how to start the design of steel columns

[nedians]

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.

 

[par060]

The manual of Steel construction from AISC has column load tables....this is a pretty good starting point

 

[Pwilliams]

A good start would be the Columns Load tables. All the columns listed in the tables satisfied noncompact section limits and kl/r = 200. If the building (assuming it is a building) is a braced frame partially restrained, the section obtained from the tables will be adequate. If it is a braced rigid frame, you will have to take into account for combined stress. Regarding the requied area method, the first value in the tables when kl/r=0 is Fy*Area.

Paul

 

[engcomp]

Have a look at "5. COLUMNS" in...

http://engs-comp.com/mathserv/project1.html

and let me know if this would answer your question?

Helmut
engcomp@pbq.com.au

 

[Gourile]

See 'Design of Steel Structures' by Bowles. It gives you an estimating formula obtained from the AISC interaction formula to guess for preliminary area and radius of jiration of the column by just knowing the load and moment.

 

[macknwo]

Hi, In the AISC specifications there is an aprox. method of W shape column design wich uses the moments applied to the column and converts them to an equivelant axial compression force using U and M factors. With the axial load you can search tables in the AISC for the column that will resist this new axial load. After you have done this you must check the column's stability using the equation given by the AISC for columns under axial and moment loads.

 

[engcomp]

To design the columns, we need a data base and a selection procedure. Process the lines that set up the data base and the code lines that set up the procedure.

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

 

[IJR]

Helmut, that stuff you wrote will get you into trouble one day, cause seems you are the pal to look at when I want to do some mathcad kind stuff. Cool notes Helmut and respects

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