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Contents | 1. Introduction to structural design | 2. Loads | 3. Wood | 4. Steel |

Introduction to reinforced concrete | Material properties | Sectional properties | Design approaches | Construction systems | Tension elements |

Concrete columns are cast into forms containing a matrix of steel reinforcement. This reinforcement is distributed just inside the perimeter of the forms in a pattern designed to confine the concrete, much like sand would be confined when placed into a steel drum. In both cases (sand in a steel drum; concrete in a steel "cage"), the ability of the material to sustain an axial compressive stress is enormously increased by the presence of the confining steel, whether or not the steel contributes directly to the support of the external load.

Two patterns of steel reinforcement are commonly used for columns: a series of square or rectangular
*ties* (Figure 5.16*a*) placed horizontally around a minimum of four longitudinal steel bars; or a continuous circular *spiral* wire (Figure 5.16*b*) wrapped around a minimum of six longitudinal bars. Tied columns are usually rectangular and spiral columns are usually circular, but either pattern of reinforcement can be used for any column cross section. In general, spiral reinforcement provides more reliable confinement of the concrete, and a more ductile type of failure than tied columns;
strength reduction factors for spiral versus tied columns take this relative safety into account. The
actual design of ties and spirals is based on fairly straight-forward guidelines, summarized in Appendix Table A-5.4. The design and analysis examples that follow do not include the calculation of tie or
spiral spacing and size.

Figure 5.16: Containment of longitudinal bars using (*a*) ties; and (*b*) spiral reinforcement

The amount of longitudinal steel in reinforced concrete columns, measured according to the ratio of steel area to gross column area (reinforcement ratio), must fall between two limiting values. The lower limit of 1% provides a minimum amount of steel to protect against tension failures due to unanticipated bending moments; the upper limit of 8% prevents overcrowding of steel bars within the concrete formwork. Because longitudinal column reinforcement is typically spliced — and therefore doubled in area — where an upper column is cast above a lower column (see Figure 5.53), it is common to limit the maximum reinforcement ratio to 4%. The reinforcement ratio is defined as:

(5.1)

where ρ* _{g}* = the reinforcement ratio of longitudinal steel area to gross area;

It is assumed in this chapter that reinforced concrete column stability is not a factor in the column's
strength; that is, the column is not slender enough for buckling to be a problem. As a general
rule of thumb, concrete columns braced against lateral misalignment ("sidesway"), with a slenderness
ratio, *KL/r*, no greater than 40, are rarely influenced by stability considerations. Taking the radius of gyration of a rectangular column as approximately equal to 0.3 times the smaller cross-sectional
column dimension, *h* (that is, assuming *r* = 0.3*h*), and taking the effective length coefficient, *K* = 1.0, we get *KL/r* = 1.0*L*/ (0.3*h*) ≤ 40. Solving for the ratio of unbraced length, *L*, to minimum cross-sectional dimension, *h*, we find that slenderness effects may typically be neglected in axially-loaded reinforced concrete columns when *L/h* ≤ 12. For slender concrete columns, other techniques must be used to account for the possibility of buckling.

For columns, at least 1½ in. of concrete is left outside the matrix of reinforcement to protect it from corrosion and to provide fire resistance (2 in. for No. 6 or larger bars if the concrete is exposed to the weather, or the earth; 3 in. for all bars if the concrete is cast directly against the earth — see Appendix Table A-5.1). For typical reinforcement sizes, the distance from the outside of the concrete column to the centerline of the longitudinal reinforcement can be taken as about 2½ in. or 3 in. (Figure 5.17).

Figure 5.17: Detail of reinforced concrete element showing approximate distance from centerline of rebar to
outside face of concrete

For a reinforced concrete column subjected to pure axial compression, the ultimate load at failure is simply the concrete strength (failure stress) times its area, plus the yield stress of the longitudinal steel rebars times their area (Figure 5.18).

Figure 5.18: Nominal stresses at failure of axially loaded reinforced concrete column

The failure strength of concrete is taken
as 85% of its cylinder strength, *f _{c}*', since the more
rapid rate of loading of the test cylinders (Figure
5.19, curve

Figure 5.19: Stress-strain diagrams for plain concrete showing (*a*) fast loading characteristic of test cylinders; and (*b*) slow loading characteristic of actual structures

The strain at which
steel longitudinal reinforcement bars yield depends
on their yield stress. For grade 60 rebars
(*f _{y}* = 60 ksi), the yield strain (stress divided by
modulus of elasticity) is 60/29,000 = 0.002. For
grade 40 (

(5.2)

where *A _{s}* is the longitudinal steel area, and

There are two strength reduction safety factors for axially-loaded reinforced concrete columns: φ is the ordinary factor, while α accounts for the possibility of non-axial loading. Both factors depend on whether the column is tied or spiral (see Appendix Table A-5.5). Combining these strength reduction factors with factored loads (typically 1.2*D* + 1.6*L* where live and dead load govern, per Appendix Table A-2.7*a*), we get equations for the design and analysis of axially-loaded reinforced concrete columns. An example of such an equation for dead load (*D*) and live load (*L*) only, where *P _{u}* is the factored or "design" load, is:

(5.3)

** Problem definition.** Assuming

Figure 5.20: Column cross-section for Example 5.1

** Solution overview.** Find concrete and steel areas; multiply by failure stresses for concrete and steel
and add together for ultimate capacity. Multiply ultimate capacity by strength reduction factors and
compare with factored loads to determine whether capacity is adequate for given loads.

*Problem solution*

**1.** From Appendix Table A-5.2, the steel area for 4 No. 9 bars, *A _{s}* = 4.00 in

**2.** The concrete area, Ac = *A _{g}* –

**3.** From Equation 5.2, the nominal capacity or failure load, *P _{n}* = 0.85

**4.** From Appendix Table A-5.2, strength reduction factors for a tied column are: φ = 0.65 and α = 0.80.

**5.** Based on Equation 5.3, check whether *P _{u}* = 1.2

**6.** In this example, all column parameters were given. However, we can still check that the column has an acceptable reinforcement ratio and that the bars fit within the cross section. Using Equation 5.1, we check that the reinforcement ratio is between 1% and 8% (that is, between 0.01 and 0.08): ρ* _{g}* =

** Problem definition.** Assuming

** Solution overview.** Use Equation 5.3 relating reduced strength to factored loads and solve for steel
area. The area of concrete within the column cross section is found by subtracting the steel area
from the gross cross-sectional dimensions; that is,

*Problem solution*

**1.** From Equation 5.3: *P _{u}* = 1.2

1.2(150) + 1.6(100) ≤ (.65)(.80)[0.85(3)(144 – *A _{s}*) + 60

340 ≤ (0.52)[367.2 – 2.55*A _{s}* + 60

653.85 ≤ 367.2 + 57.45*A _{s}*.

57.45*A _{s}* ≥ 286.65.

**2.** From Appendix Table A-5.2, choose 4 No. 10 bars with actual *A _{s}* = 5.08 in

**3.** Using Equation 5.1, check that the reinforcement ratio is between 1% and 8% (that is, between 0.01 and 0.08): ρ* _{g}* =

** Problem definition.** Assuming

** Solution overview.** Use Equation 5.3 relating reduced strength to factored loads and solve for gross
area. With the reinforcement ratio, ρ

*Problem solution*

**1.** From Equation 5.3: *P _{u}* = 1.2

*P _{u}* = 1.2

The choice of a reinforcement ratio is somewhat arbitrary; we select ρ* _{g}* = 0.04; then, with strength reduction factors, φ and α, found from Appendix Table A-5.5, we get:

1.2(150) + 1.6(125) ≤ (.75)(.85)[0.85(5)(1.00 – 0.04)*A _{g}* + 60(0.04)

380 ≤ (0.6375)[4.08*A _{g}* + 2.40

596.1 ≤ 6.48*A _{g}*.

*A _{g}* ≥ 91.99 in

**2.** From Equation 5.3: *P _{u}* = 1.2

1.2(150) + 1.6(125) ≤ (.75)(.85)[0.85(5)(78.54 – *A _{s}*) + 60

380 ≤ (0.6375)[333.8 – 4.25*A _{s}*s + 60

596.1 ≤ 333.8 + 55.75*A _{s}*.

55.75*A _{s}* ≥ 262.3.

*A _{s}* ≥ 4.71 in

**3.** From Appendix Table A-5.2, choose 6 No. 8 bars with actual *A _{s}* = 4.74 in

**4.** Using Equation 5.1, check that the reinforcement ratio is between 1% and 8% (that is, between 0.01 and 0.08): ρ* _{g}* =

The actual reinforcement ratio, ρ* _{g}* = 0.060, is much higher than our initial assumed value of ρ

© 2020 Jonathan Ochshorn; all rights reserved. This section first posted November 15, 2020; last updated November 15, 2020.