use of a coefficient, to replace the constant 0.85 as the definition for the height of the stress block shown in Fig. 4-16a. This new coefficient is defined as follows:
(a) For concrete strengths, up to and including 8000 psi,
(4-15a) (b) For 8000 psi between 8000 and 18,000 psi,
(4-15b) (c) For greater than 18,000 psi,
(4-15c) Until the ACI Code adopts a modification of the stress block shown in Fig. 4-16a, the authors recommend the use of this coefficient, when analyzing the flexural strength of columns constructed with concrete strengths exceeding 8000 psi.
4-4 ANALYSIS OF NOMINAL MOMENT STRENGTH FOR SINGLY
The assumed stress distribution is given in Fig. 4-18c. Above the neutral axis, the stress-block model from Fig. 4-16 is used to replace the actual concrete stress distribution.
The coefficient is multiplied by the depth to the neutral axis,c, to get the depth of the stress block,a. The concrete is assumed to carry no tension, so there is no concrete stress dis- tribution below the neutral axis. At the level of the steel, the stress, is assumed to be equal to the steel yield stress, This corresponds to the assumptions that the steel strain exceeds the yield strain and that the steel stress remains constant after yielding occurs (Fig. 4-7).
The final step is to go from the stress distributions to the equivalent section forces shown in Fig. 4-18d. The concrete compression force, is equal to the volume under the stress block. For the rectangular section used here,
(4-13b) The compression force in the concrete cannot be evaluated at this stage, because the depth to the neutral axis is still unknown. The tension force shown in Fig. 4-18d is equal to the tension steel area, multiplied by the yield stress, Based on the assumption that the steel has yielded, this force is known.
A key step in section analysis is to enforce section equilibrium. For this section, which is assumed to be subject to only bending (no axial force), the sum of the compres- sion forces must be equal to the sum of the tension forces. So,
(4-2) or
The only unknown in this equilibrium equation is the depth of the stress block. So, solving for the unknown value of a,
(4-16) and
(4-17) c = a
b1 a = b1c = Asfy
0.85fcœb
0.85fcœbb1c = 0.85fcœba = Asfy Cc = T
fy. As,
Cc = 0.85fcœbb1c = 0.85fcœba Cc, fy.
fs, b1
esey
(assumed)
fs fy
Cc
T
ecu 0.85fc
b
(a) Singly reinforced section. (b) Strain distribution. (c) Stress distribution. (d) Internal forces.
h d
c
Neutral axis
As
b1ca a/2
d a/2
Fig. 4-18
Steps in analysis of Mnfor singly reinforced rectangular sections.
With the depth to the neutral axis known, the assumption of yielding of the tension steel can be checked. From similar triangles in the linear strain distribution in Fig. 4-18b, the following expression can be derived:
(4-18) To confirm the assumption that the section is under-reinforced and the steel is yield- ing, show
(4-19) Once this assumption is confirmed, the nominal-section moment capacity can be cal- culated by referring back to the section forces in Fig. 4-18d. The compression force is act- ing at the middepth of the stress block, and the tension force is acting at a distance dfrom the extreme compression fiber. Thus, the nominal moment strength can be expressed as either the tension force or the compression force multiplied by the moment arm,
(4-20) For singly reinforced sections, it is more common to express the nominal moment strength using the definition of the tension force as
(4-21) This simple expression can be used for all singly reinforced sections with a rectan- gular (constant width) compression zone after it has been confirmed that the tension steel is yielding. The same fundamental process as used here to determine for singly rein- forced rectangular sections will be applied to other types of beam sections in the following parts of this chapter. However, the reader is urged to concentrate on the process rather than the resulting equations. If the process is understood, it can be applied to any beam section that may be encountered.
EXAMPLE 4-1 Calculation of for a Singly Reinforced Rectangular Section
For the beam shown in Fig. 4-19a, calculate and confirm that the area of ten- sion steel exceeds the required minimum steel area given by Eq. (4-11). The beam sec- tion is made of concrete with a compressive strength, psi, and has four No. 8 bars with a yield strength of
For this beam with a single layer of tension reinforcement, it is reasonable to assume that the effective flexural depth,d, is approximately equal to the total beam depth minus 2.5 in. This accounts for a typical concrete clear cover of 1.5 in., the diameter of the stirrup (typically a No. 3 or No. 4 bar) and half the diameter of the beam longitudinal reinforce- ment. Depending on the sizes of the stirrup and longitudinal bar, the dimension to the center of the steel layer will vary slightly, but the use of 2.5 in. will be accurate enough for most design work unless adjustments in reinforcement location are required to avoid rebar inter- ference at connections with other members. Small bars are often used in the compression
fy = 60 ksi.
fcœ = 4000 Mn
Mn
Mn Mn = Asfyad - a
2b Mn = Tad - a
2b = Ccad - a 2b
d - a/2:
es Ú ey = fy
Es = fy1ksi2 29,000 ksi es = ad - c
c becu es
d - c = ecu c
3 No. 25 bars a⫽ 151 mm b⫽ 250 mm
h⫽ 565 mm d⫽ 500 mm
(b) 20 in.
12 in.
2.5 in.
4 No. 8 bars
(a) Fig. 4-19
Beam sections for (a) Example 4-1 and (b) Example 4-1M.
zone to hold the stirrups in position, but these bars normally are ignored unless they were specifically designed to serve as compression-zone reinforcement.
1. Following the procedure summarized in Fig. 4-18, assume that the steel strain exceeds the yield strain, and thus, the stress in the tension reinforcement equals the yield strength, . Compute the steel tension force:
The assumption that will be checked in step 3. This assumption generally should be true, because the ACI Code requires that the steel area be small enough in beam sections such that the steel will yield before the concrete reaches the maximum useable compression strain.
2. Compute the area of the compression stress block so that .This is done for the equivalent rectangular stress block shown in Fig. 4-16a. The stress block con- sists of a uniform stress of 0.85 distributed over a depth which is measured from the extreme compression fiber. For psi, Eq. (4-14a) gives
Using Eq. (4-16), which was developed from section equilibrium,
3. Check that the tension steel is yielding.The yield strain is
From above, in. Now, use strain compatibility, as expressed in Eq. (4-18), to find
Clearly, exceeds so the assumption used above to establish section equilibrium is confirmed. Remember that you must make this check before proceeding to calculate the section nominal moment strength.
4. Compute .Using Eq. (4-21), which was derived for sections with constant width compression zones,
Mn
ey, es
es = ad - c
c becu = a17.5 - 5.48
5.48 b0.003 = 0.00658 c = a/b1 = 5.48
ey = fy
Es = 60 ksi
29,000 ksi = 0.00207 a = b1c = Asfy
0.85fcœb = 190 kips
0.85 * 4 ksi * 12 in. = 4.66 in.
b1 = 0.85.
fcœ = 4000
a = b1c fcœ
Cc T es 7 ey
T = Asfy = 3.16 in.2 * 60 ksi = 190 kips As = 4 No. 8 bars = 4 * 0.79 in.2 = 3.16 in.2
fy
fs
5. Confirm that tension steel area exceeds .For Eq. (4-11), there is a requirement to use the larger of or 200 psi in the numerator. In this case,
psi, so use 200 psi. Thus,
exceeds so this section satisfies the ACI Code requirement for minimum tension
reinforcement. ■
EXAMPLE 4-1M Analysis of Singly Reinforced Beams: Tension Steel Yielding—SI Units
Compute the nominal moment strength, of a beam (Fig. 4-19b) with
MPa MPa, mm, mm, and three No. 25 bars
(Table A-1M) giving Note that the difference between the
total section depth,h, and the effect depth,d, is 65 mm, which is a typical value for beam sections designed with metric dimensions.
1. Compute a(assuming the tension steel is yielding).
Therefore,
2. Check whether the tension steel is yielding.The yield strain for the reinforcing steel is
From Eq. (4-18),
Thus, the steel is yielding as assumed in step 1.
3. Compute the nominal moment strength, .From Eq. (4-21), is (where )
= 273 * 106 N-mm = 273 kN-m Mn = Asfyad - a
2b = 1530 mm2 * 420 N/mm2a500 - 151 2 b mm
= 1 N/mm2 1 MPa
Mn Mn
es = a500 mm - 178 mm
178 mm b * 0.003 = 0.00543 ey= fy
Es = 420 MPa
200,000 MPa = 0.0021 c = a/b1 = 151/0.85 = 178 mm.
= 1530 mm2 * 420 MPa
0.85 * 20 MPa * 250 mm = 151 mm a = Asfy
0.85fcœb
As = 3 * 510 = 1530 mm2.
d = 500 b = 250
fy = 420
1b1 = 0.852, Mn, fcœ = 20
As,min, As
As,min = 200 psi
fy bwd = 200 psi
60,000 psi * 12 in. * 17.5 in. = 0.70 in.2 324000 psi = 190
32fcœ
As,min
Mn = 2880 k-in. = 240 k-ft Mn = Asfyad - a
2b = 190 kipsa17.5 in. -4.66 in.
2 b
d 21.5 in.
0.85fc
block
Cc 0.85fca2/2 a
Fig. 4-20
Analysis of arbitrary cross section—Example 4-2.
4. Confirm that the tension steel area exceeds .For the given concrete strength of 20 MPa, the quantity which is less than 1.4 MPa.
Therefore, the second part of Eq. (4-11M) governs for as
exceeds so this section satisfies the ACI Code requirement for minimum tension
reinforcement. ■
EXAMPLE 4-2 Calculation of the Nominal Moment Strength for an Irregular Cross Section
The beam shown in Fig. 4-20 is made of concrete with a compressive strength, psi and has three No. 8 bars with a yield strength, This example is presented to demonstrate the general use of strain compatibility and section equilibrium equations for any type of beam section.
1. Initially, assume that the stress in the tension reinforcement equals the yield strength , and compute the tension force
The assumption that the tension steel is yielding will be checked in step 3.
2. Compute the area of the compression stress block so that .As in the prior problem, this is done using the equivalent rectangular stress block shown in Fig. 4-16a.
The stress block consists of a uniform stress of 0.85 distributed over a depth
which is measured from the extreme compression fiber. For psi, Eq. (4-14a) gives The magnitude of the compression force is obtained from equilibrium as
Cc= T = 142 kips = 142,000 lbs b1 = 0.85.
fcœ = 3000
a = b1c, fcœ
Cc T T = Asfy = 2.37 in.2 * 60 ksi = 142 kips
As = 3 No. 8 bars = 3 * 0.79 in.2 = 2.37 in.2 T Asfy: fy
fs
fy = 60 ksi.
fcœ = 3000 As,min, As
As,min = 1.4bwd
fy = 1.4 MPa * 250 mm * 500 mm
420 MPa = 417 mm2
As,min 0.252fcœ = 1.12 MPa,
As,min
By the geometry of this particular triangular beam, shown in Fig. 4-20, if the depth of the compression block is a, the width at the bottom of the compression block is also a, and the area is This is, of course, true only for a beam of this particular triangular shape.
Therefore, and
3. Check whether .This is done by using strain compatibility. The strain distribution at ultimate is shown in Fig. 4-20c. As before,
Using strain compatibility, as expressed in Eq. (4-18), calculate
Although this is close to the yield strain, it does exceed the yield strain as calculated in step 3 of Example 4-1 for Grade-60 reinforcement. Thus, the assumption made in step 1 is satisfied.
4. Compute .
wherejdis a general expression for the lever arm, i.e., the distance from the resultant ten- sile force (at the centroid of the reinforcement) to the resultant compressive force Because the area on which the compression stress block acts is triangular in this example,
acts at 2a/3 from top of the beam. Therefore,
Note:If we wanted to calculate for this section, we should base the calculation on the average width of the portion of the section that would be cracking in tension. It is not easy to determine this value, because the distance that the flexural crack penetrates into the section is difficult to evaluate. However, it would be conservative to use the width of the section at the extreme tension fiber, 24 in., in Eq. (4-11). As in the Example 4-1, 200 psi will govern for the numerator in this equation. So,
Therefore, even with a conservative assumption for the effective width of the cracked tension zone, this section has a tension steel area that exceeds the required minimum
area of tension reinforcement. ■
As,min = 200 psi
fy bwd = 200 psi
60,000 psi * 24 in. * 21.5 in. = 1.72 in.2 As,min
= 2060 lb-in. = 171 k-ft
= 2.37 in.2 * 60 ksia21.5 - 2 * 10.6 3 b in.
Mn = Asfyad - 2a 3b jd = d - 2a
3 Cc
Cc. Mn = Ccjd = Tjd
Mn
es = a21.5 - 12.4
12.4 b0.003 = 0.00220 c = a
b1 = 10.6 in.
0.85 = 12.4 in.
fs fy
a = A
142,000 1b * 2
0.85 * 3000 psi = 10.6 in.
Cc = 10.85fcœ21a2/22 a2/2.