chapt 10

Chapter 10

10-1 A five-span one-way slab is supported on 12-in.-wide beam with center-to-center spacing of 16 ft. The slab carries a superimposed dead load of 10 psf and a live load of 100

psf. Using ,desing the slab. Draw a cross section showing

the reinforcement. Use Fig. A-5 to locate the bar cut-off points.

Estimate slab thickness

Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is established later.

Table A-9:

End bay:

Interior bay:

Note that slab thickness was chosen on the basis of deflection control, since flexure and shear probably won’t govern the design (this will be checked later).

Try (this may need to be checked for deflections in the end span).

Assuming a cover of 0.75 in. and No. 4 bars as the slab reinforcement,

Compute factored loads

Considering a 1-ft wide strip of slab:

Slab self weight:

Superimposed dead load:

10-1

Total dead load:

Live load:

Factored load: Load per foot along design strip =

Note that and we can use the ACI Moment coefficients for the calculation of the positive and the negative moments.

Thickness for flexure

The maximum value for is at the first interior support since throughout. Using the appropriate moment coefficient from ACI Code Section 8.3.3,

For a reinforcing ratio of , which is a reasonable upper limit for a slab, the reinforcing index can be found from Eq. (5-20),

From Eq. (5-21) calculate the flexural resistance factor, R.

Using this value of R, the required value of can be determined using Eq. (5-22), assuming

that (will check it later).

For

i.e., min to keep is Actual will be less than 0.01 (O.K. for flexure).

Thickness for shear

The max shear is at the exterior face of the first interior support. Using the appropriate shear coefficient from ACI Code Section 8.3.3,

So, use a 7 in. slab.

Flexural reinforcement

Max

Assuming that , find the required reinforcement for a 1-ft wide strip

of slab.

10-2

Iterate to find the depth of the compression stress block and recompute the value of the required reinforcement:

Since the depth to the neutral axis, c, is less than , clearly the section is tension

controlled , and

The minimum reinforcement required by ACI Code Section 10.5.4, is

The maximum spacing of the bars is, by ACI Code Section 7.6.5,

So use No. 4 bars at 9 in.

Temperature and shrinkage steel as required by ACI Code Section 7.6.5,

and

So provide No. 4 bars at 16 in.

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the following table.

10-3

Calculation of reinforcement required in the slab.

1. 15.0 15.0

2. 62.3 62.3 62.3 62.3 62.3 62.3

3. Moment Coef.

4. 2.6 4.5 6.2 5.7 3.9 5.7 3.9

5. 0.01 0.17 0.24 0.15 0.22 0.15

6. 0.15 0.15 0.15 0.15 0.15 0.15

7. Reinforcement#4 @ 16 in.

#4 @ 9 in.

#4 @ 9 in.

#4 @ 16 in.

#4 @ 9 in.

#4 @ 16 in.

8. 0.15 0.27 0.27 0.15 0.27 0.15

Fig. S10-1.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points were located using Fig. A-5(c).

Fig. S10-1.1 Slab reinforcement detailing.

10-4

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 dead load of 20 psf and a

live load of 150 psf. Design the slab, using . Select bar

cut-off points using Fig. A-5 and draw a cross-section showing the reinforcement.

Short span clear length:

Long span clear length:

Average clear span length for first interior support:

Estimate slab thickness

Assume partitions are not sensitive to deflections. Will require recheck if sensitivity is established later.

Table A-9:

End bay:

Interior bay:

Note that slab thickness is chosen on basis of deflection control, since flexure and shear probably won’t govern the design (will be checked later).

Try

Assuming a cover of 0.75 in. and No. 4 bars as the slab reinforcement,

10-5

Compute factored loads

Considering a 1-ft wide strip of slab:

Slab self weight:

Superimposed dead load:

Total dead load:

Live load:

Factored load: Load per foot along design strip =

Note that and we can use the ACI Moment coefficients for the calculation of the positive and the negative moments (ACI Code Section 8.3.3).

Thickness for flexure

The maximum moment will occur at either:(1) exterior face of the first interior support, or(2) face of the middle support

For negative moments at the face of an interior support, ACI Code Section 8.0 defines as the average of the clear spans of the two adjacent spans. Using the appropriate moment coefficient from ACI Code Section 8.3.3,

(1)

(2)

For a reinforcing ratio of , which is a reasonable upper limit for a slab, the reinforcing index can be found from Eq. (5-20),




For

i.e., min to keep is Actual will be less than 0.01 (O.K. for flexure).

10-6

Thickness for shear

The max shear will occur in one of the two locations discussed for the maximum moments. Using the appropriate shear coefficient from ACI Code Section 8.3.3,

(1)

(2)

Flexural reinforcement

Max

Assuming that , find the required reinforcement for a 1-ft wide strip

of slab.



controlled , and

The minimum reinforcement required by ACI Code Section 10.5.4, is

The maximum spacing of the bars is, by ACI Code Section 7.6.5,

So use No. 4 bars at 7.5 in.

Temperature and shrinkage steel as required by ACI Code Section 7.6.5,

10-7

and

So provide No. 4 bars at 16 in.

The flexural reinforcement for the supports and the midspan for all the spans is calculated in the following table.

Calculation of reinforcement required in the slab.

1. 13.0 13.0 14.0 15.0 15.0 15.0

2. 61.1 61.1 70.8 81.3 81.3 81.3

3. Moment Coef.

4. 2.5 4.4 7.1 6.4 5.1 7.4 5.1

5. 0.10 0.18 0.3 0.21 0.31 0.21

6. 0.14 0.14 0.14 0.14 0.14 0.14

7. Reinforcement#4 @ 16 in.

#4 @12 in.

#4 @7.5 in.

#4 @ 12 in.

#4 @ 7.5 in.

#4 @ 12 in.

8. 0.15 0.20 0.32 0.20 0.32 0.20

Fig. S10-2.1 shows a cross-section of the slab showing the reinforcement. The bar cut-off points were located using Fig. A-5(c).

Fig. S10-2.1 Slab reinforcement detailing.

10-8

10-3 A three span continuous beam supports 6-in.-thick one-way slabs that span 20 ft center-to-center of beams. The beams have clear spans, face-to-face of 16-in.-square columns, of 27, 30, and 27 ft. The floor supports ceiling, ductwork, and lighting fixtures weighing a total of 8 psf, ceramic floor tile weighting 16 psf, partitions equivalent to a

uniform deal 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 reinforcement past points of inflection. Draw an elevation view of the beam and enough cross-sections to summarize the design.

1. Compute the trial factored load on the beam

(a) Dead load

slab self-weight:

ceiling, tile, partitions:

The beam size is not known at this stage, so it must be estimated for preliminary design purposes. Once the size of the beam has been established, the factored load will be corrected and then used in subsequent calculations. The beam size will be estimated in step 2.

(b) Live Load

The ASCE/SEI 7-05 recommendations allow live-load reductions based on tributary area multiplied by a live-load element factor, , to convert the tributary area to an influence area.

Positive moment at span AB and negative moment at exterior support, A.

10-9

Note that L shall not be less than for members supporting one floor (O.K.)

Negative moment at interior support, B.

Positive moment at span BC.

The size of the beam will be chosen on the basis of negative moment at the first interior support. For this location, the factored load on the beam, not including the beam stem below the slab, is:

The tributary width for the beam is 15 ft and the factored load from the slab per foot of beam is

Two approximate methods can be used to estimate the weight of the beam stem:(a) the factored dead load of the stem is taken as 12 to 20 percent of the other factored loads on the beam. This gives 0.43 to 0.72 kip/ft.

(b) the overall depth of beam h is taken to be 1/18 to 1/12 of the larger span, , and is taken to be . This gives the overall as 20 to 30 in., with the stem extending 14 to 24 in. below the slab, and gives as 10 to 15 in. The factored load of such sizes ranges from 0.17 to 0.45 kip/ft.

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

2. Estimate the size of the beam stem

(a) Calculate the minimum depth based on deflections.

ACI Table 9.5(a) (Table A-9) gives the minimum depths, unless deflections are checked. For partitions flexible enough to undergo some deflection, minimum depth for beam BC is

, where the span center-to-center of supports

Thus,

10-10

(b) Determine the minimum depth based on the negative moment at the exterior face of the first interior support.

The beam fits the requirements in ACI Code Section 8.3.3 and can use the moment and shear coefficients. For the support at B,

where (the average of the two adjacent spans).

Using the procedure developed in Chapter 5 for the design of singly reinforced beam sections, the reinforcement ratio that will result in a tension-controlled section can be estimated from Eq. (5-18) as,

For this reinforcement ratio, use Eq. (5-20), to find the reinforcing index,




Since columns are 16 in., try a 14 or 16 in. wide stem. Let’s try

Then,

With one layer of steel at supports, (O.K. for deflections).So, try a 14-in. wide-by-24-in. beam.

(c) Check the shear capacity of the beam

Maximum shear is at the exterior face at support B,

From ACI Code Section 11.2.1.1,

ACI Code Section 11.4.7.9 sets the maximum nominal is

Thus, (O.K. for shear)

10-11

(d) Summary

Use :

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

Weight per foot of the stem below slab

Corrected total factored load for 1st internal support moment:

Since this is less that the 4.1 kip/ft used earlier to estimate the beam size, the section chosen will be adequate.

Factored Total dead load:

Factored total loads:

(a) Positive moment at span AB and negative moment at exterior support, A.

(b) Negative moment at interior support, B.

(c) Positive moment at span BC.

4. Calculate the beam flange width for positive-moment regions

From ACI Code Section 8.12.2,

Therefore, the effective flange width is 81 in. and shown in Fig. S10-3.1

Fig. S10-3.1 Beam cross-section

10-12

5. Can we use the ACI Code Moment Coefficients?

Ratio of successive spans = ;

;

There are more than two spans; The loads are uniformly distributed.

Thus, we can use the ACI Code Moment Coefficients

6. Compute the beam moments

27.0 27.0 28.5 30.0

4.4 4.4 4.0 4.3

3210 3210 3250 3870

-200 230 -325 242

7. Design the flexural reinforcement

(a) Max negative moment (first interior support)

Max

Because the beam acts as a rectangular beam with compression in the web, we can assume that

.For , the required reinforcement for that section is,



controlled , and

10-13

The other negative moment sections have a lower design moment, so it will be conservative to

use the ratio of obtained here to quickly determine the area of tension steel required at

those other locations. That ratio is

(b) Max positive moment

Max

Because the beam acts as a T-shape beam with compression in the top flange, assume that the

compression zone is rectangular, i.e. , use .For ,

the required reinforcement that section is,


Cleary the section is tension controlled , and doing one iteration for the

negative moment section results in

The other positive moment section has a lower design moment, so it will be conservative to use

the ratio of obtained here to quickly determine the area of tension steel required at those

other locations. That ratio is

(c) Calculate the minimum reinforcement

From ACI Code Section 10.5.1, .

For 4500 psi concrete, , thus

10-14

(d) Calculate the area of steel and select the bars.

The remaining calculations are done in the following table.

-200 230 -325 242

coef. Eq. A and B 0.0114 0.0104 0.0114 0.0104

2.28 2.39 3.70 2.52

Yes Yes Yes Yes

Bars selected 4 No. 7 4 No. 72 No. 8 and

4 No. 73 No. 7 and

2 No.6provided, 2.40 2.40 3.98 2.68

Yes Yes

Note that in the negative moment regions some of the bars can be placed in the slab besides the beam and it is not necessary to check whether they will fit into the web width.

8. Check the distribution of the reinforcement

(a) Positive moment region

From ACI Code Section 10.6.4, the maximum bar spacing is

, where

Thus,

Bar spacing

It was also clear that the bar spacing is smaller than 10.3 in., since there are four bars and

(b) Negative moment region

ACI Code Section 10.6.6 says “part” of the negative moment steel shall be distributed over a width equal to the smaller of the effective flange width (81 in.) and At the interior supports, there are 6 top bars. Place the two No. 8 bars at the corners of the stirrups, two No. 7 bars over the beam web, and the other two No. 7 bars in the slab. Within a width of 34.2 in. we must place six bars. These cannot be further apart than 10.3 in. (as calculated in part a). We shall arbitrarily place two bars at 5 in. outside the web of the beam.

10-15

ACI Code Section 10.6.6 requires “some” longitudinal reinforcement in the slab outside this band. We shall assume that the shrinkage and temperature steel in the slab will satisfy this requirement.

9. Design the shear reinforcement

The shear force diagrams are calculated in the following table and shown at the bottom of the table. The shear coefficients for the supports are from ACI Code Section 8.3.3 and the coefficient fro the midspan of the beam is based on Eq. (6-26).

27 30

4.4 4.3

1.9 1.8

at support and midpsan

1.0 0.125 1.15 1.0 0.125 1.0

59.4 59.4 64.5 64.5

51.3 54

59.4 6.4 68.3 64.5 6.7 64.5

79.2 8.5 91.1 86.0 8.9 86.0

(a) Exterior end of beam AB

Because the beam is supported by a column, the critical section is located at d away from the face of the support.

Equation for :

,

Because exceeds , stirrups are required.

Try No. 3 Grade 40 stirrups double-leg stirrups with a 90o hook

10-16

Max spacing: and from ACI Code Section 11.4.5.1,

To satisfy the minimum stirrup requirement in ACI Code Section 11.4.6.3, the stirrup spacing must be,

Note that .

Thus, use 10.5 in. as maximum stirrup spacing.

The spacing required to support the shear force at the support is,

We can change the stirrup spacing to 10.5 in. when

This occurs at about 4 ft from face of support A.

We can stop the stirrups when from face of support A.

Place the first stirrup at 3 in. from support A, then 9 stirrups at 6 in. and 9 stirrups at 10.5 in.

(b) Interior end of beam AB

Equation for :

The spacing required to at this point is,

Change the stirrup spacing to 10.5 in. when .

This occurs at 5.3 ft from face of support B.We can stop the stirrups at 11.6 ft from face of support B.

10-17

Place the first stirrup at 2 in. from support B, then 15 stirrups at 4.5 in. and 9 stirrups at 10.5 in.

(c) Ends of beam BC

Equation for :

The spacing required to at this point is,

Change the stirrup spacing to 10.5 in. when .

This occurs at 5.4 ft from face of support.We can stop the stirrups at 12.8 ft from face of support.

Place the first stirrup at 2.5 in. from support B, then 14 stirrups at 4.5 in. and 9 stirrups at 10.5 in.

10. Bar cutoffs

(a) Detailing requirements:

Thus, exceeds for AB span, while governs for span BC.

The bottom and top bars have clear spacing and cover of at least and are enclosed by at least minimum stirrups. Therefore, this is Case 1 in Table 8-1 (ACI Code Section 12.2.2).

(b) Cutoffs for bottom steel

Span AB4 No. 7-Extend 2 full length into each support, cut off the other two at the positive moment point of inflection so that extra stirrups are not required.

Exterior end: From Fig. A-2, inflection point at from face of column.

Rule 3-a - Extend past the flexural cutoff point, i.e. 32.4 in.-21.5 in = 10.9 in. from face of column at A. Say 10 in.Rule 4-a - Distance from midspan to cutoff point greater than .

10-18

Rule 1-b - This is an interior beam with open stirrups. Since this is a discontinuous end use 90 deg. standard hooks on 2 No. 7 bars.

Rule 4-d – At the inflection point, the remaining steel is two No. 7, Thus,

The shear at 32.4 in. from the exterior end is,

Thus,

This exceeds - therefore, OK.

Interior end: From Fig. A-2, inflection point at from face of column

Rule 3-a - Extend bars to 33.7 in.- 21.5 in. = 12.2 in. from face of column. Use 10 in. to match other end.Rule 4-a - SatisfiedRule 1-b - This is an interior beam with open stirrups. Rule 1-b applies. Lap splice 2#7 bars from the exterior span with 2 No. 7 bars from the interior span with a Class A tension lap splice

Rule 4-d


Thus,


Span BC2 No. 8 and 3 No. 7 at midspan – Extend 2 No. 7 into supports. Cutoff 2 No. 8 and 1 No. 7 bars at the positive moment point of inflection so that extra stirrups are not required. Inflection point at

from face of column

Rule 3-a - Extend past the flexural cutoff point, i.e. 52.6 in.-22.5 in. = 30.1 in.

from face of column. Say 30 in.Rule 4-a - Satisfied.Rule 1-b – Lap splice 2#7 bars at supportRule 4-d


10-19

Thus,


Cutoffs for top steelSpan AB, exterior end, 4 No. 7

Use standard hook,

Set tail cover = 2 in., then . This is OK in a 16 in. column.

Negative moment point of inflection at

Rule 3-b - Extend d to 53.1+21.5 = 74.6 in. Cutoff at 75 in. = 6 ft-3 in. from face of column.Rule 4-b - Since 75 in. > 39.1 in. Rule 2 is satisfied. Span AB, interior end, 2 No. 8 and 4 No. 7 bars

Point of inflection at

Rule 3-b - Extend d to 77.8+21.5 = 99.3 in. Say 8 ft.-4 in. from face of columnRule 4-b – Since 100 in. > 44.7 in. Rule 2 is satisfied.

Span BC, 2 No. 8 and 4 No. 7 bars

Negative moment point of inflection at

Rule 3-b – Extend to 108.9 in. Say 9 ft- 1 in, from face of columnRule 4-b – OK

Since all cutoffs are past points of inflection, they are not in zones of flexural tension, therefore extra stirrups are not needed.

Provide 2 No. 4 top bars as stirrup support, lab splice with negative moment steel.

10-20

Fig. S10-3.2 Beam reinforcing detailing.

10-21

10-4 Repeat Problem 10-3, but cut off up to 50 percent of the negative- and positive-moment bars in each span where they are no longer required.

Bat cutoff for positive moment steel in AB span

Flexural reinforcement: 4#7 bars

Extend two bars full length and into each support, cutoff the other two where they no longer

required .

From Fig. A-2, flexural cutoff point is at from exterior end

(support A), and from interior end (support B).

Rule 3-a: Extend past the flexural cutoff point from exterior end

from interior end

Rule 4-a: Distance between midspan to cutoff point =

for #7 bar

Bar cutoff for negative moment steel

Flexural reinforcement: 2#8 and 4#7 bars

AB span, interior end

From Fig. A-2, flexural cutoff point is at

Rule 3-b: Extend past the flexural cutoff point Rule 4-b: for #7 top bar Therefore, use 55.5 in. for cutoff point.

BC span

From Fig. A-1, flexural cutoff point is at

Rule 3-b: Extend past the flexural cutoff point Rule 4-b: for #7 top bar Therefore, use 58.5 in. for cutoff point.

10-5 Explain the reason for the two live-load patterns specified in ACI Section 8.11.2

Live load patterns are arranged to maximize the moments and shears in beams and columns at various sections due to gravity loading.

10-22

chapt 10

Documents

oneway slab

slab reinforcement detailing

ft wide strip of slab

slab design stripplan12

slab design stripplanshort

required reinforcement

slab self weight

minimum reinforcement