how to start the design of steel columns
how to start the design of steel columns
(OP)
How should one start the design of steel columns , should we assume kl/r or Area required or which factor should be assumed and why? according to aisc.
When was the last time you drove down the highway without seeing a commercial truck hauling goods?
Download nowINTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS Come Join Us!Are you an
Engineering professional? Join Eng-Tips Forums!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail. Posting GuidelinesJobs |
how to start the design of steel columns
|
RE: how to start the design of steel columns
RE: how to start the design of steel columns
Paul
RE: how to start the design of steel columns
http://engs-comp.com/mathserv/project1.html
and let me know if this would answer your question?
Helmut
engcomp@pbq.com.au
RE: how to start the design of steel columns
RE: how to start the design of steel columns
RE: how to start the design of steel columns
1. Set up data base
from AISC design Table 3.1-4(A)
First a label for use in output:
STRUT = {"310UC 158","310UC 137","310UC 118","310UC 96.8"}
STRUT = {STRUT, "250UC 89.5", "250UC 72.9", "200UC 59.5"}
STRUT = {STRUT, "200UC 52.2", "200UC 46.2", "150UC 37.2"}
STRUT = {STRUT, "150UC 30.0", "150UC 23.4", "100UC 14.8"}
Then some properties (variable names are self-explanatory):
W_KGPM = {158, 137, 118, 96.8, 89.5, 72.9, 59.5, 52.2}
W_KGPM = {W_KGPM, 46.2, 37.2, 30, 23.4, 14.8}
A_MM2 = {20100, 17500, 15000, 12400, 11400, 9320, 7620}
A_MM2 = {A_MM2, 6660, 5900, 4730, 3860, 2980, 1890}
RX_MM = {139, 137, 136, 134, 112, 111, 89.7, 89.1, 88.2}
RX_MM = {RX_MM, 68.4, 67.5, 65.1, 41.1}
RY_MM = {78.9, 78.2, 77.5, 76.7, 65.2, 64.5, 51.7, 51.5}
RY_MM = {RY_MM, 51, 38.5, 38.1, 36.6, 24.5}
TANA[1:13] = 0.0 := 0.
from AISC design Table 3.1-4(B)
FY_MPA[1:5] = 280 := 280
FY_MPA[6:10] = 300 := 300
FY_MPA[11:13] = 320 := 320
KF[1:13] = 1.0 := 1.
from AISC design Table 8.1-6
PHIMX_KNM = {676, 580, 494, 422, 309, 266, 177, 154}
PHIMX_KNM = {PHIMX_KNM, 133, 83.6, 71.9, 50.7, 21.4}
PHIMY_KNM = {305, 261, 222, 187, 143, 123, 80.6, 70.3}
PHIMY_KNM = {PHIMY_KNM, 60.3, 36.9, 31.7, 21.2, 9.91}
from AISC design Table 6.1
ALPHA_B[1:13] = 0.0 := 0.
Sort data in ascending order of W_KGPM
STRUT = statsort(W_KGPM, STRUT)
PHIMX_KNM = statsort(W_KGPM, PHIMX_KNM)
PHIMY_KNM = statsort(W_KGPM, PHIMY_KNM)
A_MM2 = statsort(W_KGPM, A_MM2)
RX_MM = statsort(W_KGPM, RX_MM)
RY_MM = statsort(W_KGPM, RY_MM)
TANA = statsort(W_KGPM, TANA)
FY_MPA = statsort(W_KGPM, FY_MPA)
KF = statsort(W_KGPM, KF)
ALPHA_B = statsort(W_KGPM, ALPHA_B)
W_KGPM = statsort(W_KGPM)
2. Procedure for column design
This is called when P and LE are given. Process the whole block from PROC SelCol to END PROC. The procedure stays active for the duration of the project. It must be reprocessed if you wish to use it when you reload the project.
PROC SelCol
LAMBDA_N = (LE/RY_MM)*sqrt(KF)*sqrt(FY_MPA/250)
ALPHA_A = 2100*(LAMBDA_N-13.5)/(LAMBDA_N^2 - 15.3*LAMBDA_N + 2050)
LAMBDA = LAMBDA_N + ALPHA_A*ALPHA_B
ETA = max(0, 0.00326*(LAMBDA-13.5))
XI = ((LAMBDA/90)^2 + 1 + ETA)/(2*(LAMBDA/90)^2)
ALPHA_C = XI*(1 - sqrt(1 - (90/(XI*LAMBDA))^2))
NS = KF*A_MM2*FY_MPA
PHINC = min(PHI*ALPHA_C*NS, PHI*NS)/1000
for J = 1 to Endvalid(PHINc)+1
if J > Endvalid(PHINc) then
print
print sprint("No suitable column for P(-1) at P(0)", P, LE)
elseif PHINc[J] >= P then
print sprint("for P(-1) kN use &: P(-1) at P(0)", P, STRUT[J], PHINc[J], LE)
exit for
end if
next J
END PROC
3. Loads and sizes
**** PHI = .9
**** from B8, P = DLR[8,1]+LLR[8,1]+PLRR[8,1] := 115.1491
**** plus B9, P = P + DLR[9,1]+LLR[9,1]+PLRL[9,1] := 273.7722
**** plus B14, P = P + DLR[14,1]+LLR[14,1]+PLRL[14,1] := 694.5051
**** plus B18, P = P + DLR[18,1]+LLR[18,1] := 792.3763
**** Effective length in mm, LE = 2900
SelCol
Output by SelCol:
for 792.4 kN use 150UC 37.2: 846.8 at 2900
**** load in kN, P = P + V := 1557.2835
**** effective length in mm, LE = 2100
SelCol
Output by SelCol:
for 1557.3 kN use 200UC 52.2: 1591.4 at 2100
**** during erection, P = 2.5*13.9*12.5/4 := 108.5938
**** effective length in mm, LE = 10000
SelCol
Output by SelCol:
for 108.6 kN use 200UC 46.2: 191.4 at 10000
How do you process this little code? In any word processor, using MATHSERV
(see http://engs-comp.com/mathserv/index.html)
or, slightly modified, using any BASIC interpreter.
Let me know if this helps? Helmut
engcomp@pbq.com.au
RE: how to start the design of steel columns
Now back to nedians
Suppose nedians is not in The States:
Here is some basic stuff:
Columns, being elements mainly under compression tend to be uneconomical if not selected wisely.
1) If your column has too much axial force, then YES do your simple Force/Area stuff but here is the trap: You will have buckling problems. and buckling no matter what code you use depends on KL/r, so make sure your KL/r is not less than some recommended value, usually 200.
At the center of a building column section based on Force/Area tend to be big enough and KL/r will be very low by habit.
To utilize these columns efficiently especially when you have pin ended columns try to get them square, ie rx=ry.
2) If you have significant moment then your force/Area approach wont usually work. This is true with sway frames having less number of columns, cause as they sway they attract large moments at ends. Here you have a beam column
You gotta do the force/area+Moment/Modulus stuff as a rough guess and then check the stability equations.
Practice a lot and use the steel tables available in your location. Note that most steel sections are proportioned to work efficiently as columns (In UK they call these Universal columns(UCs)and In US they have Wide flanges) and beams (In UK they call those universal beams UBs, and also sections that have both bending stiffness and a lot of stiffness in the weaker direction to do some beam column action.
Next columns with a lot of axial load can be tubes, pipes, double angles etc.
Regs
IJR