Chapter 10 Continuous Beams and One-Way Slabs

10Continuous Beamsand One-Way Slabs

468

10-1 INTRODUCTION

In the design of a continuous beam or slab, it is necessary to consider several ultimate limitstates—failure by flexure, shear, bond, and possibly torsion—and several serviceabilitylimit states—excessive deflections, crack widths, and vibrations. Each of these topics hasbeen covered in a preceding chapter. In this chapter, continuity in reinforced concretestructures is discussed, followed by a design example showing the overall design of acontinuous beam.

10-2 CONTINUITY IN REINFORCED CONCRETE STRUCTURES

The construction of two cast-in-place reinforced concrete structures is illustrated inFigs. 10-1 to 10-3. In Fig. 10-1, formwork is under construction for the beams and slabssupporting a floor. The raised platform-like areas are the bottom forms for the slabs. Therectangular openings between them are forms for the beam concrete. The beams meet atright angles at the columns.

Once the formwork is complete and the reinforcement has been placed in theforms (Fig. 10-2), the concrete for the slabs and beams will be placed in one monolithicpour. (See ACI Code Section 6.4.7.) Following this, the columns for the next story areerected, as shown in Fig. 10-3. ACI Code Section 6.4.6 requires that the concrete in thecolumns or walls have set before the concrete in the floor supported by those columns isplaced. This sequence is required because the column concrete will tend to settle in theforms while in the plastic state. If the floor concrete had been placed, this would leave agap between the column concrete and the beam. By placing the floor concrete after thecolumn concrete is no longer plastic, any gap that formed as the concrete settled will befilled. The resulting construction joints can be seen at the bottom and top of each columnin Fig. 10-3.

As a result of this placing sequence, each floor acts as a continuous unit. Because thecolumn reinforcement extends through the floor, the columns act with the floors to form acontinuous frame.

Section 10-2 Continuity in Reinforced Concrete Structures • 469

Fig. 10-1Formwork for a beam-and-slab floor. (Photograph cour-tesy of J. G. MacGregor.)

Fig. 10-2Beam and slab reinforcementin the forms. (Photographcourtesy of J. G. MacGregor.)

Braced and Unbraced Frames

A frame is said to be a sway frame (or an unbraced frame) if it relies on frame action toresist lateral loads or lateral deformations. Thus, the frame in Fig. 10-4a is unbraced,while that in Fig. 10-4b is braced by the shear wall. If the lateral stiffness of the bracing

470 • Chapter 10 Continuous Beams and One-Way Slabs

Fig. 10-4Sway frames and bracedframes.

Fig. 10-3Construction of columns in atall building. (Photographcourtesy of J. G. MacGregor.)

element in a story exceeds 6 to 10 times the sum of the lateral stiffnesses of all thecolumns in that story, a story can be considered to be braced. ACI Code Section10.10.5.2 gives a procedure for classifying stories in a frame as sway or non-sway sto-ries. Most concrete buildings are braced by walls, elevator shafts, or stairwells. The floormembers in an unbraced frame must resist moments induced by lateral loads as well asgravity loads; in most cases, by contrast, the lateral-load moments can be ignored in the

Section 10-2 Continuity in Reinforced Concrete Structures • 471

design of beams in a braced frame. The design examples in this chapter are limited tobeams in braced frames.

Design Loads for a Continuous Floor System

The design loads are selected in accordance with ASCE/SEI 7-10, Minimum Design Loadsfor Buildings and Other Structures [10-1]. The dead loads include the weight of the floor,0.5 psf for asphalt tile or carpet, 4 psf for mechanical equipment and lighting fixtures hungbelow the floor, and 2 psf for a suspended ceiling.

ASCE/SEI 7-10 specifies the minimum live loads for offices as 50 psf and for lobbiesas 100 psf. (See Table 2-1.) The live loads for file storage rooms and computer rooms arebased on the actual usage of the room. An allowance must be made for the weight of partitions(considered to be live load in ASCE/SEI 7-10) if the floor loading is 80 psf or less. Thedesign live loads can be reduced as a function of the floor area, as discussed in Section 2-8.No live-load reduction is allowed for one-way slabs.

A second loading case required in ASCE/SEI 7-10 for office buildings is a concentrat-ed load of 2000 lb acting on a 30-in.-by-30-in. area placed on the floor to produce the maxi-mum moments or shears in the structural members. The uniform live load is assumed to notbe acting on the structure when this load acts. For most floor beams and slabs, the uniformlive loads will give greater moments and shears in the beams than the concentrated load.

Transfer of Column Loads through the Floor System

In the building shown in Fig. 10-3, the columns were built of 5000-psi concrete and the floorswere built of 3750-psi concrete. These strengths were chosen on the basis of ACI Code Sec-tion 10.12, which allows the column concrete to be stronger than the floor concrete.

The floor concrete in the vicinity of the column–beam joint must transfer axial loadsfrom the column above the joint to the column below. It is usual practice for the columnconcrete placement to be stopped at the elevation of the underside of each floor, as shownin Figs. 10-1 and 10-3, to allow the concrete to settle in the forms before the concrete in thefloor over the column is placed. ACI Code Section 10.12 allows the concrete in interior,edge, and corner columns to have a strength up to 1.4 times the strength of the concrete in thefloor system (slab and beams) without any strengthening of the floor. On the basis of tests re-ported by Biachini et al. [10-2], the effective strength of the concrete in the floor system is en-hanced by the lateral confinement from the floor surrounding the column. For larger ratios ofcolumn concrete strength to floor system concrete strength,

(a) ACI Code Section 10.12.1 allows the higher strength column concrete to beplaced in the floor under and around the column, or

(b) ACI Code Section 10.12.2 allows the design of the joint based on the lowerstrength concrete with dowels provided to make up for the lower strength concrete, or

(c) ACI Code Section 10.12.3 allows the joint region to be designed by using aneffective compressive strength of the concrete in the floor, taken as

(10-1)

In the third case, the maximum ratio is where is the specifiedstrength of the column concrete and is the specified strength of the slab concrete.f¿cs

f¿ccf¿cc>f¿cs = 2.5,

f¿ce = 0.75f¿cc + 0.35f¿cs

f¿ce,


More recent tests, reported by Ospina and Alexander [10-3], showed that the lateralconfinement of the floor concrete is reduced when slab loads stress the slab reinforcementin tension. These tests supported the use of the 1.4 strength-ratio threshold for interiorcolumns and edge columns, but suggested that the threshold strength ratio be lowered to 1.2for corner columns. The joint strengths were also sensitive to the ratio of floor thickness tocolumn width, h/c. The effective concrete strength decreased when h/c increased.

10-3 CONTINUOUS BEAMS

The three major stages in the design of a continuous beam are design for flexure, designfor shear, and design of longitudinal reinforcement details. In addition, it is necessary toconsider deflections and crack control and, in some cases, torsion. When the area supportedby a beam exceeds it is usually possible to use a reduced live load in calculatingthe moments and shears in the beam.

EXAMPLE 10-1 Design of a Continuous T-Beam

Design the floor beam B3–B4–B3 in Fig. 10-5. This beam supports its own dead loadplus the load from a 6-in. slab. The beam is supported on girders at lines A, B, C, and Dand is symmetrical about the centerline of the building. Use normal-weight concrete with

for flexural reinforcement, and for stir-rups. Base the loadings on the ASCE/SEI 7-10 recommendations and use load factors fromACI Code Section 9.2.

1. Compute the trial factored loads on the beam. Because the size of the beamstem is unknown at this stage, it is not possible to compute the final loads for use in the de-sign. For preliminary purposes, estimate the size of the beam stem. Once the size has beenestablished, the factored load will be corrected and then used in subsequent calculations.

The ASCE/SEI 7-10 recommendations allow live-load reductions based on tributaryareas multiplied by a live-load element factor, to convert the tributary area to aninfluence area. (See Section 2-8.) For a beam, this is the area between the beam in questionand the beams on either side of it, over the length of the beam. Three tributary areas will beconsidered in the design of beam B3–B4–B3, as shown in Fig. 10-6.

(a) To compute the positive moment at midspan of beam B3 and negative mo-ment at the exterior end of beam B3, it is necessary to load spans AB and CD withlive load. Because the majority of the moment in the left span results from loads onthat span, we shall take the tributary area to be the area over beam B3, extendinghalfway to the adjacent beams:

From Eq. (2-12), the reduced live load is

where the unreduced live load, is 100 psf and for a typical floor beam KLL = 2.0.Lo,

L = Loa0.25 +15

2KLLAT bAT� 33ft a15ft

2+

14.3ft

2b = 483 ft2

AT

KLL = 2,

fyt = 60,000 psify = 60,000 psifcœ = 4000 psi,

400 ft2,

Section 10-3 Continuous Beams • 473

L shall not be less than for members supporting one floor. Thus,

(b) To compute the maximum negative moment at the interior support, B, spansAB and BC should be loaded. The corresponding tributary area, is the area be-tween lines of zero shear parallel to the beam and on either side of it, as shown inFig. 10-6b.

AT � a33ft + 30ftb a15ft

2+

14.3ft

2b = 923 ft2

AT,

L = 73.2 psf.0.50Lo

= 73.2 psf

L = Loa0.25 +15

22 * 483b = 100 * 0.732

Fig. 10-5Typical floor plan—Example 10-1.


Fig. 10-6Tributary areas—Example 10-1.

and

where L shall not be less than for members supporting one floor. Thus,L = 59.9 psf.

0.50Lo

L = 100a0.25 +15

22 * 923b = 59.9 psf

KLL = 2


(c) To compute the positive moment in the center span (beam B4), span BCshould be loaded. The tributary area is shown in Fig. 10-20c. Here,

In this example, three reduced live loads were calculated, in part, to illustrate the cal-culation. To simplify design calculations, one might use a single reduced live load through-out, equal to the largest of the three reduced live loads.

Summary of unfactored reduced live loads.For negative moment at exterior end, positive moment at midspan, and shear inbeam B3 = 73.2 psf.For negative moment at support B = 59.9 psf.For positive moment at midspan of B4 = 75.5 psf.

2. Select Method of Analysis of Beam and Values of the Strength-ReductionFactor.

If the beam fits the requirements in ACI Code Section 8.3.3, we will use the momentcoefficients from that section.

In this example,

(a) there are two or more spans,

(b) the ratio of the longer clear span to the shorter clear span is approximatelywhich is less than 1.20,

(c) the loads are uniformly distributed,

(d) the “unit” (unfactored) live load does not exceed three times the “unit” (un-factored) dead load, and

(e) the members are prismatic.

So far, the beam fits all of these except item (d), which we will check later.

Use ACI Code Section 8.3.3 to compute moments and shears.

3. Select the strength-reduction factors. Because the design is based on loadcombinations from ACI Code Section 9.2, the strength reduction factors must be from ACICode Section 9.3:

(The goal will be todesign tension-controlled sections.)

The size of the stem beam B3–B4–B3 will be chosen on the basis of negative momentat the first interior support. For this location, the factored load on the beam is as follows:

(a) Dead load (not including the portion of the beam stem below the slab):

Live load (reduced): 59.9 psfTotal factored load from slab:

wu = 1.2 * 81.5 + 1.6 * 59.9 = 194 psf

Slab 75 psf

Floor, ceiling, mechanical 6.5 psf

81.5 psf

Shear f = 0.75.

For flexure in tension-controlled beam sections, f = 0.90.

33>30 = 1.10,

L = 100a0.25 +15

22 * 440b = 75.5 psf

AT � 30ft a15ft

2+

14.3ft

2b = 440 ft2


Assume that the tributary width which loads beam B3–B4–B3 extends halfway from thisbeam to the adjacent beams on each side of , or

Therefore, the factored load per foot from the slab is

It is now necessary to estimate the weight of the beam stem. Two approximate meth-ods of doing this are:

(1) The factored dead load of the stem is taken as 12 to 20 percent of the other fac-tored loads on the beam. This gives 0.34 to 0.57 kip/ft.

(2) The overall depth of beam h is taken to be 1/18 to 1/12 percent of the larger span, and is taken to be 0.5h. This gives the overall h as 22 to 33 in., with the stem extend-ing 16 to 27 in. below the slab, and gives as 11 to 17 in. The factored load from stemsof such sizes ranges from 0.22 to 0.57 kip/ft.

As a first trial, assume the factored weight of the stem to be 0.50 kip/ft. Then

say, 3.4 kips/ft.

4. Choose the actual size of the beam stem. The size of the stem is governed bythree factors: (a) deflections, (b) moment capacity at the point of maximum negativemoment, and (c) shear capacity. In addition, the overall depth of the floor should be min-imized to reduce the overall height of the building.

(a) Calculate the minimum depth based on deflections. ACI Table 9.5(a)(Table A-9) gives minimum depths, unless deflections are checked. Because the an-ticipated partitions will be flexible enough to undergo some deflections, this tablecan be used.

Beam B3 is one-end continuous, and normal-density concrete.Therefore,

where shall be taken as the span center-to-center of supports. Consequently,

(b) Determine the minimum depth based on the negative moment at theexterior face of the first interior support. Using the appropriate moment coeffi-cient from ACI Code Section 8.3.3,

for negative moment is defined as the average of the two adjacent spans.Using dimensions given in Fig. 10-5,

/n =1388 + 3242 in.

2= 356 in. = 29.7 ft

/n

Moment = -wu/n2

10

Minimum h based on deflections =33ft * 12 in. / ft

18.5= 21.4 in.

/

Minimum h =/

18.5

fy = 60,000 psi,

= 3.35 kips>ft Total trial load per foot = 2.85 + 0.50 kip>ftbw

bw

/,

14.7 ft * 194 psf = 2.85 kip>ft0.5 115ft + 14.3 ft2 = 14.7 ft

B3–B4–B3


Therefore,

We will use the procedure developed in Chapter 5 for the design of singly rein-forced beam sections with rectangular compression zones. Use Eq. (5-19) to select areinforcement ratio, that will result in a tension-controlled section.

From this, use Eq. (5-21) to find the reinforcement index,

and then, use Eq. (5-22) to calculate the flexural resistance factor, R.

Finally, this value of R can be used to determine the required value of usingEq. (5-23a).

Possible choices for b and d are

Select the second of these. With one layer of reinforcement at the supports,This exceeds the minimum h based on deflections. Try a

12-in. wide-by-24-in. deep extending 18 in. below the slab stem, with

(c) Check the shear capacity of the T-beam.

Maximum shear from beam loads at interior end of B3:

From ACI Code Section 11.2.1.1, the nominal is

ACI Code Section 11.4.7.9 sets the maximum nominal is

Thus, the absolute maximum fVn = 0.75 132.6 + 131 kips2 = 123 kips.

Vs = 82fc¿bwd = 131 kips

Vs

Vc = 2l2fc¿bwd = 32.6 kips

Vc

Vu = 1.15 wu/n>2 = 1.15 * 3.40 k/ft * 32.3ft/2 = 63.1 kips

Vu = f1Vc + Vs2d � 21.5 in.

= 23.7 in.h M 21.2 + 2.5

b = 14 in., d = 19.6 in.

b = 12 in., d = 21.2 in.

b = 10 in., d = 23.2 in.

bd2 ÚMufR

=300 k-ft * 12 in./ft

0.9 * 0.745 ksi= 5370 in.3

bd2,

= 0.213 * 4 ksi11 - 0.59 * 0.2132 = 0.745 ksi

R = vfcœ11 - 0.59v2v = r

fy

fcœ = 0.0142

60 ksi

4 ksi= 0.213

v:

r1initial2 �b1fc

œ

4fy=

0.85 * 4 ksi

4 * 60 ksi= 0.0142

r,

Mu M -3.4 * 29.72

10= - 300 kip-ft


The maximum factored due to the applied loads and dead loads is 63.1 kips.Therefore, the beam stem selected has ample capacity for shear.

(d) Summary. Use

(18 in. below slab)

assuming one layer of steel at all sections

This section is shown in Fig. 10-7.

5. Compute the dead load of the stem, and recompute the total load per foot.

Total dead load for B3–B4–B3 is:

Live load for B3–B4–B3 is:

Beam B3,

Negative moment at B,

Positive moment for B4, wL = 0.0755 ksf * 14.7 ft = 1.11 kip>ftwL = 0.0599 ksf * 14.7 ft = 0.881 kip>ftwL = 0.0732 ksf * 14.7 ft = 1.08 kip>ftwD = 0.0815 ksf * 14.7 ft + 0.225 kip>ft = 1.42 kip>ft

Weight per foot of the stem below the slab =12 * 18

144* 0.15 = 0.225 kip>ft

d � 21.5 in.,

h � 24 in.

b � 12 in.

Vu

12 in.

0.59 in.

12 in.

4.66 in.

Fig. 10-7Beam sections.


Summary of factored loads on B3–B4–B3.

Beam B3,

Negative moment at B,

Positive moment for B4,

Note that although the factored weight of the beam stem is only times the original guess, the error in the original estimate of the total factored load at B isonly 9 percent. If the final factored load per foot changed by more than about 10 percent, itmay be desirable to recalculate the beam size and load per foot.

6. Calculate the flange width for the positive-moment regions. The beam acts as aT- beam in the positive-moment regions. The effective width of the overhanging flange oneach side of the web is the smallest of (ACI Code Section 8.12.2) the following:

Therefore, the effective flange width is 81 in., as shown in Fig. 10-7a.

7. Compute the beam moments. The moments can be computed by using frameanalysis, as was discussed in Chapter 5. However, in step 2 it was shown that we could usethe ACI moment coefficients from ACI Code Section 8.3.3 if the unfactored unit live loadwas less that three times the unfactored unit dead load. From the load calculations in step5, it is clear that this condition is satisfied. Thus, the ACI moment coefficients can be usedin this design. The moment calculations are presented in Table 10-1. Because the buildingis symmetrical about its center, only a part of the beam is shown in this table.

Table 10-1, Line 1 The values of are the clear-span lengths, except that in com-puting the negative moment at supports B and C, is the average of the two adjacentspans.

Table 10-1, Line 2 The factored uniform loads differ for each span, due to the differ-ent live-load reduction factors for the various sections.

Table 10-1, Line 4 Because the exterior end of span AB is supported on a spandrelgirder, the moment coefficient at A is 1/24. The resulting moment diagram is shown inTable 10-2.

8. Design the flexural reinforcement. The reinforcement will be designed inTable 10-2. This is a continuation of Table 10-1. Prior to entering Table 10-2, however,

/n/n

bw +157 in.

2+

166 in.

2= 174 in.

bw + 218 * 62 = 12 in. + 2 * 48 in. = 108 in.

0.25/n 1based on the shorter span for simplicity2 = 0.25 * 324 in. = 81 in.

0.27>0.5 = 0.54

wu = 1.2 * 1.42 + 1.6 * 1.11 = 3.48 kip>ftwu = 1.2 * 1.42 + 1.6 * 0.881 = 3.11 kip>ftwu = 1.2 * 1.42 + 1.6 * 1.08 = 3.43 kip>ft

TABLE 10-1 Calculation of Moments for Beam B3–B4–B3—Example 10-1

32.3

3.43

3590

�1/24

�149

32.3

3.43

3590

1/14

256

29.7

3.11

2740

�274

29.7

3.11

2740

�274

27.0

3.48

2540

1/16

159

1/16

Line

2. wu, kip/ft

4. Cm �1/10 �1/10�1/11 �1/11

B3 B4

3. wu /2n

1. /n, ft

5. Mu � Cmwu /2n, kip-ft


it is necessary to compute some constants for use in subsequent calculations of area ofreinforcement.

(a) Compute the area of steel required at the point of maximum nega-tive moment (first interior support). From Eq. (5-16),

Because there is negative moment at the support, the beam acts as a rectangularbeam with compression in the web, as shown in Fig. 10-7b. Assume that

This value can be improved with one iteration using Eq. (5-17) to find the depthof the compression stress block, a:

and then reusing Eq. (5-16) with this calculated value of a:

Before proceeding, we need to confirm that this is a tension-controlled section.This could be done by calculating the strain at the level of the tension rein-forcement, and confirming that it exceeds 0.005 or by simply showing thatthe depth to the neutral axis, c, is less than 3/8 of d. Using the latter approach

et ,

As ÚMu

ffyad -a

2b =

274 k-ft * 12 in./ft

0.9 * 60 ksi * 121.5 in. - 2.32 in.2 = 3.17 in.2

a =Asfy

0.85 fcœb

=3.15 in.2 * 60 ksi

0.85 * 4 ksi * 12 in.= 4.63 in.

As �274 k-ft * 12 in./ft

0.9 * 60 ksi * 10.9 * 21.5 in.2 = 3.15 in.2

and f = 0.90j = 0.9

As ÚMu

ffy ad -a

2b �

Muffy(jd )

TABLE 10-2 Calculation of Reinforcement Required—Example 10-1

Line

5. Mu, kip-ft �149 256 �274 159

6. As coefficients 0.0116 0.0105 0.0116 0.0105

7. As(req’d), in.2 1.72 2.68 3.17 1.67

8. As > As,min Yes Yes Yes Yes

9. Bars selected 3 No. 7 4 No. 8

2 No. 6

4 No. 7 3 No. 7

10. As provided, in.2

11. bw o.k.

1.80 3.16 3.28 1.80

Yes— — Yes

�149

Mu

�274�274

�256�159


with for 4000-psi concrete and calculating a revised value ofwhich was obtained using the new required area of steel,

This is less than so the section is tension-controlled and can bedesigned using The other negative-moment sections have a lower designmoment, so it will be conservative to use the ratio of obtained here toquickly determine the area of tension steel required at those other locations.That ratio is

Lines 6 and 7 of Table 10-2 Calculate the area of steel required in negative-momentregions as

(b) Compute the area of steel required at the point of maximum positivemoment (point near the middle of the exterior span). In the positive-momentregions, the beam acts as a T-shaped beam with compression in the top flange. As-sume that the compression zone is rectangular, as shown in Fig. 10-7a. Take

(wide compression zone) for the first calculation of

Assume that a is less than then

where b is the effective flange width of 81 in.

Doing one iteration, as was done for the negative-moment section, results in and For this value of a, it is clear that c is less than

so the section is tension-controlled and can be designed usingThe other positive-moment section has a lower design moment, so it

will be conservative to use the ratio of obtained here to quickly determinethe area of tension steel required at that other location. The ratio is

The area of steel required at positive-moment regions is calculated in lines 6 and7 of Table 10-2 by using As = 0.0105Mu.

AsMu

=2.68 in.2

256 kip-ft= 0.0105 ( in.2 / kip-ft)

As/Mu

f = 0.9.3�8 d = 8.06 in.,

a = 0.59 in.,2.68 in.2As =

a =2.79 in.2 * 60 ksi

0.85 * 4 ksi * 81 in.= 0.61 in. 6 hf = 6.00 in.

a =Asfy

0.85fcœb

hf;

As �256 k-ft * 12 in./ft

0.9 * 60 ksi 10.95 * 21.5 in.2 = 2.79 in.2

As:j = 0.95

As = 0.0116Mu.

AsMu

=3.17 in.2

274 kip-ft= 0.0116 (in.2>kip-ft)

As/Mu

f = 0.9.3/8 d = 8.06 in.,

c =a

b1=

4.66 in.

0.85= 5.48 in.

13.17 in.22: Asa = 4.66 in.,b1 = 0.85


(c) Calculate the minimum reinforcement. The minimum reinforcementrequired is, by ACI Code Section 10.5.1, and given earlier as Eq. (4-11):

(ACI Eq. 10-3)

For 4000-psi concrete, the second term governs, so

Table 10-2, Line 8 in line 7 of Table 10-2 exceeds in both the positive-and negative-moment regions.

(d) Calculate the area of steel and select the bars. The remaining calcula-tions are done in Table 10-2. The areas computed in line 7 exceed in allcases. If they did not, should be used.

Lines 9 and 10 of Table 10-2 give the bars selected at each location, togetherwith their areas. The bars were selected by using Table A-4. Small bars were se-lected at the exterior support, because they have to be hooked into the support andthere may not be enough room for a standard hook on larger bars.

Line 11 of Table 10-2. Check whether the bars will fit into the beam web usingTable A-5. In the negative-moment regions, some of the bars can be placed in theslab beside the beams; hence, it is not necessary to check whether they will fit intothe web width.

9. Check the distribution of the reinforcement.

(a) Positive-moment region. It is necessary to satisfy ACI Code Section10.6.4. This will be done at the section with the smallest number of positive-moment bars: the middle of span B–C. Here, there are three No. 7 bars, asshown in Fig. 10-8a.The clear cover to the surface of the flexural steel is

The maximum bar spacing is

(9-7)

(ACI Eq. 10-4)

where (in which ). Thus,

Because and there are three bars, s is clearly smaller than 10.3 in.

(b) Negative-moment region. ACI Code Section 10.6.6 says “part” of thenegative-moment steel shall be distributed over a width equal to the smaller ofthe effective flange width (81 in.) and At each ofthe interior negative-moment regions, there are six top bars. Two of these will beplaced in the corners of the stirrups, as shown in Fig. 10-8b, two over the beamweb, and the other two in the slab. (Note that, for bars placed in the slab to be

/n>10 = 324>10 = 32.4 in.

bw = 12 in.

s = 15 - 2.5 * 1.875 = 10.3 in. … 12 in.

fy = 60 ksifs = can be taken as 0.67fy

s = 15a40,000

fsb - 2.5cc … 12a 40,000

fsb

cc = 1.5-in. cover + 0.375-in. stirrups = 1.875 in.

As,min

As,min

As,minAs

Ú 0.86 in.2

As,min Ú200 * 12 * 21.5

60,000

As,min =32fcœfy

bwd, and Ú200bwd

fy


completely effective, there should be reinforcement perpendicular to the beam inthe slab. In this case the slab reinforcement will serve this purpose.) The two barsplaced in the slab will be placed to give spacing, s, approaching the maximumallowed. The maximum spacing allowed is

Within a width of 32.4 in. we must place six bars. These cannot be farther apartthan 10.3 in. We shall arbitrarily place two bars in the slab at 5 in. outside theweb of the beam.

ACI Code Section 10.6.6 requires “some” longitudinal reinforcement in theslab outside this band. We shall assume that the shrinkage and temperature steelalready in the slab will satisfy this requirement. The final distribution of negative-moment steel is as shown in Fig. 10-8b.

10. Design the shear reinforcement. The shear-force diagrams are calculated inTable 10-3 and shown at the bottom of that table. The shear coefficients at midspan of thebeams (line 5) are based on Eq. (6-26).

= 10.3 in.= 15 - 2.5 * 1.875

s = 15a40,000

fsb - 2.5 cc

12 in.

-moment

Fig. 10-8Distribution of reinforcement.


(a) Exterior end of B3. Because this beam is supported by other beams, the crit-ical section for shear is located at the face of the support. ACI Code Section 11.4.6.1requires stirrups if where

Because kips exceeds kips, stirrups are required.Try No. 3 Grade-40 double-leg stirrups with a 90° hook enclosing a No. 7

stirrup-support bar (Fig. 10-8b). Because the required value of

does not exceed ACI Code Section 11.4.5.1 limits the maximumstirrup spacing to the smaller of d/2 and 24 in.

To satisfy the minimum stirrup area requirement in ACI Code Section 11.4.6.3,the stirrup spacing must be less than the smaller of

or

s =Avfyt

0.752f¿c # bw= 23.2 in.

s =Avfyt

50bw= 22.0 in.

d

2= 10.75 in.

42fœc bwd,

Vs = Vu�f - Vc

Vc>2 = 16.3Vu>f = 73.9

Vc2

= 16.3 kips

= 2 * 1 24000 * 12 * 21.5 = 32,600 lbs = 32.6 kips

Vc = 2l2fcœ bwdVu Ú fVc>2,

A B3 B B4 C

32.33 27.03.48

47.0

1.01.0 0.1250.125 1.01.15

55.4

73.96.98

9.3

63.8

85.0

47.0

62.6 8.0

62.6

62.6

9.3

73.9

9.3

85.0

8.08.0

5.99

62.6

47.0

3.43

55.4

1.6 � 1.08 1.6�1.11

7. Vu /f

(kips)

10-110-3


Use 10 in. as the maximum spacing.The spacing required to support the shear forces is

(6-21)

At end A, kips, and

At some point in the span, the stirrup spacing will be changed to 10 in. Com-pute where ,

Setting this value occurs at

Compute where is equal to and we can stop using stirrups:

At the exterior end of B3, use No. 3 Grade-60 double-leg stirrups: one at 3 in.,six at 6 in., and thirteen at 10 in. on centers.

(b) Interior end of B3. The shear at support B is The spacing re-quired at this point is

Change the stirrup spacing to 10 in. , where , is 61.0 kips.

occurs at

at the point where the stirrups can be stopped. This occurs at

At the interior end of B3, use No. 3 Grade-60 double-leg stirrups: one at 3 in.,twelve at 5 in., and eleven at 10 in. on centers.

x = 176 in. from end B

Vu>f = Vc>2x =

85.0 - 61.0

85.0 - 9.3* 194 = 61.5 in. from end B

s = 10 in.

s = 10 in.Vn

= 5.42 in., so use s = 5 in.

s =0.22 in.2 * 60 ksi * 21.5 in.

85.0 k - 32.6 k

85.0 kips.

x =73.9 - 16.3

73.9 - 9.3* 194 in. = 173 in.

Vc >2,Vu>fx =

73.9 - 61.0

73.9 - 9.3* 194 in. = 38.7 in. from end A

Vn = Vu �f,

= 61.0 kips

=0.22 * 60 * 21.5

10+ 32.6

Vn =Avfytd

s+ Vc

s = 10 in.Vn

= 6.87 in.—say, 6 in. on centers

s =0.22 in.2 * 60 ksi * 21.5 in.

73.9 k - 32.6 k

Vu>f = 73.9

s =Avfytd

Vu>f - Vc


(c) Ends of beam B4. The calculations for beam B4 are carried out in the samemanner as for beam B3. At each end of beam B3, use No. 3 Grade-60 double-legstirrups: one at 4.5 in. and thirteen at 10 in.

11. Check the development lengths and design-bar cutoffs.

(a) Perform the preliminary calculations.

Detailing requirements:

for No. 7, and 9 in. for No. 6.

Therefore, d always exceeds but exceeds d in beam B3.

(b) Select cutoffs for positive-moment steel in B3. The reinforcement atmidspan is four No. 8 (Table 10-2). Extend two No. 8 bar into the supports and cutoff two No. 8 where they are no longer required. The remaining bars have 50 percentof the original area; therefore, the theoretical flexural cutoff is at the point where

percent of the maximum at midspan. From Fig. A-3, this occurs atfrom the exterior end, and from

the interior end. These are points B and C in Fig. 10-9. To compute the actual cut-offpoints we must satisfy the detailing rules in ACI Code Sections 12.1, 12.2, 12.10,12.11, and 7.13. The following discussion will refer to those section.

Before doing so we shall calculate for the bottom bars.

Spacing and confinement case. The bottom bars have clear spacing and coverof at least and are enclosed by at least minimum stirrups. Therefore, this is Case1 in Table 8-1 (ACI Code Section 12.2.2). The bars are No. 8. Thus, we have

(8-12)

=60,000 * 1.0 * 1.0 * db

20 * 124000= 47.4db

/d =fyctce

20l2f¿c db

db

/d

0.26/n = 101 in.0.21 * 388 = 81.5 in.0.21/n =MuMu = 50

/n>1612db,

=324

16= 20.3 in. for B4

/n16

=388

16= 24.3 in. for B3

12db = 12 in. for No. 8, 10.5 in.

d = 21.5 in.

Actualcut-off points

Flexuralcut-off points

5�-0� 6�-6� 6�-2�

A B� F �C�B FC D E

Beam B3 Beam B4

62

101�

2 No. 8

2 No. 8Fig. 10-9Calculation of bar cut-offpoints for positive-momentsteel.


For the No. 8 bars, and thus For the No. 7 bars (BeamB4), and thus

Application of detailing rules to beam B3Some or all of each family of detailing rules apply to each region of the beam.

The detailing process starts by selecting the rules that apply in each region. It is theneasier to select the sub-rules applied in each region. Beam B3 must satisfy rules:1. Structural integrity; 2. Extension of bars into supports; 3. Effect of shear onbending moment diagrams; and 4. Anchorage, positive moment.

Structural integrity. This beam is a continuous interior beam. ACI Code Sec-tion 7.13.2.5 applies. In beam line B3–B4–B3, a tension tie is created by lapsplicing at least one-quarter of the positive-moment reinforcement, but not lessthan two bars, from span B3 with a similar amount of steel from beam B4. Thesplices will be Class B splices, at or near the supports, with a lap of The development length of a No. 8 bottom bar is 47.4 in., so

Lap splice two No. 8 bottom bars from span B3, a distance of 62 in. with twoNo. 7 bottom bars from span B4.

The two No. 7 bottom bars which will be part of the longitudinal structural in-tegrity tie must be placed inside the lower corners of U stirrups and must passthrough the column core. The longitudinal tie bars must be hooked into the supportsat the two outer ends of the beam using standard 90º hooks at the discontinuous ends.

Extension of bars into the supports. ACI Code Section 12.11.1 requires that atleast one-quarter of the positive-moment reinforcement, but not less than two barsfrom span B3 must extend a minimum of 6 in. into the supports. This requirementhas already been satisfied.

Effect of shear. The bars must extend d or 12 past the positive-momentflexural cut-off points. Extend bars to points and

(Fig. 10-9).

Anchorage. The bars must extend past the point of maximum bar stress. Themaximum bar stress occurs at midspan. By inspection, the distances frommidspan to and exceed Therefore, cut off two No. 8 bars at

from the interior face of the exterior column, and 79.5 in.,say 6 ft 6 in., from the exterior face of the interior column.

Anchorage positive moment. Bars must extend past the actual cut-off points ofadjacent bars. The actual extensions past and exceed 60 in. and 78 in., re-spectively, both of which are greater than

Anchorage at point of inflection. At positive-moment points of inflection, the barsmust satisfy

(8-21)

From Fig. A-3, the positive-moment points of inflection are at from the exterior end and from the interior end. At each of these points, the remaining steel is two No. 8 bars, . Accordingly we have

a =Asfy

0.85f¿cbe= 0.34 in.

As = 1.58 in.20.146/n = 56.6 in.

0.098/n = 38 in.

/d …MnVu

+ /a

/d = 47.4 in.C¿B¿

/d

21.5 in. = 5ft,81.5 in. -/d = 47.4 in.C¿B¿

/d

C¿B¿� 1.79 ftd � 21.5 in.

db

1.3/d = 61.6 in.1.3/d.

/d = 41.5 in.db = 0.875 in./d = 47.4 in.db = 1.0 in.


Check at the exterior end: from the shear-force diagram in Table 10-3, the at 38 in.from the exterior end is

and is the larger of d or 12db, but smaller than the actual extension (which exceeds38 in.), so

Thus,

This exceeds —therefore, o. k.Check at interior end:

Summary of the bar cut-off points for the positive-moment steel in beam B3. Thisis the end span of a continuous interior beam. Hook two No. 8 bars in the exteriorsupport and lap-splice two No. 8 bars with No. 7 bars in the interior support. Cut off twoNo. 8 bars at 5 ft from the face of the exterior column and at 6 ft 6 in. from the face ofthe interior column. These locations are shown as points and D in Fig. 10-9.

(c) Select cutoffs for the positive-moment steel in B4. This is a continuous in-terior beam. At midspan, we have three No. 7 bars. Run two No. 7 bars into the sup-ports and cut off the remaining No. 7 bar. The remaining is times the original amount. Therefore, the flexural cutoff is located at the point wherethe moment is 0.67 times the midspan moment. From Fig. A-1, this occurs at

from the face of the support (Fig. 10-9). Applyingthe rule for the effect of shear, we find that the bar must extend 21.5 in. past thispoint, to 75.7 in. from the support. Therefore, cut off one No. 7 bar at 6 ft 2 in.from the support.

The bars have a spacing and cover of at least and are enclosed by at least min-imum stirrups. Therefore, for the No. 7 bars,

Structural integrity. Lap splice the two No. 7 bars with the No. 8 bars frombeam B3. The length of a Class B lap splice for a No. 8 bar is Therefore, lap splice the bars 62 in.

Anchorage. The positive-moment points of inflection are from supports. At these sections the reinforcement is two No. 7 bars, and we have

a =Asfy

0.85f¿cbe= 0.26 in.

0.146/n = 47.3 in.

1.3/d = 61.6 in.

= 41.5 in./d = 47.4 * 0.875db

0.30/n = 0.30 * 324 = 97.2 in.

1.80 = 0.671.20>As

A, B¿, C¿,

= 53.6 in.—therefore, o.k.

MnVu

+ /a =2020

62.9+ 21.5

Vu at 56.6 in. from interior end = 62.9 kips

/d = 47.4 in.

MnVu

+ /a =2020

45.9+ 21.5 = 65.5 in.

/a = 21.5 in.

/a

Vu = 45.9 kips

Vuf

= 73.9 -38

194173.9 - 9.32 = 61.2 kips

Vu

Mn = Asfyad -a

2b = 2020 kip-in.


This exceeds —therefore, o.k. The cut-off points for the positive-moment steelare as indicated in Fig. 10-9.

(d) Select cutoffs for the negative-moment steel at the exterior end of B3.At the exterior end, there are three No. 7 bars. These must be anchored in the span-drel beam, which is 16 in. wide. The basic development lengths of standard hooksare (ACI Code Section 12.5) and book Section 8-4.

(8-18)

ForNo. 7 bar:

This can be multiplied by 0.7 if the side cover (perpendicular to the plane of thehook) exceeds 2.5 in. and for 90° hooks, if the tail cover is not less than 2 in. Thehooks in question enter the spandrel beam perpendicular to the axis of the beam andhence satisfy the side cover requirement. The tail cover can be set at 2 in. Therefore,

This can be accommodated in the spandrel beam. Therefore, anchor the top barsinto the spandrel beam with 90° standard hooks.

Extend all the top bars past the negative-moment point of inflection before cuttingthem off. From Fig. A-3, the negative-moment point of inflection is at

from the face of the exterior support (point G in Fig. 10-10). At this cut-offpoint, we must satisfy the rule for the effect of shear and that the bars extend at least

from the support face.

Effects of shear. Requires at least one-third of the bars to extend the longest of d,and past the point of inflection. governs here.

Therefore, cut off all bars at 5 ft 6 in. from the faceof the support.Anchorage. Bars must extend at least from the face of the support. Using

/d for No. 7 top bar = 1.3 * 41.5 = 54 in.

ct = 1.3,/d

41.9 + 24.3 = 66.2 in., say/n>16 = 24.3 in./n>1612db,

/d

41.9 in.0.108/n =

/dh for No. 7 bar = 16.6 * 0.7 = 11.6 in.

/dh = 16.6 in.

/dh = a0.02cefy

l2fcœ bdb

/d

MnVu

+ /a =1540

35.0+ 21.5 = 65.5 in.

Vu = 35.0 kips

Mn = Asfyad -a

2b = 1540 kip-in.

Actualcut-off points

Flexuralcut-off pointsand extensions

5�-6�

27.2�

9�-4�8�-4�2 No. 7

& 2 No. 6 4�-6�

Fig. 10-10Calculation of bar cut-offpoints for negative-momentsteel.


Because 5 ft 6 in. exceeds 54 in., this rule is satisfied. Therefore, cut off all top barsat the exterior support at 5 ft 6 in. from the face of the support.

(e) Select cutoffs for the negative-moment steel at the interior end of beamB3. The negative-moment steel at the first interior support consists of four No. 7 andtwo No. 6 bars. We will cut off two No. 7 bars when they are no longer required forflexure and extend the remaining bars (0.63 ) past the negative point of inflection.From Fig. A-3, at approximately 0.070 from theface of the support (point H in Fig. 10-10).

Effects of shear. Bars must extend past H to 48.7 in. from the face ofthe support.

Anchorage. The bars must extend from the point of maximum bar stress atthe face of the support. In step 11(d), this distance was found to be 54 in. for atop No. 7 bar.

Therefore, cut off two No. 7 bars at 4 ft 6 in. from the face of the interior support.

The remaining four bars will be extended past the negative-moment point of in-flection. From Fig. A-3, this is from the face of the support(point J in Fig. 10-10).

Effects of shear. At least one-third of bars must extend past thepoint of inflection to 111 in. from the face of the support.

Anchorage. Bars must extend past the actual cutoff Therefore,the bars must extend to 108 in. from the face of the support.

Therefore, cut off two No. 7 and two No. 6 top bars at 9 ft 4 in. from the exteri-or face of the first interior support.

(f) Select cutoffs for the negative-moment steel in beam B4. When the interi-or span of a three-span continuous beam is less than about 90 percent as long as theend spans, it can have negative moment at midspan under some loads, even thoughthis is not apparent from the ACI moment coefficients. For this reason, we will ex-tend two No. 7 top bars the full length of the beam. The remaining four bars willbe extended past the negative-moment point of inflection and will be cut off. FromFig. A-1 the negative-moment point of inflection is from the faceof the column (point K in Fig. 10-10).

Effects of shear. Says that these bars must extend by the larger of d, orpast this point to from the face of the support.

Anchorage. Says that these bars must extend by past the face of thesupport—therefore o.k.

Cut off two No. 7 top and two No. 6 bars at 8 ft 4 in. from the interior face of theinterior support, and extend the remaining two No. 7 top bars the full length ofthe interior span (see Fig. 10-10).

(g) Check shear at points where bars are cut off in a zone of flexural tension.Cutoffs and occur in zones of flexural tension. ACI Code Section12.10.5.1 requires special consideration of these regions if exceeds two-thirds of

Cutoff 5 ft from support, At this point there are No. 3 double-leg stirrups at 10 in. on centers.

Vs =0.22 * 60 * 21.5

10= 28.4 kips

Vc = 32.6 kips.Vu = 40.4 kips.B¿:

f1Vc + Vs2. Vu

H¿B¿, C¿, F¿,

/d = 54 in.

77.8 + 21.5 = 99.3 in./n>1612db,

0.24/n = 77.8 in.

H¿./d = 54 in.

/n>16 = 24.3 in.

0.224/n = 86.9 in.

/d

d = 21.5 in.

/n = 27.2 in.Mu = 0.63Mu1max2 As


Thus, is 0.88 of the shear permitted.

In step 10(a) for the design of shear reinforcement at the exterior end of beam B3,we started with a stirrup spacing of 6 in. near the support and increased to a 10-in.spacing at 39 in. from the face of the support. To satisfy ACI Code Section 12.12.5.1,we will maintain the 6-in. spacing up to and beyond cut-off point This results in2/3 at cut-off point The tighter 6-in. stirrup spacing at point should be extended at least d/2 toward the point of maximum positive momentwhich occurs near midspan, because if a flexural-shear crack initiates at it willtend to extend toward the maximum positive moment. Therefore, at the exteriorend of beam B3, use No. 3 Grade-60 double-leg stirrups: one at 3 in., twelve at6 in. (extends d/2 beyond point ), and ten at 10 in. (extends beyond the pointwhere exceeds ). The final stirrup layout for the external end of beam B3 isshown in the left portion of Fig. 10-11a. At least two of the top No. 7 bars, whichwere shown in Fig. 10-10 to be cut off at 5 ft 6 in. from the support face, should beextended to 14 ft 8 in. from the support face to hold up the stirrups, as shown by thedashed line in Fig. 10-11a.

Cutoff At this point, and thus, the same 6-in. stirrup spacingwill satisfy ACI Code Section 12.12.5.1.

Cutoff At this point, so a closer spacing of stirrups will berequired. Using a 4-in. spacing:

Vs =0.22 * 60 * 21.5

4= 71.0 kips

Vu = 47.9 kips,Hœ:

Vu = 40.9 kips,Cœ:

Vu1�2 fVc

B¿

B¿,

B¿B¿.fVn Ú VuB¿.

Vu

f1Vc + Vs2 = 0.75132.6 + 28.42 = 45.8 kips

14 ft 8 in.

(a) Spacing of stirrups for beam B3.

(b) Spacing of stirrups for beam B4.

15 ft 2 in.

3 in. 12 at 6 in. 10 at 10 in. 9 at 10 in. 6 at 6 in.4 at 4 in.

6 at 6 in. 3 in.

13 at 10 in. 13 at 10 in. 4 in.4 in.

Fig. 10-11Final layout of stirrups for Example 10-1.


and

This exceeds at the cut-off point It should be noted that cut-off point is forthe negative-moment reinforcement, so the tighter stirrup spacing used at shouldbe extended at least d/2 toward the point of maximum negative moment, which occursat the face of the support. The tighter stirrup spacing at should extend towardmidspan, as was done for cut-off point Therefore, at the interior end of beamB3, use No. 3 Grade-60 stirrups: one at 3 in., six at 6 in., four at 4 in. (starts at leastd/2 before point and extends just beyond ), six at 6 in. (extends d/2 past point

) and nine at 10 in. (extends past the point where exceeds ). The finalstirrup layout for the interior end of beam B3 is shown in the right portion of Fig. 10-11a. At least two of the top bars, which were shown in Fig. 10-10 to be cut off at9 ft 4 in. from the support face, should be extended to 15 ft 2 in. from the support faceto hold up the stirrups, as shown by the dashed line in Fig. 10-11a.

Cutoff At this point so a stirrup spacing of 10 in. will give 2/3Therefore, at each end of beam B4, use No. 3 Grade-60 double-leg

stirrups: one at 4 in. and thirteen more at 10 in. (extends beyond the point whereexceeds ). The final stirrup layout for beam B4 is shown in Fig. 10-11b.

Because two top No. 7 bars already were placed over the full length of beam B4,no changes are required to provide support for the stirrups.

12. Design web-face steel. Because d does not exceed 3 ft, no side-face steel isrequired (ACI Code Section 10.6.7).

This concludes the design. The bar detailing in step 11 of this example is quitelengthy. The process could be simplified considerably if all bars were extended past thepoints of inflection before being cut off. ■

In Example 10-1, the moments were computed by using the ACI moment coefficients.Figure 10-12 shows the elastically computed moment envelope for this beam, calculated byusing the reduced live load of 75.5 psf computed in step 1c for positive moment in beam B4(the center span). The beam was modeled for analysis with springs at the two discontinuous

Vu1�2 fVc

fVn Ú Vu .Vu = 28.2 kips,Fœ:

Vu1�2 fVcC¿

H¿H¿

B¿.C¿

H¿H¿H¿.Vu

2

3fVn = 0.667 * 0.75132.6 + 71.02 = 51.8 kips

–77.1 kip-ft[–149]

–77.1 kip-ft

–338 kip-ft[–274]

–338 kip-ft

+337 kip-ft [+256] +337 kip-ft

[+159]119 kip-ft

Fig. 10-12Elastic-moment envelopes—Example 10-1.

Section 10-4 Design of Girders • 493

ends to represent torsional stiffness of the spandrel beams. Their spring stiffness was derivedfrom Eq. (7-43).

The moments plotted in Fig. 10-12 can be compared to the moments from the ACImoment coefficients, shown in the moment diagram in the heading of Table 10-2 andgiven (in brackets) in Fig. 10-12. Moments in the end spans and at the interior supportsare relatively close between the two procedures. The negative moments at the exteriorends are directly a function of the assumed torsional stiffness of the spandrel beams.The interior-span positive moment is overestimated by the ACI moment coefficients,which, in addition, fail to predict the possibility of negative moment at midspan of theinterior span.

The lengthy example presented in this chapter appears somewhat cumbersome thefirst time through. With practice and understanding of the principles, the design of contin-uous beams and slabs becomes straightforward. The beams and slabs designed for onefloor can frequently be used many times throughout the building.

10-4 DESIGN OF GIRDERS

Girders support their own weight plus the concentrated loads from the beams they sup-port. It is customary to compute the moments and shears in the girder by assuming thatthe girder supports concentrated loads equal to the beam reactions, plus a uniform loadequal to the self-weight of the girder, plus the live load applied directly over the girder.This approach neglects two-way action in the slab adjacent to the girder.

When the flexural reinforcement in the slab runs parallel to the girder, the two-way action shown in Fig. 10-13 causes negative moments in the slab along theslab–girder connection, as indicated by the curvature of slab strip B in Fig. 10-13. Toreinforce for these moments, ACI Code Section 8.12.5 requires slab reinforcementtransverse to the girder, as shown in Fig. 10-14. The reinforcement is designed by as-suming that the slab acts as a cantilever, projecting a distance equal to the effectiveoverhanging slab width and carrying the factored dead load and live load supported bythis portion of the slab.

Frequently, edge girders are loaded in torsion by beams framing into them from thesides between the ends of the girder. This is compatibility torsion, because it exists onlybecause the free-end rotation of the beam is restrained by the torsional stiffness of the girder.In such a case, ACI Code Section 11.5.2.2 allows a reduction in the design torque, as dis-cussed in Chapter 7. If torque is present, the stirrups must be closed stirrups (ACI CodeSections 7.11.3 and 11.5.4.1).

E

DA

B

C

Fig. 10-13One-way and two-way slabaction near beam supports.


10-5 JOIST FLOORS

Long-span floors for relatively light live loads can be constructed as a series of closelyspaced, cast-in-place T-beams (or joists) with a cross section as shown in Fig. 10-15. Thejoists span one way between beams. Most often, removable metal forms referred to asfillers or pans are used to form the joists. Occasionally, joist floors are built by using clay-tilefillers, which serve as forms for the concrete in the ribs that are left in place to serve as theceiling (ACI Code Section 8.13.5).

When the dimensions of the joists conform to ACI Code Sections 8.13.1 to 8.13.3,they are eligible for less cover to the reinforcement than for beams (ACI Code Sec-tion 7.7.2(c)) and for a 10 percent increase in the shear, carried by the concrete (ACICode Section 8.13.8). The principal requirements are that the floor be a monolithic combi-nation of regularly spaced ribs and a top slab with

1. ribs not less than 4 in. in width,

2. depth of ribs not more than times the minimum web width, and

3. clear spacing between ribs not greater than 30 in.

Ribbed slabs not meeting these requirements are designed as slabs and beams.The slab thickness is governed by ACI Code Section 8.13.6.1, which requires a

thickness of not less than 2 in. for joists formed with 20-in.-wide pans and for 30-in.pans. The slab thickness and cover to the reinforcement are also a function of the fire-resistance rating required by the local building code. For a 1-hour fire rating, a coverand a 3- to slab are required. For a 2-hour rating, a 1-in. cover and a slab arerequired. These vary from code to code and as a function of the type of ceiling used. Joistfloor systems are also used for parking garages. Here, the top cover of the slab reinforce-ment and the quality of concrete used must be chosen to reduce the possibility of corrosion(ACI Code Chapter 4). Epoxy-coated or galvanized bars are often used in such applica-tions. Wheel loads on the slab and abrasion of the concrete in traffic paths should also beconsidered.

4 12 in.3 1

2 in.

34 in.

2 12 in.

3 12

Vc,

Fig. 10-14Transverse reinforcementover girders.

Section 10-5 Joist Floors • 495

Flexural reinforcement, ACI Code Section 8.13.6.2 but not less thanshrinkage reinforcement, ACI Code Section 7.12.2.

Standard hookon at leastone bottom bar,ACI Code Section7.13.2.1

At least one bottom barcontinuous or splicedwith a class B splice,ACI Code Section 7.13.2.1

(d) Bar cut-off points.

Fig. 10-15One-way-joist construction.

Standard Dimensions and Layout of One-Way Joists

The dimensions of standard removable pan forms are shown in Fig. 10-15. The forms areavailable in widths of 20 and 30 in., measured at the bottom of the ribs. Standard depths are6, 8, 10, 12, 14, 16, and 20 in. End forms have one end filled in to serve as the side form forthe supporting beams. Tapered end forms allow the width of the ends of the joists to increase,as shown in Fig. 10-15b, giving additional shear capacity at the end of the joists. The 20-in.tapered end pans reduce in width from 20 in. to 16 in. over a 3-ft length. The 30-in. pansreduce to 25 in., also over a 3-ft length. Pan forms wider than 30 in. also are available, butthe resulting floors or roofs must be designed as conventional slabs and beams (ACI CodeSection 8.13.4).


When such a floor is laid out, the rib and slab thicknesses are governed by strength,fire rating, and the available space. The overall depth and rib thickness are governed bydeflections and shear. No shear reinforcement is used in most cases. Generally, the mosteconomical forming results if the joists and supporting beams have the same depth. Fre-quently, this will involve beams considerably wider than the columns. Such a system is re-ferred to as a joist-band system. Generally, it is most economical to use untapered endforms whenever possible, and to use tapered forms only at supports having higher-than-average shears.

Although not required by the ACI Code, load-distributing ribs perpendicular to thejoists are provided at the midspan or at the third points of long spans. These have at leastone continuous No. 4 bar at the top and the bottom. The CRSI Handbook [10-4] suggestsno load-distributing ribs in spans of up to 20 ft, one at midspan for spans of 20 to 30 ft, andtwo at the third points for spans over 30 ft.

For joist floors meeting the requirements of ACI Code Section 8.3.3, the ACI mo-ment and shear coefficients can be used in design, taking as the clear span of the joiststhemselves. For uneven spans, it is necessary to analyze the floor. If the joists are sup-ported on wide beams or wide columns, the width of the supports should be modeled inthe analysis. The negative moments in the ends of the joists will be underestimated if thisis not done.

Reinforcing Details

The bar cutoffs given in Fig. 10-15d apply if the joists fit the requirements of ACI CodeSection 8.3.3. The CRSI Handbook recommends cutting off half of the top and bottom barsat shorter lengths. If this is done, the shear at the points where bars are cut off in a zone offlexural tension should satisfy ACI Code Section 12.10.5. ACI Code Section 7.13.2.1requires at least one bottom bar to be continuous or spliced over supports with a Class B ten-sion splice. At edge beams, at least one of the bottom bars in each joist must be fully anchoredinto the beam with a standard hook.

10-6 MOMENT REDISTRIBUTION

As a continuous beam is loaded beyond its service loads, the tension reinforcement willeventually yield at some section (usually a negative-moment section). With furtherloading, this section will deform as a plastic hinge, at a constant moment taken equal tothe nominal moment. Increased loads cause an increase in the sum of the positive andnegative moments in the beam. Because the moment cannot increase at the plastichinge, the increase in moments all occurs at the positive-moment hinge location. Thus,the shape of the moment diagram must shift, with more moment going to those sectionswhich are still elastic. Eventually, a mechanism forms, and the plastic capacity isreached. Procedures for carrying out inelastic analyses of concrete structures are givenin [10-5] and [10-6].

For a uniformly loaded fixed-end beam, the maximum elastic positive momentsare half the maximum negative moment As a result, approximately

twice as much reinforcement is required at the supports compared with what is required atmidspan. Sometimes, this leads to congestion of the steel at the supports. The plastic-moment distribution results in a much more variable reinforcement layout depending onthe actual plastic-moment capacities in the positive- and negative-moment regions, and

1w/2>122.1w/2>242

/n

Section 10-6 Moment Redistribution • 497

moments can be redistributed from the elastic distribution, allowing reductions in the peakmoments, with corresponding increases in the lower moments. The amount of redistribu-tion that can be tolerated is governed by two aspects. First, the hinging section must beable to undergo the necessary inelastic deformations. Because the inelastic rotational ca-pacity is a function of the reinforcement ratio, as shown in Fig. 4-11, this implies an upperlimit on the reinforcement ratio. Second, hinges should not occur at service loads, becausewide cracks develop at hinge locations.

ACI Code Section 8.4 allows the maximum moments at the supports or midspan tobe increased or decreased by not more than percent, with a maximum of 20 percent,provided that at the section where moments are being reduced, where is thetensile strain in the layer of steel closest to the tension face of the member.

These moments must have been computed via an elastic analysis, and not the ACImoment coefficients. The modified moments must be used in calculating the momentsat all critical sections, shears, and the bar cut-off points. (See ACI Code Sections 8.4.1to 8.4.3)

EXAMPLE 10-2 Moment Redistribution

Compute the design moments for the three-span beam from Example 10-1, using anelastic analysis (Fig. 10-12) and moment redistribution. Use ACI Code Sections 8.4.1through 8.4.3. Step 8 of the example will be repeated. Using

Thus, the maximum redistribution that is allowed is approximately Using the elastic moment shown in Fig. 10-12, the reduced moment is accordingly

This value exceeds the design moment (274 kip-ft) thatwas used for this section in Example 10-1. Thus, change the tension reinforcement for the sec-tion to be four No. 7 bars and four No. 5 bars . The four No. 5 bars will be1As = 3.64 in22338 = 308 ft-kips.11 - 0.0882 *

11000Pt = 8.82 percent.

= 0.00877

Pt =21.5 - 5.48

5.48* 0.003

5.48 in.,c = a>b1 = 4.66>0.85 =

PtPt Ú 0.00751000Pt

distributed into the flanges, two on each side of the web. Check whether isadequate:

Thus, the modified section is adequate.It now is necessary to compute the positive midspan moments corresponding to this

reduced value of the negative moment at the interior supports. This step is done to see if theredistribution of negative moments to midspan positive moments will cause these values toexceed those previously calculated for other loading cases (Fig. 10-12). The midspan mo-ments must correspond to the load case that produced the maximum negative moment atthe interior support. The appropriate live-load pattern would include live load on the twospans that are adjacent to the interior support. In step 5 of Example 10-1, the total factoreddistributed load, for this load case was found to be 3.11 kip/ft. An elastic analysis forwu ,

fMn = fAsfyad -a

2b = 3700 k-in. = 308 kip-ft

a =Asfy

0.85 fcœbw

=3.64 in.2 * 60 ksi

0.85 * 4 ksi * 12 in.= 5.35 in.

As = 3.64 in.2


REFERENCES

10-1 Minimum Design Loads for Buildings and Other Structures, (ASCE/SEI 7-10), American Society ofCivil Engineers, Reston, VA, 2010, 608 pp.

10-2 Bianchini, A. C., Woods, Robert E., and Kesler, C. E., “Effect of Floor Concrete Strength on ColumnStrength,” ACI Journal, Proceedings, Vol. 56, No. 11, May 1960, pp. 1149–1169.

10-3 Carlos E. Ospina and Scott D.B. Alexander, “Transmission of Interior Concrete Column Loads throughFloors,” ASCE Journal of Structural Engineering, Vol. 124, No. 6, 1998.

10-4 CRSI Design Handbook, Tenth Edition, Concrete Reinforcing Steel Institute, Schaumberg, IL, 2008.10-5 Richard W. Furlong, “Design of Concrete Frames by Assigned Limit Moments,” ACI Journal,

Proceedings, Vol. 67, No. 4, April 1970, pp. 341–353.10-6 Alan H. Mattock, “Redistribution of Design Bending Moments in Reinforced Concrete Continuous

Beams,” Proceedings, Institution of Civil Engineers, London, Vol. 13, 1959, pp. 35–46.

PROBLEMS

10-1 A five-span one-way slab is supported on 12-in.-widebeams with center-to-center spacing of 16 ft. Theslab carries a superimposed dead load of 10 psf anda live load of 60 psf. Using and

design the slab. Draw a cross sec-tion showing the reinforcement. Use Fig. A-5 tolocate bar cut-off points.

10-2 A four-span one-way slab is supported on 12-in.-wide beams with center-to-center spacing of 14, 16,16, and 14 ft. The slab carries a superimposed deadload of 20 psf and a live load of 100 psf. Design theslab, using and Select bar cut-off points using Fig. A-5 and draw across section showing the reinforcement.

10-3 A three-span continuous beam supports 6-in.-thickone-way slabs that span 20 ft center-to-center of

fy = 60,000 psi.fœc = 3500 psi

fy = 60,000 psi,fœc = 4000 psi

beams. The beams have clear spans, face-to-face of16-in.-square columns, of 27, 30, and 27 ft. Thefloor supports ceiling, ductwork, and lightingfixtures weighing a total of 8 psf, ceramic floor tile weighing 16 psf, partitions equivalent to auniform dead load of 20 psf, and a live load of 100 psf. Design the beam, using Use for flexural reinforcement and for shear reinforcement.Calculate cut-off points, extending all reinforce-ment past points of inflection. Draw an elevationview of the beam and enough cross sections tosummarize the design.

10-4 Repeat Problem 10-3, but cut off up to 50 percentof the negative- and positive-moment bars in eachspan where they are no longer needed.

fyt = 40,000 psify = 60,000 psi

fœc = 4500 psi.

this load case resulted in midspan moment values of approximately 260 kip-ft for theexterior span and 65 kip-ft for the interior span. Even if the full distributed moment, whichwas calculated above to be was added to these midspan moments,their values still would be less than the values calculated for other load cases intended tomaximize the midspan positive moments (Fig. 10-12). Thus, for this case, very little can beaccomplished by redistributing the maximum negative moments. The only change is thereduction in the maximum design moment at the interior supports ■

338 - 308 = 30 kip-ft,

Chapter 10 Continuous Beams and One-Way Slabs

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Transcript of Chapter 10 Continuous Beams and One-Way Slabs