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horizontal cold joint in one-way slab/joist construction?

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Lion06

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Nov 17, 2006
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I have a situation in one-way concrete joist construction where the final (5) trucks were held up and ended up being poured after the rest of the concrete had set. This has created a horizontal cold joint in at least one girder, and several joists as well as a vertical cold joint in several joists at the end of the span where the shear is highest.
I am trying to find information about how this is handled and to verify capacities for shear flow across this joint using in-place reinforcement for shear friction calculations.
Any guidance or references that anyone could provide would be greatly appreciated!!
 
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Usually this is a classic case of simply determining the section properties and Q value at the joint to determine the horizontal shear at the joint.

Shear = VQ/I

V is the vertical shear at the point in question
Q is the first moment of the area on one side of the horiz. joint
I is the moment of inertia of the total section.

Once you have the shear (in kips/in or SI equivalent) then you simply can use the ACI shear friction values in Chapter 11 of ACI 318.

The trick would be using the appropriate surface roughness coefficient in the shear friction formula.
 
JAE-
I considered this, but what steel would I use for shear friction? The stirrups are already being used as vertical shear steel, I can't double dip and use it for shear friction reinforcing also, can I? If I assume them to be transferring shear across that vertical joint, how will they also act to prevent the diagonal shear cracking?
 
I thought I remembered a provision of doing just what you describe but when I went searching today I couldn't find it. Could it have possibly been in a previous code and was edited out?
 
Chapter 17 of ACI addresses this. Double-dipping is permitted. I've got a similar condition, except it hasn't been constructed yet. The contractor is proposing to pour the monolithic slab and beam in separate pours, so I have to investigate the horizontal shear.
 
If your stirrups are in tension (regardless of the reason why) this will cause compression on your shear interface.

In typical perpendicular shear the tension steel is used even though it is there for bending-same principal.

There is a clause in the Australian code that covers this exact case, I will see if I can find it for you tonight.
 
JAE
For calcultaing Q we need the cross sectional area of either side of joint into the arm. What will be the arm? Is it from center of area on either side of joint into it's centroid to neutral axis. Also we assume neutral axis at geometric center of concrete beam with rebars?
 
I've always neglected the rebar. Just used the gross cross sectional properties of the beam and the effective width portion of the slab.

Q is: (Area outside the horizontal plane you are seeking shear at) x (distance from NA of that area to the NA of the total section)

 
I agree with what folks have typed so far and can only add one thing that could be forgotten: development of the stirrups. This could be a challenge depending on the cold joint elevation.
 
Section 17.5 provides a fairly straightforward analysis method, without having to calculate Q. Specify that the cold joint is roughened to 1/4", and you can get pretty decent capacity using the stirrups you already have there. There is a limit to the available capacity, beyond which you will have to design as JAE describes above.
 
i don't have the code handy, but wouldn't you have to compute the Q initially to come up with the demand stress at the critical section?

i believe that that particular section (don't shoot me if i'm wrong) states provisions on strength vs. the actual analysis method. i'll check it tommorow when i get to the office.

wouldn't it also be a good idea to provide a bonding agent such as a sika material to the existing concrete?

one can also drill into the beam web vertical bars for shear friction i believe.
 
In ACI 318-05, Chapter 17, Vnh is the nominal horizontal shear strentgh. Vu is the vertical shear from analysis?
Or we need to convert back the shear stress obtained from formula VQ/bI multiply by bv x d .
 
"Vu is factored shear force at the section considered"

"Vnh shall not be taken greater than 80bvd"

if the Vu is larger than the phi x 500bvd, you place shear friction steel.

 
Swivel63,
Appreciate your input but please read question carefully.
You answered
"Vu is factored shear force at the section considered"
Let me repeat my question here
"Vu is the VERTICAL shear from analysis?
Or we need to convert back the HORIZONTAL shear stress obtained from formula VQ/bI multiply by bv x d ."
 
if you read the code, it says "Vu is factored shear force at the section considered" and the section where it says this is the horizontal shear force section.

that would lead me to believe that the Vu they're talking about is actually a Vuh (horizontal) that you derive from Vuv (vertical) x Q / I (in kips/in).
 
I think they dumbed it down for us and we are supposed to use the vertical shear (Vu). How would you go about calculating a horizontal shear? You could get a shear flow in K/in or a shear stress in K/in^2, but unless you select an arbitrary length of beam to apply that to you can't get a horizontal shear force. Since ACI gives no guidance on what length of beam to apply the horizontal shear flow or horizontal shear stress to (and doesn't even mention it), I would think that the Vu is the vertical shear force.
 
i think that's the same problem i had when i first came across this issue. but depending on your stirrup spacing or your shear friction bar spacing, i guess you could convert it into a kip/in reading.

bv - width of cross section at contact surface
d - (same as everywhere else in the code)

however in the commentary it says that "the definition of d used in chapter 11 for determination of vertical shear strength is also appropriate when determining horizontal shear strength."

in the PCA notes it says that

"section 17.5.1 requires full tranfer of horizontal shear forces by friction at the contact surface, properly anchored ties, or both. Unless calculated in accordance with 17.5.4, the factored applied HORIZONTAL shear force Vu < phi x Vnh, where phi x Vnh is the horizontal shear strength."

it doesn't specify the length of which you would take to apply to your Vu, though.



 
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