hook within hoop

[itez]

Please see attached illustration.

For a beam that is 300mm in width and 500mm in depth. How necessarily is it to add hook within a closed hoop? How would it affect the shear resistance? For a wider width or column. It is necessary to restraint the core from dilating. But for a beam with only 300mm width (or 220mm hoop since there is less the 0.40mm concrete covers), how would it exactly affect the shear resistance when closed hoop is already provided?

 

[BAretired]

The additional bar increases the shear carried by the steel by 50%. Omitting the central bar and reducing the stirrup spacing by one third achieves the same result and is the more typical detail.

BA

 

[itez]

Did you mean the additional bar increase the shear resistance or increasing the shear itself (meaning making the shear worse)??

If it increases the shear resistance. Can you create the same result by wrapping the external beam by U shaped carbon fiber reinforced polymer?

 

[itez]

I think you meant the additional bar increased the shear resistance of the beam. Now my problem is. I'm only used to closed hoop and didn't notice the additional bar in the plan. I'm the project manager and engineer. And the beams are already casted with stirrup spacing not reduced by one third. In cases like this. How do you fix it... where you missed to add the additional bar. Would drilling the beam from bottom and adding the bars vertical at center do it? Or retrofit with carbon fiber reinforced polymer. What is usually done in the field in cases like this.

 

[BAretired]

Shear resistance of a reinforced concrete beam consists of two parts, V[sub]c[/sub] and V[sub]s[/sub]. Adding a bar as shown on your sketch increases V[sub]s[/sub] but has no effect on V[sub]c[/sub], so the total shear resistance is not increased by 50% only the resistance of the steel.

I would rather not comment on carbon fiber reinforced polymer as I have no experience with it. I presume it would increase shear resistance but would be concerned about it effectiveness during a fire.

BA

 

[itez]

I'm talking about diagonal cracks being restrained by hoops or stirrups. The number one purpose of these transverse steel is on this diagonal cracks. concrete alone would be poor in shear. Vc is diagonal shear capacity of the concrete? But it's so poor (because concrete very poor in tension which is related to the diagonal crack) that is why tranverse steel needed. So Vs (diagonal crack shear capacity of the stirrups and hoops) are much more than Vc (diagonal shear capacity of concrete alone). Do you agree? I think Vs handles 80% of the diagonal shear and Vc only 20% (?)

 

[IDS]

Do you agree? I think Vs handles 80% of the diagonal shear and Vc only 20% (?)

It depends on the steel diameter and spacing, and the concrete grade. The contribution of the concrete may be more than 20%, but certainly reducing the shear steel by 33% will significantly reduce the shear capacity.

You need to tell the engineer responsible for the design and ask for a solution.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[itez]

The designer said to decrease the live load and don't have experience with carbon fiber reinforced polymer wrapped in U-shape beneath the beam. How much would this increase the Vc of the concrete? Or V(CFRP). Anyone has used this? What else did you do to increase the shear capacity?

I suggested to the designer about drilling hole beneath the beam and inserting bars to increase the Vs. But he has not done this before. Has anyone tried this or have seen this elsewhere so I can discuss with him about this by showing it is plausible?

 

[BAretired]

The designer said to decrease the live load and don't have experience with carbon fiber reinforced polymer wrapped in U-shape beneath the beam. How much would this increase the Vc of the concrete? Or V(CFRP). Anyone has used this? What else did you do to increase the shear capacity?

Is the designer prepared to decrease the live load to the point where the existing stirrups are adequate without the central bar? I have no experience with carbon fiber reinforced polymer, but U-shape suggests you are talking about a Tee beam. I seriously doubt whether polymer could adhere to the sides of the beam enough to generate any useful shear strength.

I suggested to the designer about drilling hole beneath the beam and inserting bars to increase the Vs. But he has not done this before. Has anyone tried this or have seen this elsewhere so I can discuss with him about this by showing it is plausible?

This idea might work, but 500mm is a pretty deep beam and you would not want to drill through any existing stirrups. How would you anchor the bars...with epoxy? Would it be easier to drill from the top and anchor the bars with a nut top and bottom and forget about the epoxy? I have never done it before, but I would think it is a possibility.

BA

 

[IDS]

Drilled in bars may well be the best option. I'd suggest talking to one of the companies that produce products for post-installed shear reinforcement.

I don't know the commercial or legal arrangements on this job, but surely the original designer will have sign-off on whatever is selected as the remedial work.

Doug Jenkins
Interactive Design Services

http://newtonexcelbach.wordpress.com/

 

[dhengr]

You said.... “I'm the project manager and engineer.” It’s a shame the project manager and project engineer didn’t check the shop drawings or inspect the rebar in place, prior to the pouring of the conc. Those are both normal activities and responsibilities of those parties on a construction project. They also have to read and study the plans for the job.

BA’s method of anchoring the added shear steel with nuts and a washers would work Start with a 1.5" deep +/- cores, then drill the hole, and maybe epoxy the rod in place. Then, once the nuts are tightened the cores can be grouted full (patched) around the nuts. You will have to map all the existing steel so you don’t cut it in the drilling process. You might also want to test the actual conc. strength and tie steel strength, and the original shear steel calcs. to see if you can’t get a bit more cap’y out of what’s already there. This might allow you to use a smaller rebar and drill for the added shear steel.

They do use FRP to do what you are talking about in the way of increasing bending and shear cap’y. of conc. beams. It is a very expensive repair method, and not something that anyone without considerable experience in design, material selection, surface preparation and application should try on their own. You should contact a specialty contractor who does this kind of work if you want to know more about it.

 

[itez]

Ok. Just for my knowledge in further discussions with the designer. The ACI requires stirrup spacing of d/4. How many of you follow the d/4 code or use wider spacing? If the beam depth is 500mm. ACI requires 500/4 = 125mm or 4-5 inches spacing. The structural plan of the project has 4 inches stirrups spacing. Is 4 inches typical in your plans? Of course I'd discuss any further shear strengthening with the designer, but just need to know the ACI side on this.

 

[BAretired]

This is an international forum. We are not discussing ACI requirements. We are discussing an issue which has developed on your jobsite as a result of your failure to follow engineering drawings. We would like to help, but if you want a sensible answer to your questions, you must give us a little more information. There may be other ways of looking at the situation, but we are not clairvoyant, so help us along. What is the load on the beam and how does it relate to other beams and columns?

​

BA

 

[itez]

It is a girder and I will really put the shear bars or retrofit after the designer calculated how many spaces apart tomorrow.

Please see attached picture of the beam an hour before concrete was poured with longitudinal reinforcements in place and closed hoops (stirrups with 135" hooks). The top bars are 8 pcs of 20mm grade 60 layered in 4s and bottom bars are 4 pcs of 20mm grade 60. The beam width is 300mm and depth 500mm. The slabs are connected to it. Close hoops (stirrups) spacing is 100mm.

Do you call it a T-beam? The slabs are just connected to it with no longitudinal bars extended outside the rectangular beam. So technically it is not called a T-beam which has more reinforcements. True?

I'm wondering if carbon fiber can just be wrapped around the beam. The slabs are light at only 100mm thick and it is 4 way so other beams around it can carry it (said the designer). You agree that carbon fiber completely wrapping the beam can increase shear resistance and this can be a good solution (instead of just U-shape wrapping below the slabs)?

Also alternatively, do you know of drill machine that has least 400mm drill in length? We can see only 6 inches drill and not 16 inches drill. Is there such?

Thanks a lot.

 

[BAretired]

I don't see how carbon fiber can be wrapped around the beam, but you may want to discuss it with someone who has used that technique before. I have not.

Your photo suggests that missing the reinforcement while drilling through the beam is likely to be a challenge. You should discuss this with a concrete drilling company in your area. You may need to use GPR (Ground Penetrating Radar) to locate reinforcement before drilling.

BA

 

[dhengr]

Your original sketch sure looks different than your latest photo, but never mind that, nobody ever pays any attention to that insignificant crap in commenting on an OP. You would be surprised at how an experienced engineer makes a pretty good judgement of the problem when he/she is shown a complete and properly proportioned sketch, and is given a meaningful description of the problem. And, three bars t&b is not quite the same as 6 or 8 bars t&b in the same size beam, on this kind of problem.

I’m also curious how bldg. design and construction works in your area/country? How many layers of design engineers, project managers, and project engineers are there, none of them knowing what the other is doing or how they go about doing it? Does the design engineer ever come out and look at the construction progress? Does a design engineer have to know anything about constructibility? What is the training and testing required to be called an engineer in your country or can anyone just print up a calling card saying ‘I are an ingineer’? Why are you coming to E-Tips for a situation like this, when you should be going directly to the design engineer, who can at least look at a sketch/set of plans which aren’t so drastically in error? Those must be some pretty scary bldgs. to live in and around. We all run into problems or errors on construction jobs from time to time, but some of you guys really seem to have a corner on that market.

 

[itez]

In my country. Designers said they only calculate for reinforcement ratios and constructibility have to be handled by contractor but coordinated with designer for alteration in plans if construction is not possible. So what we do is draw shop drawing and submit to them for approval (especially for complicated design). In the case of the building. I thought it was just typical and mostly I just use U shape stirrup or hoop in small beam like 300x500mm. In the past we only put middle stirrup in big beams So I missed the drawing on the single middle stirrup and it is my mistake.

The designer said they used the most conservative seismic codes such that even if calculations would show 10 inches spacing, they would use D/4 or 500/4 or 125mm spacing and even 100mm or 4" is specified. Also the designer won't show me any calculations except if I show it to him first so I need to know something before I discuss with him later.

BA, in the picture. Do you consider it as a tee-beam and what could be the effective width of it then?

I'm solving for Vc from Vc=2*rho*b*d*sqrt(fc)
given
b=11.8112 inches (300mm)
d=19.68504 inches (500mm)
fc=4000 psi
rho=0.75
Vc==2*rho*b*d*sqrt(fc)=22,057.26 lbs.

But it is a tee-beam (is it based on the pictures?). So what should be the effective width? If it's 15" instead of 11.8", then Vc becomes 29,527.21 lbs.

Now for Vs=2*rho*As*fy*d/s
given rho=0.75, As=2(0.11), fy=40,000 psi, d=19.68504", s=4 inches
then Vs = 64,940.63 lbs

and total shear capacity = Vu = Vc + Vs = 22,057.26 + 64,940.63 = 86997.89 lbs

Shear is related to moment. If shear = 86997.89 lbs. What is the corresponding moment capacity? What is the formula to relate shear and moment? I just want to verify his work by knowing some basic formula so I'd understand his after I discuss with him because I have to shoulder the very expensive carbon fiber wraps retrofit. Thanks.

 

[hokie66]

This would be considered a tee beam, but that doesn't materially improve your shear capacity. I don't think there is a practical way of improving the shear capacity of this beam. Post-installed shear reinforcement requires anchorage at each end of verticals, and that looks impractical to achieve, based on your photo. I don't know much about carbon fibre wraps, but would be dubious of manufacturer's claims.

 

[BAretired]

BA, in the picture. Do you consider it as a tee-beam and what could be the effective width of it then?

For purposes of shear calculations, b is the width of the web which is 300mm whether it is a tee beam or a rectangular beam.

I'm solving for Vc from Vc=2*rho*b*d*sqrt(fc)
given
b=11.8112 inches (300mm)
d=19.68504 inches (500mm)
fc=4000 psi
rho=0.75
Vc==2*rho*b*d*sqrt(fc)=22,057.26 lbs.

I do not use the ACI code. I use CSA A23.3 (Canadian code). The term 'd' is the effective depth from the c.g. of steel to the compression face of the concrete. It is less than 500mm, probably about 400mm. I do not know what 'rho' is. Maybe someone else can advise.

But it is a tee-beam (is it based on the pictures?). So what should be the effective width? If it's 15" instead of 11.8", then Vc becomes 29,527.21 lbs.

If you are drilling holes and inserting bars, it makes no difference whether it is a tee-beam or not. If you are wrapping with carbon fibers, the flange will get in the way, so I can't see that being possible.

Now for Vs=2*rho*As*fy*d/s
given rho=0.75, As=2(0.11), fy=40,000 psi, d=19.68504", s=4 inches
then Vs = 64,940.63 lbs

and total shear capacity = Vu = Vc + Vs = 22,057.26 + 64,940.63 = 86997.89 lbs

Again, d < 500mm. The term 2*rho appears wrong . By CSA code, your value of Vs is too high.

In my area, steel suppliers are not stocking 40,000psi steel any more. If you order it, you will usually get 60,000psi steel. This is something you should check with your supplier.

Shear is related to moment. If shear = 86997.89 lbs. What is the corresponding moment capacity? What is the formula to relate shear and moment? I just want to verify his work by knowing some basic formula so I'd understand his after I discuss with him because I have to shoulder the very expensive carbon fiber wraps retrofit. Thanks.

There is no simple relationship between shear and moment. In a simple beam, moment is 0 at each support and maximum at midspan while shear is maximum at each support and 0 at midspan (assuming uniform load). For a continuous beam, moment is negative at the support (tension on top) and positive at midspan (tension on bottom). Shear is maximum at supports and zero somewhere near mispan.

BA

 

[itez]

Now for Vs=2*rho*As*fy*d/s
given rho=0.75, As=2(0.11), fy=40,000 psi, d=19.68504", s=4 inches
then Vs = 64,940.63 lbs

and total shear capacity = Vu = Vc + Vs = 22,057.26 + 64,940.63 = 86997.89 lbs
Again, d < 500mm. The term 2*rho appears wrong . By CSA code, your value of Vs is too high.

rho is the strength reduction factor which is 0.75 for shear calculations.. it is 0.9 for beam.
Why, what is the web reinforcement shear formula for Vs in your country? In the ACI code, it is really

Vs=2*rho*As*fy*d/s = 2*0.75*0.22*40,000*19.68504 inches / 4
Vs = 64,940 lbs

Maybe it's high value is the spacing of 4 inches. When the spacing becomes twice or 8 inches, Vs becomes half or only 32470 lbs. What is wrong in the calculation?

About the T-beam and the flank action on top. Well. I'd suggest to the designer if I can put some hole in the flank and do full carbon fiber wrap around the beam covering 1 meter from the column face. Remember stirrups only work after the diagonal cracking already occurs. It pins the diagonal cracking. So full CFRP wrap from Sika around the whole beam should prevent the diagonal cracks from fully opening and brittle failure (in addition to the existing 4 inches stirrup already present).