Design of long column in non-sway frame (ACI 318-02,05)

Moment magnification for columns in braced frames (non-sway)

For a slender column in a braced frame that is subjected to axial compression and moments.  If M₁ is the smaller and M₂ is the larger moment, the moment need to be design for magnified moment if the ratio

kl_u/r > 34 - 12(M₁/M₂)

where k is slenderness ratio, l_u is unsupported length, r is radius of gyration. k shall not be taken as 1 unless analysis shows that a lower value is justified.

M₁/M₂ is positive if the column is bent in single curve and M₁/M₂ is not to be taken less than -0.5 (kl_u/r = 40).

The design moment shall be amplified as 

M_c = d_nsM₂

where

d_ns = C_m/(1-P_u/0.75P_c)  ³ 1 is moment magnification factor  for non-sway frame,

C_m = 0.6+0.4(M₁/M₂) ³ 0.4

P_u is factored column load, and 

P_c = p²EI/(kl_u)²  is Euler's critical buckling axial load,

EI shall be taken as

EI = (0.2E_cI_g+E_s/I_se)/(1+b_d) or EI = 0.4E_cI_g/(1+b_d)

Where  is the ratio of maximum factored axial dead load to total factored load

Example: 

A 12"x12" interior reinforced concrete column is supporting a factored axail dead load of 200 kips and a factored axial live load of 150 kip. Factored column end moments are -35 kip-ft and 45 kip-ft. The column is a long column and has no sway.

Design data:

Total Factored axial: P_u = 350 kips

Factored moment: M₁ = -35 kips, M₂ = 45 kips (bent in single curve)

Compressive strength of concrete: 4000 ksi

Yield strength of steel: 60 ksi

Unsupported length of column: 10 ft

Requirement: Determine the magnified design moment

Column size: b = 12 in, h = 12 in

Gross area: Ag = 144 in².

Concrete cover: 1.5 in

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

Gross moment of inertia: I_g = (12 in)⁴/12 = 1728 in⁴.

Radius of gyration, r = Ö(1728/144) = 3.5 in or r = 0.3(12 in) = 3.6 in

Assume slenderness factor, k = 1 without detail analysis

Slenderness factor, klu/r = (1)(120 in)/3.6 in = 35 > 34-12(35/45) = 25, long column

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

Elastic modulus of steel: E_s = 29000 ksi

Assume 1% area of reinforcement: As = (0.01)(144 in²) =1.44 in².  

Assume half of the reinforcement at each side of column, distance between rebas = 12-1.5*2-0.5*2-1 = 7 in

Moment of inertia of steel reinforcement: I_se = (0.72 in²)(7/2)²*2 = 17.6 in⁴.

The ratio of factored load, b_d = 200/350 = 0.57

The flexural stiffness, EI = 0.4E_cI_g/(1+b_d)= 0.4(3605)(144)/(1+0.57) = 1.58x10⁶ kip-in². 

or EI = (0.2E_cI_g+E_s/I_se)/(1+b_d) = [0.2 (3605)(144)+(29000)(17.6)]/(1+0.57) = 1.11x10⁶ kip-in². 

The critical load, P_c = p²EI/(kl_u)²  = p²(1.58x10⁶)/(35)² = 1087 kip

Factor Cm = 0.6+0.4(35/45) = 0.911

Moment magnification factor, d_ns = C_m/(1-P_u/0.75P_c)  = (0.911)/[1-350/1087] = 1.6

The magnified design moment, M_c = 1.6 (45) = 71.8 ft-kip