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Design of steel columns (LRFD, Load & resistant factor design)

Design code: AISC LRFD 2nd edition, 1994.

Design requirements

1. Maximum factored axial compressive Force, P_u must not exceed axial strength, fP_n.

Design procedure:

1. Calculate factored axial compressive force, P_u.
2. Select a trial beam size.
3. Calculate allowable axial force, fP_n (see below)

Determine factored axial force, P_u

Load factors:

1.4D                                                                            [LRFD A4-1]

1.2D+1.6L+0.5(L_r or S or R)                                      [LRFD A4-2]

1.2D+1.6(L_r or S or R)+(0.5L or 0.8W)                     [LRFD A4-3]*

1.2D+1.3W+0.5L+0.5(L_r or S or R)                           [LRFD A4-4]*

1.2D±1.0E+0.5L+0.2S                                                [LRFD A4-5]*

0.9D±(1.3W or 1.0E)                                                  [LRFD A4-6]

where D is dead load, L is live load, L_r is roof live load, S is snow load, R is rain load, W is wind load, E is seismic load.

*Load factored for L is 1 for garage, public assembly and area over 100 psf live load.

Determine axial compressive strength,  fP_n (LRFD) for W, I, tube, & pipe section

1. Determine slenderness ratio in both axes.

K_xL_x/r_x and K_yL_y/r_y

where

L_x ,L_y are laterally unsupported lengths in X and Y direction,

r_x ,r_x are radius of gyrations in X and Y direction

K_y ,K_y are slenderness factors in X and Y direction and can be determined as

2. Calculate, slenderness parameter l_c with the larger value of  K_xL_x/r_x and K_yL_y/r_y with

l_c = (KL/rp)ÖF_y/E

where E is elastic modulus, F_y is yield strength of steel.

3. If l_c £ 1.5, then F_cr = (0.685 ^lc2) F_y    

If l_c > 1.5, then F_cr = (0.877 / l_c²) F_y

4. The axial compressive strength, fP_n = 0.85 A_gF_cr, where Ag is gross section area of member.

Example 1:

Situation: A structural column is supporting roof

Design Code: AISC LRFD 2nd edition

Roof live load: W_L = 20 psf

Roof dead load: W_D = 20 psf

Unsupported Length of column: L_ux = 15 ft, L_uy = 15 ft

Top and bottom of column is pinned

Tributary area 30 ft x 30 ft

Material: ASTM A36, yield strength, Fy = 50 ksi

Requirements:  Select a W8 beam

Solution:

Try W6x15, A = 4.43 in², r_x = 2.56 in, r_y = 1.46 in

Total column load: Pu = {[(1.2)(20)+(1.6)(20)](30)(30)+1.2(15)(15)}/1000 =47.1 kip

Slenderness factor, K_x = 1, K_y = 1

Slenderness ratio, K_xL_ux/r_x = 70.3, K_yL_uy/r_y = 123.3

Elastic modulus, E = 29000 ksi

Slenderness parameter, l_c = (123.3/3.14)Ö(50/29000) = 1.63   > 1.5

Fcr = [0.877/(1.63)²](50) = 16.5 ksi

Compressive strength of column, fP_n = 0.85(4.43)(16.5) = 62.2 kips   (O.K.)

Example 2:

Situation: In example 1 assume that the column is supported laterally at mid-height of column in minor axis,

Requirement: Select a more economical column

Solution:

Try W6x9, A = 2.68 in², r_x = 2.47 in, r_y = 0.905 in.

Lateral unsupported length, L_ux = 15 ft, L_uy = 7.5 ft

Slenderness ratio, K_xL_ux/r_x = 72.8, K_yL_uy/r_y = 99.4

Slenderness parameter, l_c = (99.4/3.14)Ö(50/29000) = 1.31   < 1.5

F_cr = [(0.685)^1.312](50) = 26 ksi

Compressive strength of column, fP_n = 0.85(2.68)(26) = 59.2 kips   (O.K.)