Column Bending angle and Failure Mode 1

[mes7a]

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?

 

[jayrod12]

Think of applying a point load to the middle of a simply supported beam, and the same magnitude of load near the end of the beam. Which deflects more? Same principle here as I see it. Except instead of an applied load, you have a reduced cross section. Essentially synonymous when discussing stresses in generalities.

 

[mes7a]

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?

 

[sponton]

it's not like that, it's the fact that when you have cracks you have increased stresses around the crack. [stress concentrates in the areas that are weaker] It's impossible to have it not stress unless you brace the whole area such that the column receives minimum lateral load [since vertical load would make the crack close itself [in the case of an angle it is questionable because of non-symmetric/non-uniform loading]

 

[BAretired]

If the column is hinged top and bottom and axially loaded, a cavity on the west side of the column will tend to make the column deflect eastward. If the properties of the materials are known, the magnitude of deflection can be calculated. Deflection will be maximum when the cavity is at mid-height.

BA

 

[mes7a]

If the column is hinged top and bottom and axially loaded, a cavity on the west side of the column will tend to make the column deflect eastward. If the properties of the materials are known, the magnitude of deflection can be calculated. Deflection will be maximum when the cavity is at mid-height.

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?

 

[racookpe1978]

IF everything were exactly positioned exactly as you have drawn it and exactly sized exactly as you have calculated it and IF every material property were exactly at nominal "textbook" strength and yield and crystal "average" size and shape and properties ...

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.

 

[BAretired]

If a cavity exists in an area which, in the absence of a cavity, would be uncompressed under a particular load case, then the cavity does not affect column strength for that load case.

BA

 

[mes7a]

Let's say you have the opposite case. See:

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.

 

[BAretired]

M = As.Fy.d
where As = area of steel on each face
Fy = yield stress
d = c/c distance between reinforcement

BA

 

[mes7a]

M = As.Fy.d
where As = area of steel on each face
Fy = yield stress
d = c/c distance between reinforcement

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?

 

[BAretired]

For intermediate bars, the formula is similar except d is the c/c distance between intermediate reinforcement and Fy may not be realized if the strain is not sufficient.

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

 

[mes7a]

For intermediate bars, the formula is similar except d is the c/c distance between intermediate reinforcement and Fy may not be realized if the strain is not sufficient.

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).

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?

 

[BAretired]

Construct an interaction diagram for the column without concrete. When the load is purely axial, the maximum value of P in round numbers is 20*300*400 = 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

 

[mes7a]

Construct an interaction diagram for the column without concrete. When the load is purely axial, the maximum value of P in round numbers is 20*300*400 = 19,200,000N or 19,200kN.

But 20*300*400 = 2,400,000 N or 2,400 kN. Not 19,200kN. Would this affect anything below?

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.

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?

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.

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??

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

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!

 

[BAretired]

But 20*300*400 = 2,400,000 N or 2,400 kN. Not 19,200kN. Would this affect anything below?

You are correct. It was past my usual bedtime and my brain was playing tricks on me. I have edited my earlier post.

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?

I arbitrarily assumed the load was centered over the compression steel. Since d=420mm, the steel is 410mm from the center.

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??

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.

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?

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

 

[BAretired]

When the eccentricity e is greater than d/2, the reaction in the 8 bars on the compression side is P(e+d/2)/d.

At failure, this means that P(e+d/2)/d = As*Fy

P = As*Fy*d/(e+d/2)​

M = P*e​

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

 

[mes7a]

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.

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?

 

[BAretired]

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.

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.

But where is the role of concrete in the tension side at the interaction diagram formula at large eccentricity?

Concrete plays no role on the tension side of a column. In that regard, it is the same as a beam.


BA

 

[mes7a]

Ok. Tnx. BAretired. In column with fixed base, if the loading is eccentric.. it seems it's not all tensions at one side. Is the following the more accurate curvature?

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.