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5.5.1

Overview

The term compression member is used to refer to columns and other members, such as walls, that are subjected primarily to compressive forces. This section focuses on the determination of the nominal axial strength Pnfor short, reinforced concrete columns

that are subjected to essentially concentric axial loads. A short column is defined as one in which slenderness effects need not be considered. Slender columns are discussed in Chap. 8.

The nominal axial strength of a short column is related to the area of the column, the compressive strength of the concrete, the area and yield stress of the longitudinal reinforcement, and the type of transverse reinforcement. These quantities and their relationship to axial strength are discussed later.

5.5.2

Maximum Concentric Axial Load

Consider a reinforced concrete column subjected to a concentric axial load P. Assume that the longitudinal reinforcement is symmetrically distributed in the section and that lateral reinforcement that meets the size and spacing requirements of ACI 7.10 is provided. The type of lateral reinforcement is not relevant to the discussion at this time. When subjected to P, the length of the column L decreases by an amount equal to the longitudinal strainε times the original length L. For a concentric axial load, ε is uniform across the section. The strains in the concrete and the steel are equal because the

concrete and the longitudinal steel are bonded together. For any givenε, it is possible to compute the stresses in the concrete and longitudinal steel using the stress–strain curves of the materials (see Chap. 2). The loads in the concrete and the longitudinal steel are equal to the stresses multiplied by the corresponding areas, and the total load that a short column can carry is the sum of the maximum loads carried by the concrete and the steel.

The maximum compressive axial load that can be resisted by the concrete Pcis equal

to the following:

Pc= 0.85 fc(Ag− Ast) (5.33)

In this equation, Ag is the gross area of the column and Ast is the total area of longitudinal reinforcement in the column; thus, (Ag− Ast) is equal to the area of the concrete. The factor 0.85 is based on the results of numerous tests.8

The maximum axial load that can be carried by the longitudinal reinforcement Ps

is equal to the area times the yield strength of the reinforcement:

Ps = fyAst (5.34)

Therefore, the maximum concentric axial load Po that can be carried by a short

column is equal to the summation of the maximum loads of the concrete and the steel:

Po = 0.85 fc(Ag− Ast)+ fyAst (5.35) Equation (5.35) forms the basis of the nominal axial strength, which is discussed next.

5.5.3

Nominal Axial Strength

In general, the maximum nominal axial strength Pn,max is equal to a constant times the concentric axial load strength Po. The constant depends on the type of transverse

reinforcement utilized in the section and accounts for any accidental eccentricities— or, equivalently, any accidental bending moments—that may exist in a compression member and were not considered in the analysis. These eccentricities can arise from unbalanced moments in the beams framing into the column, misalignment of columns from floor to floor, or misalignment of the longitudinal reinforcement in the column, to name a few.

For members with spiral reinforcement, the constant is equal to 0.85. Therefore,

Pn,max= 0.85[0.85 fc(Ag− Ast)+ fyAst] (5.36) The constant 0.85 produces nominal strength approximately equal to that from earlier Codes with the axial load applied at an eccentricity equal to 5% of the column depth.

Similarly, for members with tie reinforcement, the constant is equal to 0.80, and

Pn,maxis

In this case, the axial load is applied at an eccentricity equal to approximately 10% of the column depth.

Equations (10-1) and (10-2) in ACI 10.3.6 give the design axial strengthφ Pn= φ Pn,max for columns with spiral and tie lateral reinforcement, respectively. These equations are formed by multiplying the maximum nominal axial strength Pn,maxby the correspond- ing strength reduction factorφ. Thus,

r _{For members with spiral reinforcement}

φ Pn,max= 0.85φ[0.85 fc(Ag− Ast)+ fyAst] (5.38)

r _{For members with tie reinforcement}

φ Pn,max= 0.80φ[0.85 fc(Ag− Ast)+ fyAst] (5.39) The strength reduction factorφ is equal to 0.75 for compression-controlled sections with spiral reinforcement and 0.65 for other reinforced members including tied rein- forcement (see Section 4.3 and ACI 9.3.2.2). It was noted in Section 4.3 that the larger

φ− factor for columns with spiral reinforcement reflects the more ductile behavior of

such columns compared with columns with tied reinforcement.

5.5.4

Longitudinal Reinforcement Limits

ACI 10.9.1 prescribes the limits on the amount of longitudinal reinforcement for com- pression members, which are applicable to all such members regardless of the type of lateral reinforcement:

r _{Minimum Ast= 0.01Ag} r _{Maximum Ast= 0.08Ag}

The lower limit is meant to provide resistance to bending, which may exist even though an analysis shows that it is not present, and to reduce the effects of creep and shrinkage of the concrete under sustained compressive stresses (see Chap. 2).

The upper limit is a practical maximum for longitudinal reinforcement in terms of economy and placement of the bars: For proper concrete placement and consolidation, the size and number of longitudinal bar sizes must be chosen to minimize reinforce- ment congestion, especially at beam–column joints. If column bars are lap spliced, the maximum area of longitudinal reinforcement should not exceed 4% of the gross column area at the location of the splice.

ACI 10.9.2 also contains requirements on the minimum number of longitudinal bars in compression members and the minimum volumetric reinforcement ratio for columns with spiral reinforcement. These and other requirements are covered in Chap. 8.

The flowchart shown in Fig. 5.34 can be used to determine Pn,maxfor compression members.

Example 5.9 Determine the maximum nominal axial strength Pn,maxof the 24-in-diameter column

FIGURE5.34 Maximum nominal axial strength for compression members.

Solution The flowchart shown in Fig. 5.34 is utilized to determine Pn,maxfor this compression

member.

Step 1: Check the minimum and maximum longitudinal reinforcement limits.The minimum and maximum amounts of longitudinal reinforcement permitted in a compression member are specified in ACI 10.9.1.

Minimum: A_st= 0.01Ag= 0.01 × π × 242/4 = 4.52 in2

Maximum: Ast= 0.08Ag= 0.08 × π × 242/4 = 36.2 in2

The provided area of longitudinal reinforcement Ast= 8 × 1.00 = 8.00 in2 falls between the

FIGURE5.35 The column of Example 5.9.

Step 2: Identify the type of lateral reinforcement. Spiral reinforcement is identified in Fig. 5.35.

Step 3: Determine the maximum nominal axial strength Pn,max.For compression members

with spiral reinforcement conforming to ACI 7.10.4, Pn,maxis determined by Eq. (5.36):

Pn,max = 0.85[0.85 fc(Ag− Ast)+ fyAst]

= 0.85[0.85 × 6 × (452.4 − 8.00) + (60 × 8.00)] = 2,335 kips

Comments

The design axial strengthφ Pn,maxis equal to the strength reduction factorφ, which is 0.75 for

compression-controlled sections with spiral reinforcement [see Section 4.3 and ACI 9.3.2.2(a)], times

Pn,max:

φ Pn,max= 0.75 × 2,335 = 1,751 kips