Simpson Strap on Top Plate Connection 1

[nrguades]

Hi to all...

Greetings...

I received a comment regarding double top plate splice connection. The chord force is around 14600 lbs. Im using CMST14 with 37 - 16d nails ea side and joint in the top plate splice shall be offset at least 4 ft.

Heres my design parameters & calculation:

Design Code: IBC 2003
Lumber Type: Douglar Fir Larch, G=0.5
Lumber Grade: No. 2
Load Duration Factor = 1.6
Wet Service Factor = 1.0
Temperatur Factor = 1.0


Ft = 575 psi (Table 4a, NDS SUPP 2001)
Cross Sectional Area of 2x6 = 8.25 sq in
Size Factor = 1.3

the allowable tension capacity of 2x6 only =
575x1.3x8.25x1.6 = 9867 lbs

therefore the minimum force resisted by simpson strap =
14600 - 9867 = 4733 lbs

Capacity of CMST14 with 37 - 16d nails ea side = 6490 lbs > 4733 lbs therefore the connection for me is okey...

The comment suggested that double top plates cannot resist the specified load and we need a steel angle to reinforce the chord inorder to resist the force of 14.6 kips. He said that the steel angle will all carry the 14.6 load.

Hoping for you comments on the above issue, most specifically on the principle I used to determine the force carried by the simpson strap....

Thanks

Respectfully,
Noel

 

[UcfSE]

Who suggested the comments, an engineer, someone who should actually know what he or she is talking about? Does that person have calculations to show you are wrong and he or she is correct?

 

[RARWOOD]

The basic approach followed when using a double top plate is to assume that only one plate is effective at any given point. When this approach is used the joints are staggered at least 4'. Then the two plates are nailed toghter just beyond the splice. For example in this case the tension force in the top 2 x 6 is transfered to the bottom 2 x 6 just before the joint in the top 2 x 6; with no force being transfered across the joint in the top 2 x 6.

What you are trying to do is make both 2 x 6 top plates effective. To accomplish this you are trying to use a tension strap to carry the tension load across the joint in the top and bottom 2 x 6s. This would require you to have a tension strap across the joint when you have a break in the top 2 x 6 and a tension strap on the bottom 2 x 6 when you have a break in the bottom 2 x 6.

To simplify the example assume that the joints of the two 2 x 6 top plates line up. You then would need to transfer your tension load across the joint just as you would need to design a tension splice in you had a single soild member
3" x 6". You could do that by using a steel plate on top and bottom of the top plates then through bolting through both plates.

You made an error in assuming that you only need to resist 4733 lbs. at your splice. Following your approach you would only be able to carry 6490 x 2 = 12980 lbs.

What you want to do is design your splice to resist 7300 lbs. Then with a splice plate at every top and bottom joint, your 2 x 6's would each carry 7300 lbs.

The suggestion of using a steel angle may have been made because of the difficulty of developing a tension tie across the joint in the bottom 2 x 6. Your major problem is how to splice across the joints with out causing problems in your other framing.

An excellent reference for use in learning more about wood construction, is Donald Breyer's book "Design of Wood Structures". You could also contact Simpson directly or the APA.

 

[RARWOOD]

nrguades

I left out one important point to make both top plates effective you also have to join them togther. It would be similar to a wide flange beam with a cover plate laid on top. If the cover plate is not welded to the wide flange then it would not resist any force as the wide flange was loaded.

 

[nrguades]

UcfSe & RARSWC

Thanks for your reply...

I just have some additional clarification on RARSWC explanation. Your example is assuming that the joint on double top plates are lined up. How about if you have to offset the splice using strap at every 4 ft too, would it be theoritically correct that the weakest point for tension will be resisted by a the full capacity of a single 2x6 which is equal to 9867 lbs and the rest will be resisted by the strap which is equal to 4733 lbs?

thanks...

Respectfully,
Noel

 

[UcfSE]

You should probably think about making sure deformations are compatible with the steel strap and the 2x6. If the 2x6 is much much stiffer axially than the steel strap, it may take a disproportionate amount of load over and above what you have assumed, and vice versa. I would look for a way along that line of thinking to distribute forces between the two, for myself.

 

[RARWOOD]

I will attempt to clarify my response, although without a sketch of what you are doing, I am not sure you and I are on the same page.

I understand your problem to be the following:

You have double top plates, starting at a cross section on the left where both plates are continous, you go over 2' where the top 2x6 ends and buts up to another 2x6. You then go over 4' where the bottom 2x6 ends and buts up to another 2x6. This pattern then continues for some distance.

Lets first look at the system without any Simpson Straps. Generally what is done is that at the first splice 2' in from the left end, the top and bottom 2x6's would be nailed toghther to transfer the tension load out of the top 2x6 into the bottom 2x6. In your case the tension force would be 9867 lbs. So if you were design for a 9867 lbs load you would need enough fasteners to transfer that load from one 2x6 to the other.

Now 4' over the bottom 2x6 ends at a splice. At this location you would again attach the bottom 2x6 to the top 2x6. So as you move down the wall the tension load is alternately carried by the top 2x6 and the bottom 2x6.

Now if I understand what you are doing you are coming along and adding a Simpson strap at all the splices in the top 2x6. Now neglecting the different deformation properties of the wood, assuming that the strap is carrying 4733 across the splice. The following is what I believe would be happening.

I think you would have 14600 lbs in your top 2x6. My logic is as follows. Assume you transfer 9867 into the top 2x6 then at the first Simpson strap you transfer 4733 lbs into the strap and across the splice.

You now have 9876 lbs in the top 2x6 which at the first splice goes down into the bottom 2x6 you also have 4733 lbs in the top 2x6. Now when you go 4' over and get to the splice in the bottom 2x6, if you only have straps on top that 9876 load goes back into the top 2x6.

At this point I believe you now have 9876 lbs + 4733 lbs = 14600 lbs in your top plate which would exceed your allowable tension capactity.

What ever you try to do you have a nasty problem because of the high forces involved. I considered looking at a solid LVL as a double plate or triple plate. However the length you need to get the fastners in gets quiet large and probably causes other problems.

Most certainly you have a tough problem to solve!

 

[Cardona]

This post is probably too late for nrguades but I believe it is interesting enough that my comments may help some other members.

First thing is: IT DOES NOT WORK. We cannot disregard the fact that once the strap is properly nailed to the wood, the two pieces (steel and wood) are acting together and therefore they must be treated as a composite member, as rightly was pointed out by UcfSE above, and the different stiffness must be taken into account. This fact is considered by the manufacturers when they develop their tables. Otherwise just finding a strap that is not stressed beyond the allowable stress point would be enough.

We have two different ways to build the wood plate : a) as a two independent 2x6 members, probably with nails scattered just to keep them together. In this case each member, top and bottom, is carrying half the load, that is 7.3 kips each (how to do this is probably matter of another post), and b) as a built-up 4x6 beam (only 3 inches high) half discontinued at each splice.

In no case the proposed top only strap will work because in case a), at the first discontinuity in the lower member, it will stop carrying any load and, in case b), at every splice where there is no strap the other member will have to carry the full load; with an area of wood of 8.25 SI the stress would be 1.77 ksi. The allowable stress is 1196 psi (1.3x1.6x575), therefore the wood fails.

Treating this as a composite member then the modular ratio, that is Es/Ew, n = 15.38. The steel strap has an area of As=0.22 SI (3 inches wide by 14 ga.) so the steel’s equivalent wood area is A’s = n x As = 3.45 SI

Using straps at every splice, top and bottom, we have for case a):

Each member is carrying 7.3 kips in tension. At the point of splice (probably at the strap’s half length) we have the whole load resisted by the strap. The stress is 7.3kips/0.22SI = 32.57ksi. Unless high yield steel is used, the strap fails.

And for case b):

When the strap begins and up to the splice we have a composite beam. The area of wood Aw=16.5SI and the strap’s equivalent wood area is A’s=3.45 SI. The total load of 14.6kips is divided proportionally to these areas and the stresses are then 0.73ksi for wood and 11.26 ksi for steel.

Up to here all is fine, but when we reach the first splice half of the wood area disappears, now Aw=8.25 SI and the stresses change to 19.2 ksi for steel and 1.25 ksi for wood. This is more than the allowable 1196 psi. Wood fails.

 

[RARWOOD]

Cardona

It may be just too early in the morning, as I've just finished my first cup of coffee, but I don't follow your example at all. Perhaps after I've had a little more coffee and thought about it, your example may become clearer.

At this point I have several comments:

To the best of my knowledge, having worked for a manufacturer of lumber connectors, companies such as Simpson and USP do not consider the difference in stiffness between wood and steel, when they develop their strap capacties.

I have never seen a design exmaple or attended a seminar where the stiffness of the steel was considered in the design of a wood splice using steel straps. In every example, I've seen you determine the load that needs to be transfered and then you select a strap with that capacity or higher.

Generally the strength of a strap is limited by the number of nails in the strap. One could check some of the published table values by mutiply the allowable nail load by the number of nails. Also manufacturers do use steel steel with different physical properties for different applications.

In the application that is being disscussed my understanding is that we are talking about a splice using a light gauge steel plate across the end joint of the top plate. In this example I don't see how composite action is developed.

Thank you for your post as it is an interest topic to think about.

 

[UcfSE]

RARSWC,

Manufacturers are concerned only for the maximum capacity of a strap, not necessarily whether it can develop that maximum capacity in every conceivable application. Also, manufacturers are basing their reported strengths on the strap alone. They are not trying to find a splice capacity, for instance, that uses both a wood splice and a steel strap to supply the tensile strength of the splice.

In the example you reference at a seminar, you are finding the tensile force in the top plate and providing for the entire force with a steel strap. That is not the same thing that the OP is asking. The OP is using both wood and steel to take care of the tension. The strap along cannot handle the whole tensile force in the top plate. Therefore the relative stiffness of the two materials must be considered when finding the forces in each material. Note that this is not the same as composite action, say from a bending point of view.