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Reinforced Concree Design

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

klu/r > 34 - 12(M1/M2)

where k is slenderness ratio, lu 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 (klu/r = 40).

The design moment shall be amplified as 

Mc = dnsM2

where

dns = Cm/(1-Pu/0.75Pc ³ 1 is moment magnification factor  for non-sway frame,

Cm = 0.6+0.4(M1/M2) ³ 0.4

Pu is factored column load, and 

Pc = p2EI/(klu) is Euler's critical buckling axial load,

EI shall be taken as

EI = (0.2EcIg+Es/Ise)/(1+bd) or EI = 0.4EcIg/(1+bd)

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: Pu = 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, dt = 0.5 in, ds = 1 in

Gross moment of inertia: Ig = (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, Ec = 57Ö4000 = 3605 ksi

Elastic modulus of steel: Es = 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: Ise = (0.72 in2)(7/2)2*2 = 17.6 in4.

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

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

or EI = (0.2EcIg+Es/Ise)/(1+bd) = [0.2 (3605)(144)+(29000)(17.6)]/(1+0.57) = 1.11x106 kip-in2

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

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

Moment magnification factor, dns = Cm/(1-Pu/0.75Pc)  = (0.911)/[1-350/1087] = 1.6

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