Virtual vs Real Neutral Axis 3

[hocho]

The neutral axis in the interaction diagram doesn't really correspond to the physical location of the column. Or do they?

For example. Below the balanced point. At zero moment. The neutral axis is very small (or infinitely small). At zero moment (edit: I meant at zero axial load). It acts like a beam. Yet in a beam. The real neutral axis is the middle of the beam. So the interaction diagram neutral axis doesn't correspond to the physical location of the neutral axis. What is the formula that relates the virtual and real neutral axis (what official terms distinguish these two)? Is there a software that can show or distinguish them and plot them both?

 

[IDS]

Hocho - have a look at the spreadsheets listed here:

https://newtonexcelbach.wordpress.c...3-uls-design-of-reinforced-concrete-sections/

and here:

https://newtonexcelbach.wordpress.c...sign-functions-4-umom-and-new-maxax-function/

The ULS Design spreadsheet plots the beam cross section with the neutral axis position for any given axial load. Let me know if that is what you wanted, or if you had something else in mind.

Also I happen to be working on a spreadsheet for generating moment curvature diagrams at the moment, but it will be easy to generate an interaction diagram using the same function, so I'll look at options for how the results can be displayed.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hocho]

The axial load in a column is the difference between the compression forces in the concrete (plus compression steel), and the tension in the tension steel. When the applied loads are on the interaction diagram limit both the concrete and the steel will be at their yield stress. If this point is below the balance point, remember that the axial load is the minimum required to avoid failure, not the maximum. If the axial load is increased, with no increase in moment, the depth of the compressive stress block will increase, and the tensile strain in the steel will reduce. As the axial load is increased the concrete strain at first reduces, then when the depth of the stress block is deep enough the concrete strain starts to increase again, until it returns to the design yield strain, when the axial load crosses the upper interaction diagram line.

Btw.. something confused me above. You said "As the axial load is increased the concrete strain at first reduces, then when the depth of the stress block is deep enough the concrete strain starts to increase again"
But I know whenever axial load is increase, the concrete strain always increase.. how can it reduce initially? I'm visualizing the triangle and geometry of the strains. Thanks.

 

[IDS]

Btw.. something confused me above. You said "As the axial load is increased the concrete strain at first reduces, then when the depth of the stress block is deep enough the concrete strain starts to increase again"
But I know whenever axial load is increase, the concrete strain always increase.. how can it reduce initially? I'm visualizing the triangle and geometry of the strains. Thanks.

The combined axial load and moment are equivalent to an eccentric vertical load, with the eccentricity on the compressive side. If you apply an additional load at the column centroid it reduces the eccentricity of the resultant load, and the width of the compression region increases, so the total compression force increases, but the maximum stress reduces. It's a similar situation to a footing with a vertical load outside the middle third. If you apply an additional load on the uplift side, the pressure on the compression side will reduce, even though the total reaction force is increased.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hocho]

he combined axial load and moment are equivalent to an eccentric vertical load, with the eccentricity on the compressive side. If you apply an additional load at the column centroid it reduces the eccentricity of the resultant load, and the width of the compression region increases, so the total compression force increases, but the maximum stress reduces. It's a similar situation to a footing with a vertical load outside the middle third. If you apply an additional load on the uplift side, the pressure on the compression side will reduce, even though the total reaction force is increased.

Doug Jenkins

Thanks the great explanation. Say when you see buildings anywhere you go like in vacation. Is there a way to tell approximately where is the neutral axis location
For example. Are most columns compression region beyond the centroid at middle meaning neutral axis has mostly compressive region or is.the tension region larger. For instance. In edge columns is it possible the neutral axis has most compression region even thou axial load is not large? Is this common in building or rare? Thank you.

 

[IDS]

Hocho - I'll leave that one to others. I work on bridges and buried structures, so I don't actually know what is normal in framed building structures (but I would guess it varies a lot depending on building height and loading).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hocho]

Most concrete columns in practice fall into the compression branch of the interaction diagram. No special effort is made to be close to the balanced point. As a matter of fact, it is desirable to be as high as possible on the interaction diagram; that means the column is doing what it is intended to do, namely to carry axial load.

BA

BA. Usually how big is the percentage of moments compared to axial load did you allocate to edge columns in your gravity load only building? I know seismic need much higher allocations. Im asking for static typical framed building side columns moments where only 3 beams framed to it. What is the typical moments of these for small.houses and big houses or those you usually.build?

 

[BAretired]

For a typical medium or high rise apartment building with unit floor load w and column spacing a, the axial load is slightly less than wa²/2 per floor and the moment is approximately wa³/24 which is shared by two columns, one above and one below the floor.

A column which supports N levels including roof is carrying:

P = N.w.a²/2
and the typical moment is M = wa³/48

so the M/P ratio is a/24N

A 12 story building would have an M/P ratio of a/24 in the top floor and a/288 in the main floor. If a = 16', M/P = 8" in the top floor and 0.67" in the main floor. Ordinarily, a minimum eccentricity of one or two inches would apply in the lower floors.

BA

 

[hocho]

The axial load in a column is the difference between the compression forces in the concrete (plus compression steel), and the tension in the tension steel. When the applied loads are on the interaction diagram limit both the concrete and the steel will be at their yield stress.

Doug. If there is no concrete and only the compression steel taking the resistance at the compressive side. How do you solve for the balanced point provided the bars are of equal amount? At what moment and axial load will the strains of both the tension and compression steel be at yield at same time? Or won't it occur at all?

 

[Retrograde]

If the reinforcement is symmetric the balance point will be at pure bending (ie. zero axial load).

 

[hocho]

Retro. Is the formula to get the moment for the above case:

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

Or is it M = As.Fy.e?

where e = eccentricity from centroid to the load?

Elsewhere. I think BA seemed to believe the former. Why is it "d" and not the "e" (eccentricity)?

 

[IDS]

If the reinforcement is symmetric the balance point will be at pure bending (ie. zero axial load).

No, if the axial load is zero then the tensile force in the tension steel is equal to the compression force in the concrete plus compression steel, so the tension steel will reach yield before the compression steel, and before the concrete reaches ultimate strain (unless the steel has a very high yield strain and high %, or the reinforcement layout is very unusual).

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[hocho]

No, if the axial load is zero then the tensile force in the tension steel is equal to the compression force in the concrete plus compression steel, so the tension steel will reach yield before the compression steel, and before the concrete reaches ultimate strain (unless the steel has a very high yield strain and high %, or the reinforcement layout is very unusual).

Doug. Retro was describing my inquiry what if there was no concrete and only bars in the compression side and tension side of a column. Just hypothetical if such structure can be invented.

Then was Retro right that in concreteless column with bars only that "If the reinforcement is symmetric the balance point will be at pure bending (ie. zero axial load)."

 

[BAretired]

With no concrete, it would be a steel section.

As.Fy.d is the moment capacity with no axial load.

With axial load, the force in each set of bars would be P/2 ± M/d or P(1/2 ± e/d)

The limiting strength occurs when one or both sets of bars reaches yield.

BA

 

[hocho]

BA, before you deleted the post with the moment of Inertia (doesn't want to make it more complicated?). I already input the formula into excel and spending 2 hours trying to understand it. Also comparing those formula with your latest post.

With no concrete, you are dealing with a steel section.

M = As.Fy.d is the moment capacity of the steel reinforcement only. The axial load would be zero at that value.

With axial load, the force in the bars is P(1/A ± P.e/I where A and I refer to the properties of the steel reinforcement.

A = 2.As

I = As.d^2/2

The stress in the bars on each face, S = P/As(1/2 ± 2e/d^2)

The limiting value would be when one or both sets of bars reach yield stress.
BA

First. Let's use the figure you have computed yourself in past thread. You said in another thread that:

"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. A

If d=420mm, then M = As.Fy.d = 8(300)*400*420 = 403,000,000 n-mm OR 403Kn-M
B
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. C

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. D

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. E

BA

Here's the interaction diagram for the values you have given (for the column without concrete but bars only):

The letters corresponds to the Red Letter I included in your post above.

For a week. I am unable to plot the limit curve that you see for typical interaction diagram. Where is the curve? That's why I was asking Doug how to get the balanced point of the concreteless column.


But now the following formula may be related to it and complete the interaction diagram.

Given the above values you yourself gave.

As = 20*300M = 2400
fy = 400 MPA
d= 420mm
M = 403.2 kN-m
e = 0.21 mm
Using Axial Load of 100 kN.

A = 2.As = 4800
I (moment of inertia?) = As.d^2/2 = 2400 kN * 420 mm ^2/2 = 211680000
With axial load, the force in the bars is P(1/A ± P.e/I) = 100 * (1/4800 + 100* 0.21/211680000) = 0.020843 (plus),
0.0208234 (minus)

The stress in the bars on each face, S = P/As(1/2 ± 2e/d^2) = 100/2400 * (1/2 + 2 (0.21)/420mm^2)= 0.020833 (plus),
0.0208332 (minus)

At only 100 kN. Why is the bars already near yield strain of 0.020833 (is that the formula of stress or strain (it looks like strain)? In your earlier calculations and the interaction diagram. It is supposed to have 960 kN capacity??

Anyway. In your latest post. You mentioned a formula... solving for it in excel:

"With axial load, the force in each set of bars would be P/2 ± M/d or P(1/2 ± e/d)".. solving

P/2 ± M/d = 50.9 (plus), 49 (minus)

P(1/2 ± e/d) = 55.25 (plus), 44.75 (minus)

is this the ksi or yield stress and earlier formula the strain?

But again.. at only 100kN.. why are they in yield stress whereas in the interaction diagram figures you calculated. It has capacity of 960 kN (point C). I'm very confused. Thank you.

 

[BAretired]

hocho,
To answer the first part of your post, after writing the following post, I realized I had omitted a term:

With no concrete, you are dealing with a steel section.
M = As.Fy.d is the moment capacity of the steel reinforcement only. The axial load would be zero at that value.

With axial load, the force in the bars is P(1/A ± P.e/I where A and I refer to the properties of the steel reinforcement.

A = 2.As
I = As.d^2/2

The stress in the bars on each face, S = P/As(1/2 ± 2e/d^2)
The limiting value would be when one or both sets of bars reach yield stress.

It should have read:

With axial load, the stress in the bars is P/A ± P.e.y/I where y is d/2.
Then S = P/As(1/2 ± e/d)

Instead of correcting that post, I rewrote it expressing it in terms of bar force instead of stress.



BA

 

[BAretired]

hocho,
In the second part of your post, the problem is complicated by the presence of four intermediate bars. For pure axial load, each of the four intermediate bars carry as much load as an outside bar but for eccentric axial load, their contribution is not too clear and was neglected in my earlier post to mes7a.



BA

 

[BAretired]

Taking hocho's example,

A column has 20 bars, 8 on each side and 4 intermediate bars. Each bar has As = 300mm². A void has occurred in the concrete during pouring such that only the steel bars are acting for a short length near the bottom. The void was filled with epoxy which may have some beneficial effect, but this is ignored in the following analysis.

As = 20*300M = 2400
fy = 400 MPA
d= 420mm
M = 403.2 kN-m (this occurs with P = 0)
e = 0.21 mm
Using Axial Load of 100 kN.

f = P/A ± P.e.y/I
I = 300[2*8*(d/2)² + 2*2(d/6)²] = 1233d²
y = d/2 so y/I = 1/2466d
f = 100,000/2400 ± 100,000(0.21)/(2466*420) = 41.7 ± 0.02 = 41.7 MPa << 400

M_[P=100] = (400 - 41.7)2466d = 371e6 N-mm or 371kN-m (92% of M_[P=0])
Or, in other words, e = 3.71m when P = 100 kN.

With just the reinforcement and no concrete, there is no tension branch to the interaction diagram. The steel on the compression side will always yield first.

BA

 

[hocho]

Taking hocho's example,

A column has 20 bars, 8 on each side and 4 intermediate bars. Each bar has As = 300mm2. A void has occurred in the concrete during pouring such that only the steel bars are acting for a short length near the bottom. The void was filled with epoxy which may have some beneficial effect, but this is ignored in the following analysis.

As = 20*300M = 2400
fy = 400 MPA
d= 420mm
M = 403.2 kN-m (this occurs with P = 0)
e = 0.21 mm
Using Axial Load of 100 kN.

f = P/A ± P.e.y/I
I = 300[2*8*(d/2)2 + 2*2(d/6)2] = 1233d2
y = d/2 so y/I = 1/2466d
f = 100,000/2400 ± 100,000(0.21)/(2466*420) = 41.7 ± 0.02 = 41.7 MPa << 400

M[P=100] = (400 - 41.7)2466d = 371e6 N-mm or 371kN-m (92% of M[P=0])
Or, in other words, e = 3.71m when P = 100 kN.

With just the reinforcement and no concrete, there is no tension branch to the interaction diagram. The steel on the compression side will always yield first.

You mean at P=100 kN. Moment capacity is 371 kN-m? how can e=3.71 meters be so big.. Why is there is no tension branch to pull it on opposite side? But in the earlier computations that makes up the following interaction diagram. When P=960kN. It has moment capacity of 202 kN-m at eccentricity of 0.21.
Are they compatible points that can be included in the interaction diagram (shown in pink dot). In the following.. is the curve more of the blue one or red one.. and how to compute for the balanced point (whatever it can be?). Million thanks Ba.

 

[hocho]

After entering your formulas in Excel. I tried value Axial load of 960kN. Moment is negative. Trying 900kN. Moment capacity is 25.7kN-m at an eccentricity of 0.028 meter. But in the following where you wrote before that when the load is directly over the steel on the compression side, the capacity is 960kN. with moment capacity left of 202 kN-m at eccentricity of 210mm. You mean this isn't really the case and the real one is the 900kN and 25.7KN-m? Or maybe my excel input has typos. I'm confused. Please enlighten. Thanks.

(you wrote before
"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."

 

[BAretired]

As = 20*300M = 2400 (this is wrong) should be 6000
fy = 400 MPA
d= 420mm
M = 403.2 kN-m (this occurs with P = 0)

Using Axial Load of 960 kN.

P/A =960,000/6000 = 160MPa
This leaves 400-160 = 240MPa for bending
M.y/I = 240
M = 240*I/y = 240*2466*420 = 248e6N-mm or 248 kN-m.

e = M/P = 248/960 = 0.258m or 258mm

The point on the interaction diagram, P, M is 960, 258

My earlier calculation for 100 kN is also wrong as I used the wrong area of steel. I will correct that in a subsequent post.


BA