Design of long column in sway frame

Moment magnification for columns in unbraced frames (sway)

Determine if the frame is a sway frame

1. The frame can be assumed as non-sway if the end moment from second-order analysis not exceeds 5% of the first-order end moment.

2. The frame can be assumed as non-sway if stability index, 

Q = åP_u D_o/ V_ul_c ³ 0.05

where åP_u and V_u are the total vertical load and the story shear,  D_o is the first-order relative deflection between the top and bottom of that story due to V_u, and  l_c is the length of the column.

3. The moment need not to be design for magnified moment if the ratio

kl_u/r £ 22

where k is slenderness ratio, l_u is unsupported length, r is radius of gyration. 

Calculating magnified moments

If M₁ is the smaller and M₂ is the larger moment, the design moment shall be amplified as 

M₁ = M_1ns+d_sM_1s

M₂ = M_2ns+d_sM_2s

where M_1ns and M_2ns are moments from loadings that are not contribute to sway (i.e. gravity load), and M_2s and M_2s are moments from loading that contribute to sway (i.e. wind and seismic)

1. The magnified sway moment d_sM_s can be determined by a second-order analysis based on the member stiffness reduced for crack section.

2. The magnified sway moment can be calculated as

d_sM_s =M_s/(1-Q)  ³ M_s when 1 £ d_s £ 1.5

3. The magnified sway moment can be calculated as

d_sM_s =M_s/[1-åP_u/(0.75åP_c )]³ M_s 

where åP_u is the summation for all the factored vertical loads in a story and åP_c = åp²EI/(kl_u)²  is the summation of critical bucking loads for all the columns  in a story from non-sway frame.

Limitations

1. The ratio of second-order lateral deflection to the first-order lateral deflection based on factored dead and live loads plus lateral load shall not exceed 2.5.

2. The value of Q shall not exceed 0.6.

3. The magnification factor d_s shall not exceed 2.5.

Example: 

A reinforced concrete moment frame has four 18"x18" reinforced concrete columns.

Design data:

Factored column axial loads:

Live load: P_L1 = 150 kips

Dead load: P_D1 = 200 kips

Lateral load: P_W1 = 40 kips

Column 2 & 3 (interior columns)

Live load:  P_L2 = 300 kips

Dead load: P_D2 = 400 kips

Lateral load: P_W2 = 20 kips

Factored column moments:

Live load moment: M_L11 = -25 ft-kips, M_L12 = 40 ft-kips

Dead load moment: M_D11 = -35 ft-kips, M_D12 = 45 ft-kips

Lateral load moment: M_W11 = -75 ft-kips, M_W12 = 80 ft-kips

Column 2 & 3 (interior columns)

Live load moment: M_L21 = -15 ft-kips, M_L22 = 25 ft-kips

Dead load moment: M_D21 = -30 ft-kips, M_L22 = 40 ft-kips

Lateral load moment: M_W21 = -65 ft-kips, M_W22 = 75 ft-kips

Compressive strength of concrete: f_c' = 4000 psi

Yield strength of steel: f_y = 60 ksi

Unsupported length of column: l_u = 10 ft

Requirement: Calculate magnification design moments for column 2 and 3.

1. Calculate total factored loads

Column 2 & 3: P_u = (P_L2+P_D2+P_W2) = 720 kips

All column:

åP_u = (P_L1+P_D1+P_W1+P_L2+P_D2+P_W2) = 2220 kips

Column size: b = 18 in. h = 18 in.

Gross area of column: Ag = bh = 324 in².

Assume concrete cover: 1.5 in for interior column

Assume #4 ties, d_t = 0.5 in and #8 bars, d_s = 1 in, for vertical reinforcement

Calculate gross moment of inertia of column: I_g = bh³/12 = 8748 in⁴.

Radius of gyration, r = ÖI_g/A_g = 5.2 in

Assume slenderness factor, k =2 for moment frame with side-sway, 

the slenderness ratio, k l_u/r = 46. > 22 (long column)

Young's modulus of concrete, E_c = 57 (Ö4000) = 3605 ksi

Elastic modulus of steel: E_s = 29000 ksi

Assume 1% of reinforcement area, A_s = 0.01 A_g = 3.24 in².

Assume that reinforcement are placed equally at each side of the column, the distance between vertical reinforcement is

18-2(1.5)-2(0.5)-1 = 12 in

the moment of inertia of reinforcement, I_se = As (13/2)² = 137 in⁴.

Factor, b_d = P_D2/(P_D2+P_L2+P_W2) = 0.556

The stiffness factor, EI = (0.2E_cI_g+E_s/I_se)/(1+b_d) = 6.61x10⁶ kip-in².

or EI = 0.4E_cI_g/(1+b_d) = 8.11x10⁶ kip-in².

Calculate magnification factor for non-sway moment:

P_c = p²EI/(kl_u)²  = 1390 kips.

Assume C_m = 1 (Transverse load at top and bottom of column)

d_ns = C_m/[1-P_u/(0.75P_c )] = 3.235

Calculate magnification factor fo sway moment:

Euler's critical buckling axial load of all columns, åP_c = p²EI/(kl_u)²  x 4 = 5558 kips.

The moment magnification factor, d_s = 1/[1-åP_u/(0.75åP_c )] = 2.14

Magnified design for column 2 and 3, M_u2 = d_ns (M_L22+M_D22)+d_s (M_W2) = 370.7 ft-kip