Slender Columns - PowerPoint Presentation

Slide1

Slender Columns

1Slide2

Slender Columns When

a column bends or deflects laterally an amount

Δ

, its axial load will cause an increased column moment equal to PΔ. This moment will be superimposed onto any moments already in the column. Should this PΔ moment be of such magnitude as to reduce the axial load capacity of the column significantly, the column will be referred to as a slender column.

Section 10.10.1 of the Code states that the design of a compression member should, desirably, be based on a theoretical analysis of the structure that takes into account the effects of axial loads, moments, deflections, duration of loads, varying member sizes, end conditions, and so on. If such a theoretical procedure is not used, the Code (10.10.2) provides an approximate method for determining slenderness effects. This method, which is based on the factors just mentioned for an “exact” analysis, results in a moment magnifier δ, which is to be multiplied by the larger moment at the end of the column, and that value is used in design. If bending occurs about both axes, δ is to be computed separately for each direction and the values obtained multiplied by the respective moment values.

2Slide3

NONSWAY AND SWAY FRAMES

For this discussion it is necessary to distinguish between frames without

side sway and those

with side sway. In the ACI Code these are referred to respectively as nonsway frames and sway frames. The columns in nonsway frames must be designed according to Section 10.12 of the Code, while the columns of sway frames must be designed according to Section 10.13. As a result, it is first necessary to decide whether we have a nonsway frame or a sway frame.

It must be realized that we rarely find a frame that is completely braced against swaying or one that is completely unbraced against swaying. Therefore, it is a delicate decision.3Slide4

NONSWAY AND SWAY FRAMES

The question may possibly be resolved by examining the lateral stiffness of

the bracing elements for the story in question. You may observe that a particular column is located in a story where there is such substantial lateral stiffness provided by bracing members, shear walls, shear trusses, and so on that any lateral deflections occurring will be too small to affect the strength of the column appreciably. You should realize while examining a particular structure that there may be some nonsway stories and some sway stories. If we cannot tell by inspection whether we have a

nonsway frame or a sway frame, the Code provides two ways of making a decision. First, in ACI Section 10.11.4.1, a story in a frame is said to be a nonsway one if the increase in column end moments due to second-order effects is 5% or less of the first-order end moments.4Slide5

NONSWAY AND SWAY FRAMES

The second method presented by the Code for determining whether a particular

frame is

braced or unbraced is given in the Code (10.11.4.2). If the value of the so-called stability index is ≤ 0.05, the Commentary states that the frame may be classified as a nonsway one. (Should Vu be equal to zero, this method will not apply.)

5Slide6

NONSWAY AND SWAY FRAMES

Despite these suggestions from the ACI, the individual designer is going to have

to make decision as to what is adequate bracing and what is not, depending on the presence of structural walls and other bracing items. For the average size reinforced concrete building, load eccentricities and slenderness values will be small and frames will be considered to be braced. Certainly, however, it is wise in questionable cases to err on the side of the unbraced.

6Slide7

SLENDERNESS EFFECTS

The slenderness of columns is based on their

geometry and on their lateral bracing. As their slenderness increases, their bending stresses increase, and thus buckling may occur. Reinforced concrete columns generally have small slenderness ratios. As a result, they can usually be designed as short columns without strength reductions because of slenderness. If slenderness effects are considered small, then columns can be considered “short” and can be designed according to Chapter 10. However, if they are “slender,” the moment for which the column must be designed is

increased or magnified. Once the moment is magnified, the column is then designed according to Chapter 10 using the increased moment.7Slide8

SLENDERNESS EFFECTS

Several items involved in the calculation of slenderness ratios are

discussed. These include unsupported column lengths, effective length factors, radii of gyration, and the ACI Code requirements. The ACI Code (10.10.2.1) limits second-order effects to not more than 40% of first-order effects.Unsupported Lengths The length used for calculating the slenderness ratio of a column, lu

, is its unsupported length. This length is considered to be equal to the clear distance between slabs, beams, or other members that provide lateral support to the column. If haunches or capitals (see Figure 16.1) are present, the clear distance is measured from the bottoms of the capitals orhaunches.8Slide9

SLENDERNESS EFFECTS

To calculate the slenderness ratio of a particular column, it is necessary to estimate its

effective length. This is the distance between points of zero moment in the column. For this initial discussion it is assumed that no sidesway or joint translation is possible. Sidesway or joint translation means that one or both ends of a column can move laterally with respect to each other. If there were such a thing as a perfectly pinned end column, its effective length

would be its supported length, as shown in Figure 11.1(a). The effective length factor k is the number that must be multiplied by the column’s unsupported length to obtain its effective length. For a perfectly pinned end column, k = 1.0.Effective Length Factors9Slide10

SLENDERNESS EFFECTS10Slide11

SLENDERNESS EFFECTS

Columns with different end conditions have entirely different effective lengths.

For instance, if there were such a thing as a perfectly fixed end column, its points of inflection (or points of zero moment) would occur at its one-fourth points, and its effective length would be lu/2, as shown in Figure 11.1(b). As a result, its k value would equal 0.5.

Obviously, the smaller the effective length of a particular column, the smaller its danger of buckling and the greater its load-carrying capacity. Figure 11.1(c) shows a column with one end fixed and one end pinned. The k factor for this column is approximately 0.70.Effective Length Factors11Slide12

SLENDERNESS EFFECTS

The concept of effective lengths is simply a mathematical method of taking

a column—whatever its end and bracing conditions—and replacing it with an equivalent pinned end-braced column. A complex buckling analysis could be made for a frame to determine the critical stress in a particular column. The k factor is determined by finding the pinned end column with an equivalent length that provides the same critical stress. The k factor procedure is a method of making simple solutions for complicated frame-buckling problems.

Effective Length Factors12 Reinforced concrete columns serve as parts of frames, and these frames are sometimes braced and sometimes unbraced. A braced frame is one for which sidesway or joint translation is prevented by means of bracing, shear walls, or lateral support from adjoining structures.Slide13

SLENDERNESS EFFECTS

An

unbraced

frame does not have any of these types of bracing supplied and must depend on the stiffness of its own members to prevent lateral buckling. For braced frames, k values can never be greater than 1.0, but for unbraced frames, the k values will always be greater than 1.0 because of sidesway.

Effective Length Factors13 An example of an unbraced column is shown in Figure 11.2(a). The base of this particular column is assumed to be fixed, whereas its upper end is assumed to be completely free to both rotate and translate. The elastic curve of such a column will take the shape of the elastic curve of a pinned-end column of twice its length. Its effective length will therefore equal 2lu, as shown in the figure. In Figure 11.2(b), another unbraced column case is illustrated.Slide14

SLENDERNESS EFFECTS

The

bottom of this column is connected to beams that provide resistance to rotation but not enough to be considered a fixed end. In most buildings, partial rotational restraint is common, not pinned or fixed ends. Section 11.4 shows how to evaluate such partial restraint. For the case shown in Figure 11.2(b), if the beam at the bottom is flexible compared with the column, the k factor approaches infinity. If it is very stiff, k approaches 2.Effective Length Factors

14 The code (10.10.6.3) states that the effective length factor is to be taken as 1.0 for compression members in frames braced against sidesway unless a theoretical analysis shows that a lesser value can be used.Slide15

SLENDERNESS EFFECTS

Effective Length Factors

15

Should the member be in a frame not braced against sidesway, the value of k will be larger than 1.0 and must be determined with proper consideration given to the effects of cracking and reinforcing on the column stiffness. ACI-ASCE Committee 441 suggests that it is not realistic to assume that k will be less than 1.2 for such columns; therefore, it

seems logical to make preliminary designs with k equal to or larger than that value.Slide16

DETERMINING K FACTORS WITH ALIGNMENT CHARTS

The preliminary procedure used for estimating effective lengths involves the use of the

alignment charts shown in Figure 11.3.1,2. The chart of part (a) of the figure is applicable to braced frames, whereas the one of part (b) is applicable to unbraced frames. To use the alignment charts for a particular column, ψ factors are computed at each end of the column. The ψ

factor at one end of the column equals the sum of the stiffness [Σ(EI/l)] of the columns meeting at that joint, including the column in question, divided by the sum of all the stiffnesses of the beams meeting at the joint. Should one end of the column be pinned, ψ is theoretically equal to ∞, and if fixed, ψ = 0. Since a perfectly fixed end is practically impossible to have, ψ is usually taken as 1.0 instead of 0 for assumed fixed ends. When column ends are supported by, but not rigidly connected to a footing, is theoretically infinity, but usually is taken as about

10 for practical design

.

16Slide17

SLENDERNESS EFFECTS

Effective Length Factors

17Slide18

DETERMINING K FACTORS WITH ALIGNMENT CHARTS

One of the two

ψ values is called ψA and the other is called ψB. After these values are computed, the effective length factor k is obtained by placing a straightedge between ψA and ψB. The point where the straightedge crosses the middle nomograph

is k. It can be seen that the ψ factors used to enter the alignment charts and thus the resulting effective length factors are dependent on the relative stiffnesses of the compression and flexural members. If we have a very light flexible column and large stiff girders, the rotation and lateral movement of the column ends will be greatly minimized. The column ends will be close to a fixed condition, and thus the ψ values and the resulting k values will be small. Obviously if the reverse happens—that is, large stiff columns framing into light flexible girders—the column ends will rotate almost freely, approaching a pinned condition. Consequently, we will have large ψ and k

values.

18Slide19

DETERMINING K FACTORS WITH ALIGNMENT CHARTS

19Slide20

DETERMINING K FACTORS WITH ALIGNMENT CHARTS

To calculate the

ψ values it is necessary to use realistic moments of inertia. Usually, the girders will be appreciably cracked on their tensile sides, whereas the columns will probably have only a few cracks. If the I values for the girders are underestimated a little, the column k factors will be a little large and thus on the safe side.

Several approximate rules are in use for estimating beam and column rigidities. One common practice of the past for slenderness ratios of up to about 60 or 70 was to use gross moments of inertia for the columns and 50% of the gross moments of inertia for the beams. In ACI Section 10.11.1, it is stated that for determining ψ values for use in evaluating k factors, the rigidity of the beams may be calculated on the basis of 0.35Ig to account for cracking and reinforcement, while 0.70Ig may be used for compression members. This practice is followed for the examples in this chapter. Other values for the estimated rigidity of walls and flat plates are provided in the same section.

20Slide21

DETERMINING K FACTORS WITH EQUATIONS

Instead

of using the alignment charts for determining

k values, the ACI Commentary (R10.12.1) provides an alternate method that involves the use of relatively simple equations. These equations, which were taken from the British Standard Code of Practice, are particularly useful with computer programs. For braced compression members, an upper bound to the effective length factor may be taken as the smaller value determined from the two equations to follow in which ψA

and ψB are the values just described for the alignment charts (commonly called the Jackson and Moreland alignment charts as described in footnote 2 of this chapter). ψmin is the smaller of ψA and ψB.21Slide22

DETERMINING K FACTORS WITH EQUATIONS

The value of

k for unbraced compression members restrained at both ends may be determined from the appropriate one of the following two equations, in which ψm is the average of ψA and ψB: The value of the effective length factor of

unbraced compression members that are hinged at one end may be determined from the following expression, in which ψ is the value at the restrained end:

22Slide23

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

After this section, the remainder of

the lecture is devoted to an approximate design procedure wherein the effect of slenderness is accounted for by computing moment magnifiers which are multiplied by the column moments. A magnifier for a particular column is a function of its factored axial load Pu and its critical buckling load Pc. Before moment magnifiers can be computed for a particular structure, it is

necessary to make a first-order analysis of the structure. The member section properties used for such an analysis should take into account the influence of axial loads, the presence of cracked regions in the members, and the effect of the duration of the loads. Instead of making such an analysis, ACI Code 10.11.1 permits use of the following properties for the members of the structure. These properties may be used for both nonsway and sway frames.23Slide24

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

24Slide25

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

There

is a major difference in the behavior of columns in

nonsway or braced frames and those in sway or unbraced frames. In effect, each column in a braced frame acts by itself. In other words, its individual strength can be determined and compared to its computed factored loads and moments. In an unbraced or sway frame, a column will probably not buckle individually but will probably buckle simultaneously with all of the other columns on the same level. As a result, it is necessary in a sway frame to consider the buckling strength of all the columns on the level in question as a unit. For a compression member in a nonsway frame, the effective slenderness ratio kl

u /r is used to determine whether the member is short or slender. For this calculation, lu is the unbraced length of the member. The effective length factor k can be taken as 1.0 unless an analysis provides a lesser value. The radius of gyration r is equal to 0.25 times the diameter of a round column and 0.289 times the dimension of a rectangular column in the direction that stability is being considered. The ACI Code (10.11.2) permits the approximate value of 0.30 to be used in place of 0.289, and this is done herein. For other sections the value of r will have to be computed from the properties of the gross sections.25Slide26

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

For

nonsway frames, slenderness effects may be ignored if the following expression is satisfied:

In this expression, M1 is the smaller factored end moment in a compression member. It has a plus sign if the member is bent in single curvature (C-shaped) and a negative sign if the member is bent in double curvature (S-shaped). M2 is the larger factored end moment in a compression member, and it always has a plus sign. In this equation the term (34 - ) shall not be taken larger than 40, according to ACI Code 10.12.2.

26Slide27

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

Should

klu/r for a particular column be larger than the applicable ratio, we will have a slender column. For such a column the effect of slenderness must be considered. This may be done by using approximate methods or by using a theoretical second-order analysis that takes into account the effect of deflections. If klu/r >

100, a theoretical second order analysis must be made (Code 10.11.5). A second-order analysis is one that takes into account the effect of deflections and also makes use of a reduced tangent modulus. The equations necessary for designing a column in this range are extremely complicated, and, practically, it is necessary to use column design charts or computer programs. Fortunately, most reinforced concrete columns have slenderness ratios less than 100.27Slide28

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

The design of slender columns is appreciably more complicated than the design of

short columns. As a result, it may be wise to give some consideration to the use of certain minimum dimensions so that none of the columns will be slender. In this way they can be almost completely avoided in the average-size building. If k is assumed equal to 1.0, slenderness can usually be neglected in braced frame columns if lu/h

is kept to 10 or less on the first floor and 14 or less for the floors above the first one. To determine these values it was assumed that little moment resistance was provided at the footing–column connection and the first-floor columns were assumed to be bent in single curvature. Should the footing–column connection be designed to have appreciable moment resistance, the maximum lu/h value given above as 10 should be raised to about 14 or equal to the value used for the upper floors.Avoiding Slender Columns28Slide29

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

Should we have an

unbraced frame and assume k = 1.2, it is probably necessary to keep lu/h to 6 or less. So for a 10-ft clear floor height it is necessary to use a minimum h of about 10 ft/6

= 1.67 ft = 20 in. in the direction of bending to avoid slender columns. Example 11.1 illustrates the selection of the k factor and the determination of the slenderness ratio for a column in an unbraced frame. For calculating I/L values let us use 0.70 times the gross moments of inertia for the columns, 0.35 times the gross moments of inertia for the girders, and the full lengths of members center to center of supports.Avoiding Slender Columns29Slide30

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

Example 11.1

1. Using

the alignment charts of Figure 11.3, calculate the effective length factor for column AB of the unbraced frame of Figure 11.4. Consider only bending in the plane of the frame.2. Compute the slenderness ratio of column AB. Is it a short or a slender column? The maximum permissible slenderness ratio for a

short unbraced column is 22. Assume single curvature with M1 = M2 = 75 ft-k.Solution(a) Effective Length Factor for Column AB

30Slide31

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

31Slide32

FIRST-ORDER ANALYSES USING SPECIAL MEMBER PROPERTIES

32Slide33

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

When

a column is subjected to moment along its

unbraced length, it will be displaced laterally in the plane of bending. The result will be an increased or secondary moment equal to the axial load times the lateral displacement or eccentricity. In Figure 11.5 the load P causes the column moment to be increased by an amount PΔ. This moment will cause o to increase a little more, with the result that the PΔ moment will increase, which in turn will cause a further increase in d and so on until equilibrium is reached.

33Slide34

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

We

could take the column moments, compute the lateral deflection, increase the

moment by PΔ, recalculate the lateral deflection and the increased moment, and so on. Although about two cycles would be sufficient, this would still be a tedious and impractical procedure. It can be shown that the increased moment can be estimated very well by multiplying the primary moment by 1/(1 - P/Pc), where P

is the axial load and Pc is the Euler buckling load π²EI/(klu)². In Example 11.2 this expression is used to estimate the magnified moment in a laterally loaded column. It will be noted that in this problem the primary moment of 75 ft-k is estimated to increase by 7.4 ft-k. If we computed the deflection due to the lateral load, we would get 0.445 in. For this value PΔ = (150)(0.445) = 66.75 in.-k = 5.6 ft-k. This moment causes more deflection, which causes more moment, and so on.

34Slide35

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

Example 11.2

Solution

1. Compute the primary moment in the column shown in Figure 11.6 due to the lateral 20-k load.2. Determine the estimated total moment, including the secondary moment due to lateral de- flection using the appropriate magnification factor just presented

. E = 3.1 X 10³ ksi. Assume k = 1.0 and lu = 15 ft.Figure 11.635Slide36

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

Example 11.2

Solution

36Slide37

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

As we have seen, it is possible to calculate approximately the increased moment

due to lateral deflection by using the (1- P/Pc) expression. In ACI Code 10.12.3 the factored design moment for slender columns with no sway is increased by using the following expression, in which Mc is the magnified or increased moment and M2 is the larger factored end

moment on a compression member: Should our calculations provide very small moments at both column ends, the Code provides an absolutely minimum value of M2 to be used in design. In effect, it requires the computation of a moment based on a minimum eccentricity of 0.6 + 0.03h, where h is the overall thickness of the member perpendicular to the axis of bending.37Slide38

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

A moment magnifier

δ, is used to estimate the effect of member curvature or lateral deflection in a column in a nonsway frame. It involves a term Cm, which is defined later in this section. The determination of the moment magnifier δns

involves the following calculations:38for normal-weight concrete (see Section 1.11 for other densities).Slide39

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

4

.

Ise = moment of inertia of the reinforcing about the centroidal axis of the section. (This value equals the sum of each bar area times the square of its distance from the centroidal axis of the compression member.)5.

The term βdns accounts for the reduction in stiffness caused by sustained axial loads and applies only to nonsway frames. It is defined as the ratio of the maximum factored sustained axial load divided by the total factored axial load associated with the same load combination. It is always assumed to have a plus sign and is never permitted to exceed 1.0.6. Next it is necessary to compute EI. The two expressions given for EI in the Code were developed so as to account for creep, cracks, and so on. If the column and bar sizes have already been selected, or estimated, EI can be computed with the following expression, which

is

particularly satisfactory for columns with high steel

percentages.

39Slide40

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

The

alternate expression for

EI that follows is probably the better expression to use when steel percentages are low. Notice also that this expression will be the one used if the reinforcing has not been previously selected.7. The Euler buckling load is computed:40Slide41

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

8. For

some moment situations in columns,

the amplification or moment magnifier expression provides moments that are too large. One such situation occurs when the moment at one end of the member is zero. For this situation the lateral deflection is actually about half of the deflection in effect provided by the amplification factor. Should we have approximately equal end moments that are causing reverse curvature bending, the deflection at mid-depth and

the moment there are close to zero. As a result of these and other situations, the Code provides a modification factor (Cm) to be used in the moment expression that will result in more realistic moment magnification.41Slide42

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

For braced frames without transverse loads,

C

m can vary from 0.4 to 1.0 and is determined with the expression as given below. For all other cases it is to be taken as 1.0. (Remember the sign convention: M1 is positive for single curvature and is negative for reverse curvature, and M2 is always positive.) Should M

2,min as computed with ACI Equation 10-14 be larger than M2, the value of Cm above shall either be taken as equal to 1.0 or be based on the ratio of the computed end moments M1/M2 (ACI Section 10.12.3.2). Example 11.3 illustrates the design of a column in a nonsway frame.42Slide43

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

Example 11.3

Solution

43The tied column of Figure 11.7 has been approximately sized to the dimensions 12 in. × 15 in. It is to be used in a frame braced against sidesway. The column is bent in single

curvature about its y-axis and has an lu of 16 ft. If k = 0.83, fy = 60,000 psi, and f’c = 4000 psi, determine the reinforcing required. Consider only bending in the plane of the frame. Note also that the unfactored dead axial load PD is 30 k, and concrete is normal weight.Slide44

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

44Slide45

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

Example 11.3

45Slide46

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

46Slide47

MAGNIFICATION OF COLUMN MOMENTS IN NONSWAY FRAMES

47

Since

Kn and Rn are between the radial lines labeled fs/fy = 1.0 and

ϵt = 0.005 on the interaction diagrams, the φ factor is permitted to be increased from the 0.65 value used. An area of reinforcing of only 1.80 in.2 is found to be sufficient. This significant reduction occurs because of the increased φ factor in this region. Most columns are designed for multiple load combinations, and the designer must be certain that the column is able to resist all of them. Often there are some columns with high axial load and low moment, such as 1.2D + 1.6L, and others with low axial load and high moment, such as 0.9D + 1.6E. The first of these is likely to have a φ

factor of 0.65.

The second

, however, is more likely to be eligible for the increase in the

φ

factor.Slide48

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

Tests

have shown that even though the lateral deflections in

unbraced frames are rather small, their buckling loads are far less than they would be if the frames had been braced. As a result, the buckling strengths of the columns of an unbraced frame can be decidedly increased (perhaps by as much as two or three times) by providing bracing. If a frame is unbraced against sidesway, it is first necessary to compute its slenderness ratio. If k

lu/r is less than 22, slenderness may be neglected (ACI 10.13.2). For this discussion it is assumed that values > 22 are obtained. When sway frames are involved, it is necessary to decide for each load combination which of the loads cause appreciable sidesway (probably the lateral loads) and which do not. The factored end moments that cause sidesway are referred to as M1s and M2s, and they must be magnified because of the P

Δ

effect.

The other end moments, which do

not cause

appreciable

sidesway

, are

M

1

ns

and

M

2

ns

. They are determined by first-order

analysis and

will not have to be magnified.

48Slide49

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

The Code (

10.10.7) states that the magnified moment δs can be determined by any one of the following three methods.1. The moment magnifier may be calculated with the equation given below in which Q is the stability index previously presented. Should the computed value of δs be greater than 1.5, it will be

necessary to compute δs by ACI Section 10.10.7.4 or by a second- order analysis.49Slide50

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

In

this last equation

ΣPu is the summation of all the vertical loads in the story in question, and ΣPc is the sum of all the Euler buckling loads (Pc = π²El/(klu)²) for all of the sway resisting columns in the story with k

values determined as described in ACI Section 10.10.7.2. This formula reflects the fact that the lateral deflections of all the columns in a particular story are equal, and thus the columns are interactive.502. With the second method and the one used in this chapter, the magnified sway moments may be computed with the following expression:Slide51

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

Whichever

of the preceding methods is used to determine the

δs values, the design moments to be used must be calculated with the expressions that follow. Sometimes the point of maximum moment in a slender column will fall between its ends. The ACI Commentary (R10.10.2.2) says the moment magnification for this case may be evaluated using the procedure described for nonsway frames (ACI Section 10.10.6). Example 11.4 illustrates the design of a slender column subject to sway.

51Slide52

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

Example 11.4

52

Select reinforcing bars using the moment magnification method for the 18 in. × 18 in. unbraced column shown in Figure 11.8 if lu = 17.5 ft, k = 1.3, fy = 60 ksi, and

f’c = 4 ksi. A first-order analysis has resulted in the following axial loads and moments:The loading combination assumed to control with sidesway is ACI Equation 9.6.

The loading combination assumed to control for the case with no

sidesway

is ACI Equation

9.2 (Section

4.1 of this text).Slide53

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

Example 11.4

53

Note that ACI Equation 9.3 or 9.4 may also control for sidesway, but in this case it is unlikely.

SolutionSlide54

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES

Example 11.4

54

SolutionSlide55

MAGNIFICATION OF COLUMN MOMENTS IN SWAY FRAMES55Slide56

ANALYSIS OF SWAY FRAMES

The frame of Figure 11.9 is assumed to be

unbraced

in the plane of the frame. It supports a uniform gravity load, wu, and a short-term concentrated lateral load, Pw. As a result, it is necessary to consider both the moments resulting from the loads that do not cause appreciable sidesway and the loads that do. It will, therefore, be necessary to compute both δ and δs values, if the column proves to be slender.

56Slide57

57Slide58

ANALYSIS OF SWAY FRAMES The

M

s

values are obviously caused by the lateral load in this case. The reader should realize, however, that if the gravity loads and/or the frame are unsymmetrical, additional Ms or sidesway moments will occur. If we have an unbraced frame subjected to short-term lateral wind or earthquake loads, the columns will not have appreciable creep (which would increase lateral deflections and, thus, the P

moments). The effect of creep is accounted for in design by reducing the stiffnessEI used to calculate Pc and thus δs by dividing EI by 1 + βdns , as specified in ACI Section 10.10.6.1. Both the concrete and steel terms in ACI Equation 10-14 are divided by this value. In the case of sustained lateral load, such as soil backfill or water pressure, ACI Section 10.10.4.2 requires that the moments of inertia for compression members in Section 11.6 be divided by (1 + βds ). The term βds is the ratio of the maximum factored sustained shear within a story to the maximum factored shear in that story for the same load combination.

58Slide59

ANALYSIS OF SWAY FRAMES

The beam and columns of the frame have been tentatively sized as shown in the

figure. In

Example 11.5 the frame is analyzed for each of the conditions specified in ACI Section 9.2 using 1.3W instead of 1.6W. In the example, the magnification factors δ and δs are computed for each of the loading conditions and used to compute the magnified moments. Notice in the solution that different k values are used for determining

δ and δs. The k for the δ calculation is determined from the alignment chart of Figure 11.3(a) for braced frames, whereas the k for the δs calculation is determined from the alignment chart of Figure 11.3(b) for unbraced frames.59 To illustrate the computation of the magnified moments needed for the design of a slender column in a sway frame, the simple frame of Figure 11.9 has been chosen. Slide60

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

60Determine the moments and axial forces that must be used for the design of column CD of the unbraced frame of Figure 11.9. Consider only bending in the plane of the frame. The assumed member sizes shown in the figure are used for the analyses given in the problem.

fy = 60,000 psi and f’c = 4000 psi. For this example, the authors considered the load factor cases of ACI Equations 9-2, 9-4, and 9-6. For other situations, other appropriate ACI load factor equations will have to be considered.Slide61

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

61Slide62

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

62Slide63

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

63Slide64

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

64Slide65

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

65Slide66

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

66Slide67

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

67Slide68

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

68Slide69

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

69Slide70

ANALYSIS OF SWAY FRAMES

Example 11.5

Solution

70