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```					T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Columns

Portland Cement Association
Page    1   of   9

The following examples illustrate the           Materials
design methods presented in the article         • Concrete: normal weight (150 pcf), ¾-in.
“Timesaving Design Aids for Reinforced            maximum aggregate, f′c = 5,000 psi
Concrete, Part 3: Columns and Walls,” by        • Mild reinforcing steel: Grade 60 (fy =
David A. Fanella, which appeared in the           60,000 psi)
November 2001 edition of Structural
noted, all referenced table, figure, and        • Floor framing dead load = 80 psf
equation numbers are from that article.         • Superimposed dead loads = 30 psf
The examples presented here are for             • Live load = 100 psf (floor), 20 psf (roof)
columns.
Building Data
Examples for walls are available on our         • Typical interior bay = 30 ft x 30 ft
Web page: www.portcement.org/buildings.         • Story height = 12 ft-0 in.

Example 1                                       The table below contains a summary of the
axial loads due to gravity. The total
In this example, an interior column at the      factored load Pu is computed in accordance
1st floor level of a 7-story building is        with Sect. 9.2.1, and includes an estimate
designed for the effects of gravity loads.      for the weight of the column. Live load
Structural walls resist lateral loads, and      reduction is determined from ASCE 7-98.
the frame is nonsway.                           Moments due to gravity loads are negligible.

Floor   DL (psf)    LL (psf)   Red. LL (psf)    Pu (kips) Cum. Pu (kips)
7        80         20          20.0            142           142
6       120        100          50.0            238          380
5       120        100           42.7           227          607
4       120        100          40.0            223          830
3       120        100          40.0            223        1,053
2       120        100          40.0            223        1,276
1       120        100          40.0            223        1,499
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Columns

Portland Cement Association
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Use Fig. 1 to determine a preliminary size      For a 22 x 22 in. column at the 1st floor
for the tied column at the 1st floor level.     level:

Assuming a reinforcement ratio ρg =             Pu /Ag = 1,499/484 = 3.10 ksi
0.020, obtain Pu /Ag ≈ 3.0 ksi (f′c = 5 ksi).
From Fig. 1, required ρg = 0.026, or
Since Pu = 1,499 kips, the required Ag =
1,499/3.0 = 499.7 in.2                          As = 0.026 x 22 x 22 = 12.58 in.2

Try a 22 x 22 in. column (Ag = 484 in.2)        Try 8-No. 11 bars (As = 12.48 in.2)
with a reinforcement ratio ρg greater than
0.020.                                          Check Eq. (10-2) of ACI 318-99:

Check if slenderness effects need to be         φPn(max) = 0.80φ[0.85f’c (Ag – Ast) + fy Ast]
considered.
φPn(max) = 1,542 kips > 1,499 kips O.K.
Since the column is part of a nonsway
frame, slenderness effects can be               From Table 1, 5-No. 11 bars can be
neglected when the unsupported column           accommodated on the face of a 22-in. wide
length is less than or equal to 12h, where      column with normal lap splices and No. 4
h is the column dimension (Sect. 10.12.2).      ties. In this case, only 3-No. 11 bars are
provided per face.
12h = 12 x 22 = 264 in. = 22 ft > 12 ft
story height, which is greater than the         Use 8-No. 11 bars (ρ = 2.58%).
unsupported length of the column.
Therefore, slenderness effects can be           Determine required ties and spacing.
neglected.
According to Sect. 7.10.5.1, No. 4 ties are
Use Fig. 1 to determine the required area       required when No. 11 longitudinal bars are
of longitudinal reinforcement.                  used.
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Columns

Portland Cement Association
Page   3   of   9

22 − 2 1.5 + 0.5 +
According to Sect. 7.10.5.2, spacing of                                         1.41 
                  
ties shall not exceed the least of:
Clear space =                     2  − 1.41
2
16 long. bar diameters = 16 x 1.41                        = 6.885 in.
16 long. bar diameters = 22.6 in.
Since the clear space between longitudinal
48 tie bar diameters = 48 x 0.5               bars > 6 in., cross-ties are required per
48 tie bar diameters = 24 in.                 Sect. 7.10.5.3.

Least column dimension = 22 in. (governs)     Reinforcement details are shown below.

Check clear spacing of longitudinal bars:     See Sect. 7.8 for additional          special
reinforcement details for columns.

8-No. 11

22″

No. 4 ties @ 22″

22″
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Columns

Portland Cement Association
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Example 2                                      • Point 2 (fs1 = 0)

In this example, a simplified interaction         Layer 1:
diagram is constructed for an 18″ x 18″                          d
1 − C 2 1 = 1 − 1 (1) = 0
tied column reinforced with 8-No. 9 Grade                        d1
60 bars (ρg = 8/182 = 0.0247). Concrete
compressive strength = 4 ksi.                     Layer 2:
1 − C2 2 = 1 − 1 
d            9.00 
        = 0.42
Use Fig. 3 to determine the 5 points on                          d1         15.56 
the interaction diagram.
Layer 3:
• Point 1: Pure compression
1 − C2 3 = 1 − 1 
d             2.44 
        = 0.84
d1         15.56 
′
φPn(max) = 0.80 φA g [ 0.85 fc
′
+ ρ g ( fy − 0.85 fc )]            Since 1 – C2 (d3 /d1) > 0.69, the steel in
layer 3 has yielded.
= 0.56 × 18 2 [( 0.85 × 4 )
+ 0.0247 (60 − (0.85 × 4 ))]        Therefore, set 1 – C2 (d3 /d1) = 0.69 to
= 871 kips                            ensure that the stress in the bars in
layer 3 is equal to 60 ksi.
1.5″ (typ.)
d3 = 2.44″

18″
No. 3 tie
d2 = 9.00″

3-No. 9
d1 = 15.56″

18″

2-No. 9

3-No. 9
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             n                    di         • Point 3 (fs1 = -0.5fy)
φPn = φ C 1d 1b + 87 ∑ A si  1 − C 2
                 
            i=1                   d1 

Layer 1:
d
= 0.70 {(2.89 × 15.56 × 18)                              1 − C 2 1 = 1 − 1.34 (1) = −0.34
d1
+ 87[(3 × 0) + (2 × 0.42)
+ (3 × 0.69)]}                                       Layer 2:
= 0.70 (809 .4 + 253 .2)                                        d
1 − C 2 2 = 1 − 1.34     9.00  = 0.23
       
= 744 kips                                                       d1               15.56 

Layer 3:
              βd 
φMn = φ 0.5C 1d 1b h − 1 1 
1 − C 2 3 = 1 − 1.34 
                                                 d                2.44 
               C2                                                              = 0.79
d1            15.56 
           d i  h      
 − di   / 12
n
+ 87 ∑ A si  1 − C 2
                       
i=1                d 1  2
                    Use 0.69

= 0.70{[( 0.5 × 2.89 × 15.56 × 18)                                          n                 di  
φPn = φ C 1d 1b + 87 ∑ A si  1 − C 2
              
            i=1                d1 

0.85 × 15.56  
×  18 −
                   
          1.00     
+ 87[(3 × 0)(9 − 15.56)                              = 0.70 {(2.15 × 15.56 × 18)
+ (2 × 0.42)(9 − 9)                                    + 87[(3 × -0.34 ) + (2 × 0.23)
+ (3 × 0.69)(9 − 2.44 )]} / 12                         + (3 × 0.69)]}
= 0.70(602 .2 + 131 .4 )
= 0.70 (1, 932 .1 + 1, 181 .4 ) / 12                        = 514 kips
= 182 ft - kips
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              βd                              Layer 3:
φMn = φ 0.5C 1d 1b h − 1 1 

               C2                              1 − C 2 3 = 1 − 1.69 
d

2.44 
 = 0.74
d1            15.56 
           d i  h      
 − di   / 12
n
+ 87 ∑ A si  1 − C 2
                                    Use 0.69
i=1                d 1  2
       
             n              d 
φPn = φ C 1d 1b + 87 ∑ A si  1 − C 2 i  
= 0.70{[( 0.5 × 2.15 × 15.56 × 18)                                              
        d1 

            i=1
0.85 × 15.56  
×  18 −
                   
          1.34                            = 0.70 {(1.71 × 15.56 × 18)
+ 87[(3 × -0.34 )(9 − 15.56)                     + 87[(3 × -0.69) + (2 × 0.02)
+ (2 × 0.23)(9 − 9)                              + (3 × 0.69)]}
+ (3 × 0.69)(9 − 2.44 )]} / 12                 = 0.70 ( 478 .9 + 3.5) = 338 kips

= 0.70 (2, 447 .8 + 1, 763 .5) / 12                                       βd 
φMn = φ 0.5C 1 d 1b h − 1 1 

= 246 ft - kips
                C2 
        d  h         
+ 87 ∑ A si  1 − C 2 i  − d i   / 12
n
• Point 4 (fs1 = -fy)                                                                            
i=1             d 1  2
       
Layer 1:
d
1 − C 2 1 = 1 − 1.69 (1) = −0.69                         = 0.70{[( 0.5 × 1.71 × 15.56 × 18)
d1                                                                0.85 × 15.56  
×  18 −
                   
          1.69     
Layer 2:                                                          + 87[(3 × -0.69)(9 − 15.56)
1 − C 2 2 = 1 − 1.69 
d               9.00 
        = 0.02                           + (2 × 0.02)(9 − 9)
d1            15.56 
+ (3 × 0.69)(9 − 2.44 )]} / 12
= 0.70 (2, 436 .3 + 2, 362 .8) / 12
= 280 ft - kips
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• Point 5: Pure bending                        fs2 = E s ε s2
= 29, 000 × ( −0.0038 ) = −108 .8 ksi
Use iterative procedure to determine            > -60 ksi, use fs2 = −60 ksi
φMn.

Try c = 4.0 in.                             Ts2 = As2 fs2 = 2 × ( −60) = −120 kips

c − d1 
ε s 1 = 0.003 

c − d1 
                    ε s 3 = 0.003         
 c                                          c 
4 − 2.44 
= 0.003 

4 − 15.56 
                       = 0.003           
     4                                          4    
= −0.0087                                   = 0.0012

fs 1 = E s ε s 1                            fs 3 = E s ε s 3
= 29, 000 × (−0.0087 ) = −251 .4 ksi        = 29, 000 × 0.0012 = 33 .9 ksi
> - 60 ksi, use fs1 = −60 ksi
C s3 = A s 3 fs 3 = 3 × 33 .9 = 102 kips
Ts1 = As 1 fs 1 = 3 × (−60) = −180 kips
′
C c = 0.85 fc ab
c − d1                         = 0.85 × 4 × (0.85 × 4 ) × 18
ε s2 = 0.003 
        
 c                              = 208 kips
4 − 9
= 0.003 
        
 4                           Total T = (-180) + (-120) = -300 kips
= −0.0038
Total C = 102 + 208 = 310 kips

Since T ≈ C, use c = 4.0 in.
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Mns 1 = Ts 1  − d 1 
h                                Compare simplified interaction diagram to
        
2                                interaction diagram generated from the
PCA computer program PCACOL.
= (−180 ) − 15.56  / 12
18
       
2      
= 98.4 ft - kips                          The comparison is shown on the next page.
As can be seen from the figure, the
comparison between the exact (black line)
Mns 2 = Ts2  − d2 
h
                                  and simplified (red line) interaction
2                                 diagrams is very good.
= (−120 ) − 9  / 12
18
      
2     
=0

Mns 3 = C s 3  − d 3 
h
       
2      
= 102  − 2.44  / 12
18
         
2        
= 55.8 ft - kips

3
Mn = 0.5C c (h − a) + ∑ Mnsi
i=1
= [ 0.5 × 208 × (18 − 3.4 )] / 12 + 154 .2
= 280 .7 ft - kips

φMn = 0.9 × 280 .7 = 253 ft - kips
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T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Beams and One-Way Slabs

Portland Cement Association
Page   1   of   6

The following example illustrates the         Example Building
design methods presented in the article
“Timesaving Design Aids for Reinforced        Below is a partial plan of a typical floor in a
Concrete, Part 1: Beams and One-way           cast-in-place reinforced concrete building.
Slabs,” by David A. Fanella, which            The floor framing consists of wide-module
appeared in the August 2001 edition of        joists and beams. In this example, the
Structural Engineer magazine. Unless          beams are designed and detailed for the
otherwise noted, all referenced table,        combined effects of gravity and lateral
figure, and equation numbers are from         (wind) loads according to ACI 318-99.
that article.

30′-0″                30′-0″                 30′-0″
32′-6″
32′-6″

18″x18″ (typ.)       24″x 24″ (typ.)
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T MESAV NG DES GN A DS
Beams and One-Way Slabs

Portland Cement Association
Page   2   of   6

Design Data                                                       15       
L = L o  0.25 +



           K LL A T   
Materials
• Concrete: normal weight (150 pcf), ¾ -
From Table 4.2 of ASCE 7-98, KLL = live
in. maximum aggregate, f′c = 4,000 psi     load element factor = 2 for interior beams
• Mild reinforcing steel: Grade 60 (fy =
60,000 psi)                                AT = tributary area = 32.5 x 30 = 975 ft2

Loads                                         KLLAT = 2 x 975 = 1,950 ft2 > 400 ft2
• Joists (16 + 4 x 6 + 66) = 76.6 psf
½
• Live load = 100 psf                        L = L o  0.25 +

 = 0.59L
           1,950 

o
• Wind loads: per ASCE 7-98
Since the beams support only one floor, L
Gravity Load Analysis                         shall not be less than 0.50Lo.

The coefficients of ACI Sect. 8.3 are         Therefore, L = 0.59 x 100 = 59 psf.
utilized to compute the bending moments
and shear forces along the length of the      Total factored load wu:
beam. From preliminary calculations, the
beams are assumed to be 36 x 20.5 in.         wu = 1.4(76.6 + 23.7 + 30) + 1.7(59)
Live load reduction is taken per ASCE 7-         = 282.7 psf
98.                                              = 282.7 x 32.5/1,000 = 9.19 klf

36 × 20.5
× 150                 Factored reactions per ACI Sect. 8.3:
Beam weight =   144           = 23.7 psf
32.5
Neg. Mu at ext. support = wuln2/16
Live load reduction per ASCE 7-98 Sect.                               = 9.19 x 28.252/16
4.8.1:                                                                = 458.4 ft-kips
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Beams and One-Way Slabs

Portland Cement Association
Page   3    of   6

Pos. Mu at end span = wuln2/14                Design for Flexure
= 9.19 x 28.252/14
= 523.9 ft-kips           Sizing the cross-section

Neg. Mu at int. col.    = wuln2/10*           Per ACI Table 9.5(a), minimum thickness =
= 9.19 x 28.1252/10   l/18.5 = (30 x 12)/18.5 = 19.5 in.
= 726.9 ft-kips
Since joists are 20.5 in. deep, use 20.5-in.
Pos. Mu at int. span = wuln2/16               depth for the beams for formwork economy.
= 9.19 x 282/16
= 450.3 ft-kips          With d = 20.5 – 2.5 = 18 in., solve Eq. (2)
for b using maximum Mu along span (note:
gravity moment combination governs):
Vu at exterior col.     = wuln/2
= 9.19 x 28.25/2
bd2 = 20Mu
= 129.8 kips
b = 20 x 726.9/182 = 44.9 in. > 36 in.
This implies that using a 36-in. wide beam,
Vu at interior col.     = 1.15wuln/2          ρ will be greater than 0.5ρmax.
= 1.15 x 129.8
= 149.3 kips          Check minimum width based on ρ = ρmax
(see Chapter 3 of the PCA publication
Wind Load Analysis                            Simplified Design of Reinforced Concrete
Buildings of Moderate Size and Height for
As noted above, wind forces are computed      derivation):
per ASCE 7-98. Calculations yield the
following reactions:                          bd2 = 13Mu
b = 13 x 726.9/182 = 29.2 in. < 36 in.
Mw = ± 90.3 ft-kips                           This implies that ρ will be less than ρmax.
Vw = 6.0 kips

*Average of adjacent clear spans              Use 36 x 20.5 in. beam.
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Beams and One-Way Slabs

Portland Cement Association
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Required Reinforcement

Beam moments along the span are
summarized in the table below.

End Span     Interior span
(ft-kips)      (ft-kips)
Exterior negative         -211.2          
Interior negative       -335.0         -301.8
Exterior negative         -95.6           
Live (L)            Positive                  109.3          94.0
Interior negative         -151.7       -136.7
Exterior negative        ± 90.3           
Wind (W)             Positive                                 
Interior negative        ± 90.3        ± 90.3
Exterior negative     -458.4           
1          1.4D + 1.7L        Positive                523.9         450.3
Interior negative      -726.9        -654.9
Exterior negative      -228.5          
-458.8
2    0.75(1.4D + 1.7L + 1.7W) Positive                392.8          337.7
Interior negative     -660.3          -376.1
-430.0         -606.3
Exterior negative        -72.7          
-307.5
3         0.9D + 1.3W         Positive                  217.3        186.8
Interior negative       -418.9        -154.2
-184.1       -389.0
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Beams and One-Way Slabs

Portland Cement Association
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Eq. (6) is used to determine the required            ensure that the number of bars chosen
reinforcement, which is summarized in the            conform to the code requirements for cover
table below. Tables 1 and 2 are utilized to          and spacing.

Location                    Mu (ft-kips)     As (in.2)*   Reinforcement
Exterior negative            -458.8          6.37          8-No. 8
End Span     Positive                      523.9          7.28          10-No. 8
Interior negative            -726.9         10.10          13-No. 8
Interior Span Positive                       450.3          6.25          8-No. 8
* A s = M u /4d

Min. A s = 3 4,000 × 36 × 18/60,000 = 2.05 in. 2

= 200 × 36 × 18/60,000 = 2.16 in. 2 (governs)

Max. A s = 0.0214 × 36 × 18 = 13.87 in. 2

For example, at the exterior negative                Vu = 1.4D + 1.7L = 149.3 kips (governs)
location in the end span, the required As =          Vu at d from face = 149.3 – 9.19(18/12)
Mu/4d = 458.8/(4 x 18) = 6.37 in.2 Eight                                 = 135.5 kips
No. 8 bars provides 6.32 in.2 (say OK;                                         ′
Max. (φVc + φVs ) = φ10 fc b w d = 348.4 kips
less than 1% difference). From Table 1, the
′
φVc = φ2 fc b w d = 69.7 kips
minimum number of No. 8 bars for a 36-
in. wide beam is 5. Similarly, from Table 2,         Required φVs = 135.5 – 69.7 = 65.8 kips
the maximum number of No. 8 bars is 16.
Therefore, 8-No. 8 bars are adequate.                From Table 4, No. 5 U-stirrups at d/3
provides φVs = 94 kips > 65.8 kips.
Design for Shear                                     Length over which stirrups are required =
[149.3 – (69.7/2)]/9.19 = 12.45 ft from
face of support.
Shear design is illustrated by determining
the requirements at the exterior face of
Use No. 5 stirrups @ 6 in.
the interior column.
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T MESAV NG DES GN A DS
Beams and One-Way Slabs

Portland Cement Association
Page    6   of   6

Reinforcement Details                                         required bar lengths due to wind effects.
For overall economy, it may be worthwhile to
The figure below shows the reinforcement                      forego the No. 5 bars and determine the
details for the beam. The bar lengths are                     actual bar lengths per the above ACI
computed from Fig. 8-3 of the PCA                             sections.
publication Simplified Design of Reinforced
Concrete Buildings of Moderate Size and                       Since the beams are part of the primary
Height. In lieu of computing the bar                          lateral-load-resisting system, ACI Sect.
lengths in accordance with ACI Sects.                         12.11.2 requires that at least one-fourth of
12.10 through 12.12, 2-No. 5 bars are                         the positive moment reinforcement extend
provided within the center portion of the                     into the support and be anchored to
span to account for any variations in                         develop fy in tension at the face of the
support.

1′-6″    7′-1″                                      9′-6″      2′-0″

A
8-No. 8            2-No. 5                13-No. 8              Class A tension splice

20.5″
Standard
hook (typ.)                                                                                       5-No. 8
3-No. 8          7-No. 8              A    3′-6″             3′-6″    3-No. 8
15-No. 5 @ 9″                    26-No. 5 @ 6″
2″                                                                   2″
6″                               30′-0″

13-No. 8                                                                      4½″

No. 5 U-stirrups                                                              16″

10-No. 8

1½″ clear (typ.)                  36″
Section A-A
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T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   1   of   7

The following example illustrates the                Example Building
design methods presented in the article
“Timesaving Design Aids for Reinforced               Below is a partial plan of a typical floor in a
Concrete, Part 2: Two-way Slabs,” by                 cast-in-place reinforced concrete building. In
David A. Fanella, which appeared in the              this example, an interior strip of a flat
October 2001 edition of Structural                   plate floor system is designed and detailed
Engineer magazine. Unless otherwise                  for the effects of gravity loads according
noted, all referenced table, figure, and             to ACI 318-99.
equation numbers are from that article.

20′-0″              20′-0″            20′-0″
24′-0″

Design strip
24′-0″

20″x 20″ (typ.)                       24″x 24″ (typ.)
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   2   of   7

Design Data                                   From Fig. 2, d/c1 ≈ 0.39

Materials                                     d = 0.39 x 20 = 7.80 in.
• Concrete: normal weight (150 pcf), ¾-
in. maximum aggregate, f′c = 4,000 psi     h = 7.80 + 1.25 = 9.05 in.
• Mild reinforcing steel: Grade 60 (fy =
60,000 psi)                                Try preliminary h = 9.0 in.

• Live load = 50 psf                         Use Fig. 3 to determine if the Direct Design
Method of ACI Sect. 13.6 can be utilized to
Minimum Slab Thickness                        compute the bending moments due to the
Longest clear span ln = 24 – (20/12) =
• 3 continuous spans in one direction,
22.33 ft
more than 3 in the other O.K.
From Fig. 1, minimum thickness h per ACI      • Rectangular panels with long-to-short
span ratio = 24/20 = 1.2 < 2 O.K.
Table 9.5(c) = ln/30 = 8.9 in.
• Successive span lengths in each
direction are equal O.K.
Use Fig. 2 to determine h based on shear
requirements at edge column assuming a        • No offset columns O.K.
9 in. slab:                                   • L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.
• Slab system has no beams N.A.
wu = 1.4(112.5 + 30) + 1.7(50) = 284.5 psf
Since all requirements are satisfied, the
A = 24 x [(20 + 1.67)/2] = 260 ft2            Direct Design Method can be used.

A/c12 = 260/1.672 = 93.6
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   3   of   7

Total panel moment Mo in end span:                      For simplicity, use Mo = 282.2 ft-kips for all
spans.
w u l 2 l 2 0.285 × 24 × 18.167 2
Mo =           n =
Division of the total panel moment Mo into
8                8
negative and positive moments, and then
= 282 .2 ft - kips                                  column and middle strip moments, involves
the direct application of the moment
Total panel moment Mo in interior span:                 coefficients in Table 1.

w u l 2 l 2 0.285 × 24 × 18.0 2
Mo =           n =
8               8
= 277 .0 ft - kips

Slab                       End Spans                      Int. Span
Moments
(ft-kips)     Ext. neg.      Positive        Int. neg.    Positive

Total
73.4          146.7           197.5         98.8
Moment
Column
73.4           87.5           149.6         59.3
Strip
Middle
0            59.3           48.0          39.5
Strip
Note: All negative moments are at face of support.
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   4     of   7

Required slab reinforcement.

Mu           b*       d**          As†     Min. As‡                   +
Span Location                                                                 Reinforcement
(ft-kips)      (in.)     (in.)       (in.2)    (in.2)
End Span
Ext. neg.      73.4         120       7.75       2.37        1.94        12-No. 4
Column
Positive       87.5         120       7.75       2.82        1.94        15-No. 4
Strip
Int. Neg.     149.6         120       7.75       4.83        1.94        25-No. 4
Ext. neg.       0.0         168       7.75        ---        2.72        14-No. 4
Middle
Positive       59.3         168       7.75       1.91        2.72        14-No. 4
Strip
Int. Neg.      48.0         168       7.75       1.55        2.72        14-No. 4
Interior Span
Column
Positive       59.3         120       7.75        1.91       1.94        10-No. 4
Strip
Middle
Positive       39.5         168       7.75        1.27       2.72        14-No. 4
Strip
*Column strip width b = (20 x 12)/2 = 120 in.
*Middle strip width b = (24 x 12) – 120 = 168 in.
**Use average d = 9 – 1.25 = 7.75 in.
†As = Mu /4d where Mu is in ft-kips and d is in inches
‡Min. As = 0.0018bh = 0.0162b; Max. s = 2h = 18 in. or 18 in. (Sect. 13.3.2)
+
For maximum spacing: 120/18 = 6.7 spaces, say 8 bars
168/18 = 9.3 spaces, say 11 bars

Design for Shear                                          Check slab reinforcement at exterior column
for moment transfer between slab and
Check slab shear and flexural strength at                 column.
edge column due to direct shear and
unbalanced moment transfer.                               Portion of total unbalanced moment
transferred by flexure = γfMu
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   5   of   7

b1 = 20 + (7.75/2) = 23.875 in.                                     Provide the required 8-No. 4 bars by
concentrating 8 of the column strip bars
b2 = 20 + 7.75 = 27.75 in.                                          (12-No. 4) within the 47 in. slab width over
the column.
b1 /b2 = 0.86
Check bar spacing:
From Fig. 5, γf = 0.62*
For 8-No. 4 within 47 in. width: 47/8 =
γfMu = 0.62 x 73.4 = 45.5 ft-kips                                   5.9 in. < 18 in. O.K.

Required As = 45.5/(4 x 7.75) = 1.47 in.2                           For 4-No. 4 within 120 – 47 = 73 in. width:
73/4 = 18.25 in. > 18 in.
Number of No. 4 bars = 1.47/0.2 = 7.4,
say 8 bars                                                          Add 1 additional bar on each side of the
47 in. strip; the spacing becomes 73/6 =
Must provide 8-No. 4 bars within an                                 12.2 in. < 18 in. O.K.
effective slab width = 3h + c2 = (3 x 9) +
20 = 47 in.                                                         Reinforcement details at this location are
shown in the figure on the next page (see
Fig. 6).

∗

∗
The provisions of Sect. 13.5.3.3 may be utilized; however, they are not in this example.
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page    6   of   7

5′-6″

3-No. 4
Column strip – 10′-0″

3′-11″

8-No. 4
1′-8″

3-No. 4

Check the combined shear stress at the                   moment transferred by eccentricity of
inside face of the critical transfer section.            shear must be 0.3Mo = 0.3 x 282.2 =
84.7 ft-kips (Sect. 13.6.3.6).
V  γ M
vu = u + v u
Ac   J/c                                             γv = 1 – γf = 1 – 0.62 = 0.38

Factored shear force at edge column:                     c2 /c1 = 1.0

Vu = 0.285[(24 x 10.83) – (1.99 x 2.31)]                 c1 /d = 20/7.75 = 2.58
Vu = 72.8 kips
Interpolating from Table 7, f1 = 9.74 and
When the end span moments are                            f2 = 5.53
determined from the Direct Design
Method, the fraction of unbalanced                       Ac = f1 d2 = 9.74 x 7.752 = 585.0 in.2
T II M E S A V II N G D E S II G N A II D S
T MESAV NG DES GN A DS
Two-Way Slabs

Portland Cement Association
Page   7   of   7

J/c = 2f2d3 = 2 x 5.53 x 7.753 = 5,148 in.3                        bo /d = [(2 x 23.875) + 27.75]/7.75 = 9.74
βc = 1
72, 800 0.38 × 84.7 × 12, 000
vu =           +
585 .0        5, 148                                       φvc = 215 psi > vu = 199.4 psi OK
v u = 124 .4 + 75 .0 = 199 .4 psi
Reinforcement Details
Determine allowable shear stress φvc from
The figures below show the reinforcement
Fig. 4b:
details for the column and middle strips.
The bar lengths are determined from
bo /d = (2b1 + b2)/d
Fig. 13.3.8 of ACI 318-99.
1′-8″ 5′-6″                          5′-6″      2′-0″ 5′-6″
3′-8″        3′-8″
12-No. 4

14-No. 4                                      13-No. 4

Standard
hook (typ.)
6″       2-No. 4        13-No. 4                   Class A tension splice

Column strip
20′-0″

1′-8″ 4′-0″                             4′-0″ 2′-0″ 4′-0″

14-No. 4                                       14-No. 4
Standard
hook (typ.)                                                       6″

7-No. 4
6″       7-No. 4        7-No. 4                          7-No. 4
3′-0″
Middle strip
20′-0″

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