4-8 ANALYSIS OF FLANGED SECTIONS
1. Determine for a beam T-section that is part of a continuous floor system
Consider the portion of the continuous floor system shown in Fig. 4-45 and the central floor beam spanning in the horizontal direction. The beam sections corresponding to sec-tion lines A–A and B–B in Fig. 4-45 are given in Figs. 4-46 and 4-47, respectively. The lim-its given in ACI Code Section 8.12 for determining the effective width of the compression flange for a beam section in a continuous floor system are
be … bw + 2¢10 ft - bw
It should be noted that for a floor system with a uniform spacing between beams, the third limit defined above should always result in a value equal to the center-to-center spac-ing between the beams. The first limit governs for this section, so in the followspac-ing parts of this example it will be assumed that be = 72in.
A
A
B
B
10 ft
Slab thickness ⫽ 5 in.
24 ft
10 ft
Fig. 4-45
Continuous floor system for Example 4-5.
be⫽ 72 in.
2.5 in.
5 in.
24 in.
12 in.
3.5 in.
6 No. 7 bars 2 No. 8 bars
⫹
Fig. 4-46
Section A–A from continuous floor system in Fig. 4-45.
be⫽ 72 in.
2.5 in.
5 in.
24 in.
12 in.
2.5 in.
3 No. 7 bars
3 No. 5 bars 3 No. 5 bars
3 No. 8 bars
Fig. 4-47
Section B–B from continuous floor system in Fig. 4-45.
For parts (1) and (2) use the following material properties:
2. For the T-section in Fig. 4-46, calculate and . For the given section, and
For a typical floor system, midspan sections are subjected to positive bending, and sections near the end of the span are subjected to negative bending. The beam section in Fig. 4-46 represents the midspan section of the floor beam shown in Fig. 4-45 and thus is subjected to positive bending. The tension reinforcement for this section is provided in two layers. The minimum spacing required between layers of reinforcement is 1 in. (ACI Code Section 7.6.2). Thus, the spacing between the centers of the layers is approximately 2 in.
Assuming the section will include a No. 3 or No. 4 stirrup, it is reasonable to assume that the distance from the extreme tension edge of the section to the centroid of the lowest layer of steel is approximately 2.5 in. So, the distance from the tension edge to the centroid of the total tension reinforcement is approximately 3.5 in. Thus, the effective flexural depth, d, and the distance from the top of the section (compression edge) to the extreme layer of tension reinforcement, can be calculated to be
Calculation of Assume this is a Case 1 analysis and assume that the tension steel is yielding For section equilibrium, use Eq. (4-16) with substi-tuted for b, giving
This is less than as expected. This value also is less than so we can ignore the compression reinforcement for the analysis of This is a very common result for a T-section in positive bending. For such beams with large compression zones, compression steel is not required for additional moment strength. The compression steel in this beam section may be present for reinforcement continuity requirements (Chapter 8), to reduce deflections (Chapter 9), or to simply support shear reinforcement.
The depth to the neutral axis, c, which is equal to will be approximately equal to 1 in. Comparing this to the values for d and it should be clear without doing calculations that the tension steel strain, easily exceeds the yield strain (0.00207) and the strain at the level of the extreme layer of tension reinforcement, easily exceeds the limit for tension-controlled sections (0.005). Thus, and we can use Eq. (4-21) to calculate as
Check of Tension is at the bottom of this section, so it is clear that we should use in Eq. (4-11). Also, is equal to 190 psi, so use 200 psi in the numerator:
As,min = 200 psi
0.8514 ksi2172 in.2 = 0.88 in.
be
3. For the T-section shown in Fig. 4-47, calculate and . Because this section is subjected to negative bending, flexural tension cracking will develop in the top flange and the compressive zone is at the bottom of the section. Note that ACI Code Section 10.6.6 requires that a portion of the tension reinforcement be distributed into the flange, which coincidentally allows all the negative-moment tension reinforcement to be placed in one layer. Thus, assume that the No. 5 bars in the flange are part of the tension reinforce-ment. So, for the given section,
Using assumptions similar to those used in prior examples, is approximately equal to 2.5 in.
and is approximately equal to the total beam depth, h, minus 2.5 in., i.e., 21.5 in.
Calculation of Because this is a doubly reinforced section, we initially will assume the tension steel is yielding and use the trial-and-error procedure described in Section 4-7 to find the neutral axis depth, c.
Try
Because we should decrease c for the second trial.
Try
With section equilibrium established, we must confirm the assumption that the tension steel is yielding. Because for this section, we can confirm that this is a tension-controlled section in the same step. Using Eq. (4-18):
Clearly, the steel is yielding and this is a tension-controlled section
So, using use Eq. (4-34) to calculate as
5.1 in. b0.003 = 0.00965 d = dt
5.5 in. b10.0032 = 0.00164 c d/4 « 5.5 in.
Calculation of As discussed in Section 4-8, the value of for beam sec-tions with a flange in the tension zone is a function of the use of that beam. The beam section for this example is used in the negative bending zone of a continuous, statically indeterminate floor system. Thus, the minimum tension reinforcement should be calculated using as was done in part (2) of this example. Using an effective depth, d, of 21.5 in., and noting that is less than 200 psi, the following value is calculated using Eq. (4-11):
If the beam section considered here was used as a statically determinate cantilever beam subjected to gravity loading (all negative bending), then the term should be replaced with the smaller of or For this section, is the smaller value, so for such a case, the value of would be
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EXAMPLE 4-6 Analysis of a T-Beam with the Neutral Axis in the Web
Compute the positive moment strength and for the beam shown in Fig. 4-48. Assume that the concrete and steel strengths are 3000 psi and 60 ksi, respec-tively. Also assume the beam contains No. 3 stirrups as shear reinforcement, which are not shown in Fig. 4-48.
1. Compute . Assume this beam is an isolated T-beam in which the flange is used to increase the area of the compression zone. For such a beam, ACI Code Section 8.12.4 states that the flange thickness shall not be less than one-half the width of the web and that the effective flange width shall not exceed four times the width of the web. By observation, the given flange dimensions satisfy these limits. Thus,