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Reinforced Concrete Design

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 = åPu Do/ Vulc ³ 0.05

where åPu and Vu are the total vertical load and the story shear,  Do is the first-order relative deflection between the top and bottom of that story due to Vu, and  lc is the length of the column.

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

klu/r £ 22

where k is slenderness ratio, lu 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 = M1ns+dsM1s

M2 = M2ns+dsM2s

where M1ns and M2ns are moments from loadings that are not contribute to sway (i.e. gravity load), and M2s and M2s are moments from loading that contribute to sway (i.e. wind and seismic)

1. The magnified sway moment dsMs 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

dsMs =Ms/(1-Q)  ³ Ms when 1 £ ds £ 1.5

3. The magnified sway moment can be calculated as

dsMs =Ms/[1-åPu/(0.75åPc )]³ M

where åPu is the summation for all the factored vertical loads in a story and åPc = åp2EI/(klu) 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 ds shall not exceed 2.5.

Example: 

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

Design data:

Factored column axial loads:

Column 1 & 4 (Exterior columns)

Live load: PL1 = 150 kips

Dead load: PD1 = 200 kips

Lateral load: PW1 = 40 kips

Column 2 & 3 (interior columns)

Live load:  PL2 = 300 kips

Dead load: PD2 = 400 kips

Lateral load: PW2 = 20 kips

Factored column moments:

Column 1 & 4 (Exterior columns)

Live load moment: ML11 = -25 ft-kips, ML12 = 40 ft-kips

Dead load moment: MD11 = -35 ft-kips, MD12 = 45 ft-kips

Lateral load moment: MW11 = -75 ft-kips, MW12 = 80 ft-kips

Column 2 & 3 (interior columns)

Live load moment: ML21 = -15 ft-kips, ML22 = 25 ft-kips

Dead load moment: MD21 = -30 ft-kips, ML22 = 40 ft-kips

Lateral load moment: MW21 = -65 ft-kips, MW22 = 75 ft-kips

Compressive strength of concrete: fc' = 4000 psi

Yield strength of steel: fy = 60 ksi

Unsupported length of column: lu = 10 ft

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

1. Calculate total factored loads

Column 2 & 3: Pu = (PL2+PD2+PW2) = 720 kips

All column:

åPu = (PL1+PD1+PW1+PL2+PD2+PW2) = 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, dt = 0.5 in and #8 bars, ds = 1 in, for vertical reinforcement

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

Radius of gyration, r = ÖIg/Ag = 5.2 in

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

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

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

Elastic modulus of steel: Es = 29000 ksi

Assume 1% of reinforcement area, As = 0.01 Ag = 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, Ise = As (13/2)2 = 137 in4.

Factor, bd = PD2/(PD2+PL2+PW2) = 0.556

The stiffness factor, EI = (0.2EcIg+Es/Ise)/(1+bd) = 6.61x106 kip-in2.

or EI = 0.4EcIg/(1+bd) = 8.11x106 kip-in2.

Calculate magnification factor for non-sway moment:

Pc = p2EI/(klu) = 1390 kips.

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

dns = Cm/[1-Pu/(0.75Pc )] = 3.235

Calculate magnification factor fo sway moment:

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

The moment magnification factor, ds = 1/[1-åPu/(0.75åPc )] = 2.14

Magnified design for column 2 and 3, Mu2 = dns (ML22+MD22)+ds (MW2) = 370.7 ft-kip