Compression Steel

Section 10.5.1 states that the minimum amount of reinforcing required equals the larger of the two values that follow:

A_{s min}=3 f_c^

f_y b_wd (ACI Equation 10-3)

or

A_{s min} =200b_wd f_y

For statically determinate members with their flanges in tension, b_w in the above expres-sion is to be replaced with either 2b_w or the width of the flange, whichever is smaller.

5.6 L-Shaped Beams

The author assumes for this discussion that L beams (i.e., edge T beams with a flange on one side only) are not free to bend laterally. Thus they will bend about their horizontal axes and will be handled as symmetrical sections, exactly as with T beams.

For L beams, the effective width of the overhanging flange may not be larger than one-twelfth the span length of the beam, six times the slab thickness, or one-half the clear distance to the next web (ACI 8.12.3).

If an L beam is assumed to be free to deflect both vertically and horizontally, it will be necessary to analyze it as an unsymmetrical section with bending about both the horizontal and vertical axes. An excellent reference on this topic is given in a book by MacGregor.¹

5.7 Compression Steel

The steel that is occasionally used on the compression sides of beams is called compression steel, and beams with both tensile and compressive steel are referred to as doubly reinforced beams. Compression steel is not normally required in sections designed by the strength method because use of the full compressive strength of the concrete decidedly decreases the need for such reinforcement, as compared to designs made with the working-stress design method.

Occasionally, however, space or aesthetic requirements limit beams to such small sizes that compression steel is needed in addition to tensile steel. To increase the moment capacity of a beam beyond that of a tensilely reinforced beam with the maximum percentage of steel [when(_t = 0.005)], it is necessary to introduce another resisting couple in the beam. This is done by adding steel in both the compression and tensile sides of the beam. Compressive steel increases not only the resisting moments of concrete sections but also the amount of curvature that a member can take before flexural failure. This means that the ductility of such sections will be appreciably increased. Though expensive, compression steel makes beams tough and ductile, enabling them to withstand large moments, deformations, and stress reversals such as might occur during earthquakes. As a result, many building codes for earthquake zones require that certain minimum amounts of compression steel be included in flexural members.

Compression steel is very effective in reducing long-term deflections due to shrinkage and plastic flow. In this regard you should note the effect of compression steel on the long-term deflection expression in Section 9.5.2.5 of the code (to be discussed in Chapter 6 of this text).

Continuous compression bars are also helpful for positioning stirrups (by tying them to the compression bars) and keeping them in place during concrete placement and vibration.

1Wight, J. K. and MacGregor, J. G., 2011, Reinforced Concrete Mechanics and Design, 6th ed. (Upper Saddle River, NJ:

Pearson Prentice Hall), pp. 165–168.

Tests of doubly reinforced concrete beams have shown that even if the compression concrete crushes, the beam may very well not collapse if the compression steel is enclosed by stirrups. Once the compression concrete reaches its crushing strain, the concrete cover spalls or splits off the bars, much as in columns (see Chapter 9). If the compression bars are confined by closely spaced stirrups, the bars will not buckle until additional moment is applied. This additional moment cannot be considered in practice because beams are not practically useful after part of their concrete breaks off. (Would you like to use a building after some parts of the concrete beams have fallen on the floor?)

Section 7.11.1 of the ACI Code states that compression steel in beams must be enclosed by ties or stirrups or by welded wire fabric of equivalent area. In Section 7.10.5.1, the code states that the ties must be at least #3 in size for longitudinal bars #10 and smaller and at least #4 for larger longitudinal bars and bundled longitudinal bars. The ties may not be spaced farther apart than 16 bar diameters, 48 tie diameters, or the least dimension of the beam cross section (code 7.10.5.2).

For doubly reinforced beams, an initial assumption is made that the compression steel yields as well as the tensile steel. (The tensile steel is always assumed to yield because of the ductile requirements of the ACI Code.) If the strain at the extreme fiber of the compression concrete is assumed to equal 0.00300 and the compression steel, A^_s, is located two-thirds of the distance from the neutral axis to the extreme concrete fiber, then the strain in the compression steel equals ²₃× 0.003 = 0.002. If this is greater than the strain in the steel at yield, as say 50,000/(29 × 10⁶) = 0.00172 for 50,000-psi steel, the steel has yielded. It should be noted that actually the creep and shrinkage occurring in the compression concrete help the compression steel to yield.

Sometimes the neutral axis is quite close to the compression steel. As a matter of fact, in some beams with low steel percentages, the neutral axis may be right at the compression steel. For such cases, the addition of compression steel adds little, if any, moment capacity to the beam. It can, however, make the beam more ductile.

When compression steel is used, the nominal resisting moment of the beam is assumed to consist of two parts: the part due to the resistance of the compression concrete and the balancing tensile reinforcing, and the part due to the nominal moment capacity of the compression steel and the balancing amount of the additional tensile steel. This situation is illustrated in Figure 5.13. In the expressions developed here, the effect of the concrete in compression, which is replaced by the compressive steel, A^_s, is neglected. This omission will cause us to overestimate M_n by a very small and negligible amount (less than 1%). The first of the two resisting moments is illustrated in Figure 5.13(b).

M_n1= A_s1f_y

 d−a

2



M_n1 = A_s1f_y

(

)

M_n = M_n1 + M_n2 M_n2 = A_s'f_s'(d – d') = A_s2f_y (d – d')

F I G U R E 5.13 Doubly reinforced beam broken into parts.

McCormac c05.tex V2 - January 9, 2013 11:07 P.M. Page 129

5.7 Compression Steel 129

The second resisting moment is that produced by the additional tensile and compressive steel(As2 and A^_s), which is presented in Figure 5.13(c).

M_n2= A^sf_y(d − d^)

Up to this point it has been assumed that the compression steel has reached its yield stress. If such is the case, the values of A_s2 and A^_s will be equal because the addition to T of A_s2f_y must be equal to the addition to C of A^_sf_y for equilibrium. If the compression steel has not yielded, A^_s must be larger than A_s2, as will be described later in this section.

Combining the two values, we obtain M_n = As1f_y

The addition of compression steel only on the compression side of a beam will have little effect on the nominal resisting moment of the section. The lever arm, z, of the internal couple is not affected very much by the presence of the compression steel, and the value of T will remain the same. Thus, the value M_n = Tz will change very little. To increase the nominal resisting moment of a section, it is necessary to add reinforcing on both the tension and the compression sides of the beam, thus providing another resisting moment couple.

Examples 5.7 and 5.8 illustrate the calculations involved in determining the design strengths of doubly reinforced sections. In each of these problems, the strain, f_s^, in the com-pression steel is checked to determine whether or not it has yielded. With the strain obtained, the compression steel stress, f_s^, is determined, and the value of A_s2 is computed with the following expression:

A_s2f_y= A^sf_s^

In addition, it is necessary to compute the strain in the tensile steel,_t, because if it is less than 0.005, the value of the bending,φ, will have to be computed, inasmuch as it will be less than its usual 0.90 value. The beam may not be used in the unlikely event thatt is less than 0.004.

To determine the value of these strains, an equilibrium equation is written, which upon solution will yield the value of c and thus the location of the neutral axis. To write this equation, the nominal tensile strength of the beam is equated to its nominal compressive strength. Only one unknown appears in the equation, and that is c.

Initially the stress in the compression steel is assumed to be at yield(fs^= fy). From Figure 5.14, summing forces horizontally in the force diagram and substitutingβ1c for a leads to

A_sf_y= 0.85fc^β1cb+ A^sf_y c= (A_s − A^_s)f_y

0.85f_c^β₁b

Referring to the strain diagram of Figure 5.14, from similar triangles

^s =c− d^ c (0.003)

If the strain in the compression steel _s^ > _y= f_y/E_s, the assumption is valid and f_s^ is at yield, f_y. If_s^ < _y, the compression steel is not yielding, and the value of c calculated above is not correct. A new equilibrium equation must be written that assumes f_s^< f_y.

A_sf_y= 0.85fc^β1cb+ A^s

²'s

c d'

²t

0.003

C_c = 0.85f'_c ab C'_s = A'_sf'_s = A_s2f_y

T = A_sf_y A'_s

As d

strain internal forces F I G U R E 5.14 Internal strains and forces for doubly reinforced rectangular beam.

The value of c determined enables us to compute the strains in both the compression and tensile steels and thus their stresses. Even though the writing and solving of this equation are not too tedious, use of the Excel spreadsheet for beams with compression steel makes short work of the whole business.

Examples 5.7 and 5.8 illustrate the computation of the design moment strength of doubly reinforced beams. In the first of these examples, the compression steel yields, while in the second, it does not.

Example 5.7

Determine the design moment capacity of the beam shown in Figure 5.15 for which f_y= 60,000 psi and f_c^ = 3000 psi.

S O L U T I O N

Writing the Equilibrium Equation Assuming f_s^ = f_y A_sf_y= 0.85fc^bβ1c+ A^sf_y

(6.25 in.²) (60 ksi)= (0.85) (3 ksi) (14 in.) (0.85c) + (2.00 in.²) (60 ksi)

2 #9 (2.00 in.²)

4 #11 (6.25 in.²)

27 in.

in.

3 in.

14 in.

21¹₂ d' = 2 in.¹₂

F I G U R E 5.15 Beam cross section for Example 5.7.

McCormac c05.tex V2 - January 9, 2013 11:07 P.M. Page 131

5.7 Compression Steel 131

c= (6.25 in.²− 2.00 in.²) (60 ksi)

(0.85) (3 ksi) (0.85) (14 in.) = 8.40 in.

a= β₁c= (0.85) (8.40 in.) = 7.14 in.

Computing Strains in Compression Steel to Verify Assumption that It Is Yielding

^s= c− d^

c (0.003)= 8.40 in.− 2.5 in.

8.40 in. (0.003)= 0.00211

_y= f_y

E_s = 60,000 psi

29,000,000 psi= 0.00207 < _s^ ∴ f_s^= f_yas assumed.

Note: Example 5.8 shows what to do if this assumption is not correct.

A_s2= A^_sf_s^ Then the design moment strength is

φM_n= φ

Compute the design moment strength of the section shown in Figure 5.16 if f_y= 60,000 psi and f_c^= 4000 psi.

S O L U T I O N

Writing the Equilibrium Equation Assuming f_s^ = f_y A_sf_y= 0.85fc^bβ1c+ A^sf_y

(5.06 in.²) (60 ksi)= (0.85) (4 ksi) (14 in.) (0.85c) + (1.20 in.²) (60 ksi) c= (5.06 in.²− 1.20 in.²) (60 ksi)

(0.85) (4 ksi) (0.85) = 5.72 in.

a= β1c= (0.85) (5.72 in.) = 4.86 in.

Computing Strains in Compression Steel to Verify Assumption that It Is Yielding

^_s= c− d^

c (0.003)= 5.72 in.− 2.5 in.

5.72 in. (0.003)= 0.00169

y= f_y

E_s = 60,000 psi

29,000,000 psi= 0.00207 > ^s ∴ fs^= fyas assumed Since the assumption is not valid, we have to use the equilibrium equation that is based on f_s^ not yielding.

Compute strains, stresses, and steel areas

_s^ = Then the design moment strength is

φMn= φ