T T T I I I M M M E E E S S S A A A V V V I I I N N N G G G D D D E E E S S S I I I G G G N N N A A A I I I D D D S S S Columns Portland Cement Association Page 1 of 9 The following examples illustrate the design methods presented in the article “Timesaving Design Aids for Reinforced Concrete, Part 3: Columns and Walls,” by David A. Fanella, which appeared in the November 2001 edition of Structural Engineer magazine. Unless otherwise noted, all referenced table, figure, and equation numbers are from that article. The examples presented here are for columns. Examples for walls are available on our Web page: www.portcement.org/buildings . Example 1 In this example, an interior column at the 1 st floor level of a 7-story building is designed for the effects of gravity loads. Structural walls resist lateral loads, and the frame is nonsway. Materials •Concrete: normal weight (150 pcf), ¾-in. maximum aggregate, f ′ c = 5,000 psi •Mild reinforcing steel: Grade 60 (f y = 60,000 psi) Loads •Floor framing dead load = 80 psf •Superimposed dead loads = 30 psf •Live load = 100 psf (floor), 20 psf (roof) Building Data •Typical interior bay = 30 ft x 30 ft •Story height = 12 ft-0 in. The table below contains a summary of the axial loads due to gravity. The total factored load P u is computed in accordance with Sect. 9.2.1, and includes an estimate for the weight of the column. Live load reduction is determined from ASCE 7-98. Moments due to gravity loads are negligible. Floor DL (psf) LL (psf) Red. LL (psf) P u (kips) Cum. P u (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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TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Columns
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 1 o f 9
The following examples illustrate the
design methods presented in the article
“Timesaving Design Aids for Reinforced
Concrete, Part 3: Columns and Walls,” by
David A. Fanella, which appeared in the
November 2001 edition of StructuralEngineer magazine. Unless otherwise
noted, all referenced table, figure, and
equation numbers are from that article.
The examples presented here are for
columns.
Examples for walls are available on our
Web page: www.portcement.org/buildings .
Example 1
In this example, an interior column at the1st floor level of a 7-story building isdesigned for the effects of gravity loads.Structural walls resist lateral loads, andthe frame is nonsway.
• Floor framing dead load = 80 psf• Superimposed dead loads = 30 psf
• Live load = 100 psf (floor), 20 psf (roof)
Building Data
• Typical interior bay = 30 ft x 30 ft
• Story height = 12 ft-0 in.
The table below contains a summary of the
axial loads due to gravity. The totalfactored load Pu is computed in accordancewith Sect. 9.2.1, and includes an estimatefor the weight of the column. Live loadreduction is determined from ASCE 7-98.Moments due to gravity loads are negligible.
Floor DL (psf) LL (psf) Red. LL (psf) Pu (kips) Cum. Pu (kips)
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Columns
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 2 o f 9
Use Fig. 1 to determine a preliminary sizefor the tied column at the 1st floor level.
Assuming a reinforcement ratio ρg =
0.020, obtain Pu /Ag ≈ 3.0 ksi (f′c = 5 ksi).
Since Pu = 1,499 kips, the required Ag =
1,499/3.0 = 499.7 in.
2
Try a 22 x 22 in. column (Ag = 484 in.2)
with a reinforcement ratio ρg greater than0.020.
Check if slenderness effects need to beconsidered.
Since the column is part of a nonswayframe, slenderness effects can beneglected when the unsupported columnlength is less than or equal to 12h, whereh is the column dimension (Sect. 10.12.2).
12h = 12 x 22 = 264 in. = 22 ft > 12 ftstory height, which is greater than theunsupported length of the column.
Therefore, slenderness effects can beneglected.
Use Fig. 1 to determine the required areaof longitudinal reinforcement.
For a 22 x 22 in. column at the 1st floorlevel:
Pu /Ag = 1,499/484 = 3.10 ksi
From Fig. 1, required ρg = 0.026, or
As= 0.026 x 22 x 22 = 12.58 in.2
Try 8-No. 11 bars (As = 12.48 in.2)
Check Eq. (10-2) of ACI 318-99:
φPn(max) = 0.80φ[0.85f’c (Ag – Ast) + fy Ast]
φPn(max) = 1,542 kips > 1,499 kips O.K.
From Table 1, 5-No. 11 bars can beaccommodated on the face of a 22-in. widecolumn with normal lap splices and No. 4ties. In this case, only 3-No. 11 bars areprovided per face.
Use 8-No. 11 bars (ρ = 2.58%).
Determine required ties and spacing.
According to Sect. 7.10.5.1, No. 4 ties arerequired when No. 11 longitudinal bars areused.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Columns
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 8 o f 9
kipsft8
12652
18)80
d2
hT 1s
=
−
−
0
122
18)20
d2
hT
2s
=
−
−
kipsft5
124
2
18 02
d2
hC 3s
=
−
−
kipsft80
2542]808
M3
1nsi
=
+
+ ∑=
kipsft5380n =
Compare simplified interaction diagram tointeraction diagram generated from thePCA computer program PCACOL.
The comparison is shown on the next page.As can be seen from the figure, thecomparison between the exact (black line)and simplified (red line) interactiondiagrams is very good.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Beams and One-Way Slabs
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 1 o f 6
The following example illustrates the
design methods presented in the article
“Timesaving Design Aids for Reinforced
Concrete, Part 1: Beams and One-way
Slabs,” by David A. Fanella, which
appeared in the August 2001 edition of
Structural Engineer magazine. Unless
otherwise noted, all referenced table,
figure, and equation numbers are from
that article.
Example Building
Below is a partial plan of a typical floor in acast-in-place reinforced concrete building.The floor framing consists of wide-modulejoists and beams. In this example, thebeams are designed and detailed for the
combined effects of gravity and lateral(wind) loads according to ACI 318-99.
The coefficients of ACI Sect. 8.3 areutilized to compute the bending momentsand shear forces along the length of thebeam. From preliminary calculations, thebeams are assumed to be 36 x 20.5 in.Live load reduction is taken per ASCE 7-98.
psf3.7eighteam32.5
15014420.56== ×
×
Live load reduction per ASCE 7-98 Sect.4.8.1:
+=
TL
150.25o
From Table 4.2 of ASCE 7-98, KLL = liveload element factor = 2 for interior beams
AT = tributary area = 32.5 x 30 = 975 ft2
KLLAT = 2 x 975 = 1,950 ft2 > 400 ft2
o.59L1,950
150.25o =
+=
Since the beams support only one floor, Lshall not be less than 0.50Lo.
For example, at the exterior negativelocation in the end span, the required As =Mu/4d = 458.8/(4 x 18) = 6.37 in.2 EightNo. 8 bars provides 6.32 in.2 (say OK;less than 1% difference). From Table 1, theminimum number of No. 8 bars for a 36-in. wide beam is 5. Similarly, from Table 2,the maximum number of No. 8 bars is 16.Therefore, 8-No. 8 bars are adequate.
Design for Shear
Shear design is illustrated by determiningthe requirements at the exterior face ofthe interior column.
Vu = 1.4D + 1.7L = 149.3 kips (governs)Vu at d from face = 149.3 – 9.19(18/12)
= 135.5 kips
kips48.40cax. w =′=+φ φφ
kips9.7c w =′=φ φ
Required φVs = 135.5 – 69.7 = 65.8 kips
From Table 4, No. 5 U-stirrups at d/3
provides φVs = 94 kips > 65.8 kips.Length over which stirrups are required =[149.3 – (69.7/2)]/9.19 = 12.45 ft fromface of support.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Beams and One-Way Slabs
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 6 o f 6
Reinforcement Details
The figure below shows the reinforcementdetails for the beam. The bar lengths arecomputed from Fig. 8-3 of the PCApublication Simplified Design of Reinforced
Concrete Buildings of Moderate Size and
Height. In lieu of computing the barlengths in accordance with ACI Sects.12.10 through 12.12, 2-No. 5 bars areprovided within the center portion of thespan to account for any variations in
required bar lengths due to wind effects.For overall economy, it may be worthwhile toforego the No. 5 bars and determine theactual bar lengths per the above ACIsections.
Since the beams are part of the primarylateral-load-resisting system, ACI Sect.12.11.2 requires that at least one-fourth ofthe positive moment reinforcement extendinto the support and be anchored todevelop fy in tension at the face of thesupport.
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Two-Way Slabs
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 1 o f 7
The following example illustrates the
design methods presented in the article
“Timesaving Design Aids for Reinforced
Concrete, Part 2: Two-way Slabs,” by
David A. Fanella, which appeared in the
October 2001 edition of StructuralEngineer magazine. Unless otherwise
noted, all referenced table, figure, and
equation numbers are from that article.
Example Building
Below is a partial plan of a typical floor in acast-in-place reinforced concrete building. Inthis example, an interior strip of a flatplate floor system is designed and detailedfor the effects of gravity loads according
TTT III MMM EEE SSS AAA VVV III NNN GGG DDD EEE SSS III GGG NNN AAA III DDD SSS
Two-Way Slabs
P o r t l a n d C e m e n t A s s o c i a t i o n
P a g e 3 o f 7
Total panel moment Mo in end span:
kipsft82
8
1678485
8
wM
2n
o
.
..
=
××==
ll
Total panel moment Mo in interior span:
kipsft77
8
08485
8
wM
2n
o
.
..
=
××==
ll
For simplicity, use Mo = 282.2 ft-kips for allspans.
Division of the total panel moment Mo intonegative and positive moments, and thencolumn and middle strip moments, involvesthe direct application of the momentcoefficients in Table 1.
*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 ShearCheck slab shear and flexural strength atedge column due to direct shear andunbalanced moment transfer.
Check slab reinforcement at exterior columnfor moment transfer between slab andcolumn.
Portion of total unbalanced momenttransferred by flexure = γfMu