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?

 

[hocho]

When I changed the As to 6000 from 2400 in excel.. at P=100kN. M= 397 kN-m at e = 3.97 mtr instead of M=371 kN-m at e = 3.71 mtr. It seems the changes in moments is not much.. why? But if the axial load increase.. there is more effect in the moments from the increase in As. How come?

How can one tell if there is tension branch or not to the interaction diagram (why did you say there isn't?). Also if there is concrete contribution, will eccentricity be lesser?

Looking at the strain diagram in the following:

If there is no concrete contribution.. the strain of the compression bars will be more.. but the tension side will be equally more from the seasaw effect. So there seems to be tension side. But how do you get the balanced point of the interaction diagram for this concreteless column?

 

[BAretired]

Getting back to the 100 kN axial load
P/A = 100,000/6000 = 16.7 MPa
This leaves 400 - 16.7 = 383 MPa for bending

M = 383*I/y = 383*2466*420 = 397e6N-mm or 397 kN-m
and e = M/P = 3.97 m

The interaction curve is actually a straight line from P = 2400, M = 0 down to P = 0, M = 414. The pure moment value is slightly higher than we had before because it takes into account intermediate bars which the earlier estimate omitted since their contribution was thought to be small.

There is no tension branch and no balanced point. Failure occurs when the bars on the compression side reach yield stress.


BA

 

[hocho]

Getting back to the 100 kN axial load
P/A = 100,000/6000 = 16.7 MPa
This leaves 400 - 16.7 = 383 MPa for bending

M = 383*I/y = 383*2466*420 = 397e6N-mm or 397 kN-m
and e = M/P = 3.97 m

The interaction curve is actually a straight line from P = 2400, M = 0 down to P = 0, M = 414. The pure moment value is slightly higher than we had before because it takes into account intermediate bars which the earlier estimate omitted since their contribution was thought to be small.

There is no tension branch and no balanced point. Failure occurs when the bars on the compression side reach yield stress.

When the bars on the compression side reach yield stress.. what is the strain in the tension side.. below yield strain? I can't seem to draw it by means of the strain triangular geometric relationship. The following diagram is wrong.. isn't it?

When there is no concrete.. the balance point won't shift left and right.. and will only stay at the middle?

Or don't you use any strain diagram but based it on the Moment of Inertia? Btw.. since I-beams (wide flange) are pure steel.. don't they also have balanced point? What other steel use the same computations as the concreteless column with just bars?

 

[BAretired]

A balanced point does not apply to steel shapes. The balanced point for a reinforced concrete column is not located anywhere on the column.

The balanced point on an interaction diagram for a concrete column is defined as a strain of 0.003 on the compression face occurring simultaneously with a yield strain on the tensile steel. Geometry will determine where the neutral axis occurs. It is the point of zero strain. It varies according to the yield strength of the steel. A maximum strain of 0.003 is somewhat arbitrarily set and in fact there are some who believe it should be set a little higher.

Ordinarily, for a steel shape the ultimate moment is defined as M = Z.Fy where Z is the plastic modulus. In the above, I have used the relationship M = I.Fy/y which is the same as S.Fy where S is the section modulus. The reason for doing this is that, in order to use the plastic modulus, the strain in the outer bars would have to be three times yield strain or, in the case of 400 MPa (60 ksi) steel, would be in the order of 0.006 which seemed too high.

BA

 

[BAretired]

When the bars on the compression side reach yield stress.. what is the strain in the tension side.. below yield strain? I can't seem to draw it by means of the strain triangular geometric relationship.

Depends on the eccentricity. When e = 0 (P is maximum), the strain on all bars is Fy/E. When e is infinite (P=0) the strain is equal but opposite sign. For all other values of e, the strain is between those limits.

BA

 

[hocho]

A balanced point does not apply to steel shapes. The balanced point for a reinforced concrete column is not located anywhere on the column.

The balanced point on an interaction diagram for a concrete column is defined as a strain of 0.003 on the compression face occurring simultaneously with a yield strain on the tensile steel. Geometry will determine where the neutral axis occurs. It is the point of zero strain. It varies according to the yield strength of the steel. A maximum strain of 0.003 is somewhat arbitrarily set and in fact there are some who believe it should be set a little higher.

"Balanced point" can have 2 meanings. One where in the interaction diagram it is point where axial load and moment fail at same time. When I mentioned "balanced points" in the column. I'm referring to the neutral axis near the middle where it is balancing between the left tension side and right compression forces of both concrete and steel. Here one can describe it as tipping in the balance. But then even if the neutral axis is not on the centroid.. one can say the column is tipping in the balance at the point of the neutral axis between tension and compression.. this is why the resultant is zero.. maybe I should avoid using the terms "Balanced point" in the column but a "point of balance" instead?

Ordinarily, for a steel shape the ultimate moment is defined as M = Z.Fy where Z is the plastic modulus. In the above, I have used the relationship M = I.Fy/y which is the same as S.Fy where S is the section modulus. The reason for doing this is that, in order to use the plastic modulus, the strain in the outer bars would have to be three times yield strain or, in the case of 400 MPa (60 ksi) steel, would be in the order of 0.006 which seemed too high.

But note our concreteless column with rebars at sides are not really I-beams. It's not whole section steel I-beam. Our case is like treating the rebars as creating a force field that around it that created the column. Here is it right to analyze the column as I-beam when there is really no steel connecting the bars in our concreteless column but just bars?

Depends on the eccentricity. When e = 0 (P is maximum), the strain on all bars is Fy/E. When e is infinite (P=0) the strain is equal but opposite sign. For all other values of e, the strain is between those limits.

This is what I've been trying to visualize for weeks. You said lately there is no tension branch in the interaction diagram. Yet when the compression side strain is increasing and yielding.. .. the tension side strain would also increase and yield. So how to reconcile this with the statement there is no tension branch in the interaction diagram. Maybe because of the presence of concrete in the normal case, the compression side has more resistance so it is not yet yielding while the tension side already yielding. But in the concreteless case.. both are yielding? But it seems the tension side would yield first and break before the compressive side. Not other way around. Have you got any drawing software.. can you please draw the strain diagram for the concreteless column where the strain is in between those limits?

Many thanks!

 

[BAretired]

But note our concreteless column with rebars at sides are not really I-beams. It's not whole section steel I-beam. Our case is like treating the rebars as creating a force field that around it that created the column. Here is it right to analyze the column as I-beam when there is really no steel connecting the bars in our concreteless column but just bars?

The bars are surrounded by concrete above and below the void so that they cannot move relative to one another. If the gap was large, the bars could buckle below yield stress, but the gap is not very large, so they would tend to act like short steel columns. Also, the bars are surrounded by epoxy material which would would tend to prevent buckling. We are making the assumption that the bars are capable of carrying the full yield stress before failure. That may or may not be a valid assumption but to me it appears reasonable.

Further to your request, I include a sketch showing four strain diagrams for various combinations of P and M. In Fig. 1 the load is maximum and the moment is zero. The green line indicates an equal compressive strain for each of the four layers of steel.

In Fig. 2, the eccentricity is small and failure takes place when the right hand bar reaches yield. The remaining bars are in compression but not at yield.

In Fig. 3, the eccentricity is increased to the point where the left hand bar is in tension, but well below yield.

In Fig. 4, the eccentricity is infinite, the moment is maximum and the load is zero. The left outside bars reach yield point in tension; simultaneously, the right outside bars reach the yield point in compression. The intermediate bars are strained but only to one third of the yield strain.

I hope this helps.

BA

 

[hocho]

The bars are surrounded by concrete above and below the void so that they cannot move relative to one another. If the gap was large, the bars could buckle below yield stress, but the gap is not very large, so they would tend to act like short steel columns. Also, the bars are surrounded by epoxy material which would would tend to prevent buckling. We are making the assumption that the bars are capable of carrying the full yield stress before failure. That may or may not be a valid assumption but to me it appears reasonable.

Further to your request, I include a sketch showing four strain diagrams for various combinations of P and M. In Fig. 1 the load is maximum and the moment is zero. The green line indicates an equal compressive strain for each of the four layers of steel.

In Fig. 2, the eccentricity is small and failure takes place when the right hand bar reaches yield. The remaining bars are in compression but not at yield.

In Fig. 3, the eccentricity is increased to the point where the left hand bar is in tension, but well below yield.

In Fig. 4, the eccentricity is infinite, the moment is maximum and the load is zero. The left outside bars reach yield point in tension; simultaneously, the right outside bars reach the yield point in compression. The intermediate bars are strained but only to one third of the yield strain.

I hope this helps.

Thanks very much for them. I finally understood what you meant. Do ALL steel columns such as I-beam (wide flange) or even HSS also have the same flat curve? So the reason concrete column has the unique curve with tension branch is simply because the concrete compressive block is giving more capacity to the compression side and also pressing on the tension side and decreasing the press as the axial load decrease below balanced point giving lesser moment capacity.. and this is what the concrete interaction diagram curve is all about.. For engineers who grow up familiar only with steel columns.. the concrete curve would not be a normal one. Generally do steel columns also need interaction diagram even if the curve is flat?

By the way. In Canada.. why is your 300M bars only 300mm.. elsewhere it's 314m (area of 20mm diameter).. also why is your yield stress 400MPA. Normal is 414MPA.. why didn't you use the value 414MPA? Just for ease of calculations? Actual yield strain of the bars are tested at 470MPA.. so inputting them in excel.. when the axial load is 960K.. moment is 328kN-m capacity instead of 248kN.. with increase of 80kN.

 

[BAretired]

Thanks very much for them. I finally understood what you meant. Do ALL steel columns such as I-beam (wide flange) or even HSS also have the same flat curve?

They would if capacity was calculated in the same way. But for steel columns, the ultimate capacity is calculated using the plastic modulus; this means that strains are not artificially limited to 0.003 or any such number. Also, in the case of steel I-beams or wide flange columns, the slenderness plays a much larger role than it does for stocky concrete sections. In many cases, we cannot come close to the yield point before buckling occurs. Interaction diagrams are normally not constructed for steel columns.

So the reason concrete column has the unique curve with tension branch is simply because the concrete compressive block is giving more capacity to the compression side and also pressing on the tension side and decreasing the press as the axial load decrease below balanced point giving lesser moment capacity.. and this is what the concrete interaction diagram curve is all about.. For engineers who grow up familiar only with steel columns.. the concrete curve would not be a normal one. Generally do steel columns also need interaction diagram even if the curve is flat?

The unusual shape of an interaction diagram is largely due to the properties of concrete; concrete does not have a well defined yield point, cannot resist tensile stress and cannot be compressed beyond a relatively small strain. In short, it is not an elastic material and that gives rise to the strange shape of the interaction diagram. Steel columns do not normally require interaction diagrams.

By the way. In Canada.. why is your 300M bars only 300mm.. elsewhere it's 314m (area of 20mm diameter).. also why is your yield stress 400MPA. Normal is 414MPA.. why didn't you use the value 414MPA? Just for ease of calculations? Actual yield strain of the bars are tested at 470MPA.. so inputting them in excel.. when the axial load is 960K.. moment is 328kN-m capacity instead of 248kN.. with increase of 80kN.

The bars in Canada were made slightly different than the nominal dimension, some larger, some smaller so that the area could be easily remembered, 200 mm² for 15M, 300 for 20M, 500 for 25M etc.

Before we switched to metric sizes, the yield was 60,000 psi which is 414 MPa. Most local fabricators provide a minimum guaranteed yield strength of 60,000 psi today but they are not legally required to do so; the specified yield for 400 Grade steel is 400 MPa so that is what Canadian engineers use in design. It provides a small increase in safety factor but the cost difference is not too significant.

BA

 

[hocho]

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 wa^2/2 per floor and the moment is approximately wa^3/24 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. What part of structural books did you get the formulas above. I was asking for edge column moment estimate. I couldn't find the formulas in the column sections in the books. But computing for your formulas.. the output doesn't look right:

unit floor load w = 100 psf pounds per square feet
a = 20 feet
axial load per floor = wa^2/2 = 100*20^2/2 = 20,000 kips (or 20,000 x 4.448 = 88,960 Kilonewton
moment shared by 2 columns = wa^3/24 = 100*20^3/24 = 33333 kips-ft (or about 33333 x 4.448 / 3.28 = 45,203 kN-m.

something is wrong.. values too much.

Anyway. You mentioned the M/P ratio or eccentricity is lower in the ground floor.. This is because the moments in the column above can make it straight and less tendency to bend versus when the column above can translate in any direction as in top floor column.

The bottom line of what it seems to be saying is.. moments from unbalanced loading in edge columns in ground floor are not as high as top floor. Therefore it is the seismic moments induced in the ground floor edge column that would dominate than edge column gravity moments, isn't it.

 

[BAretired]

wa^2/2 = 20,000# (not kips) or 89 kN

The M/P ratio is lower in the ground floor because P is larger; it is the sum of all the upper story loads. M from unbalanced loading is approximately equal in all floors. In most tall buildings, lateral forces are resisted by shear walls, elevator shafts or moment resisting frames, not by simple bending of columns.

We are fortunate in Alberta that we don't get significant seismic events. We don't get typhoons either but we do design for some wind forces. In the Philippines, I would expect lateral forces resulting from seismic events and wind load to be much more of a concern than gravity load.

BA

 

[hocho]

wa^2/2 = 20,000# (not kips) or 89 kN

The M/P ratio is lower in the ground floor because P is larger; it is the sum of all the upper story loads. M from unbalanced loading is approximately equal in all floors. In most tall buildings, lateral forces are resisted by shear walls, elevator shafts or moment resisting frames, not by simple bending of columns.

We are fortunate in Alberta that we don't get significant seismic events. We don't get typhoons either but we do design for some wind forces. In the Philippines, I would expect lateral forces resulting from seismic events and wind load to be much more of a concern than gravity load.

Ok. Thanks for all assistance. To summarize and for my conclusion of it all after 2 years of trying to understand the behavior of epoxy repair in column voids.

Instead of the designed 4 storey with concrete roof and all concrete walls. We would just make it 3 storey with very light roof and very light wall. That's like reducing it by 2 solid storeys (building only half the intended floor). This should produce lower seismic base shear. This should hopefully put the seismic load to within 890kN axial load capacity of the bars only column section and 328kN-m moment capacity when load is directly over the compression bars (when seismic moment is more.. then axial capacity would be less.. that's where the epoxy may help (see below)). Edge column already used 33kN from the unbalanced load, there would be 297 kN-m capacity left of the moments in the pure reinforcement column alone. If you'll add the contribution of epoxy. Then one has to use the interaction diagram formula that includes the bars and epoxy acting together, right? It's then capacity of 863 kN axial load and 488kN-m moment capacity at balanced point with eccentricity of 565mm. Improving from the sole bars only column capacity of 960kN and 328 kN-M when load is directly at compression bars with eccentricity of 0.258 meters.

reviewing the computations of the interaction diagram of the epoxy.

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. 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 (psi) 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 in addition to the bars.. 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.

Notice that at the balanced point of the epoxy or even normal concrete column. The tension bars and compression bars cancel out in the formula of axial load which is Pn= 0.85 fc' ab + As'fs' - As fs.
This means at eccentricity of 565mm (see 2 paragraph above). The compression bars axial capacity really cancel out to the tension bars? This is the part that still confusing me a bit when I tried to reconcile both the contribution of bars only and epoxy. Does it mean if there is no epoxy or concrete, the behavior of the bars only and contribution of the bars at compressive side to axial load is different than with either concrete or epoxy? Or if same.. it means at eccentricity of 565mm.. there is almost no axial capacity of the bars? Or is it because the bars only column can't reach balanced point that it won't happen? Meaning before the tension side yield.. the compression side is already yielded and even rupturing later on. Just to confirm this one.

Thanks a lot again BA!

 

[BAretired]

I don't think a Whitney stress block is appropriate for epoxy. I don't know its properties but I'm guessing that the material acts elastically throughout the range of strains under consideration. That would mean the stress variation in the epoxy is triangular in shape with a zero value at the neutral axis and a maximum at the compression face of column.

BA

 

[hocho]

I don't think a Whitney stress block is appropriate for epoxy. I don't know its properties but I'm guessing that the material acts elastically throughout the range of strains under consideration. That would mean the stress variation in the epoxy is triangular in shape with a zero value at the neutral axis and a maximum at the compression face of column.

BA

Ok. For years I always just wanted to focus on concrete structures and not steel structures to avoid more understanding and spending time studying it. But this epoxy repair in void is finally making me think of steel especially when you said the bars in compression can act like steel. After spending some time today on it (til my eyes strain reading many materials). Just some quick question about Section Modulus and Moment of Inertia (I know their basics but just verifying my understanding). I know it is one properties that concrete structures don't have or focus. My questions is (don't worry I won't drag it long just the following part). For example. The steel

from

http://www.engineeringtoolbox.com/american-wide-flange-steel-beams-d_1319.html

W12x30 has Moment of Inertia of 38.6 and Section Modulus of 6.2
W16x26 has Moment inertia of 38.4 and section modulus of 3.5.
(notice the shorter W12X30 has slightly bigger moment of inertia and bigger section modulus.
Using L/25 for W16x26 and say 10 meters span
If L/25 = 10 meters /25 = 400mm (16")
So the depth of W16x26 is optimal for 10 meters span for less deflection. But the shorter W12x30 has less depth and yet it's moment of inertia and section modulus is larger than the W16x26. Does it mean when put in a 10 meter span. The W12x30 would be as good or even better than the W16x26 because of bigger section modulus and slightly bigger moment of inertia too? Would they deflect about the same at maximum load when put in identical 10 meter span? Since the Moment of Inertia is almost the same, or would the smaller W12x30 deflect bit less since it has higher moment of inertia and section modulus too than the W16x26 under the same 10 meters span? Just this important question. I know I'll read the rest of concept in text books in weeks ahead but need to know just this one now. Thanks.

 

[BAretired]

W12x30 has Moment of Inertia of 38.6 and Section Modulus of 6.2
W16x26 has Moment inertia of 38.4 and section modulus of 3.5.

Size------------- d - - - b - - - t - - - - Area - - W- - - Ix- - - Iy- - Wx- - Wy

W 16 x 26 -- 15.69 -- 5.5 -- 0.250 -- 7.68 -- 26 -- 301 -- 9.6 -- 38.4 -- 3.5

W 12 x 30 -- 12.34 -- 6.52 -- 0.26 -- 8.8 - - 30 -- 238 -- 20.3 -- 38.6 -- 6.2

I copied two lines of the table you provided. The justification leaves something to be desired, but you can see that Ix is about 26% greater for the W16 than the W12. The strength of the two shapes is almost identical shown by Wx (section modulus). In Canada, we normally use the symbol Sx to denote section modulus about the X axis.

The beams will carry nearly the same moment if they are continuously braced against lateral torsional buckling (LTB). If they are unbraced against LTB, the W12x30 will do much better, particularly for longer spans.

Deflection is inversely proportional to the moment of inertia Ix, so the W16x26 will deflect 79% of the W12x30 deflection.


BA

 

[hocho]

Size------------- d - - - b - - - t - - - - Area - - W- - - Ix- - - Iy- - Wx- - Wy

W 16 x 26 -- 15.69 -- 5.5 -- 0.250 -- 7.68 -- 26 -- 301 -- 9.6 -- 38.4 -- 3.5

W 12 x 30 -- 12.34 -- 6.52 -- 0.26 -- 8.8 - - 30 -- 238 -- 20.3 -- 38.6 -- 6.2

I copied two lines of the table you provided. The justification leaves something to be desired, but you can see that Ix is about 26% greater for the W16 than the W12. The strength of the two shapes is almost identical shown by Wx (section modulus). In Canada, we normally use the symbol Sx to denote section modulus about the X axis.

The beams will carry nearly the same moment if they are continuously braced against lateral torsional buckling (LTB). If they are unbraced against LTB, the W12x30 will do much better, particularly for longer spans.

Deflection is inversely proportional to the moment of inertia Ix, so the W16x26 will deflect 79% of the W12x30 deflection.

Size----------------- d - - - b - - - t - - - - Area - - W- - - Ix- - - Iy- - Wx- - Wy

W 14 x 22 ----13.74 --- 5 ------ 0.230 -- 6.49 -- 22 -- 199 -- 7 -- 29.0 -- 2.8

W 10 x 45 ---10.10 -- 8.020 -- 0.350 -- 13.3 -- 45 -- 248 -- 53.4 -- 49.1 -- 13.3

In the above example. The shorter w10x45 has both bigger section modulus and moment of inertia compared to the deeper w 14 x 22.

1) first question, why manufacturer a deeper wide flange when the moment of inertia and section modulus are smaller than a less deep one? loading requirement?

2) from say L/25... if L=350" or 29.166 feet.. then w14x22 fulfill it because L/25 = 350/25 = 14". Now my question is. What if your use the w10x45 on the same 29.166 feet span. What would deflect more.. the w14x22 or the w10x45? I'm asking because I want to know if its possible the w10x45 can deflect less even if its only for 250" or 20.83 feet if its use on the 29.166 feet span used by the w14x22?

2)If the answer is yes to question 2.. and i'm expecting it (please confirm).. can you consider the span to depth ratio deflection some kind of tension stress? this deflection is supposed or calculated to occur at maximum load isn't it. So if you load is just one half.. you can use a smaller size wide flange even if it's only half the required span/depth ratio?

 

[BAretired]

Size----------------- d - - - b - - - t - - - - Area - - W- - - Ix- - - Iy- - Wx- - Wy

W 14 x 22 -----3.74 --- 5 ------ 0.230 -- 6.49 -- 22 -- 199 -- 7 -- 29.0 -- 2.8

W 10 x 45 ---10.10 -- 8.020 -- 0.350 -- 13.3 -- 45 -- 248 -- 53.4 -- 49.1 -- 13.3

In the above example. The shorter w10x45 has both bigger section modulus and moment of inertia compared to the deeper w 14 x 22.

1) first question, why manufacturer a deeper wide flange when the moment of inertia and section modulus are smaller than a less deep one? loading requirement?

A W14x22 is only half the weight of a W10x45 so it is much cheaper as a beam. So the reason is economy. The W10x45 is not usually used as a beam although it can be in situations where clearance is an issue. It is more suitable as a column because of its much larger Iy value.

2) from say L/25... if L=350" or 29.166 feet.. then w14x22 fulfill it because L/25 = 350/25 = 14". Now my question is. What if your use the w10x45 on the same 29.166 feet span. What would deflect more.. the w14x22 or the w10x45? I'm asking because I want to know if its possible the w10x45 can deflect less even if its only for 250" or 20.83 feet if its use on the 29.166 feet span used by the w14x22?

The W10x45 would deflect less, actually 80% of the deflection of the W14x22 but would cost twice as much in material.

2)If the answer is yes to question 2.. and i'm expecting it (please confirm).. can you consider the span to depth ratio deflection some kind of tension stress? this deflection is supposed or calculated to occur at maximum load isn't it. So if you load is just one half.. you can use a smaller size wide flange even if it's only half the required span/depth ratio?

The span to depth ratio is not a tension stress or any other kind of stress. Economical design requires that a beam having the least weight be selected provided it satisfies all of the requirements, namely strength and deflection. The span to depth ratio is a measure of deflection when the beam is stressed to a particular stress under service loads.

BA

 

[hocho]

I don't think a Whitney stress block is appropriate for epoxy. I don't know its properties but I'm guessing that the material acts elastically throughout the range of strains under consideration. That would mean the stress variation in the epoxy is triangular in shape with a zero value at the neutral axis and a maximum at the compression face of column.

What is the resultant then of this epoxy compressive block? In the concrete one, the resultant is the average of the concrete compressive block (the C).. or you mean one must not even use resultant in the epoxy block? Or how to compute for it.. how do you handle this say the stress variation in the epoxy is really triangular in shape.. For concrete, it's like more of slight sloped square? I'm trying to compute for this epoxy axial and moment contribution. Thanks.

 

[BAretired]

What you can do is construct a transformed section where the epoxy is transformed to the equivalent of a steel section. This is done by taking the ratio of the modulus of elasticity for the two materials. If we assume that E = 29,000,000 psi for steel and 450,000 psi for epoxy, the modular ratio is 64.4. A 20" width of epoxy is considered equivalent to a strip of steel with a width of 20/64.4 = 0.31" and a dimension of 10" in the other direction starting at the middle of the column and ending at the edge of where the concrete should be if it were present.

We now have an unsymmetrical section consisting of 20 bars located as previously discussed plus an eccentrically located 10" x 0.31" steel plate. The center of gravity of the transformed section, the transformed area, the transformed moment of inertia and the transformed section modulus can all be calculated. With those properties, P and M can be calculated for various eccentricities and an interaction diagram can be drawn.

The maximum stress in the epoxy is the maximum stress in the steel plate divided by the modular ratio, 64.4. The compression attributed to the epoxy is the average pressure multiplied by the area of epoxy. It will be located at the center of gravity of the stress distribution which is easily calculated because the stress distribution is linear.

BA

 

[hocho]

BA.. I'd not dig thru slab and remove the sleeve as it seems fixed and it's not possible with tenant on top of it. So I planned to just remove the waterproof topping above second floor where a roof will be put to lessen the load opposite the epoxy void to avoid compression in the lower S shape of the moment. While trying out test column and beams in Etabs to see how lessening the loading in the second floor would affect the ground floor. I noticed something unexpected. If the 2nd floor has no load or fewer load. The moments in the ground floor is larger. See:

If there is load in the 2nd floor top. The moments in the ground floor got lower (only 2.75kN in the lower left.. notice this is just test column and beams and not my building etab file which is with my designer). See:

Notice the moment in the lower left (in ground floor) decrease from 6.26kN to 2.75kN (when load of 2nd floor top is added). This is really true.. ah? what is this principle called... Equilibrium? Where in the book is this mentioned. Because if true it means if I removed the waterproof topping above the second floor. It would do the opposite of what I intended.. it would increase the moments in the ground floor.. so maybe I shouldn't touch it and leave it as it is. Hmm....

Any profound thing about this with regards to Neutral Axis?