Recall, the general design strength equation for flexure is
(4-1b) where is the strength reduction factor. For beams, the factor is defined in ACI Code Section 9.3.2 and is based on the expected behavior of the beam section, as represented by the moment–curvature curves in Figs. 4-11 and 4-12. Because of the monolithic nature of reinforced concrete construction, most beams are part of a continuous floor system, as shown in Fig. 1-6. If a beam section with good ductile behavior was overloaded accidentally, it would soften and experience some plastic rotations that would permit loads to be redis- tributed to other portions of the continuous floor system. This type of behavior essentially
f f
fMn Ú Mu fcœ
fy
rb = 0.85b1fcœ
fy ¢ 87,000 87,000 + fy≤ Es = 29,000,000
ecu
rb = 0.85b1fcœ fy ¢e ecu
cu + ey≤ rb = As1bal2
bd = 0.85b1fcœ fy * b
bd * ¢e ecu
cu + ey≤d
rb, As1bal2 = 1
fy30.85fœcbb1c1bal24 As
As1bal2fy = 0.85fœcbb1c1bal2 T1bal2 = Cc1bal2
Mn. Cc
Mn
d dt
c
(a) Beam section. (b) Strain distribution.
es
et
ecu 0.003
Fig. 4-22
Definitions for dtand et.
creates an increased level of safety in the structural system, so a higher is permitted for beams designed to exhibit ductile behavior. For beams that exhibit less ductile behavior, as indicated for the sections with larger tension steel areas in Fig. 4-11, the ability to redis- tribute loads away from an overloaded section is reduced. Thus, to maintain an acceptable level of safety in design, a lower is required for such sections.
Until 2002, the ACI Code defined only a single for the design of reinforced concrete beam sections, but the behavior was controlled by limiting the permitted area of tension reinforcement. The design procedure was to keep the reinforcement ratio, less than or equal to 0.75 times the balanced reinforcement ratiodefined in Eq. (4-25). This procedure, which is still permitted by Appendix C of the ACI Code, is easy to apply to singly reinforced rectangular sections, but becomes more complicated for flanged sections and sections that use compression reinforcement. When the same criteria is applied to beam sections that contain both normal reinforcement and prestressing tendons, the defin- ition for the permitted area of tension reinforcement becomes quite complex.
Another method for controlling the ductility of a section is to control the value of tension strain reached at the level of the tension reinforcement when the extreme concrete compression fiber reaches the maximum useable compression strain, i.e., at nominal strength conditions(Fig. 4-18b). Requiring higher tension strains at the level of tension steel is a universal method for controlling the ductility of all sections, as initially discussed by Robert Mast [4-14]. Starting with the 2002 edition, this is the procedure used in Chapters 9 and 10 of the ACI Code to control section ductility, and thus, specify the corresponding values for the strength-reduction factor,
Definitions of Effective Depth and Distance to Extreme Layer of Tension Reinforcement
Theeffective depth,d, is measured from the extreme compressive fiber to the centroidof the longitudinal reinforcement. This is the distance used in calculations of the nominal moment strength, as demonstrated in prior examples. To have consistency in controlling tension strains for a variety of beam and column sections, the ACI Code defines a distance, which is mea- sured from the extreme compression fiber to the extreme layer of tension reinforcement, as shown in Fig. 4-22a. The strain at this level of reinforcement, is defined as the net strain at the extreme layer of tension reinforcement at nominal-strength conditions, excluding strains due to effective prestress, creep, shrinkage, and temperature. For beam sections with more than one layer of reinforcement, will be slightly larger than the strain at the centroid of the tension reinforcement, as shown in Fig. 4-22b. The ACI Code uses the strain to define the behavior of the section at nominal conditions, and thus, to define the value of f.
et es,
et
et,
dt, f.
r, f-value
f-value
f-value
Definitions of Tension-Controlled and Compression-Controlled Sections
A tension-controlled section has a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile steel, is greater than or equal to 0.005. For Grade-60 reinforcement with a yield strength the tensile yield strain is The tension-controlled limit strain of 0.005 was chosen to be approximately 2.5 times the yield strain of the rein- forcement, giving a moment–curvature diagram similar to that shown in Fig. 4-11 for the section with an area of tension reinforcement equal to 4.50 in.2The strain diagram corre- sponding to the tension-controlled limit (TCL) is demonstrated in Fig. 4-23b, with the depth from the extreme compression fiber to the neutral axis defined as c(TCL). From the strain diagram it can be shown that
(4-26) Clearly, if a calculated value of cislessthat the strain, will exceed0.005. Thus, when analyzing the nominal flexural strength of a beam section, demonstrating that the depth to the neutral axis obtained from section equilibrium is less than as given in Eq. (4-26), will be one method to verify that the section is tension-controlled.
Acompression-controlled sectionhas a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile steel, is less than or equal to the yield strain. For beams with Grade-60 reinforcement and beams with prestressed reinforcement, ACI Code Section 10.3.3 permits the use of 0.002 in place of the yield strain. A beam section with this amount of tension reinforcement would exhibit a moment–curvature relationship sim- ilar to that shown in Fig. 4-11 for the section with the largest steel area. The strain diagram corresponding to the compression-controlled limit (CCL) is demonstrated in Fig. 4-23c, with the depth from the extreme compression fiber to the neutral axis defined as c(CCL).
From the strain diagram it can be shown that
(4-27) Clearly, if a calculated value of cisgreaterthat the strain will be lessthan 0.002.
Atransition-zone sectionhas a tension-reinforcement area such that when the beam reaches its nominal flexural strength, the net tensile strain in the extreme layer of tensile
et 3/5dt,
c1CCL2 = 3
5dt = 0.60dt
1ey = 0.002072 et,
3/8dt, et,
3/8dt, c1TCL2 = 3
8dt = 0.375dt
ey = 60/29,000 = 0.00207.
fy = 60 ksi, et,
dt
c(TCL)
c(CCL)
(a) Beam section. (b) Strain distribution at tension-controlled limit.
(c) Strain distribution at compression-controlled limit.
ecu⫽ 0.003 ecu⫽ 0.003
et⫽ 0.002 et⫽ 0.005
Fig. 4-23
Strain distributions at tension-controlled and compression-controlled limits.
steel, is between 0.002 and 0.005. A beam section with this amount of tension rein- forcement would exhibit a moment–curvature relationship in between those shown in Fig. 4-11 for sections with tension steel areas of 4.50 and 6.50 in.2
Because tension-controlled sections demonstrate good ductile behavior if overloaded, they are analyzed and designed using a strength-reduction factor, of 0.9. Because of their brittle behavior if overloaded, compression-controlled sections are analyzed and designed with equal to 0.65. (Note:This is the value for beams with standard stirrup-tie reinforce- ment similar to that shown in Fig. 4-19. As will be discussed in Chapter 11, for column sections with spiral reinforcement, the value of is 0.75 if the section is compression- controlled).
The variation of the strength-reduction factor, , as a function of either the strain,et, or the ratio,c/dt, at nominal strength conditions is shown in Fig. 4-24. For beam or column sections that are either compression-controlled (et … 0.002) or tension-controlled (et Ú 0.005), the value of is constant. When analyzing a transition-zone section with stirrup-tie (or hoop) transverse reinforcement, the value of varies linearly between 0.65 and 0.90 as a function of either etorc/dt, as given in Eqs. (4-28a) and (4-28b), respectively.
(4-28a) (4-28b) For a transition-zone section with spiral transverse reinforcement (column section), the variation of the value of as a function of either etorc/dtis given in Eqs. (4-29a) and (4-29b), respectively.
(4-29a) (4-29b) In the prior examples, the values of now can be calculated. In all three examples, there was only one layer of tension steel, so is equal to For the rectangular beams in Examples 4-1 and 4-1M, the value of esexceeded 0.005, so the f-valuewould be 0.9. For
es. et
f
f = 0.75 + 0.15a 1 c/dt - 5
3b f = 0.75 + 1et - 0.0022 * 50 f
f = 0.65 + 0.25a 1 c/dt - 5
3b f = 0.65 + 1et - 0.0022 * 250
3 f
f
f f f
f, et,
0.600 0.65
0.75 0.90
0.002 0.005
0.375
c/dt Transition zone
Stirrup-tie Spiral
Eq. (4-29a and b)
Eq. (4-28a and b)
Compression- controlled
Tension- controlled f
et Fig. 4-24
Variation of -factor with andc/dtfor spiral and stirrup- tie transverse reinforcement.
et f
the triangular beam section in Example 4-2, the value of was 0.00220. Using that as the value for in Eq. (4-28a) results in a of 0.67 (the authors recommend using only two significant figures for ).
Upper Limit on Beam Reinforcement
Prior to the 2002 edition of the ACI Code, the maximum-tension steel area in beams was limited to 0.75 times the steel area corresponding to balanced conditions (Fig. 4-21). In the latest edition of the ACI Code (ACI 318-11), Section 10.3.5 requires that for reinforced concrete (nonprestressed) beam sections (stated as members with axial com- pressive load less than ) the value of at nominal flexural strength conditions shall be greater than or equal to 0.004. This strain value was selected to approximately correspond to the former ACI Code requirement of limiting the tension steel area to 0.75 times the balanced-tension steel area. A beam section with a tension steel area resulting in at nominal conditions would have a higher value than a beam section with a lower-tension steel area that resulted in (the tension-controlled limit) at nominal strength conditions. However, because there are different for these two beam sections, the resulting values of for the two sections will be approxi- mately equal.
The rectangular beam section in Fig. 4-25 will be used to demonstrate the change in the reduced nominal moment strength, as the amount of tension-reinforcing steel is increased. Table 4-2 gives the results from a series of moment strength calcula- tions for constantly increasing values for the reinforcement ratio, The corresponding steel areas are given in the second column of Table 4-2, and the depth to the neutral axis,c, obtained from Eqs. (4-16) and (4-17) are given in the third column. The beam section represented by the last row in Table 4-2 is over-reinforced and a strain-compatibility procedure is required to establish equilibrium and find the corresponding depth to the neutral axis,c. The details of this analysis procedure will be discussed at the end of this subsection.
Values for which are equal to for a single layer of reinforcement, are obtained from Eq. (4-18) and then used to determine the corresponding values of the strength reduc- tion factor, If is greater than or equal to 0.005 (signifying a tension-controlled section), is set equal to 0.9. For values between 0.005 and 0.002, Eq. (4-28a) is used to calculate the corresponding to three significant figures for this comparsion.
For the largest in Table 4-2 (last row), the calculated value of is equal to the compression-controlled limit of 0.002, so fwas set equal to 0.65. Finally, Eq. (4-20) was
et r-value
f-value et f
et f.
es et,
r. fMn,
fMn
f-values et = 0.005
Mn et = 0.004
et 0.10fœcAg
f
f-value et
es
14 in.
22 in.
19.5 in.
f⬘c⫽ 4000 psi fy⫽ 60 ksi As
Fig. 4-25
Typical beam section with as a variable.
As
00 100 200 300 400 500 600
Moment strength, k-ft
700
0.005 0.010 0.015
Reinforcement ratio, r
0.020 0.025 0.030 0.035
et⫽ 0.005
et⫽ 0.004 et⫽ 0.002 Nominal moment
Reduced nominal moment
Fig. 4-26
Relationship between and values for Mnand fMn.
r
TABLE 4-2 Relationship between Reinforcement Ratio and Nominal Moment Strength c (in.)
0.005 1.37 2.03 0.0258 0.900 128 115
0.010 2.73 4.05 0.0115 0.900 243 218
0.015 4.10 6.08 0.00662 0.900 347 312
0.0181 4.93 7.31 0.0050 0.900 404 364
0.0207 5.64 8.36 0.0040 0.817 449 367
0.025 6.83 10.1 0.00278 0.715 519 371
0.0285 7.78 11.5 0.00207 0.656 568 372
0.030 8.19 11.7 0.00200 0.650 575 374
fMn1k-ft2 Mn1k-ft2
f et
As1in.22 r
used to calculate the nominal moment strength, which was multiplied by to get the values of given in the last column of the Table 4-2.
Some interesting results can be observed in the plots of versus and versus in Fig. 4-26. There is an almost linear increase in and for increasing values of up to the point where the tension strain, reaches the tension-controlled limit of 0.005.
Beyond this point, continues to increase almost linearly for increasing values of but the value of tends to stay almost constant due the decrease in the value of obtained from Eq. (4-28a). This is a very important result that dimishes the significance of the limit set on in ACI Code Section 10.3.5 The author believes that the important limit for the amount of tension steelto use in the design of beam sections will be to keep
1et Ú 0.0042.
et
f fMn
r, Mn
et,
r fMn
Mn
fMn Mn
r fMn
Mn, f
at or above the tension-controlled limit of 0.005, because Fig. 4-26 clearly shows that beyond this point it is not economical to add more tension steel to the section. Thus, for the flexural design procedures discussed in Chapter 5, the authors will always check that the final section design is classified as a tension-controlled section and thus, the always will be 0.90.
One final point of interest in Fig. 4-26 occurs in the plot of versus for a steel area larger that the balanced steel area given in Eq. (4-25). This section (last row of values in Table 4-2) is referred to as being over-reinforced, but the value for does not increase for this larger area of tension steel because the concrete compression zone will start to fail before the steel reaches its yield stress. Thus, the values for the compression force, and therefore the tension force,T, tend to stay relatively constant. Exact values for the steel stress and strain can be determined using the fundamental procedure of satisfying section equilibrium (Eq. (4-2)) and strain compatibility (Eq. (4-18)). Then the section nominal moment strength,
can be calculated by the more general expression in Eq. (4-20). For over-reinforced sec- tions, the nominal moment strength will tend to decrease as more tension steel is added to the section, because the moment arm will decrease as is increased. An analysis of an over-reinforced beam section is presented as Beam 3 in the following example.
EXAMPLE 4-3 Analysis of Singly Reinforced Rectangular Beams
Compute the nominal moment strengths, and the strength reduction factor, for three singly reinforced rectangular beams, each with a width and a total height As shown in Fig. 4-27 for the first beam section to be analyzed, a beam normally will have small longitudinal bars in the compression zone to hold the stirrups (shear reinforcement) in place. These bars typically are ignored in the calculation of the section nominal moment strength. Assuming that the beam has in. of clear cover and uses No. 3 stirrups, we will assume the distance from the tension edge to the centroid of the lowest layer of tension reinforcement is 2.5 in.
Beam 1: The tension steel area,
1. Compute a, c, and (same as for single layer of reinforcement).As before, assume that the tension steel is yielding, so Using Eq. (4-16), which was developed from section equilibrium for a rectangular compression zone,
= 4.00 in.2 * 60 ksi
0.85 * 4 ksi * 12 in. = 5.88 in.
a = b1c = Asfy 0.85fcœb
fs = fy. Et
Es 4.00 in.2
As4 (1.00 in.2) fcœ4000 psi and fy60 ksi.
112
h = 20 in.
b = 12 in.
f, Mn,
As 1d - a/22
Mn,
Cc, Mn
Mn r
f-value
1et Ú 0.0052, et
20 in.
12 in.
d⫽ 17.5 in.
4 No. 9 bars
Fig. 4-27
Section used for Beams 1 and 2 of Example 4-3.
For is equal to 0.85. Thus, and using strain compatibility as expressed in Eq. (4-18), find
This exceeds the yield strain for Grade-60 steel ( previously calculated), so the assumption that the tension steel is yielding is confirmed.
2. Compute the nominal moment strength, .As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones for
3. Confirm that tension steel area exceeds .Although this is seldom a problem with most beam sections, it is good practice to make this check. The expression for is given in Eq. (4-11) and includes a numerator that is to be taken equal to but not less than 200 psi. As was shown in Example 4-1, the value of 200 psi governs for beams constructed with 4000 psi concrete. Thus,
Clearly, for this section satisfies the ACI Code requirement for minimum tension reinforcement.
4. Compute the strength reduction factor, and the resulting value of . As stated previously, for a single layer of tension reinforcement, is equal to which was calculated in step 1. Because is between 0.002 and 0.005, this is a transition-zone section. Thus, Eq. (4-28a) is used to calculate
Then,
Beam 2: Same as Beam 1, except that .As shown in Fig. 4-12, chang- ing the concrete compressive strength will not produce a large change in the nominal moment strength, but it does increase the ductility of the section. Thus, increasing the concrete compressive strength might change the beam section in Fig. 4-27 from a transition-zone sectionto a tension-controlled section.
1. Compute a, c, and .Again, assume that the tension steel is yielding, so For this compressive strength, Eq. (4-14b) is used to determine that
Then, using Eq. (4-16),
= 4.00 in.2 * 60 ksi
0.85 * 6 ksi * 12 in. = 3.92 in.
a = b1c = Asfy 0.85fcœb
b1 = 0.75.
fs = fy.
Es
fcœ 6000 psi fMn = 0.87 * 291 k-ft = 253 k-ft f = 0.65 + 10.00459 - 0.0022250
3 = 0.87 f:
et
es, et
FMn
F, As
As,min = 200 psi
fy bwd = 200
60,000 * 12 in. * 17.5 in. = 0.70 in.2
32fcœ, As,min
As,min Mn = 34 90 k-in. = 291 k-ft
Mn = Asfyad - a
2b = 4.0 in.2 * 60 ksia17.5 in. - 5.88 in.
2 b
Mn ey = 0.00207, es = ad - c
c becu = a17.5 - 6.92
6.92 b0.003 = 0.00459 c = a/b1 = 6.92 in., fcœ = 4000 psi, b1
Thus, and using strain compatibility as expressed in Eq. (4-18), find
This exceeds the yield strain for Grade-60 steel confirming the assump- tion that the tension steel is yielding.
2. Compute the nominal moment strength, . As in Example 4-1, use Eq. (4-21), which applies to sections with rectangular compression zones:
3. Confirm that tension steel area exceeds .For this beam section with 6000 psi concrete, the value of exceeds 200 psi, and will govern in Eq. (4-11). Thus,
Again, for this section easily satisfies the ACI Code requirement for minimum tension reinforcement.
4. Compute the strength reduction factor, , and the resulting value of . As before, is equal to which was calculated in step 1. This beam section is clearly a tension-controlled section, so Then,
Beam 3: Same as Beam 1, except increase tension steel to six No. 9 bars in two layers (Fig. 4-28).For this section, will be larger than and will be calculated using the distance to the extreme layer of tension reinforcement, Assuming the same cover and size of stirrup, as used for din Beams 1 and 2. The value of dfor this section involves a centroid calculation for the six No. 9 bars. ACI Code Section 7.6.2 requires a clear spacing between layers of reinforcement greater than or equal to 1 in. Thus, we can assume that the second layer of steel (two bars) is one bar diameter plus 1 in. above the lowest layer,—or a total of 2.5 in. ⫹ from the extreme tension fiber. A simple calculation is used to find the distance from the bottom of the beam to the centroid of the tension reinforcement,g, and then find the value of
1. Compute a, c, and . Again, assume that the tension steel is yielding, so Then, using Eq. (4-16):
= 6.00 in.2 * 60 ksi
0.85 * 4 ksi * 12 in. = 8.82 in.
a = b1c = Asfy 0.85fcœb fs = fy.
es
d = h - g = 20 in. - 3.21 in. L 16.8 in.
g = 4.0 in.2 * 2.5 in. + 2.0 in.2 * 4.63 in.
6.0 in.2 = 3.21 in.
d = h - g.
1.128 in. + 1 in. L 4.63 in.
dt = 17.5 in.,
dt. es et
fMn = 0.9 * 311 k-ft = 280 k-ft 111 percent increase from beam 12 f = 0.9.
es, et
FMn
F As
As,min = 32fcœ
fy bwd = 326000
60,000 * 12 in. * 17.5 in. = 0.81 in.2 32fcœ
As,min
Mn = 3730 k-in. = 3 11 k-ft 17 percent increase from Beam 12 Mn = Asfyad - a
2b = 4.0 in.2 * 60 ksia17.5 in. -3.92 in.
2 b
Mn
1ey = 0.002072, es = ad - c
c becu = a17.5 - 5.23
5.23 b0.003 = 0.00704 c = a/b1 = 5.23 in.,
As for Beam 1, Thus, and using strain compatibility, as expressed in Eq. (4-18), find
This is less than the yield strain for Grade-60 steel so the assumption that the tension steel is yielding is not confirmed. Because the tension steel is not yielding, this is referred to as an over-reinforcedsection, and the previously developed procedure for cal- culating the nominal moment strength does not apply. A procedure that enforces strain compatibility and section equilibrium will be demonstrated in the following.
2. Compute the nominal moment strength, by enforcing strain compatibility and section equilibrium.Referring to Fig. 4-18, we must now assume that the steel stress,
is an unknown but is equal to the steel steel strain, multiplied by the steel modulus, Strain compatibility as expressed in Eq. (4-18) still applies, so the steel stress and thus the tension force can be expressed as a function of the unknown neutral axis depth,c.
Similarly, the concrete compression force can be expressed as a function of the neutral axis depth,c.
Enforcing equilibrium by setting we can solve a second degree equation for the unknown value of c. The solution normally results in one positive and one negative value forc; the positive value will be selected. Using all of the given section and material prop- erties and recalling that ksi and the resulting value for c is 10.1 in. Using this value, the authors obtained
An average value of kips will be used to calculate Then, using calculate using the more general expression in Eq. (4-20).
Mn = 4350 k-in. = 363 k-ft Mn = Tad - a
2b = 348 kipsa16.8 in. - 8.59 in.
2 b
Mn 0.85 * 10.1 in. = 8.59 in.,
a = b1c = Mn.
T = Cc = 348
T = 346 kips Cc = 350 kips ecu = 0.003, Es = 29,000
T = Cc,
Cc = 0.85fcœbb1c T = Asfs = AsEses = AsEsad - c
c becu
Es. es,
fs,
Mn,
1ey = 0.002072, es = ad - c
c becu = a16.8 - 10.4
10.4 b 0.003 = 0.00186 c = a/b1 = 10.4 in.
b1 = 0.85.
20 in.
12 in.
d⫽ 16.8 in.
dt⫽ 17.5 in.
6 No. 9 bars
Fig. 4-28
Section used for Beam 3 of Example 4-3.