is
bending at great moments. The compressive stress block or resultant C=0.85 fc` a b is related to moments only for the
capacity of the P above and below the balanced point of the interaction diagram. Right? This is what is very tricky to
understand. I'm realizing this because when the neutral axis is below the balanced point value.. the C resultant must
be larger.. instead the amount is smaller.. so it is only related to the moment interaction diagram.
3. But then you still have to deal with the real compression section when the column is bending at greater moments.
Here there is a point when the left side of the column is no longer in touched (like in tension). Here one can use some
idea in the derivation of the interaction diagram. Let's pretend the left side of column is no longer in compression but
only the 0.5x0.2 mtr right portion in compression (so all the axial load is taken up by this). Is this possible? see again:
For my column concern above. Let' say 0.5x0.2 mtr section is replaced with epoxy and composing of 8 20mm bars (314 mm^2 area)
And the column is bending so much only it is taking the compression portion (the other side in tension). Then.
strain = 0.003
Epoxy modulus = 450ksi
stress = 1350 PSI or 9.3 MPA
epoxy area = 0.5x0.2 = 0.1 sq. mtr
P = 9.3 MPA x 0.1 sq.m = 930 kN
so axial load capacity of 0.5x0.2 mtr epoxy section is 930 kN.
the 8 bars contribution is
strain = 0.0021
bars Modulus = 29,000ksi
stress = 60900
bars area = 0.002512
P (of bars only) = 1054 kN.
BA.. 2 yrs ago I was computing like the above but my understanding wont be complete without computing for
the bending stress of the composite concrete epoxy. At that time my understanding of moment is less and at
least more now.
You mentioned earlier about the column with only bars present and no concrete and the bars prevent from buckling
by magic. There you mentioned about eccentricity of 210mm and moment of "d=420mm, then M = As.Fy.d =
8(300)*400*420 = 403,000,000 n-mm OR 403Kn-M (bars here is 300mm instead of 300 in Canada but that will do
for rounding off). And you said "When the load is directly over the steel on the compression side,
those bars will carry it all by themselves. Their capacity is 8*300*400 = 960kN. That gives you another
point on the diagram, i.e. 960kN at an eccentricity of 210mm which means a moment of 960(0.21) = 202 kN-m.
At that point, the remainder of the bars, twelve in total, will be doing very little work.".
Now the capacity of the epoxy is 930 kN. Can you use similar concept where the eccentricity is 210mm
so the moment is 930 (0.21) = 195.3 kN-m?
Or more accurate. Since the center of the 0.5x0.2 mtr portion at right side is 0.15 meter away from
center of column Then moment is 930 (0.15) = 139.5 kN-m?
If can't apply the same concept as your example of the concreteless bars. Then how do you compute for
the moment capacity of the epoxy section with 930 kN P capacity? How do you convert it to moment
capacity?
After learning this. Last thing is to consider the centroid has moved to the left side of the harder
concrete (against epoxy) and it has natural inkling for bending stress. Here just give me a clue how
to compute for it and I'll read the book for another day. Thanks so much BA! :)
RE: Column Bending angle and Failure Mode
the compression block, the compression bars and the tension bars. They are interrelated and the
following formulas and computations seem to relate them all and it may actually be able
to solve for the moment capacity of the epoxy section.
fs (tension steel) = strain Es (d-c)/c
fs' (compression steel)= strain Es (c-d')/c
C (compressive resultant) = C = 0.85fc'ab
cb (neutral axis balance failure) = d (strain concrete ultimate/(strain concrete ultimate + strain bar ultimate)
a= 0.85 cb
Pn= 0.85 fc' ab + As'fs' - As fs
M = Pn e = 0.85 fc' ab (h/2 - a/2) + As' fs' (h/2 - d') + As fs (d-h/2)
To relate it to epoxy. I'll use stress 1350 Psi (0.003 strain x 450ksi (epoxy)) instead of concrete
4000 Psi. This is not incorrect right.. because let's treat the entire compression block to be composed
of epoxy
given:
epoxy strain 0.003 (although it can be pushed higher but let's use it as standard meantime)
steel unit strain is 60/29,000=0.0021,
column dimension 19.685" x 19.685 " (0.5 x 0.5 mtr)
area steel = 8 x 0.46 = 3.68
from excel input of formulas and values
cb (neutral axis) = 10.10885"
a (stress block depth)= 8.59"
fs' = 65.48 ksi but <= 60 ksi
C = 0.85 x 1.35 x 8.59" x 19.685" = 194 kips = 863 kN
Pn = 863 + 3.68 x 60 - 3.68 x 60 = 194 kips = 863 kN
Mn = 4318.946 in-kip = 359.9 ft-kip = 488 kN.m
This means with the epoxy as compressive block.. axial load capacity is 863 kN and Moment capacity is 488 kN.m.
eccentricity is 22.25" or 565mm.
Since the compression block is 0.2 mtr or about 8 inches.. then the entire compression block is really epoxy
I tried changing strain since epoxy has greater strain-stress block and change fs (tension) up to tensile strength
and the capacity seems to increase (ductile) but first is the above method correct BAretired or spanton?
Thanks a whole lot!
RE: Column Bending angle and Failure Mode
BA
RE: Column Bending angle and Failure Mode
Because of the word wrap problem. Let's continue in this thread BAretired, sponton and others
http://www.eng-tips.com/viewthread.cfm?qid=395487
Btw.. in the computations above. I entered all formulas in excel from the book and I crosscheck it via the book examples to see
that I entered the formulas correctly and it produces final Pn and Mn, e for different neutral axis values like in the book.
RE: Column Bending angle and Failure Mode
sponton, I've been pondering on what you said above. Yes. I'm talking about moment unbalance from column at sides of building because of
beams framing into it at one side only.. it has nothing to do with actual eccentricity.. only virtual.. virtual in the sense that
you can only relate moments to axial load via eccentricity in formulas. If no virtual eccentricity.. how else would you able
to compute moments and axial loads and interrelate them?
In actual columns. What is the maximum virtual eccentricity at most by having unbalance beams framing on one side of it only? what you think?
RE: Column Bending angle and Failure Mode
There is something odd about the interaction diagram. At balanced point and using statics and resultant..
Pb = 0.85fc'ab + As'fs' - Asfs
And As'fs' is same value to AsFs but opposite reactions. In other words the opposite reaction
literally cancel out. So at balanced point. How can you still rely on the bars to resist the
axial load at the compression side when the tension side reactions cancel it out??