Determine for a beam T-section that is part of a continuous floor system

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–AandB–Bin 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

b_e … b_w + 2¢^{10 ft - bw}

2 ≤ = 10 ft = 120 in.

b_e … b_w + 218h_f2 = 12 in. + 1615 in.2 = 92 in.

b_e … /

4 = 24 ft 112 in./ft2

4 = 72 in.

b_e

b_e, 2bw

b_w b_w

b_e, b_w A_s,min As,min

f = 0.65.

1e_t 6 0.0022,

f.

e_t

e_t, d_t,

f,

f

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 following parts of this example it will be assumed that b_e = 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.

b_e⫽ 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

SectionA–Afrom continuous floor system in Fig. 4-45.

b_e⫽ 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

SectionB–Bfrom 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 dand 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:

A_s,min = 200 psi

f_y b_wd = 200 psi

60,000 psi112 in.2120.5 in.2 = 0.82 in.² 6 A_s1o.k.2 32f_c^œ

b_w

As,min

fM_n = 0.9 * 361 = 325 k-ft M_n = 4330 k-in. = 361 k-ft M_n = A_sf_yad - a

2b = 13.60 in.²2160 ksi2a20.5 in. -0.88 in.

2 b

M_n f = 0.9,

e_t, e_s,

d_t,

a/b₁, M_n.

d¿, h_f,

a = A_sf_y

0.85f_c^œb_e = 13.60 in.²2160 ksi2

0.8514 ksi2172 in.2 = 0.88 in.

b_e 1e_s Ú e_y2.

1a … h_f2 FM_n

d_t = 24 in. - 2.5 in. = 21.5 in.

d = 24 in. - 3.5 in. = 20.5 in.

d_t,

A_s^œ = 210.79 in.²2 = 1.58 in.² 610.60 in.²2 = 3.60 in.²

A_s=

A_s,min FM_n

f_c^œ = 4000 psi 1b₁ = 0.852 and f_y = 60 ksi

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 cfor 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

fM_n = 0.9 * 401 = 361 k-ft

M_n = 3420 k-in. + 1400 k-in. = 4820 k-in. = 401 k-ft M_n = C_cad - a

2b + C_s1d - d^œ2 = 177 k * 19.3 in. + 73.7 k * 19.0 in.

M_n a = b₁c = 0.85 * 5.1 in. = 4.34 in.,

0.0052.

1e_t7 1e_s7e_y=0.002072

e_s1=e_t2 = d - c

c e_cu = a21.5 in. - 5.1 in.

5.1 in. b0.003 = 0.00965 d = d_t

T = 254 kips C_s + C_c = 251 kips C_c = 177 kips

C_s = 73.7 kips f_s^œ = 44.4 ksi 1…f_y2 e_s^œ = 0.00153

c 5.1 in.

T 6 C_c + C_s,

T = A_sf_y = 4.23 in.² * 60 ksi = 254 kips

C_c = 0.85f_c^œb_wb₁c = 0.85 * 4 ksi * 12 in. * 0.85 * 5.5 in. = 191 kips C_s = A_s^œ1f_s^œ - 0.85f_c^œ2 = 1.80 in.²147.5 ksi - 3.4 ksi2 = 79.3 kips f_s^œ = E_se_s^œ = 29,000 ksi * 0.00164 = 47.5 ksi 1…f_y2

e_s^œ = ac - d¿

c be_cu = a5.5 in. - 2.5 in.

5.5 in. b10.0032 = 0.00164 c d/4 « 5.5 in.

FM_n d = d_t

d¿ A_s^œ = 3 * 0.60 in.² = 1.80 in.²

A_s = 3 * 0.79 in.² + 6 * 0.31 in.² = 4.23 in.² A_s,min FM_n

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,

2. Computed.As in the prior example with two layers of tension reinforcement,

assume as shown in Fig. 4-48a.

3. Computea.Assume this is a Case 1 analysis and thus, the compres- sion zone will be rectangular. Accordingly,

Becauseais greater than the thickness of the flange our assumption that the compression zone is rectangular is wrong, and our calculated value of aisincorrect. It therefore is necessary to use the Case 2 analysis procedure discussed in Section 4-8 and artificially break the section into two beams shown as beam F and beam W in Fig. 4-48c and d, respectively.

4. Analysis of for the flanged section with .The compression force in beam F is given by Eq. (4-36) as

The compression force in beam W is given by Eq. (4-37) as C_cw = 0.85f_c^œb_wa

C_cf = 0.85f_c^œ1b_e - b_w2h_f = 0.85 * 3 ksi118 in. - 10 in.2 * 5 in. = 102 kips a>h_f

M_n

1h_f = 5 in.2, a = A_sf_y

0.85f_c^œb_e = 4.74 in.² * 60 ksi

0.85 * 3 ksi * 18 in. = 6.20 in.

1a … h_f2, d L h - 3.5 in. = 24.5 in.,

b_e = 18 in.

be

A_s,min fM_n

A_s,min = 200 psi

f_y 12b_w2d = 1.72 in.² 6 A_s1o.k.2 A_s,min

2b_w b_e.

2b_w

b_w A_s,min = 200 psi

f_y b_wd = 200 psi

60,000 psi112 in.2121.5 in.2 = 0.86 in.² 6 A_s1o.k.2 32f_c^œ

b_w, A_s,min

A_s,min

Of course, the depth of Whitney’s stress block,a, is the major unknown for this sec- tion analysis procedure. It is found by setting the tension force,

equal to the sum of the compression forces, as was done to derive Eq. (4-38):

With this value of a, Before calculating we must confirm that

the tension steel is yielding. Using Eq. (4-18) can be

used to calculate the tension steel strain as

This clearly exceeds the yield strain (0.00207), so the assumption that the tension steel is yielding is confirmed. It should be noted that the distance to the extreme layer of tension steel, d_t, will exceed d for this section, so the strain at the extreme layer of

e_s = d - c

c e_cu = a24.5 in. - 8.42 in.

8.42 in. b0.003 = 0.00573 c = a/b₁ = 7.15/0.85 = 8.42 in.,

M_n, C_cw = 182 kips.

a = T - C_cf

0.85f_c^œb_w = 284 k - 102 k

0.85 * 3 ksi * 10 in. = 7.15 in.

60 ksi = 284 kips,

T = A_sf_y = 4.74 in.²* Fig. 4-48

Beam sections for Example 4-6.

tension steel, will exceed Thus, exceeds 0.005, making this a tension-controlled section with equal to 0.9. Using Eq. (4-39) to calculate

5. Check whether .Assuming this beam is continuous over several spans, we can use for this calculation. For a concrete strength of 3000 psi, note that is less than 200 psi, so use 200 psi in Eq. (4-11), giving

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