Analysis of 40-year-old bar joist in HWVZ

[Streng72]

I have been asked to verify the structural stability for a 40-year-old bar joist in the Miami area. I am analyzing the roof joist using a computer model based on measurements taken in the field. The bottom chord is a double bar (3/4” dia each), the 28”-9” long top cord is a 2L 2x2x1/4, and the webs are solid 7/8” dia. bars. Per field observations there are three rows of cross bridging, which seems alrigth for this span length. I have assumed that all joints are pined and that the two end supports in the joist are fixed. I am using Fy=36 ksi (I am not very sure about that Fy yet).

After applying all the loading the bottom chord fails in compression due to wind uplift. I took the compression force and run manual calcs to determine the unsupported length which gives me about 47 inches. The max compression force is about 6.2 kips.
I am inclined to suggest they provide additional bridging at 4.5 feet o.c. (the bar joist ‘bays’ are 1.5 feet wide. I am assuming there is some fixity at the points where the bridging connects to the joist then I could apply K=0.8)

My question is: Does the bridging effectively restrain the bottom chord so that the KL=0.8 x 4.5’ is satisfactory?. Shouldn’t I use K=0.8?

Any input will be greatly appreciated.

 

[BAretired]

Where or what is HWVZ?

BA

 

[Streng72]

Oops) my mistake... I meant High-Velocity Hurricane Zones (HVHZ)

 

[BAretired]

In such areas, I would use K = 1.0 although 0.9 is probably justifiable by code.

BA

 

[SteelPE]

I'm not sure I quite understood what you said. Are you indicating that the calculated maximum unsupported length for compression of the bottom chord was around 47" for the 6.2k load you have.... and you are considering adding bridging at 54" (4.5 feet) o.c.? Shouldn't the spacing of the new bridging be less that 47" o.c.?

I must be missing something.

 

[SteelPE]

In looking at the SJI requirements, they calculate the compressive strength of the chord based upon l/r for k series joists. Therefore I would believe that they would use k=1.0 for your calculation.

 

[BAretired]

Bottom chord > 2-3/4" dia. bars...A = 0.88 in^2.
Max compression = 6.2 kips, so stress = 7.02 ksi.
L/r = 54/.375 = 144 if rods are separate.

Cr/A = 71.5 MPa (NBC) or about 10.4 ksi
Cr = 9.1 kips (factored) or about 6.1 kips unfactored.

If the two bars are welded together at 1.5' centers, L/r is much larger, so where is the problem?

Check your numbers again.



BA

 

[Streng72]

Thanks BAretired and SteelPE for your responses.

I was asking about how appropiate would be to assume K sligthly less than one 1. They could provide bridging spaced at either 3' or 4'-6". My calcs using K=1 give me the calculated spacing at 3'-11".

The real problem is that the bottom chord is very weak (I = 0.05 in4) but I don't want to suggest any retrofit to the joist, I feel more comfortable by adding lateral restraints to the bottom chord instead.

From what you two suggested I understand I should stick to the k=1 and provide bridiging every 3 feet.

 

[BAretired]

Streng, your post came only a minute past mine, so I expect you had not read it when you posted.

If the two bottom bars are welded together at 1.5' centers, I don't believe you have a problem with 4.5' center bracing. Perhaps the original bracing is adequate.

BA

 

[BAretired]

Please review your calculation of I.

BA

 

[Streng72]

Thanks again BAretired.

My concern (or confusion) is that the program tells me that the bottom chord compressive stress exceeds F'e (Euler Buckling). I am rechecking the computer model, which is basically a truss type joist with pinned ends everywhere. I placed vertical rollers in three nodes along the bottom chord to simulate the bridging. However, I just noticed that that I did not enter Lbyy nor Lbzz for the bottom chord element which is modeled as a one single member throughout (supposedly the program would calculate the unbraced length based on the nodes along the element).

I was thinking the bridging would work as an element that "breaks" KL/r.

My inertia calcs are:
Ixx = 2 [pi*(D/2)^4]/4 with actual D=0.85" (not 0.75" sorry)

 

[MiketheEngineer]

Both end supports are fixed?? Can you justify that?? And the joints are not truly pinned - maybe not totally fixed - but I would think that since they are welded - it would be close??

 

[BAretired]

So A = 2(0.5675) = 1.1349
and Ixx = 0.0512 in^4
and Iyy = 0.0512 + 2(0.5675)0.85/2 = 0.53357 in^4
and ryy = [√](0.5336/1.1349 = 0.47

L/ryy = 54/0.47 = 115

Cr/A = 99 MPa or 14.3 ksi
Cr = 14.3* 1.135 = 16.3 kips (factored load)

Safety Factor = 16.3/6.2 = 2.6 (no problem).

BA

 

[BAretired]

Sorry, forgot to take the root in ryy expression:

ryy = ?(0.5336/1.13490) = 0.47 = 0.685

L/ryy = 54/0.47 = 115 = 54/0.685 = 79

Cr/A = 99 MPa or 14.3 ksi = 149 MPa or 21.6 ksi
Cr = 14.3* 1.135 = 16.3 kips (factored load) = 21.6*1.135 = 24.5 kips

Safety Factor = 16.3/6.2 = 2.6 (no problem) 24.5/6.2 = 3.96 (still no problem).

BA

 

[Streng72]

BAretired... Thanks a lot for your insights !!!
What is your exact refeence to look for Cr/A=149Mpa ???

 

[BAretired]

Streng72,

My reference is the [!]Handbook of Steel Construction 2004 published by CISC[/!] (Canadian Institue of Steel Construction). I don't have a copy of the AISC handbook, but I imagine it contains a similar set of tables in Imperial units.

If not, it will certainly contain a formula to calculate the factored compressive resistance for a member taking into account the slenderness ratio.

BA

 

[BAretired]

Streng72,

To be complete, the Lx/rxx should also be checked. It would be 18/(0.85/4) = 84.7 which is a little more than Ly/ryy, so Cr/A = 140 MPa or 20.3 ksi. This means that the factored load is 20.3*1.1349 = 23.0 kips. If you are using ASC, the allowabel compression would be 23/1.5 by my code = 15.3 kips. I believe your code uses a higher safety factor, so you would need to adjust for that, but in any event your value of 6.2 kips is no problem.

BA

 

[BAretired]

ASC should read ASD (allowable strength design).

BA

 

[Brad805]

BA, you are a heck of a good guy. I am not so sure I am going to sit around and do calcs when I retire.

Nothing to add, sorry.

Brad

 

[BAretired]

Well Brad, I suppose I should get a hobby, but so far this is it. What I don't understand is how all of the guys who have not retired have time to respond in such detail to some of these questions. They must stay up late finishing their work.

BA