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For a slender column in a braced frame that is subjected to axial
compression and moments. If M_{1} is
the smaller and M_{2} is
the larger moment, the moment need to be design for magnified moment
if the ratio

kl_{u}/r >
34 - 12(M_{1}/M_{2})

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.

M_{1}/M_{2 }is
positive if the column is bent in single curve and M_{1}/M_{2 }is
not to be taken less than -0.5 (kl_{u}/r = 40).

The design moment shall be amplified as

M_{c} = d_{ns}M_{2}

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(M_{1}/M_{2}) ³
0.4

P_{u} is factored
column load, and

P_{c} = p^{2}EI/(kl_{u})^{2 }is
Euler's critical buckling axial load,

EI shall be taken as

EI = (0.2E_{c}I_{g}+E_{s}/I_{se})/(1+b_{d}) or
EI = 0.4E_{c}I_{g}/(1+b_{d})

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

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: M_{1} =
-35 kips, M_{2} =
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 in^{2}.

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 in^{4}.

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 in^{2})
=1.44 in^{2}.

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 in^{2})(7/2)^{2}*2 = 17.6 in^{4}.

The ratio of factored load, b_{d} =
200/350 = 0.57

The flexural stiffness, EI = 0.4E_{c}I_{g}/(1+b_{d})=
0.4(3605)(144)/(1+0.57) = 1.58x10^{6} kip-in^{2}.

or EI
= (0.2E_{c}I_{g}+E_{s}/I_{se})/(1+b_{d})
= [0.2 (3605)(144)+(29000)(17.6)]/(1+0.57) = 1.11x10^{6} kip-in^{2}.

The critical load, P_{c} = p^{2}EI/(kl_{u})^{2 }= p^{2}(1.58x10^{6})/(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