I have an architect who wants a 40' clear span opening in a concrete wall of a residence. The wall is actually one side of a concrete box: there is a reinforced concrete deck on top of it that is an exterior patio. I have 4' of depth above the opening and plenty of wall beyond it at both ends. I can design the beam, but I am concerned about creep. I've never done a 40' opening in a cast-in-place wall. I would like a column at the centre, and he has admitted that he is having issues with his overhead door guys being able to make a 40' wide door, but he really wants this opening. Any thoughts? I attached a low-resolution elevation.
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Long span concrete beam
- Thread starter shobroco
- Start date
slickdeals
Structural
Can be done. Since the beam is fully integral with the wall on both sides, the positive moment demand reduces tremendously. You also have a good compression flange because of the deck.
I say go for it. The key would be to design for the lateral loads (weak axis) from wind loads on the garage door.
I say go for it. The key would be to design for the lateral loads (weak axis) from wind loads on the garage door.
TXStructural
Structural
At this length, you could certainly have one-piece, continuous bars top and bottom on your deep beam. With good concrete, you should get good long term performance. The top of the opening can be cambered to offset initial deflection and creep. Live load deflection should not be much of an issue on something this deep. Since the top of the wall/beam is not braced (the concrete deck possibly adds torsion, without providing top bracing) you may want or need to thicken the upper wall.
Be sure you check and reinforce against shear. Since this is exposed, you could get cracks at service loads. Placing WWR on each face may help maintain appearance and improve long term performance. The added cost will be minimal on a single unit.
Be sure you check and reinforce against shear. Since this is exposed, you could get cracks at service loads. Placing WWR on each face may help maintain appearance and improve long term performance. The added cost will be minimal on a single unit.
With top and bottom reinforcement in the beam section, creep will get pretty close to zero. Check your elastic deformation and call it a day. If the door is sensitve to vertical deflections, they should provide a double channel with a vertical expansion gap.
Teguci,
Equal top and bottom reinforcement does not give zero creep. It may give a reduction in creep depending on the amount of compression transferred to the concrete to the compression reinforcement.
It will give a reduction in shrinkage and, if the section is uncracked there may be very little shrinkage deflection with equal top and bottom reinforcement in this case, but I doubt it wouuld be uncracked, so this is not likely either.
In a building like this, the visual deflection will be very obvious. It is importent to get it right and not make unconservative and illogical assumptions to save a little work.
Equal top and bottom reinforcement does not give zero creep. It may give a reduction in creep depending on the amount of compression transferred to the concrete to the compression reinforcement.
It will give a reduction in shrinkage and, if the section is uncracked there may be very little shrinkage deflection with equal top and bottom reinforcement in this case, but I doubt it wouuld be uncracked, so this is not likely either.
In a building like this, the visual deflection will be very obvious. It is importent to get it right and not make unconservative and illogical assumptions to save a little work.
- Thread starter
- #6
The architect is now saying that the door will be a horizontal bifolding hangar-type door, so technically some vertical deflection at the centre won't be a problem, but as rapt says, it will be a big visual problem. Thanks for all of the comments, you've brought up the things I have to consider.
TXStructural
Structural
The easy solution is to use more than minimum tension and compression reinforcement (no problem with this if you use similar quantities top and bottom. Also, be sure you use at least 0.003 T&S vertical steel and sufficient horizontal steel spread out across the face (see my WWR suggestion above). The 0.0030 comes from the T&S requirements of ACI 350 (environmental structures) for watertight structures. Obviously minimizing cost is not the primary concern, performance is. You want a stiff element and optimizing for material use is not the way to go here.
rapt,
A couple sources regarding creep:
"Reinforced Concrete"-James MacGregor - Chapter 3 - "Creep" and Chapter 5 - "Reasons for Providing Compression Reinforcement" includes a time deflection diagram comparing with and without compression reinf.
ACI 318 - 9.5.2.5 - Equation for calculating long term creep effects with compression steel.
After 60 days, the concrete beam with equal compression/tension reinforcement will have close to zero additional deflection. Conceivably, you could design the beam considering just the steel reinforcement as taking the longitudinal stresses (steel doesn't creep or shrink) and consider the concrete only as a matrix.
I've been thinking about this a little more. When the concrete flexes, it will crack. Everytime it rains the flexural cracks will appear over the garage door. There will be cracks at the bottom middle and at the top sides. A strategy to compete against this may be to use small diameter bars (or welded wire reinf around the skin) close to the surface to distribute the flexural cracks into many smaller cracks.
A couple sources regarding creep:
"Reinforced Concrete"-James MacGregor - Chapter 3 - "Creep" and Chapter 5 - "Reasons for Providing Compression Reinforcement" includes a time deflection diagram comparing with and without compression reinf.
ACI 318 - 9.5.2.5 - Equation for calculating long term creep effects with compression steel.
After 60 days, the concrete beam with equal compression/tension reinforcement will have close to zero additional deflection. Conceivably, you could design the beam considering just the steel reinforcement as taking the longitudinal stresses (steel doesn't creep or shrink) and consider the concrete only as a matrix.
I've been thinking about this a little more. When the concrete flexes, it will crack. Everytime it rains the flexural cracks will appear over the garage door. There will be cracks at the bottom middle and at the top sides. A strategy to compete against this may be to use small diameter bars (or welded wire reinf around the skin) close to the surface to distribute the flexural cracks into many smaller cracks.
Teguci,
Firstly, the code section you are referring to is referring to both creep and shrinkage, not just creep.
Do not believe everything you read! Do the shrinkage and creep calculations (not using the ACI code multiplier, I mean do a real creep and shrinkage analysis) and you will see the real effects. I tried to summarize them for you in the earlier post. Here is a longer version.
Creep deflection will never be completely eliminated using equal top and bottom reinforcement as there will still be compression stress in the concrete in the compression zone, so there will be creep deflection. The only way you can get no creep deflection is to get rid of all of the compression stress in the concrete. As this is not possible, there will be significant creep deflection.
Shrinkage deflection depends on the relative amounts of top and bottom reinforcement, but also depends on the stress profile through the section and the neutral axis depth.
- If there is no bending stress in a rectangular section (no possible in a 40' span beam) and you provide equal top and bottom reinforcement, there will be no shrinkage deflection. If it is a T beam, the top reinforcement will be much closer to the centroid than the bottom reinforcement so you would need even more compression face reinforcement to make this work. But this is not possible as there will be bending stress!
- Otherwise, the deflection caused by shrinkage warping is basically dependant on the relative distances of the reinforcement from the neutral axis and the relative areas of reinforcement. If the areas are equal and the distances from the neutreal axis are equal, there is no warping deflection. If for both top and bottom steel, the product of area * distance from neutral axis is equal, there will be no warping deflection.
This is never the case in RC members. The compression reinforceemnt distance from the neutral axis is much less than the tension reinforcement distance from the NA, so there is a nett bending induced and a downward deflection from shrinkage warping.
As well as this, the more reinforcement you put into the section increases the amount of restraint causing more tension stress in the member causing more cracking so you alos get more deflection from this.
The ACI simplified formulawill never result in 0 long term deflection anyway. With .5% reinforcement (rho = .005), the long term component of deflection
- at 3 months would be 1 / (1 + 50 * .005) = 1/1.25 = .8. So total deflection = 1 + .8 = 1.8 * short term deflection
- at 2 years = 2 / (1 + 50 * .005) = 1.6. So total deflection = 1 + 1.6 = 2.6 * short term deflection
Firstly, the code section you are referring to is referring to both creep and shrinkage, not just creep.
Do not believe everything you read! Do the shrinkage and creep calculations (not using the ACI code multiplier, I mean do a real creep and shrinkage analysis) and you will see the real effects. I tried to summarize them for you in the earlier post. Here is a longer version.
Creep deflection will never be completely eliminated using equal top and bottom reinforcement as there will still be compression stress in the concrete in the compression zone, so there will be creep deflection. The only way you can get no creep deflection is to get rid of all of the compression stress in the concrete. As this is not possible, there will be significant creep deflection.
Shrinkage deflection depends on the relative amounts of top and bottom reinforcement, but also depends on the stress profile through the section and the neutral axis depth.
- If there is no bending stress in a rectangular section (no possible in a 40' span beam) and you provide equal top and bottom reinforcement, there will be no shrinkage deflection. If it is a T beam, the top reinforcement will be much closer to the centroid than the bottom reinforcement so you would need even more compression face reinforcement to make this work. But this is not possible as there will be bending stress!
- Otherwise, the deflection caused by shrinkage warping is basically dependant on the relative distances of the reinforcement from the neutral axis and the relative areas of reinforcement. If the areas are equal and the distances from the neutreal axis are equal, there is no warping deflection. If for both top and bottom steel, the product of area * distance from neutral axis is equal, there will be no warping deflection.
This is never the case in RC members. The compression reinforceemnt distance from the neutral axis is much less than the tension reinforcement distance from the NA, so there is a nett bending induced and a downward deflection from shrinkage warping.
As well as this, the more reinforcement you put into the section increases the amount of restraint causing more tension stress in the member causing more cracking so you alos get more deflection from this.
The ACI simplified formulawill never result in 0 long term deflection anyway. With .5% reinforcement (rho = .005), the long term component of deflection
- at 3 months would be 1 / (1 + 50 * .005) = 1/1.25 = .8. So total deflection = 1 + .8 = 1.8 * short term deflection
- at 2 years = 2 / (1 + 50 * .005) = 1.6. So total deflection = 1 + 1.6 = 2.6 * short term deflection
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