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shearwall-multiple hold downs
- Thread starter umkorotk
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- #4
msquared48
Structural
Why would you want to spread the uplift?
I would be concerned with not utilizing the full capacity of either hold down if they are placed too close together. And if they are spread further apart, some of the leverage is lost.
As for the difference in the results, perhaps Josh Plum could comment on that. You might reconsider posting this in the RISA Technologies forum if you do not get the answer you seek. Josh is in there more and is the resident expert. You would have to red flag this string first before doing that though.
Mike McCann
MMC Engineering
I would be concerned with not utilizing the full capacity of either hold down if they are placed too close together. And if they are spread further apart, some of the leverage is lost.
As for the difference in the results, perhaps Josh Plum could comment on that. You might reconsider posting this in the RISA Technologies forum if you do not get the answer you seek. Josh is in there more and is the resident expert. You would have to red flag this string first before doing that though.
Mike McCann
MMC Engineering
JoshPlumSE
Structural
I don't trust the numbers shown on that print out very much. If there were to be a difference between the two hold downs, I would expect the outer hold down to take the greater load. Sigma = Mc/I after all....
It's always a little problematic putting in two perfectly rigid supports so close to each other. Therefore, you might either do a hand calc (like AELLC suggests), or replace the perfectly rigid reaction with a more flexible spring support.
The real behavior of a shear wall is actually a lot different than that model. You have continuous shear resistance along the length of the wall. You have continuous compressive vertical resistance along the wall. Heck, you probably have something close to contiunous tensile resistance as well. The NDS stiffness and deflection equations assume this type of continous resistance as well....we just don't rely on it for design purposes. Instead, we create a pretend force couple in the chords to resist the total moment. At least that's the current "standard of care" in the industry. If you do something else (however rational) then you probably want it to be more conservative than what the hand calc method would give.
It's always a little problematic putting in two perfectly rigid supports so close to each other. Therefore, you might either do a hand calc (like AELLC suggests), or replace the perfectly rigid reaction with a more flexible spring support.
The real behavior of a shear wall is actually a lot different than that model. You have continuous shear resistance along the length of the wall. You have continuous compressive vertical resistance along the wall. Heck, you probably have something close to contiunous tensile resistance as well. The NDS stiffness and deflection equations assume this type of continous resistance as well....we just don't rely on it for design purposes. Instead, we create a pretend force couple in the chords to resist the total moment. At least that's the current "standard of care" in the industry. If you do something else (however rational) then you probably want it to be more conservative than what the hand calc method would give.
Interestingly, the results are on the left are similar to (½ total shear)(height)/(distance between holddowns).
½(10000 lbs)(10)/10=5000 lbs
½(10000 lbs)(10)/8=6250 lbs
What are the horizontal reaction?
If the holddowns are on the same chord member, taking 1/2 the load to each holddown is reasonable. If each holddown is on a separate chord, more analysis may be warranted. I would be interested to see the result if the tension holddowns are modeled as springs. For a spring constant, I would start with k= hold down capacity/ holddown slip.
½(10000 lbs)(10)/10=5000 lbs
½(10000 lbs)(10)/8=6250 lbs
What are the horizontal reaction?
If the holddowns are on the same chord member, taking 1/2 the load to each holddown is reasonable. If each holddown is on a separate chord, more analysis may be warranted. I would be interested to see the result if the tension holddowns are modeled as springs. For a spring constant, I would start with k= hold down capacity/ holddown slip.
at times I think we are being generally over-conservative on the hold down design. When you've got standard anchor bolts holding your bottom plate to the stem wall at some regular spacing you have more of a continuous tensile resistance, and definitely continuous compressive resistance.
- Thread starter
- #15
thanks for comments and feedback everyone
I have about 11 kips uplift in 3 story condo shearwall. Capacity of 7/8" hilti HY200 in 8" conc wall about 6kips
that's the reason I'm trying to have 2 hold downs on separate chords(16" apart).
I have about 11 kips uplift in 3 story condo shearwall. Capacity of 7/8" hilti HY200 in 8" conc wall about 6kips
that's the reason I'm trying to have 2 hold downs on separate chords(16" apart).
Many times the loss of leverage is pretty small compared to what you gain in uplift capacity. If you had a second holdown 16" away your moment arm changes by 8" but you double your holdown resistance. What I have found is like others have said, you have to be careful that you pay attention to what is happening at the foundation.
CELinOttawa
Structural
There is a difference - Just look at the solution of a moment about the out of plane axis. In practice this is frequently discarded. Meet the minimum separating distance and ensure you can actually engage sufficient dead load.
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