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Shear Behavior at Midspan of RC Beams 1

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p8psk

Structural
Jun 28, 2018
24
In shear and moment diagrams of reinforced concrete beams... you have continuous shear from midspan to support as in...

RXqaYa.png


in actual...

TndEC2.jpg


You see, in midspan.. there is more strength in the beam... and even if shear forms.. it would be more vertical or more of flexural failure... therefore can you put stirrups starting D (equal to depth) away from midspan for concentrated load?

If you could.. then why does the shear and moment diagram have continuous shear from midspan to support yet in actual the diagonal shear only forms at the middle of them.. how do you explain difference in the behavior?

Thank you.
 
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p8psk said:
It's not a moot point since thousands of buildings have been built with minimal stirrups at midspan. They use the generic stirrup spacings of 1 rebar at 2" from support, 7 rebars at 4", 7 rebars at 6", and rest is one rebar for every 10" spacing irregardless of beam sizes. Thei reasoning being the D/4 is satisfied near support. So when it reaches the middle.. there is only spacing of 10" even for concentrated midspan load. I saw structural plans of many buildings this way and seldom do they use uniform stirrups to match the shear force diagram.

p8psk said:
For concentrated load at midspan and constant shear diagram from support to midspan.. uniform stirrup is needed. No problem about it. But some just didn't follow it reasoning that no shear failure could occur at midspan and it's flexural failure only.

The case you are describing - a single concentrated load on a beam - is uncommon. That's why you might not see the stirrups spaced according to that sort of shear force diagram on the plans you've seen. Most building beams, which are built for people rather than heavy equipment, are governed by distributed loads. The weight of the beam itself is distributed; the weight of the slab (or other floor system) supported by the beam is distributed; the live load varies but is enveloped for simplicity by a uniform load allowance.

Design codes do consider a concentrated load but that is for robustness of small elements. Consider a beam 8m long supporting 8m width of floor. The distributed live load on that beam would be more than 150kN in most cases, compared with a concentrated design load of 5 or 10kN (order of magnitude). Then add in 300+kN of (distributed) dead load and it's clear that the governing loads are distributed.

As BAretired mentioned, there is a minimum requirement for shear reinforcement. If the beam was sized to control deflection, it may be that the minimum is used along the whole length of the beam (ie no variation), but the governing load was still distributed.


Going back to your first question about whether stirrups are needed right at the location of the midspan concentrated load, I suggest reading about 'truss analogy' and strut-and-tie design. These explain how stirrup requirements (and longitudinal reinforcement) are related to shear force and bending moment diagrams.
 
The case you are describing - a single concentrated load on a beam - is uncommon. That's why you might not see the stirrups spaced according to that sort of shear force diagram on the plans you've seen. Most building beams, which are built for people rather than heavy equipment, are governed by distributed loads. The weight of the beam itself is distributed; the weight of the slab (or other floor system) supported by the beam is distributed; the live load varies but is enveloped for simplicity by a uniform load allowance.

Design codes do consider a concentrated load but that is for robustness of small elements. Consider a beam 8m long supporting 8m width of floor. The distributed live load on that beam would be more than 150kN in most cases, compared with a concentrated design load of 5 or 10kN (order of magnitude). Then add in 300+kN of (distributed) dead load and it's clear that the governing loads are distributed.

As BAretired mentioned, there is a minimum requirement for shear reinforcement. If the beam was sized to control deflection, it may be that the minimum is used along the whole length of the beam (ie no variation), but the governing load was still distributed.


Going back to your first question about whether stirrups are needed right at the location of the midspan concentrated load, I suggest reading about 'truss analogy' and strut-and-tie design. These explain how stirrup requirements (and longitudinal reinforcement) are related to shear force and bending moment diagrams.

A single concentrated load on the beam occurs when secondary beam frames into it at midspan. For example.

R38vGo.jpg


This is the common stirrups spacing I saw in structural designers drawing. Many designers neglect to use uniform stirrups spacing which I will surely do in any design. In evaluating those made in the above manner. I was wondering what would happen if cracks form near midspan between the stirrups very steep that falls in the red mark (small horizontal line either side of midspan). Is it shear or flexural cracks? BA convinced me it can be considered flexural crack (crack that occurs with the red mark bottom to top). However if you can argue it is considered shear crack.. please give your reason. Thanks.
 
p8psk,

1) Hairline cracks are often difficult to see in a beam. You may not notice them just walking through the building but that does not mean they are not there. Stirrups cannot provide Vs unless they are stressed and if they are stressed, they will elongate. If the stirrups elongate, the concrete in contact with them will crack. That is a very basic fact about reinforced concrete. I'm not sure why we are arguing about it.

2) The shear force for the 6m long girder above is not constant. The load consists of a concentrated load at midspan plus a uniform load for the self weight of the girder plus the dead and live load of an unknown width of slab bearing on the girder. For that load distribution, the stirrup spacing appears reasonable.

BA
 
p8psk,

1) Hairline cracks are often difficult to see in a beam. You may not notice them just walking through the building but that does not mean they are not there. Stirrups cannot provide Vs unless they are stressed and if they are stressed, they will elongate. If the stirrups elongate, the concrete in contact with them will crack. That is a very basic fact about reinforced concrete. I'm not sure why we are arguing about it.

2) The shear force for the 6m long girder above is not constant. The load consists of a concentrated load at midspan plus a uniform load for the self weight of the girder plus the dead and live load of an unknown width of slab bearing on the girder. For that load distribution, the stirrup spacing appears reasonable.

I'm not arguing about why there are cracks.. but arguing that these should be common yet even when I climb up the beams in many buildings.. i can't see any cracks.. even hairline.. according to Dolan in page 75:

"If the section were to remain uncracked, the tensile stress in the concrete would now be twice its previous value, that is, 864 psi. Since this exceeds by far the modulus of rupture of the given concrete (475 psi), cracks will have formed and the analysis must be adapted consistent with Fig 3.5. Equation (3.5), with the known quantities b,n, and As inserted, gives the distance to the neutral axis kd= 7.6 in., or k= 7.6/23 = 0.33. From Eq. (3.13), j=1 - 0.33/3= 0.89. With these values the steel stress is obtained from Eq. (3.8) as fs = 22,300 psi, and the maximum concrete stress from Eq. (3.10) as fc= 1390 psi."

The above meant that even with rebars, the modulus of rupture of concrete would still be approx 475 psi. Therefore at very low loads corresponding to stress just above 2870 psi in the tension bars, the concrete at tension should already crack.. yet majority of buildings didn't have any flexural cracks.. Does it mean their loads are so very low that they don't exceed the concrete tension modulus of rupture of 475 psi (or corresponding steel stress/strain of 2870 psi)? But how could this be when it's carrying traffic and many loads above.

Of course I understood when crack occurs, the tension bars take over until it reaches nomimal moment. No problem about this.
 
The attached PDF file by the American Society of Concrete Contractors discusses cracks in structural concrete. The article confirms that most concrete members crack under service load and the crack width increases with time.

BA
 
 https://files.engineering.com/getfile.aspx?folder=15391e8e-966c-48d4-b289-7ccc2ca7d229&file=Cracks_in_Structural_Concrete.pdf
Concrete cracks if placed in tension. One object of reinforcement is to control the width of the cracks, with the amount of required control dependent on the type of exposure of the structure. If you can't see the cracks, the reinforcement is doing its job.
 
Concrete cracks if placed in tension. One object of reinforcement is to control the width of the cracks, with the amount of required control dependent on the type of exposure of the structure. If you can't see the cracks, the reinforcement is doing its job.

I need figures. You mean after concrete reaches the modulus of rupture of about 475 psi, there would be no crack (even hairline) if the reinforcement is doing its job?

I think the right way to think is after reaching the modulus of rupture, there would be hairline crack that won't widen because the reinforcement is doing its job.

But again, in most buildings and malls. Even if you climb up the beam and inspect. There is not even hairline crack. The mystery is how does this occur where the modulus of rupture of 475 psi is reached, yet no crack?? unless they haven't reach the modulus of rupture? But how you design beam that has service load yet doesn't reach the modulus of rupture??

Please reply using figures and technical terms like the above and not general description which can be vague or prone to misinterpretation. Thank you.
 
No, thank you. I will leave your investigation and numbers to you and your mentors. Vague? Maybe to you, not to me.

But back to your beam with a point load, which now we learn is a secondary beam at the same level. The load needs to be delivered to the top of the supporting beam, so the stirrups at that point are hanging reinforcement.
 
No, thank you. I will leave your investigation and numbers to you and your mentors. Vague? Maybe to you, not to me.

But back to your beam with a point load, which now we learn is a secondary beam at the same level. The load needs to be delivered to the top of the supporting beam, so the stirrups at that point are hanging reinforcement.

But according to Macgregor.. the provision is waived if the shear, Vu2, at the end of the supported beam is less than 3 sqrt (fc bw2 dw)... quoting:

"The additional hanger reinforcement,Ah, is placed in the supporting beam to intercept
45° planes starting on the shear interface at one-quarter of the depth of the supported
beam, above its bottom face and spreading down into the supporting beam,as shown by
the 45 degrees dashed lines in Figs. 6-41a and 6-42a . These provisions can be waived if the shear,Vu2, at the end of the supported beam is less than 3 sqrt (fc)* bw2* dw) because inclined cracking is not fully developed for this value of shear force."

So if Vu2 is below 3 sqrt (fc)* bw2* dw) then the load no longer needs to be delivered to the top of the supporting beam, right? Or are the 2 cases separate?
 
I am not going to try to interpret the Macgregor text in that regard, but maintain that provision of hanging reinforcement is required. Good detailing practice sometimes overrules the niceties of specific design provisions.
 
If no compression fan forms in the supported beam.. then the loading would be uniform top to bottom of the supporting beam? Can others confirm what is the case?

According to Macgregor. This occurs due to the compression fan. (?)

zfpkzZ.jpg
 
I think the following solves the puzzles of why in spite of ACI lack of provision for hanger stirrups.. millions of beams don't come crushing down...

See illustration:

q3bDaP.jpg


This thread shows that for inclined crack that forms vertically or very steep within 'd' from midspan (within the red lines/region) where concentrated load or secondary beam frames into primary girder.. it is still flexural crack and not shear cracks.. therefore no stirrups needed at midspan and 'd' away from it. However all structural drawings either have uniform stirrups in the main girder or ending as "rest of stirrups 6 inches". This means those stirrups that end up in the midspan indirectly functions as hanger stirrups.
ACI not requiring omission of midspan stirrups saves the day because it indirectly serves the function of hanger stirrups! (at least for moderate load.. but for very huge secondary beam load.. the uniform stirrups spacing at 4" throughout saves the day)

All agree?
 
No, I don't. In my opinion, what saves the day, if hanging reinforcement is not provided in the girder, is that the structure chooses another load path. The stirrups in the secondary beam act as hangers, transferring the load up to the top of the beam and slab and then into the girder. Maybe just another way of looking at it. But that only works if the slab is of decent thickness.
 
No, I don't. In my opinion, what saves the day, if hanging reinforcement is not provided in the girder, is that the structure chooses another load path. The stirrups in the secondary beam act as hangers, transferring the load up to the top of the beam and slab and then into the girder. Maybe just another way of looking at it. But that only works if the slab is of decent thickness.

For girder with concentrated load such as secondary beam framing into it. Are you saying that some designers provide close uniform stirrups but only up to:

TjiVgu.jpg


I know the following is what should be done providing 4" spacing of stirrups throughout the entire girder and I will always do it (no problem about it or not arguing against it):

WvWl1C.jpg


I just want to know if some designers really do the former in the US or Canada. Or all automatically use the latter (which is how it should be done).
 
Not sure what some people do, as I have seen a lot of strange things. Your second picture looks like what I do, so we agree on that.
 
Here is another thread on the same topic:
thread507-347413

In order to omit stirrups for distance 'd' each side of point load, the load must be positioned on top of the girder.

In the case of a secondary beam framing in to the side of a girder, perhaps a hanger designed to carry the full reaction should be used to ensure that the load is on top of the girder. So far as I am aware, that is not common practice. Normal practice has been to run stirrups through as shown on the second diagram above. Perhaps we need to discuss this topic further.

BA
 
In order to omit stirrups for distance 'd' each side of point load, the load must be positioned on top of the girder.

In the case of a secondary beam framing in to the side of a girder, perhaps a hanger designed to carry the full reaction should be used to ensure that the load is on top of the girder. So far as I am aware, that is not common practice. Normal practice has been to run stirrups through as shown on the second diagram above. Perhaps we need to discuss this topic further.

But if you used the second diagram which is the standard or normal practice...

WvWl1C.jpg


Then all such primary-secondary beam joint already automatically have hanger reinforcement even if the designer didn't originally intent it.... is it not?
 
anyway, the most important question we must answer is...

if the secondary beam doesn't go beyond a certain Vu shear loading and there is no compression fan... is the secondary beam loading still towards the bottom portion of the girder and hanger still required? Most of the books says it is a certainty only if the secondary beam has crack and compression fan.. but what if there is no crack.

If in spite of lack of crack and a side loaded beam is treated as a bottom loading as most of the shear comes in at or near the bottom even no crack. Then most authors are wrong.

Hope we can tackle and answer this once and for all.
 

p8psk said:
Then all such primary-secondary beam joint already automatically have hanger reinforcement even if the designer didn't originally intent it.... is it not?

Well yes, it has some hanger reinforcement but Vs does not provide enough to carry the entire reaction. Vc is still required in addition to Vs. If the secondary beams were suspended below the girder, a hanger would be required to carry the full reaction.

BA
 
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