Analysis of 40-year-old bar joist in HWVZ
Analysis of 40-year-old bar joist in HWVZ
(OP)
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.
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.






RE: Analysis of 40-year-old bar joist in HWVZ
BA
RE: Analysis of 40-year-old bar joist in HWVZ
RE: Analysis of 40-year-old bar joist in HWVZ
BA
RE: Analysis of 40-year-old bar joist in HWVZ
I must be missing something.
RE: Analysis of 40-year-old bar joist in HWVZ
RE: Analysis of 40-year-old bar joist in HWVZ
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
RE: Analysis of 40-year-old bar joist in HWVZ
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.
RE: Analysis of 40-year-old bar joist in HWVZ
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
RE: Analysis of 40-year-old bar joist in HWVZ
BA
RE: Analysis of 40-year-old bar joist in HWVZ
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)
RE: Analysis of 40-year-old bar joist in HWVZ
RE: Analysis of 40-year-old bar joist in HWVZ
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
RE: Analysis of 40-year-old bar joist in HWVZ
ryy = √(0.5336/1.13490) =
0.47= 0.685L/ryy =
54/0.47 = 115= 54/0.685 = 79Cr/A =
99 MPa or 14.3 ksi= 149 MPa or 21.6 ksiCr =
14.3* 1.135 = 16.3 kips (factored load)= 21.6*1.135 = 24.5 kipsSafety Factor =
16.3/6.2 = 2.6 (no problem)24.5/6.2 = 3.96 (still no problem).BA
RE: Analysis of 40-year-old bar joist in HWVZ
What is your exact refeence to look for Cr/A=149Mpa ???
RE: Analysis of 40-year-old bar joist in HWVZ
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
RE: Analysis of 40-year-old bar joist in HWVZ
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
RE: Analysis of 40-year-old bar joist in HWVZ
BA
RE: Analysis of 40-year-old bar joist in HWVZ
Nothing to add, sorry.
Brad
RE: Analysis of 40-year-old bar joist in HWVZ
BA
RE: Analysis of 40-year-old bar joist in HWVZ
A = 2(pi*0.85^2/4) = 2(0.5675) = 1.1349
Ixx = 0.0512 in^4
Iyy = 0.0512 + 2(0.5675)(0.85/2)^2 = 0.256 in^4
ryy = √(0.256/1.1349) = 0.475
Ly/ryy = 54/0.475 = 114
Lx/rxx = 18/(.85/4) = 84.7 (so Ly controls)
Cr/A = 100 MPa or 14.5 ksi
Cr = 14.5* 1.1349 = 16.4 kips (factored load)
Safety Factor = 16.4/6.2 = 2.65 (no problem).
I think this is correct, Streng721, but it is best if you check it for yourself.
BA
RE: Analysis of 40-year-old bar joist in HWVZ
Brad
RE: Analysis of 40-year-old bar joist in HWVZ
Many thanks for taking the time to do all these calcs. You REALLY helped me out (my first time analyzing an old bar joist with a PC model).
With these hand calcs I was able to figure out what was wrong with the computer model. That's why I always prefer to run hand calculations over PC models. Many times you don't know if you have one parameter wrong in the computer that yields just the wrong output.
Thanks again !
RE: Analysis of 40-year-old bar joist in HWVZ
BA