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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