Shear Behavior at Midspan of RC Beams 1

[p8psk]

In shear and moment diagrams of reinforced concrete beams... you have continuous shear from midspan to support as in...

in actual...

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.

 

[BAretired]

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?

Yes, at midspan, the cracks tend to form vertically from the bottom and terminate at the underside of the compression block. The compression block resists shear near midspan. I believe you could start stirrups D away from midspan if the concentrated load was the only load acting and if it meant a significant cost saving. However, the self weight of the beam negates that assumption and the cost saving is minimal.

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?

You mean diagonal tension, not diagonal shear. An elemental cube subjected to pure vertical shear requires horizontal shear top and bottom in order to maintain equilibrium. The resultant stress, called the principal stress, acts at 45^o to the horizontal (diagonal tension). Diagonal tensile cracks may appear but are resisted by vertical or sloping stirrups and by bent bars.

At midspan, the moment cracks the concrete in the bottom of the beam and puts tension in the bottom reinforcement. It compresses the concrete in the top of the beam; the compression block carries most if not all of the shear.

This is quite a different behavior than that of a beam made of isotropic material.





BA

 

[p8psk]

Thanks BAretired...

Supposed you had a beam with diagonal crack right at midspan (either purposedly built that way in lab to study the behavior or it just happens). See illustration:

Would the diagonal compression block at the top of the midspan still work like a beam without crack? Like a normal tension crack with compression block engaged?

I understood that if the diagonal crack occurs at the sides.. or bonafide shear failure.. the longitudinal bar below would yield and the crack would widen or the compression block would crash.. But in the case of the diagonal crack forming right at midspan.. how would the behavior differs to one forming at the sides?

 

[BAretired]

A smooth diagonal control joint formed on purpose at midspan might behave differently than a natural crack as slip along the joint could occur.

A natural diagonal crack at midspan would be unusual. It could be caused by shrinkage or temperature strain when end supports are restrained against horizontal movement. In that case, aggregate interlock would tend to make the beam behave as if uncracked.

I believe there have been instances involving very long continuous concrete beams with rigid end restraints where shear failure occurred unexpectedly at very low shear stress but I was unable to find examples on the internet.

BA

 

[dik]

Your stirrup spacing does not match your SFD... expect to have uniformly spaced stirrups across the beam. Often beams are loaded with dual symmetric loads to provide a zero shear area at mid span, and pure flexure.

A smooth 'pour joint' does not provide for the aggregate interlock that BA mentions.

Dik

 

[hokie66]

Construction joints are often placed at midspan or wherever the minimum shear is judged to be. But the joints should not be inclined. Because of uncertainty and conservatism, my practice has been to include some big dowels at middepth of beam construction joints.

 

[dik]

Hokie... different practice... at midspan, only with keys...

Dik

 

[hokie66]

Yeah, I never use keys in suspended floors. Or floors on the ground, for that matter. Hate them.

 

[p8psk]

The following beam is just example from the internet. I'd like to know at what angles (or approximately the number in blue) of the inclined crack at midspan can it be considered or behaved already as a shear diagonal crack and not flexural crack?

 

[jayrod12]

Here's the thing with reinforced concrete. Generally the typical proportions lead to bending being the origin of failure for loading near the beam centre, regardless of magnitude of load. Once the beam gets sufficiently short enough to be shear governed, this discussion is almost a moot point since it's the cost of a single or maybe 2 stirrups.

 

[p8psk]

Here's the thing with reinforced concrete. Generally the typical proportions lead to bending being the origin of failure for loading near the beam centre, regardless of magnitude of load. Once the beam gets sufficiently short enough to be shear governed, this discussion is almost a moot point since it's the cost of a single or maybe 2 stirrups.

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. This is why I want to understand the behavior of possible inclined cracks at midspan . Like what angles or steepness can you say it's already shear crack vs just flexural crack. Again:

 

[dik]

hokie... not for slabs, but for beams... usually face centred type...

Dik

 

[BAretired]

p8psk,

It is common practice to space stirrups to match the shear force diagram. The shear strength of a beam is composed of two parts, the contribution of the concrete, Vc and the contribution of the stirrups, Vs. With some exceptions, a minimum area and maximum spacing of stirrups is required where the factored shear, Vf exceeds Vc/2. Beyond that, stirrups could be omitted by code, although many engineers simply provide stirrups throughout the span for good measure.

There is no reason why stirrup spacing should vary as you indicated if Vf is constant or nearly constant from support to midspan. In that case, stirrup spacing would be constant throughout the beam. Variable stirrup spacing is frequently found when the load is more or less evenly distributed.

BA

 

[p8psk]

p8psk,

It is common practice to space stirrups to match the shear force diagram. The shear strength of a beam is composed of two parts, the contribution of the concrete, Vc and the contribution of the stirrups, Vs. With some exceptions, a minimum area and maximum spacing of stirrups is required where the factored shear, Vf exceeds Vc/2. Beyond that, stirrups could be omitted by code, although many engineers simply provide stirrups throughout the span for good measure.

There is no reason why stirrup spacing should vary as you indicated if Vf is constant or nearly constant from support to midspan. In that case, stirrup spacing would be constant throughout the beam. Variable stirrup spacing is frequently found when the load is more or less evenly distributed.
BA

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. I was asking what degree or angles of inclined cracks at midspan before you could consider it as a shear failure and not flexural failure (please see illustration above and kindly address directly the concern). If it's always flexural failure for any steep cracks at midspan, then those who didn't use uniform uniforms for constant shear can't be blamed.. is it not. Of course in my design i'll use uniform stirrups, no problem about it. But I'm just concerned of those who didn't.

 

[BAretired]

In your illustration above, the depth of beam is about 1/6 of the span. That is considerably deeper than most beams in practice. You are assuming that there is some magical angle which determines whether a failure is a shear failure or flexural failure. You are also assuming that cracks will be straight lines pointing toward the concentrated load. And you seem to be assuming that cracks in the concrete represent failure of the beam. None of these assumptions is accurate, so your question has no single answer.

The photograph courtesy of J. G. MacGregor which you included in an earlier post is clearly not showing a beam loaded at midspan. It appears to be a partial photo of a beam with two (possibly symmetrical) concentrated loads. The portion to the right is called the shear span; the portion to the left lies between the two applied loads. There is another shear span to the left of the load not shown.

In your case, the shear span is the distance from the concentrated load to the support. Again, there are two shear spans. Cracks may be caused by flexure, diagonal tension or shear/flexure. Vertical cracks are caused by flexure. Sloping cracks are caused by diagonal tension or shear/flexure. As may be expected, cracks occur approximately normal to the principal tensile stress.

The bottom reinforcement must be adequate to resist the moment at all points on the beam. The maximum moment on your beam is PL/4. Some engineers might calculate the area of steel needed at midspan and run it full length. Others may vary the steel area because the moment is not constant across the span; it is only PL/8 at the quarter point.

The shear is constant over the entire span. If stirrups are needed at 4" o/c near the supports, then stirrups are needed at 4"o/c throughout the span with the possible exception of distance 'd' each side of the concentrated load. To use a 6" or 10" spacing in the central region would not be sufficient shear reinforcement.

BA

 

[p8psk]

The shear is constant over the entire span. If stirrups are needed at 4" o/c near the supports, then stirrups are needed at 4"o/c throughout the span with the possible exception of distance 'd' each side of the concentrated load. To use a 6" or 10" spacing in the central region would not be sufficient shear reinforcement.
BA

Thanks BA. In short. Any crack that occurs *within* (meaning from bottom to up even inclined inside) 'd' each side of the concentrated load is flexural crack, right? Illustration:

The red inclined crack occurs within 'd' from midspan concentrated load.. so can we say any crack within 'd' is always flexural crack and not shear crack?

 

[BAretired]

Sounds right to me.

BA

 

[p8psk]

Sounds right to me.

Thanks BA.

In the case of regular inclined shear cracks at the sides.
I want to ask about design methodology.

Supposed your factored shear forces (1.2 DL + 1.6 LL) is 20 kips (or about 90 kN). Should you find concrete sizes that will have Vc that completely enclose it.. that is.. finding concrete dimension with computed Vc of say 25 kips or 100 kN? Or do you make the concrete sizes smaller such that for example Vc is only half or 10 kips (only 45 kN) and the rest taken up by stirrups?

If the latter is the case, then when your load approach half of the factored load.. there should be diagonal cracks already where the stirrups will finally be engaged.. or do you design it such a way that you don't expect any diagonal cracks? This means the concrete Vc is never reach even by 1.2 DL + 1.6LL by using large dimensions concrete?

 

[BAretired]

The size of a concrete beam is determined by many factors. Rarely is shear one of them, although it could be in the case of a short, heavily loaded beam. Architectural headroom often determines the maximum depth of a beam. Sometimes mechanical ducts conflict with beams. Changes to beam size may be requested by others for non-structural reasons.

When the structural engineer is given free rein to choose the dimensions of a beam, he will usually start by choosing a trial depth which falls within code recommendations for deflection control and a trial width which simplifies form work for the contractor and which does not result in congested reinforcement. To keep form work simple, he will try to use as few different beam sizes as possible. This is a matter of engineering judgment and obviously there will be instances where a different size is dictated by the geometry of the structure.

Some beams are architecturally oversized and lightly loaded; if Vf < Vc/2, stirrups could be omitted according to code but many engineers, including me, use minimum stirrups anyway.

Many beams fall into the category where Vf > Vc/2 which means that minimum stirrups are required by code. The factored shear resistance, Vr = Vc + Vs. The maximum value of Vr is limited by code. I always found that it was prudent to stay well below the upper limit permitted by code in order to maintain a reasonable stirrup spacing but some engineers push it to the limit (another matter for engineering judgment).

Cracking of concrete is normal, and I don't think we can design beams in such a way as to avoid all cracks, diagonal or otherwise. Long continuous beams and slabs are subject to cracking as a result of temperature change and shrinkage. Beams designed completely in accordance with the code can develop cracks; it's the nature of the beast.

BA

 

[p8psk]

BA,

You mean it is common to have inclined diagonal cracks, but in most buildings and malls I saw, I haven't seen a single inclined diagonal crack.. maybe they were purposedly overdesigned? How many percentage of buildings you have built was there inclined diagonal cracks? If minimal.. can you give example of the Vc and loading values you made? Because even at service load.. Vc should be reached and cracks occurred and Vs engaged.. but I have never seen this in most buildings. So please give me actual figures or values you used. Thanks.