12-1

Chapter 12

12-1 A hinged end column 18-ft tall supports unfactored loads of 100 kips dead load and

60 kips live load. These loads are applied at an eccentricity of 3 in. at bottom and 5

in. at the top. Both eccentricities are on the same side of the centerline of the

column. Design a tied column with at least three bars per face using

and .

Factored loads and moments

1.2 100 kips +1.6 60 kips 216 kipsuP

1 2/ 0.6M M (Note that the column is bent in single curvature)

Estimate column size

Assume 0.015g

2uP 216 kips

(trial ) = 110 in.0.4 4 ksi + 60 ksi 0.0150.4

g

c y g

Af f

Choose a column cross section of 12 in. x 12 in.

Check whether the column is slender

1.0 (18 12) in.60

(0.3 12 in.)

uk

r

1

2

34 12 34 12 0.6 26.8M

M

60uk

r > 1

2

34 12 26.8M

M

The column is quite slender, increase column size to 16 in. x 16 in.

Are moments greater than the minimum?

2,min 2(0.6 0.03h) =216 kips 0.6 in. 0.03 16 in. 233 k-in. < uM P M

2use M

Compute EI

0.4

1

c g

d

E IEI

357,000 4000 3605 10 psicE

4

416 in.

5460 in.12

gI

factored dead load 1.2 100 kips0.56

all factored load 216 kipsd

12-2

3 4

9 20.4 3605 10 psi 5460 in.

5.05 10 lb-in.1 0.56

EI

Magnified moment

2c nsM M

1.0

1

mns

u

c

C

P

P

1

2

0.6 0.4 0.6 0.4 0.6 0.84m

MC

M

2 9 223

2 2 2

5.05 10 lb-in.1070 10 lb =1070 kips

1.0 18 12 in.c

EIP

k

0.841.15

216 kips1

0.75 1070 kips

ns

Select reinforcement

Assume # 8 bars for longitudinal reinforcement, # 3 bars for the ties, and a clear concrete

cover of 1.5 in.

16 in. 2 1.5 in. 2 0.375 in. 1 in.0.70

16 in.

2

216 kips0.84 ksi

(16 in.)

n uP P

bh bh

Assume E-type reinforcement

Fig.A-9a shows

Fig.A-9b shows

. Use . 20.01 2.56 in.st gA A

Use 6 bars # 6 22.64 in.stA for the column of 16 in. 16 in.

12-2 Repeat Problem 12-1, but with the top eccentricity to the right of the centerline and

the bottom eccentricity to the left.

Factored loads and moments

216 kipsuP

1 2/ 0.6M M (Note that the column is bent in double curvature)

12-7

Select reinforcement

Assume # 8 bars for longitudinal reinforcement, # 3 bars for ties, and a clear concrete

cover of 1.5 in.

16 in. 2 1.5 in. 2 0.375 in. 1 in.0.70

16 in.

Assume E-type reinforcement

Fig. A-10a yields

Fig. A-10b yields

Use 8 #8 bars ( ) for the column of 16 in. 16 in.

12-4 Use the ACI moment-magnifier method to redesign the columns in the main floor of

Example 12-3 assuming that the floor-to-floor height of the first story is 16 ft 0 in.

rather than 18 ft 0 in. Also, assume the lateral wind forces are 15 percent larger

than those used in Example 12-3.

The floor plan and elevation is shown in Fig. S12-1

The following design shall be performed for a typical interior frame, for instance the fram

along line 2. The design of the columns will follow the following steps:

A. Calculate loads

B. Calculate the beam and column properties and modulus of elasticity

C. Select preliminary column size

D. Check with gravity load case

E. Check with gravity plus wind load case

F. Finalize the design of column reinforcement

A. Calculate loads for a typical interior frame

Dead load from roof

Distributed loads on beams

3 2 3

2 2

18 in. 24 in.6 in.0.150 k/ft 0.025 k/ft 10 0.150 k/ft

12 in/ft 144 in /ft

1.45 k/ft

DL

Distributed load on girders (beams supporting other beams)

2 2

18 in. 18 in. 30 in0.025 0.150 0.6 k/ft

12 in./ft 144 in /ftDL

Concentrated on the interior columns

(

)

12-8

A B C D

1

2

3

4

5

6

32' 30' 32'

20'

20'

20'

20'

20'

11' 6"

11' 6"

11' 6"

11' 6"

11' 6"

16' 0"

Roof

5 th floor

4 th floor

3 th floor

2 th floor

Ground

floor

Slab thickness: 6"; Column size: 18" x 18"; Beam size: 18" x 30"

a) Plan

b) Section 1-1

11

N

Fig. S12-1

12-9

Concentrated on the exterior columns

(

)

Dead load from other floors

Distributed loads on beams

3 2 3

2 2

18 in. 24 in.6 in.0.150 k/ft 0.020 k/ft 10 0.150 k/ft

12 in/ft 144 in /ft

1.40 k/ft

DL

Distributed load on girders (beams supporting other beams)

2 2

18 in. 18 in. 30 in0.020 0.150 0.593 k/ft

12 in./ft 144 in /ftDL

Concentrated on the interior columns

(

)

Concentrated on the exterior columns

(

)

For each column of 11.5 ft, a weight of 3.88 kips is added for the column self-weight.

For each column of 16.0 ft, a weight of 5.40 kips is added for the column self-weight.

Live load from roof

Concentrated live load on the interior columns

(

)

Concentrated live load on the exterior columns

(

)

Live load from other floors

Concentrated live load on the interior columns

(

)

Concentrated live load on the exterior columns

(

)

B. Calculate the beam and column properties and modulus of elasticity

Column: 218 in. 18 in. 324 in.A

4

418 in.

0.7 0.7 6120 in12

c gI I

12-10

Beam:

Effective flange width 18 in. 8 6 in. 66 in.

26 66 18 24 828 inA

3

418 in. 30 in.

0.7 beam-web 0.7 28350 in.12

b gI I

Note that the selection of rigid end zones follows Example 12-3.

Modulus of elasticity 357000 4000 3600 10 psiE

C. Select preliminary column size

Columns are sized based on the gravity load 1.2 1.6 0.5 rD L L . In this load combination, live

load can be reduced. From separate analyses of dead load, live load from the roof floor, and live

load from the other floors, the axial loads of the columns in the ground floor are shown in the

Table P12-1.

The axial load in the exterior column from the live load from the other floors then can be reduced

with a reduction factor as follows:

150.25 0.46

4 4 16.75 20

The axial load in the interior column from the live load from the other floors then can be reduced

with a reduction factor as follows:

150.25 0.40

4 4 31 20

The calculation of reduced axial live load and factored load is shown in Table P12-1.

Table P12-1

All unit are in kips

Exterior

column

Interior

column

Dead load 285 487

Live load from the roof floor 10.0 18.5

Live load from the other floors 107 198

Reduced live load from the other floors 48.8 79.2

Total factored load 425 720

Assume 0.015g

Exterior column

( )

( )

( ) ( )

12-11

Interior column

( )

( )

( ) ( )

Based on the result of Example 12-3, select a column cross section of 18 in. 18 in. ( 2324 in.gA ) for both exterior and interior columns.

D. Check with gravity load case 1.2 1.6 0.5 rD L L

1. Is the story being designed sway or non-sway? In order to answer this question, we need

to calculate the stability index u oh

us c

PQ

V

.In order to have the terms from the same

analysis, we need to analyze a frame with an arbitrary lateral load of 20 kips applied at

the 2nd

floor level in conjunction with the factored dead load and live load as shown in

Fig. S12-2. In order to take into account the live load reduction, an average of live load

reduction factor for exterior and interior columns (0.43) is multiplied with the live load

factor 1.6, yielding 0.69. Therefore, the load combination used is:

1.2 0.69 0.5 20 k lateral loadrD L L .

A structural analysis gives the following results:

0.127 in.oh 414 722 723 418 2277 kipsuP

16 ft 192 in.c

2277 kips 0.127 in.0.075 0.05

20 kips 192 in.Q

The first story is a sway story.

Note that the 2277 kipsuP does not differ significantly from

( )

2. Are the columns slender?

1.2 (192 30) in.36 22

(0.3 18 in.)

uk

r

The columns are slender.

3. Compute the factored axial loads and moments from a first-order frame analysis. As

explained in Example 12-3, the unfactored moments for exterior columns can be

determined based on the live load pattern shown in Fig. S12-3 while those for interior

columns based on the live load pattern shown Fig. S12-4. After a structural analysis is

made, live load reduction factors will be applied. All results and calculation are shown in

Table P12-2.

12-12

Factored dead load and live load plus arbitrary lateral load to evaluate stability index, Q

20 kips

Fig. S12-2

All span loaded with live load Fig. S12-3

12-13

Staggered live load pattern Fig. S12-4

Table P12-2

(Forces in kips Moments in k-ft) Exterior Column Interior Column

DP 285 487

reducedLP 48.8 79.9

LrP 10 18.5

topDM 37.0 -4.9

bottomDM -36.2 5.5

top, reducedLM 20.9 0.67 14 18.2 0.46 8.4

bottom, reducedLM 20.5 0.67 13.7 2.8 0.46 1.3

The factored load on exterior and interior columns are as follows:

Exterior column

2top 1.2 37 1.6 14 66.8 k-ftM M

1bottom 1.2 36.2 1.6 13.7 65.4 k-ftM M

12-14

Interior column

2top 1.2 4.9 1.6 8.4 19.3 k-ftM M

1bottom 1.2 5.5 1.6 1.3 8.68 k-ftM M

4. Find ns for the exterior and interior column?

1.0

1

mns

u

c

C

P

P

Exterior column

1

2

65.40.6 0.4 0.6 0.4 ( ) 0.21

66.8m

MC

M

2

2Pc

EI

k

0.2

1

c g s se

dns

E I E IEI

48750 in.gI

3600 ksicE

4150 in.seI

( )

6 25.88 10 kip-inEI

2 6 22

2 2

5.88 10 kip-in2210 kips

1.0 162 in.c

EIP

k

66.8 k-ftcM

Interior column

1

2

8.680.6 0.4 0.6 0.4 ( ) 0.42

19.3m

MC

M

( )

Since (int)dns does not change significantly, EI and Pcwill

remain essentially the same.

67.8 k-ftcM

12-15

5. Check initial column sections for gravity load case

Exterior column

1.9

0.1118

e

h

0.015g

Fig. A-9b yields

Interior column

0.015g

Because reading from the graph may not be accurate given the two

values are so close, we need to select reinforcement for the column and

check its capacity against the demand. Select 8 bars #8, Example 12-3

shows , OK.

E. Check with gravity plus wind load case 1.2 1.6 0.5 rD W L L

Wind loads are given in Fig. S12-5

1. Calculation of the stability index u oh

us c

PQ

V

Similar to the gravity load case, we need to do one single analysis with the wind load plus

gravity load case. To take into account of the live load reduction, an average live load

reduction factor of 0.43 will be multiplied with the live load factor 0.5, yielding 0.22.

Therefore, the load combination becomes 1.2 1.6 0.215 0.5D W Lr L . A structural

analysis yields the following results:

0.357 in.oh

352 635 633 385 2005 kipsuP

16 ft 192 in.c

4.46 3 6.61 8.49 32.9usV kips

2005 kips 0.357 in.

0.113 0.0532.9 kips 192 in.

Q

1 1

1.131 1 0.113

sQ

The first story is a sway story.

12-16

Wind load

4.46 kips

6.61 kips

6.61 kips

6.61 kips

8.49 kips

Fig. S12-5

2. Factored axial loads and moments

A structural analysis of the frame subjected to the wind load (without the load factor) yields

the following results, as shown in Table P12-3.

Table P12-3

Forces in kips

Moments in k-ft Exterior Column Interior Column

WP 10.4 0.5

topWM 46.2 65

bottomWM -46 -64.6

Exterior column

( )

2 2 2ns s sM M M

2 1.2 0.5 1.2 37 0.5 14 51.4 k-ftns D LM M M

2 1.6 1.6 46.2 73.9 k-fts WM M

2 51.4 1.13 73.9 135M k-ft

Interior column

12-17

( )

2 2 2ns s sM M M

2 1.2 0.5 1.2 4.9 0.5 8.4 10.1ns D LM M M k-ft

2 1.6 1.6 65 104 k-fts WM M

2 10.1 1.13 104 128M k-ft

Note that the 2005 kipsuP does not differ significantly from

2 387 632 2038 kipsuP

3. Check column sections for axial loads and moments

Exterior column

0.01g

Fig. A-9b yields

At this point, we can select 8 bars #6 ( 23.52 in. , 0.011st gA ) for the exterior columns.

Interior column

0.02g

Conclusion: The cross section of exterior and interior columns is 18 in. x 18 in. Use 8 bars #6 for

exterior columns, and 8 bars #8 for interior columns.

11-15

Now select the bars:

28#6 bars 3.52 in

26#7 bars 3.6 in

Select #6 bars, spaced evenly around the column.

From the interaction diagrams, the extreme tensile layer of steel is in tension, but only requires a

Class A splice as long as not all of the bars are spliced at the same location. Since, in reality, it is

likely that all bars will be spliced in the same plane, specify a Class B splice regardless. Therefore,

from Table A-13, a splice of length 33 in is required.

As transverse reinforcement, we have selected a #3 spiral since the longitudinal bars are not larger

than #10 bars. Now select the pitch of the spiral:

22

2 2

3 in

0.375 in 60 ksi1.81 in

0.45 ' / 1 0.45 15 in 5 ksi 254 in /177 in 1

sp yt

c c g ch

d fs

D f A A

Use an 18 in diameter column reinforced with 8 #6 bars. Use 33 in lap splices, and #3 spirals with

a pitch of 1.75 in as transverse reinforcement.

11-9 Design a cross section and reinforcement to supports Pu = 450 kips, Mux = 100 k-ft,

and Muy = 130 k-ft. Use and .

Although the strain compatibility method (shown in example 11-5) is the most

theoretically correct method for designing columns for biaxial loading, it is seldom used in design.

Here we will use two more common methods for designing a column, the equivalent eccentricity

method and the Bresler reciprocal load method. Any method outlined in section 11-7 is

appropriate for the solution to this problem.

Equivalent eccentricity method:

First select the dimensions of the trial section, assuming 0.015 :

2

,

450 k230 in 15.2 in by 15.2 in

0.40 4 ksi 60 ksi 0.0150.40 '

ug trial

c y g

PA

f f

Since biaxial moments are also applied to the section, select an 18 in square section.

Assuming #3 ties and

#9 bars:

18 in 2 1.5 in 0.375 in 0.56 in

0.7318 in

Now compute ex, ey, and eox.

130 k-ft 12 in/ft

3.47 in450 k

uy

x

u

Me

P

100 k-ft 12 in/ft

2.67 in450 k

uxy

u

Me

P

11-16

Since our trial section is square,yx

x y

ee

2

450 k0.347 0.4

' 4 ksi 18 in

u

c g

P

f A

40,000 psi

0.5' 100,000 psi

yu

c g

fP

f A

2

450 k 60,000 psi 40,000 psi0.5 0.847

100,000 psi4 ksi 18 in

0.847 2.67 in 18 in

3.47 in 5.73 in18 in

y x

ox x

y

ee e

Therefore, we can design our column for: 450 kuP , and

450 k 5.73 in 2580 k-in 215 k-ftoy u oxM P e

Now use the interaction diagrams in Appendix A to determine g .

2

450 k1.39 ksi

324 in

n u

g g

P P

A A

2

2580 k-in0.442 ksi

18 in 324 in

oyn

g g

MM

hA hA

Since 0.73 0.75 we can use Fig A-9b without interpolating for this design.

So use 0.023g

Finally, we can compute the area of steel required to reinforce this section, and select the bars:

2 20.023 324 in 7.45 inst g gA A

Select 8 #9 bars, with 3 bars along each face. Select ties and splice lengths as appropriate.

We can check this solution using the Bresler reciprocal load method. Remember that we have an

18 in square column reinforced with 8 #9 bars.

2

2

8 in0.0247

324 in

stg

g

A

A

18 in 2 1.5 in 0.375 in 0.56 in

0.7318 in

ComputenxP , the factored load capacity corresponding to and x ge .

130 k-ft 12 in/ft

3.47 in450 k

uy

x

u

Me

P

, and

3.47 in0.193

18 in

x

x

e .

From Fig. A-9b, 1.9 ksinxP

bh

Therefore 616 knxP

Compute nyP , the factored load capacity corresponding to and y ge .

11-17

100 k-ft 12 in/ft

2.67 in450 k

uxy

u

Me

P

, and

2.67 in0.148

18 in

x

x

e .

From Fig. A-9b, 2.2 ksinxP

bh

Therefore 713 knyP

From equ. 11-7 calculate noP :

2 2 2

3 ' 0.85 4 ksi 324 in 8 in 60 ksi 8 in 1560 kno c g st y stP k f A A f A

1 1 1 1 1 1 1 1

616 k 713 k 0.65 1560 ku n nx ny noP P P P P

490 kuP , so our section is sufficient for the loading defined here.

5-44

5-17 For the one-was slab shown in Fig. P5-7, assume the maximum negative moment at

support c is -3.3 kip-ft/ft, and the maximum factored positive moment at midspan

point b is 2.4 kip-ft/ft.

(a) Using the given slab thickness of 6 in, determine the required reinforcement

size and spacing at both of these locations to satisfy ACI Code flexural

strength requirements. Be sure to check the ACI Code requirements for

minimum flexural reinforcement in slabs.

Negative Moment 3.3 k-ft 39.6 k-inuM :

Assume 0.75 in of cover will be provided. This results in a d = 5 in.

From Eq. (5-16):

239.6 k-in

ft ft 0.16 inftft 0.9 60 ksi 0.9 5 in

2

u

s

y

MA

af d

With this estimate, iterate once to have a better estimate of the lever arm jd

From Eq. (5-17): 20.16 in 60 ksi

0.24 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

239.6 k-in

ft ft in0.15 ft0.24 inft

0.9 60 ksi 5 in -2 2

u

s

y

MA

af d

Minimum reinforcement:

2,min in0.0018 0.0018 12 in 6 in 0.13

ft ftsA

bh

Strength requirements govern here.

Maximum reinforcement spacing is limited to 3h or 18 in, which is the same value for this 6 in deep

slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the same

here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs.

If we select #3 bars at 8 in, the result is:

2 22 12 in in in0.11 in 0.165 0.15 ft ft ft8 in

sA OK

Check that strength is satisfied:

2 0.24 in 0.9 0.165 in 60 ksi 5 in 43.5 k-in

2 2n s y

aM A f d

k-ft k-ft 3.62 3.3 ft ftn uM M OK

5-45

Remember to calculate the strain in the extreme layer of tension steel to verify that assuming

is valid.

- 5 in - 0.24 / 0.85 in0.003 0.050 0.005

0.24 / 0.85 int cu

d c

c OK

Therefore the designer is permitted to use for this beam design.

Positive Moment 2.4 k-ft 28.8 k-inuM :

Assume 0.75 in of cover will be provided. This results in a d = 5 in.

From Eq. (5-16):

228.8 k-in

ft ft 0.12 inftft 0.9 60 ksi 0.9 5 in

2

u

s

y

MA

af d

With this estimate, iterate once to have a better estimate of the lever arm jd

From Eq. (5-17): 20.12 in 60 ksi

0.17 in0.85 ' 0.85 4 ksi 12 in

s y

c

A fa

f b

From Eq. (5-16):

228.8 k-in

ft ft in0.11 ft0.17 inft

0.9 60 ksi 5 in -2 2

u

s

y

MA

af d

Minimum reinforcement:

2,min in0.0018 0.0018 12 in 6 in 0.13

ft ftsA

bh

Minimum requirements govern here.

Maximum reinforcement spacing is limited to 3 h or 18 in, which is the same value for this 6 in

deep slab. Also, we must check reinforcement spacing for crack control. Since fy and cc are the

same here as in Example 5-7, the maximum spacing for crack control is 12 in. This governs.

If we select #3 bars at 10 in, the result is:

2 22 12 in in in0.11 in 0.132 0.13 ft ft ft10 in

sA OK

Check that strength is satisfied:

2 0.17 in 0.9 0.132 in 60 ksi 5 in 35.0 k-in

2 2n s y

aM A f d

k-ft k-ft 2.9 2.4 ft ftn uM M OK

9.0

9.0

5-46

Remember to calculate the strain in the extreme layer of tension steel to verify that assuming

is valid.

- 5 in - 0.17 / 0.85 in0.003 0.072 0.005

0.17 / 0.85 int cu

d c

c OK

Therefore the designer is permitted to use for this beam design.

(b) At both locations, determine the required bar size and spacing to be provided

in the transverse direction to satisfy ACI Code section 7.12.2 requirements for

minimum shrinkage and temperature reinforcement.

Since the positive flexural region was controlled by temperature and shrinkage reinforcement, the

reinforcement specified there would suffice in the transverse direction at all locations.

So, use #3 bars at 10 in. Placement near the top or bottom of the slab makes no difference here, so

specify that bars are to be placed wherever is easiest.

(c) For both locations provide a sketch of the final design of the slab section.

Fig. S5-17 Cross-section of final design for negative and positive bending regions, respectively

Note that other selections of As may also be correct. If all checks are satisfied, without being

unreasonably conservative, the design may be considered adequate.

9.0

9.0

12in.

6in.

#3@8in.

12in.

6in.

#3@10in.

10-5

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 100 psf. Design the slab, using and

. Select bar cut-off points using Fig. A-5 and draw a cross-section

showing the reinforcement.

slab design strip

Plan

Short span clear length: 12 in.14 ft. 13 ft.in.

12ft

n

Long span clear length: 12 in.16 ft. 15 ft.in.

12ft

n

Average clear span length for first interior support:

,

13 ft. 15 ft.14 ft.

2n avg

Estimate slab thickness

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

later.

Table A-9:

End bay: 13 12Min 6.50 in.24 24

nh

Interior bay: 15 12Min 6.43 in.28 28

nh

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 6.5 in.h

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

0.56.5 0.75 5.5 in.2

d

10-6

Compute factored loads

Considering a 1-ft wide strip of slab:

Slab self weight: 6.5

150 81.25 psf12

Dsw

Superimposed dead load: 20 psfDiw

Total dead load: 81.25 20 101.25 psfDw

Live load: Factored load: Load per foot along design strip =

3L Dw w , so 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 n as the

average of the clear spans of the two adjacent spans. Using the appropriate moment coefficient

from ACI Code Section 8.3.3,

( )

( )

For a reinforcing ratio of 0.01 , which is a reasonable upper limit for a slab, the reinforcing

index can be found from Eq. (5-21),

0.01 600000.171

3500

From Eq. (5-22) calculate the flexural resistance factor, R.

0.171 3500 1 0.59 0.171 538 psiR

Using this value of R, the required value of 2bd can be determined using Eq. (5-23a), assuming

that 0.9 (will check it later).

For , √

i.e., min d to keep 0.01 is Actual 5.5 in.d will be less than 0.01 (O.K. for

flexure).

Thickness for shear

The max shear uV 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,

10-7

'0.75 2 0.75 2 3500 12 5.5 5860 lbs/ft ok for shearc c wV f b d

Flexural reinforcement

Max -

Assuming that 0.95 and 2

s y

ad d

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

of slab.

( )

( )

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 3 8 of d , the section is tension controlled

, 0.9s y .

The minimum reinforcement required by ACI Code Section 10.5.4, is 2

,min 0.0018 0.0018 12 6.5 0.14 in. /ftsA bh

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

max

3 19.5 in.

18 in

hs

Select No. 4 bars at 9 in.

⁄

⁄

Temperature and shrinkage steel as required by ACI Code Section 7.12.2,

2,min 0.0018 0.14 in. /ftsA bh and max

5 32.5 in.

18 in

hs

So provide No. 4 bars at 16 in. 2

20.2 in. in.12 0.15 in.ft ft16 in.

sA

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

following table.

10-8

Calculation of reinforcement required in the slab.

1. n 13.0 13.0 14.0 15.0 15.0 15.0

2. 2 kips-ftu nw 47.7 47.7 55.3 63.5 63.5 63.5

3. Moment Coef. 1 24 1 14 1 10 1 11 1 16 1 11 1 16

4. kips-ft/ftuM 2.0 3.4 5.5 5.0 4.0 5.8 4.0

5. 2 reqd. in. /ftsA 0.09 0.14 0.23 0.17 0.25 0.17

6. 2,min in. /ftsA 0.14 0.14 0.14 0.14 0.14 0.14

7. Reinforcement #4 @

16 in.

#4 @

16 in.

#4 @

9 in.

#4 @

11 in.

#4 @

9 in.

#4 @

11 in.

8. 2 provided in. /ftsA 0.15 0.15 0.27 0.18 0.27 0.18

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.

16”

9”

11”

13-9

The length of the critical perimeter is 2 21.6 21.6 86.4 in.ob

Now, cV is to be taken as the smallest of the following. From Eq. (13-24),

' 14 4 1 4000 86.4 5.6 122 kips 1000c c oV f b d

For Eq. (13-25), 1.0 (since column is square). Therefore,

'4 12 2 4 1 4000 86.4 5.6 184 kips1000c c oV f b d

For Eq. (13-26), 40s for this interior column. Therefore,

' 40 5.6 12 2 1 4000 86.4 5.6 140 kips100086.4

sc c o

o

dV f b d

b

Therefore, the smallest values is 122 kipscV , so 0.75 122 91.5 kips > Vc uV and the slab

is OK in two-way shear.

13-6 Assume the slab described in Problem 13-5 is supported on 10 in. 24 in. columns.

Check two-way shear at a typical interior support. Assume unbalanced moments

are negligible.

1. Compute the factored uniform load.

(

)

2. See the solution to problem 13-5 for one-way shear calculations.

3. Check two-way shear

Punching shear is critical on a rectangular section located at 2d away from the column face, as

shown in Fig. S13-6.2. The critical perimeter is 29.6 in. by 15.6 in. The average d value for

determining the shear strength of the slab is 5.6 in.d

(a) Compute uV on the critical perimeter for two-way shear.

(

)

(b) Compute cV for the critical section.

The length of the critical perimeter is 2 29.6 15.6 90.4 in.ob

Now, cV is to be taken as the smallest of the following. From Eq. (13-24),

' 14 4 1 4000 90.4 5.6 128 kips 1000c c oV f b d

13-10

Fig. S13-6.2 Critical section for two-way shear at interior column.

For Eq. (13-25), 24

2.410

(since column is 10 in. by 24 in.). Therefore,

'4 4 12 2 1 4000 90.4 5.6 117 kips10002.4

c c oV f b d

For Eq. (13-26), 40s for this interior column. Therefore,

' 40 5.6 12 2 1 4000 90.4 5.6 143 kips100090.4

sc c o

o

dV f b d

b

Therefore, the smallest values is 117 kipscV , so 0.75 117 88 kips > Vc uV and the slab is

OK in two-way shear.

13-7 The slab shown in Fig. P13-7 supports a superimposed dead load of 25 psf and a live

load of 60 psf. The slab extends 4 in. past the exterior face of the column to support

an exterior wall that weighs 400 lbs/ft of length of wall. The story-to-story height is

9.5 ft. Use 4500-psi concrete and Grade-60 reinforcement.

(a) Select slab thickness.

Determine the thickness to limit deflections.

From Table 13-1, the minimum thicknesses of the four typical slab panels are as follows:

Panel 1-2-A-B (corner; treat as exterior), and panels 2-3-A-B and 1-2-B-C (exterior)

13-11

Maximum 20 12 16 224 in.n

Minimum 224

7.47 in.30 30

nh

Panel 2-3-B-C (interior)

Maximum 224 in.n

Minimum 224

6.78 in.33 33

nh

Try 8.0 in.h

Check the thickness for shear. We should check the shear at columns A2 and B2

The tributary area for column A2 is cross-hatched in Fig. S 13-7.1 The factored uniform load can

be calculated as:

81.2 150 25 1.6 60 246 psf

12uq

Note that if the area of any of the panels exceeded 2400 ft , it would be possible to reduce the live

load before factoring it.

Fig. S 13-7.1 Initial critical shear perimeters and tributary areas for column A2.

The critical shear perimeter is located at 2d away from the interior column face and 4 in. from

the exterior column face, as shown in Fig. S 13-7.1. In the following calculation for the factored

shear force transmitted to column A2, the shear force multiplier of 1.15 required for the first

interior support will be applied directly to the appropriate tributary lengths. Then,

8 0.75 0.5 6.75 in.avgd (assuming 34

in. clear cover and No. 4 bars as slab

reinforcement).

22.75 2 23.38 69.5 in.ob

12 22.75 23.38

246 9 1.15 9 10 1.2 400 9 1.15 9 60700 lbs 61 kips12 144

uV

13-12

From Eq.(13-25),

16

116

4

2 6.0 4

(does not govern)

From Eq. (13-26),

30s , for an exterior slab-column connection

30 6.75

2 2 4.91 4.069.5

s

o

d

b

(does not govern)

Thus, using Eq. (13-24):

' 14 0.75 4 1 4500 69.5 6.75 94.4 kips > V1000c c o uV f b d

Note: 0.65 0.75u cV V

For this ratio, ACI Code Section permits modification of f for moment transfer about an axis

parallel to the edge of the slab. With that information and because this ratio is below 0.8, the slab

thickness at this connection should be sufficient for checking shear and moment transfer about an

axis perpendicular to the edge of the column. Shear check for column B2 follows the same

procedure as for column A2. Thus, use an 8-in. slab. Final shear checks will be made in part (c)

after completing the flexural design of the slab.

(b) Use the direct design method to compute moments, and then design the

reinforcement for the column and middle strips associated with column line

2.

Because there is no edge beams, 0f

Compute moments in the slab strip along column line 2

A2 B2 C2

1 (ft) 20.0 20.0

(ft)n 18.67 18.67

2 (ft) 18.0 18.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 196 196

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -51 +102 -137 -127 +69 -127

13-13

Compute moments in the slab strip along column line 1

A1 B1 C1

1 (ft) 20.0 20.0

(ft)n 18.67 18.67

2 (ft) 10.0 10.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 109 109

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -28 +57 -76 -71 +38 -71

Wall load (kip/ft) 0.48 0.48

Wall 2

8

wall no

qM

21 21

Moments from wall

(kip-ft) -5.5 11 -15 -14 +7 -14

Distribute the negative and positive moments to the column and middle strips and design the

reinforcement.

In each panel, the column strip extends 1 20.25 min , 0.25 18 12 54 in. on each side

of the column lines. The total width of the column strip is 2 54 in.= 108 in. 9 ft . The width of

the middle strip is 9 ft. The edge strip has a width of 54 in. 12 in. 66 in. 5.5 ft .

Place the steel in the long direction close to the surface of the slab. Try No. 4 bars. Thus,

8 0.75 0.25 7.0 in.d

Compute trial sA required at the section of maximum moment (column strip at B2). The largest

uM is 102.3 kip-ft. Assuming that 0.95jd ,

2102.3 12,000(trial) 3.42 in.

0.9 60,000 0.95 7.0sA

Compute a and check whether the section is tension controlled:

3.42 60,0000.82 in.

0.85 4500 5.5 12a

0.82

1.00 in.0.825

c

Clearly, the section is tension-controlled; therefore, 0.9 .

Compute the value of jd :0.82

7.0 6.59 in.2

jd

Assuming that a is constant for all sections (conservative assumption), compute a constant for

computing sA :

2 12,000(in. ) 0.0337 (kip-ft)

0.9 60,000 6.59

us u

MA M

(Eq. A)

13-14

The values of sA required in the following table are computed from Eq. (A).

From ACI Code Section 13.3.1,

,min 0.0018sA bh for Grade-60 reinforcement. Maximum bar spacing is 2h (ACI Code

Section 13.3.2), but not more than 18 in. (ACI Code Section 7.12.2.2). Therefore maximum is

16 in.

Edge column strip:

2,min 0.0018 5.5 12 8 0.95 in.sA

Minimum number of bar spaces5.5 12

4.116

Therefore, the minimum number of bars is 5.

Other strips:

2,min 0.0018 9 12 8 1.56 in.sA

The minimum number of bars is 8.

Division of moment to column and middle strip: north-south strips

Edge

Column Strip

Middle

Strip

Column

Strip

Middle

Strip

Column

Strip

Strip Width, ft 9.0 9.0 9.0 9.0 5.5

Exterior Negative Moments A1 A2 A3

Slab moment (kip-ft) -28 -51 -51 Moment Coef. 1.0 0.0 0.0 1.0 0.0 0.0 1.0

Distributed moments to

strips -28 0.0 0.0 -51 0.0 0.0 -51

Wall moment (kip-ft) -5.5 Total strip moment (kip-ft) -33.5 0.0 -51 0.0 -51

Required 2 (in. )sA 1.13 0.0 1.72 0.0 1.56

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 6 #4 9 #4 9 #4 9 #4 9 #4

2 (in. )sA provided 1.20 1.80 1.80 1.80 1.80

End Span Positive

Moments

Slab moment (kip-ft) 57 102 102 Moment Coef. 0.6 0.4 0.2 0.6 0.2 0.2 0.6

Distributed moments to

strips 34.2 22.8 20.4 61.2 20.4 20.4 61.2

Wall moment (kip-ft) 11 Total strip moment

(kip-ft) 45.2 43.2 61.2 40.8 61.2

Required 2 (in. )sA 1.52 1.45 2.06 1.37 2.06

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #4 10 #4 10 #4 10 #4 10 #4 2 (in. )sA provided 1.60 2.00 2.00* 2.00 2.00*

13-15

First Interior Negative

Moments B1 B2 B3

Slab moment (kip-ft) -76 -137 -137 Moment Coef. 0.75 0.25 0.125 0.75 0.125 0.125 0.75

Distributed moments to

strips -57 -19 -17.1 -102.3 -17.1 -17.1 -102.3

Wall moment (kip-ft) -15 Total strip moment

(kip-ft) -72 -36.1 -102.3 -34.2 -102.3

Required 2 (in. )sA 2.43 1.22 3.45 1.15 3.45

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #5 9 #4 11 #5 9 #4 11 #5 2 (in. )sA provided 2.48 1.80 3.41* 1.80 3.41*

Interior Positive Moments Slab moment (kip-ft) 38 69 69

Moment Coef. 0.6 0.4 0.2 0.6 0.2 0.2 0.6 Distributed moments to

strips 22.8 15.2 13.8 41.1 13.8 13.8 41.1

Wall moment (kip-ft) 7 Total strip moment (kip-

ft) 29.8 29 41.1 27.6 41.1

Required 2 (in. )sA 1.00 0.98 1.36 1.04 1.36

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 6 #4 9 #4 9 #4 9 #4 9 #4 2 (in. )sA provided 1.20 1.80 1.80 1.80 1.80

Interior Negative Moments C1 C2 C3

Slab moment (kip-ft) -71 -127 -127 Moment Coef. 0.75 0.25 0.125 0.75 0.125 0.125 -0.75

Distributed moments to

strips -53.2 -17.8 -15.9 -95.2 -15.9 -15.9 -95.2

Wall moment (kip-ft) -14 Total strip moment (kip-

ft) 67.2 -33.7 -95.2 -31.8 -95.2

Required 2 (in. )sA 2.26 1.14 3.21 1.07 3.21

Minimum 2 (in. )sA 0.95 1.56 1.56 1.56 1.56

Selected Steel 8 #5 9 #4 11 #5 9 #4 11 #5 2 (in. )sA provided 2.48 1.80 3.41 1.80 3.41

*As(provided) < As(required) is o.k. because adjacent positive moment regions are over-designed

and some moment redistribution can occur in ductile slabs.

(c) Check two-way shear and moment transfer at columns A2 and B2. Neglect

unbalanced moments about column line 2.

Column A2

13-16

The critical perimeter is at 2d from the face of the column, where d is the average depth. At all

exterior ends, the reinforcement is No. 4 bars and 6.75 in.avgd The shortest perimeter results

from the section shown in Fig. S13-7.2 and the perimeter dimensions are,

1

2

20 in. 2 23.38 in.

16 in. 22.75 in.

b d

b d

For moments about the z z axis,

2 23.38 5.69 23.38 27.86 in.

2 23.38 5.69 22.75 5.69ABy

Therefore, 7.86 in. and 15.52 in.AB CDc c

For moments about the w w axis,

22.7511.38 in.

2CB ADc c

As calculated in part (a), 61 kipsuV .

For slabs designed by the direct-design method, the moment transferred from the slab to the

column axis z-z is 0.3 oM , and using the moments calculated from part (b),

0.3 0.3 196 58.8 kip-ftoM (acting about the centroid of the shear perimeter).

Fig. S 13-7.2 Critical section- ColumnA2

From part (b), we found that the unbalanced moment due to the wall moments is 7 kip-ft and

assuming that the loads acts at 2 in. from the edge of the slab,

23.38 2.0 7.86 13.52 in. from centroid

The total moment to be transferred is,

13.52

58.8 7 50.9 kip-ft12

z zM

13-17

Note that the unbalanced moment about column line 2 w wM is neglected as stated in the

problem.

From Eq. (13-32), calculate the fraction of moment transferred by flexure,

1 2

1 10.60

2 21 1 23.38 22.75

3 3

f

b b

ACI Code Section 13.5.3.3 allows f to be increased to 1.0 if u cV V and the resulting is less

than 0.375 b within a width of 2 3c h centered in the column. From part (a), 0.65u cV V .

Therefore, take 1.0f and check the reinforcement required.

Width effective for flexure2 3 16 3 8 40 in.c h

Effective width 2 2 12 2tc c c c (second expression governs)

Effective width 16 2 16 48 in.

Assume that 0.95jd d , with 7.0 in.d

250.9 12,0001.70 in.

0.9 60,000 0.95 7.0sA

The steel provided is 9 No. 4 in a column strip width of 9 ft 108 in. , or roughly 13.5 on centers.

The bars within the 40 in. effective width can be used for the moment transfer. Place four

column-strip bars into this region and add 5 No. 4 bars, giving 21.8 in.sA in the effective width.

Recompute sA ,

1.80 60,000

0.71 in.0.85 4500 40

a

2 50.9 12,000(in. ) 1.70 kip-ft

0.710.9 60,000 7

2

sA

(steel chosen OK)

The reinforcement ratio is,

1.8

0.006440 7

sA

bd

and from Eq. (4-24),

0.85 0.825 4500 0.003

0.031160,000 0.003 0.00207

b

,

and thus, 0.375 0.0117 0.0064b and we can use 1.0f . As a result, it is not necessary to

transfer any of the moment about z-z axis by eccentric shear stresses.

Column B2

The critical perimeter is shown in Fig. S 13-7.3 and the centroidal axes pass through the centers

of the sides.

2 22.75 22.75 91 in.ob

13-18

The factored shear force transmitted to column B2 is,

22.75 22.75

246 9 1.15 9 10 1.15 10 101500 lbs 102 kips144

uV

Fig. S 13-7.3 Critical section- Column B2

From Eq.(13-25),

4

2 6.0 4

(does not govern)

From Eq. (13-26),

40s , for an interior slab-column connection

40 6.75

2 2 4.97 4.091

s

o

d

b

(does not govern)

Thus, using Eq. (13-24):

' 14 0.75 4 1 4500 91 6.75 124 kips > V1000c c o uV f b d

Since, 0.82 0.4u cV V no adjustment will be permitted in the ratio of unbalanced moment

resisted by eccentric shear.

The moment about x-x axis to be transferred comes from part (b) and is the difference between

the negative moments on the two sides of column B2, i.e. , 137 127 10 kip-ftu x xM .

From Eq. (13-32), calculate the fraction of moment transferred by flexure (x-x axis),

1 2

1 10.6

2 21 1 1

3 3

f

b b

The torsional moment of inertia can be calculated from Eq. (13-34),

23 3

1 1 122 2

12 12 2c

b d db bJ b d

Where 1 26.75 in. and 22.75 in.d b b Thus, 454150 in.cJ

By inspection, the reinforcement that is already in the slab is adequate for moment transfer.

13-19

From Eq. (13-30) and neglecting unbalanced moment about column line 2 (i.e. about axis y-y),

(shear transfer) 1 0.6 10 4 kip-ft 48,000 lb-in.u uM M

Then,

48,000 11.38

12.1 psi45,150

u

c

M c

J

So,

102,000 124,000

(max) 12.1 166 psi 12.1 psi=178 psi 202 psi91 6.75 91 6.75

u c

Thus, the shear is OK at this column.

13-8 Refer to the slab shown in Fig. P13-7 and the loadings and material strengths given

in Problem 13-7.

(a) Select slab thickness.

Problems 13-7 and 13-8 refer to the same flat-slab. As a result, the thickness of the slab was

chosen in part (a) of problem 13-7. Thus, use an 8 in. thick slab.

(b) Use the direct design method to compute moments, and then design the

reinforcement for the column and middle strips associated with column line

A.

Because there is no edge beams, 0f

Compute moments in the slab strip along column line A

A1 A2 A3

1 (ft) 18.0 18.0

(ft)n 16.67 16.67

2 (ft) 11.0 11.0

(ksf)uq 0.25 0.25 2

2 (kip-ft)

8

u no

qM 96 96

Moment Coef. -0.26 0.52 -0.70 -0.65 0.35 -0.65

Moments (kip-ft) -25 +50 -67 -62 +34 -62

Wall load (kip/ft) 0.48 0.48

Wall 2

8

wall no

qM

16.7 16.7

Moments from wall

(kip-ft)

-4.3 8.7 -11.7 -10.9 +5.8 -10.9

Compute moments in the slab strip along column line B

B1 B2 B3

1 (ft) 18.0 18.0

(ft)n 16.67 16.67

2 (ft) 20.0 20.0

(ksf)uq 0.25 0.25