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9 CHAPTER 9: COLUMNS 9.1 First-Order versus Second-Order Analysis A first-order analysis is based on the initial geometry of the structure, assuming elastic behavior. On the other hand, second-order analysis considers the influence of lateral drift, cracking, member curvature, shrinkage and creep on the forces in the structure.

9.2 Sway and Nonsway Frames 9.2.1 Nonsway Frames Structural frames whose joints are restrained against lateral displacement by attachment to rigid elements or bracing are called “nonsway frames”, shown in Figure 9.1. According to ACI Code 10.10.5.1 a column in a structure is nonsway if the increase in column end moments due to second-order effects does not exceed 5 percent of the firstorder end moments. Moreover, ACI Code 10.10.5.2 assumes a story within a structure is nonsway if: Q=

∑ Pu ∆o

(9.1)

Vus lc

is less than or equal to 0.05, where Q is the stability index which is the ratio of secondary moment due to lateral displacement and primary moment,

∑P

u

is the total factored

vertical load in the story, Vus is the factored horizontal story shear, l c is length of column measured center-to-center of the joints in the frame, and ∆ o is the first-order relative deflection between the top and bottom of that story due to Vus .

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2

Figure 9.1: Nonsway frame 9.2.2 Sway Frames Structural frames, not attached to an effective bracing element, but depend on the bending stiffness of the columns and girders to provide resistance to lateral displacement are called “sway frames”, shown in Figure 9.2.

Figure 9.2: Sway frame

9.3 Effective Length Factor “k” of Columns of Rigid Frames The effective length of column is the length of a column hinged at both ends and having the same buckling load as the column being considered. Thus, the effective length factor k, is the ratio of the effective length to the original length of column.

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The ACI Code R10.10.6.3 recognizes the Jackson and Moreland Alignment Charts, shown in Figure 9.3, to estimate the effective length factor k for a column of constant cross section in a multibay frame. The effective length factor k is a function of the relative stiffness at each end of the column. In these charts, k is determined as the intersection of a line joining the values of ψ at the two ends of the column. The relative stiffness of the beams and columns at each end of the column ψ is given by Eq. (9.2) ψ=

∑E ∑E

c

I c / lc

b

I b / lb

(9.2)

where, l c = length of column center-to-center of the joints l b = length of beam center-to-center of the joints

E c = modulus of elasticity of column concrete E b = modulus of elasticity of beam concrete I c = moment of inertia of column cross section about an axis perpendicular to the plane of buckling being considered. I b = moment of inertia of beam cross section about an axis perpendicular to the plane of buckling being considered.

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4

(a) (b) Figure 9.3: Alignment chart; (a) nonsway frames; (b) sway frames

∑

indicates a summation of all member stiffnesses connected to the joint and lying in the

plane in which buckling of the column is being considered. Consider the two-story frame shown in Figure 9.4. To determine the effective length factor k for column EF,

Figure 9.4: Two-story frame

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ψE =

5

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Ec ( I DE / H1 ) + Ec ( I EF / H 2 ) Eb ( I BE / L1 ) + Eb ( I EH / L2 )

and ψF =

Ec ( I EF / H 2 ) Eb ( I CF / L1 ) + Eb ( I FI / L2 )

ACI Code 10.10.6.3 specifies that for columns in nonsway frames, the effective length factor k should be taken as 1.0, unless analysis shows that a lower value is justified. The ACI Code 10.10.4.1 specifies that the influence of cracking along the length of the member, presence of axial loads, and effects of duration of loads be taken into consideration when calculating k by using reduced values of moment of inertia as follows: Beams -------------------------------- 0.35 I g Columns------------------------------ 0.70 I g Uncracked Walls -------------------- 0.70 I g Cracked Walls ----------------------- 0.35 I g where I g is the gross moment of inertia. As an alternate to using alignment charts to determine k, the following simplified equations are used for computing the effective length factors for nonsway and sway frame members. For columns in nonsway frames, the smaller of Eq. (9.3) and Eq. (9.4) is used k = 0.7 + 0.05 (ψ A + ψ B ) ≤ 1.0

(9.3)

k = 0.85 + 0.05 ψ min ≤ 1.0

(9.4)

where ψ A and ψ B are the values of ψ at the two ends of the column, and ψ min is the smaller of the two values. For columns in sway frames restrained at both ends, k is taken as for ψ m < 2 k=

20 − ψ m 1 + ψm 20

for ψ m ≥ 2

(9.5)

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k = 0.9 1 + ψ m

(9.6)

where ψ m is the average of ψ values at the two ends of the column. For columns in sway frames hinged at one end, k is taken as k = 2.0 + 0.3 ψ

(9.7)

where ψ is the values at the restrained end of the column.

9.4 The ACI Procedure for Classifying Short and Slender Columns According to ACI Code 10.10.1, columns can be classified as short when their effective slenderness ratios satisfy the following criteria: For nonsway frames k lu ≤ 34 − 12 (M1 / M 2 ) ≤ 40.0 r

(9.8)

or For sway frames k l u / r ≤ 22

(9.9)

Furthermore, compression members are considered braced against sidesway when bracing elements have a total stiffness, resisting lateral movement of that story, of at least 12 times the gross stiffness of the columns within the story. where k = effective length factor l u = unsupported length of member, defined in ACI Code 10.10.1.1 as clear distance between floor slabs, beams, or other members capable of providing lateral support, as shown in Figure 9.5.

Figure 9.5: Unsupported length of member

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r = radius of gyration associated with axis about which bending occurs. For rectangular cross sections r = 0.30 h, and for circular sections, r = 0.25 h as specified by ACI Code 10.10.1.2. h = column dimension in the direction of bending. M 1 = smaller factored end moment on column, positive if member is bent single curvature, negative if bent in double curvature. M 2 = larger factored end moment on column, always positive. Chart 9.1 summarizes the process of column design as per the ACI Code.

Column Design Sway frame

klu ≤ 22 r

kl 22 < u ≤ 100 r

klu > 100 r

Non-sway frame

Neglect slenderness (short) Moment magnification method (slender)

Exact P-∆ analysis (slender)

kl u ≤ 34 − 12 ( M 1 / M 2 ) r

100 ≥

M  klu > 34 − 12 1  r  M2 

klu > 100 r

Chart 9.1: Column Ddesign Example (9.1): The frame shown in Figure 9.6 consists of members with rectangular cross sections, made of the same strength concrete. Considering buckling in the plane of the figure, categorize column bc as long or short if the frame is:

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a. Nonsway b. Sway.

Figure 9.6: Frame and loads on column bc Solution: a. Nonsway: For a column to be short, k lu ≤ 34 − 12 (M1 / M 2 ) ≤ 40.0 r l u = 400 − 30 − 30 = 340 cm k is conservatively taken as 1.0. kl/r =

1 (340 ) = 32.38 0.3 (35)

34 − 12 (M1 / M 2 ) = 34 − 12 (− 27 / 40) = 42.1 taken as 40 > 32.38 i.e., column is classified as being short. b. Sway: The column is classified as being short when k l u / r ≤ 22

[0.7 (30)(35)3 / 12 (400)] ψc = = 0.406 [0.35 (30)(60)3 / 12 (900)] + [0.35 (30)(60)3 / 12 (750)]

8

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3 3 [ 0.7 (30 )(35) / 12 (400 ) ] + [ 0.7 (30 )(40) / 12 (450)] ψb = = 0.945 [0.35 (30)(60)3 / 12 (900)] + [0.35 (30)(60)3 / 12 (750)]

Using the appropriate alignment chart, k = 1.14, and

k l u 1.14 (340 ) = = 36.91 > 22 r 0.3 (35)

i.e., column is classified as being slender, or long. ψ could have been evaluated using Eq. (9.5) ψm = k=

0.406 + 0.945 = 0.675 2

20 − 0.675 1 + 0.675 = 1.25 20

9.5 Short Columns Subjected to Axial force and Bending Generally, columns are subjected to axial forces in addition to some bending moments. These moments are generally due to: §

Unsymmetrically placed floors, as shown in Figure 9.7.

Figure 9.7 Unsymmetrically placed floors §

Lateral loading such as wind or earthquake loads, shown in Figure 9.8.

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Figure 9.8: Rigid Frame §

Loads from eccentric loading such as crane loads acting on corbels.

§

End restraints resulting from monolithic action between floor beams and columns.

§

Accidental eccentricity resulting from column misalignment or other execution deficiencies.

9.5.1 Interaction Diagram Unlike pure axial or bending loading, where unique strength exists for a particular section, combined axial load and bending result in an infinite number of strength combinations evaluated by using equilibrium equations and compatibility of strains. When these strength combinations are plotted, a strength curve called “Interaction Diagram” is produced. To plot an interaction diagram for a particular cross section, at least five strength combinations are required, as shown in Figure 9.9. All strength combinations located in the area under the curve are possible safe strength capacities while combinations located outside the curve are failure cases. §

Point “A”:

This point on the curve represents pure axial compression capacity of the column cross section, where the eccentricity e is equal to zero. The nominal axial capacity PnA is given by Eq. (9.10) PnA = 0.85 f c′ (Ag − Asg ) + Asg f y where Asg is total column reinforcement.

(9.10)

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Figure 9.9: Strength interaction diagram §

Point “C”:

This point on the curve represents pure flexural capacity of the column cross section analyzed as doubly reinforced section, where the eccentricity e is equal to infinity. The nominal flexural capacity M nC is given by Eq. (9.11) M nC = C c (d − a / 2 ) + C s (d − d ′) §

(9.11)

Point “B”:

This point is characterized by its maximum bending strength and represents a balanced failure of the column section. Crushing of the concrete occurs simultaneously with yielding of the reinforcement, or ε c = ε cu = 0.003 , and ε t = ε y . §

Point “D”:

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Points along the curve between points A and B are characterized by compression failure of the section. Failure is initiated by crushing of the concrete before the initiation of yielding of the reinforcement, or ε c = ε cu = 0.003 , and ε t < ε y . The eccentricity e is smaller than the eccentricity at balanced failure eb , where the eccentricity increases when moving from point A towards point B along the interaction curve. §

Point “E”:

Points along the curve between points B and C are characterized by tension failure of the section. Failure is initiated by yielding of the reinforcement, or ε c = ε cu = 0.003 , and ε t ≥ 0.005 . The eccentricity e is larger than the eccentricity at balanced failure eb , where the eccentricity increases when moving from point B towards point C along the interaction curve. §

Point “F”:

This point on the curve represents pure axial tension capacity of the column cross section where the eccentricity e is equal to zero. The nominal axial capacity PnF is given by Eq. (9.12) PnF = Asg f y

(9.12)

where Asg is total column reinforcement. 9.5.2 Modes of Failure Three modes of failure are possible for columns subjected to axial force plus bending. 9.5.2.1 Balanced Failure Figure 9.10 shows a rectangular section subjected to an axial load with eccentricity adjusted so as to produce a balanced failure using the principle of static equivalence.

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Figure 9.10: Balanced failure The distance to the neutral axis from the extreme compression fiber is given as  6120  d xb =   6120 + f  y  

(9.13)

The compressive force resisted by concrete is given by C cb = 0.85 fc′ β 1 xb b The tensile force resisted by reinforcement in the tension side is Tb = As f y The compressive force resisted by reinforcement on the compression side of the cross section is C sb = As′ ( f y − 0.85 f c′ )

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In the last of the previous equations, it is assumed that compression reinforcement does  x −d′  yield, or ε s′ ≥ ε y . This can be easily checked from similar triangles, or ε s′ = 0.003  b x b   If yielding of the reinforcement does not occur, where ε s′ < ε y , the compressive force in compression reinforcement is given as C sb = As′ ( f s − 0.85 f c′ ) where the stress in the compression reinforcement is evaluated using Hook’s law, or f s = ε s′ E s From equilibrium of forces in the vertical direction Pnb = C cb + C sb − Tb

(9.14)

Substituting C cb , C sb , and Tb in Eq. (9.14), one gets Pnb = 0.85 fc′ β 1 xb b + As′ ( f y − 0.85 fc′ ) − As f y

(9.15)

From equilibrium of moments, by taking the moments about the centroid of the cross section, Pb eb = M nb = Ccb (d − a / 2 − d ′′) + Csb (d − d ′ − d ′′) + Tb d ′′ and eb =

(9.16)

M nb Pnb

The balanced strain condition represents the dividing point between compression-controlled sections and the transition zone of the strength interaction diagram. 9.5.2.2 Tension Failure Figure 9.11shows a rectangular section subjected to an axial load with eccentricity chosen larger than eb .

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Figure 9.11: Tension failure The compressive force resisted by concrete is given by C c = 0.85 fc′ β 1 x b where

0.003 x = and x = 0.375 d 0.003 + 0.005 d

The tensile force resisted by reinforcement in the tension side is T = As f y The compressive force resisted by reinforcement on the compression side of the cross section is Cs = As′ ( f y − 0.85 fc′) when ε s′ ≥ ε y .

If yielding of the reinforcement does not occur, where ε s′ < ε y , the compressive force in compression reinforcement is given as Cs = As′ ( fs − 0.85 fc′) where the stress in the compression reinforcement is evaluated using Hook’s law, or f s = ε s′ E s From equilibrium of forces in the vertical direction Pn = Cc + C s − T

(9.17)

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Substituting C c , C s , and T in Eq. (9.17), one gets Pn = 0.85 f c′ β 1 x b + As′ ( f y − 0.85 f c′ ) − As f y

(9.18)

From equilibrium of moments, Pn (e + d' ' ) = C c (d − a / 2 ) + C s (d − d' ) is used to evaluate e The nominal flexural strength M n = Pn e . 9.5.2.3 Compression Failure When the nominal compression strength Pn exceeds the balanced nomial compression strength Pnb , or when the eccentricity e is less than eb or when ε t at the extreme layer of steel at the face opposite the maximum compression force is less than ε y , the section is compression controlled. Figure 9.12 shows a rectangular section subjected to an axial load with eccentricity chosen smaller than eb .

Figure 9.12: Compression failure The compressive force resisted by concrete is given by

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C c = 0.85 f c′ a b Since the reinforcement does not yield, the tensile force resisted by reinforcement in the tension side is  d −x  β d −a  T = As fs = As Es ε s = As Es ε cu   = As (6120) 1   x   a  The compressive force resisted by reinforcement on the compression side of the cross section is C s = As′ ( f y − 0.85 f c′ ) for ε s′ ≥ ε y . From equilibrium of forces in the vertical direction Pn = C c + C s − T

(9.19)

Substituting C c , C s , and T in Eq. (9.18), one gets  β d −a  Pn = 0.85 f c′ a b + As′ ( f y − 0.85 f c′ )− As (6120 )  1  a  

(9.20)

From equilibrium of moments, by taking the moments about the axial load C c (d − a / 2 − e − d ′′) + C s (d − e − d ′ − d ′′) + T (d ′′ + e ) = 0 Cc (− d + a / 2 + e + d ′′) + Cs (− d + e + d ′ + d ′′) − T (d ′′ + e ) = 0

(9.21)

Substituting C c , C s , and T in Eq. (9.21), one gets 0.85 fc′ a b (− d + a / 2 + e + d ′′) + As′ ( f y − 0.85 fc′) (− d + e + d ′ + d ′′) β d −a  − As (6120 )   (e + d ′′) = 0  a 

(9.22)

Eq. (9.22) which is a cubic equation in terms of a , can be written in the following form: Aa 3 + B a 2 + C a + D = 0 where, A= 0.425 f c′ b B = 0.85 f c′ b (e + d ′′ − d ) C = As′ ( f y − 0.85 f c′ )(e + d ′ + d ′′ − d ) + As (6120 )(d ′′ + e )

(9.23)

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D = − As (6120) β 1 (d ′′ + e ) d Eq. (9.23) can be solved using Newton-Raphson iteration method, or any available mathematical software. When this iteration method is used, f (a ) = Aa 3 + B a 2 + C a + D

(9.24)

and the first derivative of this function is given by f ′ (a ) = 3 Aa 2 + 2 B a + C

(9.25)

Assume a trial value for a , named a o The first iteration value a1 is given as a1 = a o −

f (a o ) f ′ (a o )

(9.26)

A second iteration is evaluated using a1 evaluated from Eq. (9.26). Repeat for more iterations until you reach a converged value for a . Then, a is substituted in Eq. (9.20) to get Pn . The nominal flexural strength is evaluated by multiplying Pn by e . Example (9.6): For the column cross section shown in Figure 9.13, plot a nominal strength interaction diagram, using five strength combinations at least. Use f c′ = 250 kg / cm 2 , f y = 4200 kg / cm 2 , and E s = 2.04 (10) kg / cm 2 . 6

F Figure 9.13: Column cross section Solution:

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d = 55 –6.25 = 48.75 cm d ′ = 6.25 cm d ′′ = (55.0 – 6.25 –6.25)/2 = 21.25 cm εy = §

4200

2.04 (10)

6

= 0.002

Point “A”:

This is a case of pure axial compression load, the nominal axial load is Pn =

§

1 [0.85 (250) (55 × 35 − 39.27) + 39.27 (4200)]= 565.65 tons 1000

Point “B”:

The strength combination at this point corresponds to a balanced failure.   6120  (48.75) = 28.91 cm xb =   6120 + 4200  a b = 0.85 (28.91) = 24.573 cm Ccb = 0.85 (250)(24.573)(35) / 1000 =182.76 tons  4200  Tb = 19.64   = 82.49 tons  1000   28.91 − 6.25  ε s′ = 0.003   = 0.0023 > ε y  28.91  C sb =

19.64 (4200 − 0.85 × 250 ) = 78.31 tons 1000

From equilibrium of forces in the vertical direction, Pnb = 182.76 + 78.31 − 82.49 = 178.58 tons From equilibrium of moments, 182.76 (48.75 − 24.57 / 2 − 21.25) + 78.31 (48.75 − 6.25 − 21.25) 100 100 82.49 (21.25) = 61.97 t.m + 100

M nb =

19

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The eccentricity causing balanced failure eb = §

20

61.97 = 0.347 m 178.58

Point “E”:

x 0.003 = and d 0.003 + 0.005 x = 0.375 d = 0.375 (48.75 ) = 18.281 cm  18.281 − 6.25  ε s = 0.003   = 0.00197 ≅ ε y  18.281   0.66 − 6.25  ε s′ = 0.003   = − 0.0254  0.66  Cc = 0.85 (250 )(0.85 )(18.281)(35 ) / 1000 = 115 .57 tons  4200  T = 19.64   = 82.49 tons  1000  Cs =

19.64 (4200 − 0.85 × 250 ) = 78.31 tons , assuming that ε s′ ≥ ε y 1000

From equilibrium of forces in the vertical direction, Pn = Cc + C s − T Pn = 115 .57 + 78.31 − 82.49 = 113 .39 tons From equilibrium of moments, Pn (e + d' ' ) = C c (d − a / 2 ) + C s (d − d' ) 0.85 (18.281)   113 .39 (e + 21.25 ) = 115 .57  48.75 −  + 78.31 (48.75 − 6.25 ) 2   and e= 51.15 cm  51.15  M n = 111 .39   = 56.97 t .m  100  §

Point “D”:

The eccentricity is set at 0.25 m (smaller than 0.347 m) in order to locate a compression failure strength combination.

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A = 0.425 (250 )(35) = 3718.75 kg / cm B = 0.85 (250 )(35)(25 + 21.25 − 48.75) = − 18593.75 kg C = 19.64 (4200 − 0.85 × 250)(25 + 6.25 + 21.25 − 48.75) + 19.64 (6120 )(21.25 + 25) = 5852781.37 kg.cm D = − 19.64 (6120)(0.85)(21.25 + 25)(48.75) = − 2.3035529 (10 ) kg .cm 2 8

f (a ) = 3718.75 a 3 − 18593.75 a 2 + 5852781.37 a − 2.3035529 (10)

8

f ′ (a ) =11156 .25 a 2 − 37187.5 a + 5852781.37 Let a o = 25 cm as a first trial f (25) = − 37551380.75 f ′ (25) = 11895750 .12 The first iteration value a 1 is given as a 1 = 25 −

f (25) 37551380.75 = 25 + = 28.15 cm f ′ (25) 11895750 .12

f (28.15) = 2619416.5 f ′ (28.15) = 13646417 a 2 = 28.15 −

f (28.15) 2619416.5 = 28.15 − = 27.96 cm f ′ (28.15) 13646417

f (27.96 ) = 37237.75 f ′ (27.96 ) = 13534547 a 3 = 27.96 −

f (27.96) 37237.75 = 27.96 − = 27.96 cm f ′ (27.96 ) 13534547

 27.96 / 0.85 − 6.25  ε s′ = 0.003   = 0.0024 > ε y  27.96 / 0.85  Pn = 0.85 (250 )(27.96 )(35) / 1000 +

O.K

19.64 (4200 − 0.85 × 250) 1000  6120   0.85 (48.75) − 27.96  − 19.64   = 228.32 tons 27.96  1000   

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22

M n = 228.32 (0.25) = 57.08 t.m §

Point “F”:

This is a case of pure axial tension load, the nominal axial load is Pn = Asy f y = 39.27 (4200 / 1000 ) =164.93 tons . §

Point “C”:

This is a case of pure bending moment where the nominal flexural strength is calculated as follows: Cc = 0.85 (250)(0.85)( x)(35) / 1000 = 6.32 x tons  4200  T = 19.64   = 82.49 tons  1000  Cs =

19.64 (4200 − 0.85 × 250 ) = 78.31 tons , assuming that ε s′ ≥ ε y 1000

From equilibrium of forces in the vertical direction, 6.32 x + 78.31 − 82.49 = 0 , and x = 0.66 cm  0.66 − 6.25  ε s′ = 0.003   = − 0.0254 , which means that the compression reinforcement is  0.66  stressed in tension, a contradiction to the equilibrium equation. Cs =

  19.64  751.23   x − 6.25  6120   − 0.85× 250 = 116 .023 −  tons  1000  x   x   

From equilibrium of forces, 6.32 x + 116 .023 −

751.23 = 82.49 x

6.32 x 2 + 33.533 x − 751.23 = 0 x= x=

− 33.533 ±

(33.533)2 + 4 (6.32 )(751.23) 2 (6.32)

− 33.533 ± 141.829 12.64

x = 8.568 cm, and the other root is negative (rejected)

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23

 8.568   8.568   M n = 6.32    48.75 − 0.85   2   100   751.23   48.75 − 6.25   + 116.023 −  = 36.472 t.m  8.568   100   Figure 9.14 shows the resulting strength interaction diagram for the given cross section.

Figure 9.14: Strength interaction diagram 9.5.3 Design Interaction Diagrams The design interaction diagram for a tied or spirally-reinforced column is evaluated by carrying out three modifications on the nominal strength interaction curve, shown in Figure 9.15 as follows: a. All nominal strength combinations on the curve are multiplied by the strength reduction factor Φ . This factor is equal to 0.65 for tied columns and 0.75 for spirally reinforced columns.