time saving design

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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.

Materials

• Concrete: normal weight (150 pcf), ¾-in.

maximum aggregate, f′c = 5,000 psi

• Mild reinforcing steel: Grade 60 (fy =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 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)

7 80 20 20.0 142 142

6 120 100 50.0 238 3805 120 100 42.7 227 607

4 120 100 40.0 223 830

3 120 100 40.0 223 1,0532 120 100 40.0 223 1,276

1 120 100 40.0 223 1,499




Columns


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.




Columns


P a g e 3 o f 9

According to Sect. 7.10.5.2, spacing ofties shall not exceed the least of:

16 long. bar diameters = 16 x 1.4116 long. bar diameters = 22.6 in.

48 tie bar diameters = 48 x 0.548 tie bar diameters = 24 in.

Least column dimension = 22 in. (governs)

Check clear spacing of longitudinal bars:

in. 85

42

2

4152

spacelear

=

−

+

=

Since the clear space between longitudinalbars > 6 in., cross-ties are required perSect. 7.10.5.3.

Reinforcement details are shown below.

See Sect. 7.8 for additional specialreinforcement details for columns.

22″

22″

8-No. 11

No. 4 ties @ 22″




Columns


P a g e 4 o f 9

Example 2

In this example, a simplified interaction

diagram is constructed for an 18″ x 18″ tied column reinforced with 8-No. 9 Grade

60 bars (ρg = 8/182 = 0.0247). Concretecompressive strength = 4 ksi.

Use Fig. 3 to determine the 5 points onthe interaction diagram.

• Point 1: Pure compression

kips 71

))]50247 )5(86

)]5

f50

2

c

cmax)

=

××

′

′

• Point 2 (fs1 = 0)

Layer 1:

0d

dC

1

12 =

Layer 2:

4256500

ddC

122 =−

Layer 3:

84565

44

d

dC

1

32 =

−

Since 1 – C2 (d3 /d1) > 0.69, the steel inlayer 3 has yielded.

Therefore, set 1 – C2 (d3 /d1) = 0.69 toensure that the stress in the bars inlayer 3 is equal to 60 ksi.

d 3

=

2 .

4 4 ″

18″ 1 .

5 ″

( t y p .

)

d 2

=

9 .

0 0 ″

d 1

=

1 5 .

5 6 ″

1 8 ″

No. 3 tie

3-No. 9

2-No. 9

3-No. 9




Columns


P a g e 5 o f 9

kips44)53090

)]}9

)22)[(3

)8659(0

d

dC7

n

1 1

i2i

=+

×

×

×

−∑

=

{

kipsft82

1281320

12]}4(9

)(22

15.56))(9[(3

00

565518

)8659(0

122

h

d

dC7

C

dh

in

1 1

i2i

2

11

=+

−

−

−

×

−

×

−

−

β−

∑=

• Point 3 (fs1 = -0.5fy)

Layer 1:

344d

dC

1

12 −

Layer 2:

2356500 4

ddC

122 =−

Layer 3:

79565

44 4

d

dC

1

32 =

−

Use 0.69

kips14

)31020

)]}9

)324[(3

)8655(0

d

dC7

n

1 1

i2i

=

+

×

×

×

−∑

=




Columns


P a g e 7 o f 9

• Point 5: Pure bending

Use iterative procedure to determine

φMn.

Try c = 4.0 in.

0087

4

565 03

cd 03 11

−

−

=

−

=

kips800

ksi0fseksi0

ksi51087009

E

11

s1

1

−

−

−

ε

0038

4

9 03

c

d 03 1

2

−

−

=

−

=

kips 200

ksi0fseksi0

ksi08038009

E

22

s2

2

−

−

−

ε

00124

44 03

c

d 03

13

=

−

=

−

=

kips 023

ksi3012009

E

33

3

=

=

ε

kips 08

1855

ab5 c

=

×

′

Total T = (-180) + (-120) = -300 kips

Total C = 102 + 208 = 310 kips

Since T ≈ C, use c = 4.0 in.




Columns


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.




Columns


P a g e 9 o f 9




Beams and One-Way Slabs


P a g e 1 o f 6

The following example illustrates the



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.

30′-0″ 30′-0″ 30′-0″

3 2 ′ - 6 ″

3 2 ′ - 6 ″

18″x18″ (typ.) 24″x 24″ (typ.)






P a g e 2 o f 6

Design Data

Materials

• Concrete: normal weight (150 pcf), ¾ -

in. maximum aggregate, f′c = 4,000 psi

• Mild reinforcing steel: Grade 60 (fy =60,000 psi)

Loads

• Joists (16 + 4½ x 6 + 66) = 76.6 psf

• Superimposed dead loads = 30 psf

• Live load = 100 psf

• Wind loads: per ASCE 7-98

Gravity Load Analysis

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.

Therefore, L = 0.59 x 100 = 59 psf.

Total factored load wu:

wu = 1.4(76.6 + 23.7 + 30) + 1.7(59)= 282.7 psf= 282.7 x 32.5/1,000 = 9.19 klf

Factored reactions per ACI Sect. 8.3:

Neg. Mu at ext. support = wuln2/16

= 9.19 x 28.252/16= 458.4 ft-kips






P a g e 3 o f 6

Pos. Mu at end span = wuln2/14

= 9.19 x 28.252/14= 523.9 ft-kips

Neg. Mu at int. col. = wuln2/10*

= 9.19 x 28.1252/10= 726.9 ft-kips

Pos. Mu at int. span = wuln2/16

= 9.19 x 282/16= 450.3 ft-kips

Vu at exterior col. = wuln/2= 9.19 x 28.25/2= 129.8 kips

Vu at interior col. = 1.15wuln/2= 1.15 x 129.8= 149.3 kips

Wind Load Analysis

As noted above, wind forces are computedper ASCE 7-98. Calculations yield thefollowing reactions:

Mw = ± 90.3 ft-kipsVw = 6.0 kips

*Average of adjacent clear spans

Design for Flexure

Sizing the cross-section

Per ACI Table 9.5(a), minimum thickness =

l/18.5 = (30 x 12)/18.5 = 19.5 in.

Since joists are 20.5 in. deep, use 20.5-in.depth for the beams for formwork economy.

With d = 20.5 – 2.5 = 18 in., solve Eq. (2)for b using maximum Mu along span (note:gravity moment combination governs):

bd2 = 20Mu b = 20 x 726.9/182 = 44.9 in. > 36 in.

This implies that using a 36-in. wide beam,ρ will be greater than 0.5ρmax.

Check minimum width based on ρ = ρmax (see Chapter 3 of the PCA publicationSimplified Design of Reinforced Concrete

Buildings of Moderate Size and Height forderivation):

bd2 = 13Mu b = 13 x 726.9/182 = 29.2 in. < 36 in.

This implies that ρ will be less than ρmax.

Use 36 x 20.5 in. beam.






P a g e 4 o f 6

Required Reinforcement

Beam moments along the span aresummarized in the table below.

Load Case LocationEnd Span(ft-kips)

Interior span(ft-kips)

Exterior negative -211.2 Positive 241.4 207.5ead (D)

Interior negative -335.0 -301.8

Exterior negative -95.6 Positive 109.3 94.0ive (L)


Exterior negative ±90.3 Positive ind (W)

Interior negative ±90.3 ±90.3

No. Load CombinationExterior negative -458.4 Positive 523.9 450.31.4D + 1.7L


-228.5 xterior negative

-458.8

Positive 392.8 337.7

-660.3 -376.1

2 0.75(1.4D + 1.7L + 1.7W)

Interior negative

-430.0 -606.3-72.7 xterior negative

-307.5Positive 217.3 186.8

-418.9 -154.2

3 0.9D + 1.3W

Interior negative

-184.1 -389.0






P a g e 5 o f 6

Eq. (6) is used to determine the requiredreinforcement, which is summarized in thetable below. Tables 1 and 2 are utilized to

ensure that the number of bars chosenconform to the code requirements for coverand spacing.

Location Mu (ft-kips) As (in.2)* Reinforcement

Exterior negative -458.8 6.37 8-No. 8

Positive 523.9 7.28 10-No. 8nd Span

Interior negative -726.9 10.10 13-No. 8Interior Span Positive 450.3 6.25 8-No. 8

2s

2

2s

u

in.3.8786.0214ax.

(governs)n..168/60,000600

in..058/60,0006,000in.

4d

=××=

=××=

=××=

=

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.

Use No. 5 stirrups @ 6 in.






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.

Section A-A

1′-6″ 2′-0″

20.5″

30′-0″

7′-1″ 9′-6″

8-No. 8 2-No. 5 13-No. 8

3-No. 8 7-No. 8 3-No. 8 5-No. 8

6″

Standardhook (typ.)

3′-6″ 3′-6″

Class A tension splice

2″ 15-No. 5 @ 9″ 26-No. 5 @ 6″

2″

36″

4½″

16″

13-No. 8

10-No. 8

No. 5 U-stirrups

1½″ clear (typ.)




Two-Way Slabs


P a g e 1 o f 7

The following example illustrates the



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

to ACI 318-99.

20′-0″ 20′-0″ 20′-0″

2 4 ′ - 0 ″

2 4 ′ - 0 ″

20″x 20″ (typ.) 24″x 24″ (typ.)

D e s i g n s t r i p




Two-Way Slabs


P a g e 2 o f 7

Design Data

Materials• Concrete: normal weight (150 pcf), ¾-

in. maximum aggregate, f′c = 4,000 psi• Mild reinforcing steel: Grade 60 (fy =

60,000 psi)

Loads• Superimposed dead loads = 30 psf• Live load = 50 psf

Minimum Slab Thickness

Longest clear span ln = 24 – (20/12) =22.33 ft

From Fig. 1, minimum thickness h per ACI

Table 9.5(c) = ln/30 = 8.9 in.

Use Fig. 2 to determine h based on shearrequirements at edge column assuming a9 in. slab:

wu = 1.4(112.5 + 30) + 1.7(50) = 284.5 psf

A = 24 x [(20 + 1.67)/2] = 260 ft2

A/c12 = 260/1.672 = 93.6

From Fig. 2, d/c1 ≈ 0.39

d = 0.39 x 20 = 7.80 in.

h = 7.80 + 1.25 = 9.05 in.

Try preliminary h = 9.0 in.

Design for Flexure

Use Fig. 3 to determine if the Direct DesignMethod of ACI Sect. 13.6 can be utilized tocompute the bending moments due to thegravity loads:

• 3 continuous spans in one direction,

more than 3 in the other O.K.• Rectangular panels with long-to-short

span ratio = 24/20 = 1.2 < 2 O.K.• Successive span lengths in each

direction are equal O.K.• No offset columns O.K.• L/D = 50/(112.5 + 30) = 0.35 < 2 O.K.• Slab system has no beams N.A.

Since all requirements are satisfied, theDirect Design Method can be used.




Two-Way Slabs


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.

End Spans Int. Spanlab

Moments(ft-kips) Ext. neg. Positive Int. neg. Positive

TotalMoment

73.4 146.7 197.5 98.8

ColumnStrip

73.4 87.5 149.6 59.3

Middle

Strip 0 59.3 48.0 39.5

Note: All negative moments are at face of support.




Two-Way Slabs


P a g e 4 o f 7

Required slab reinforcement.

Span LocationMu

(ft-kips)b*

(in.)d**(in.)

As†(in.2)

Min. As‡(in.2) Reinforcement+

End SpanExt. neg. 73.4 120 7.75 2.37 1.94 12-No. 4Positive 87.5 120 7.75 2.82 1.94 15-No. 4

olumnStrip

Int. Neg. 149.6 120 7.75 4.83 1.94 25-No. 4Ext. neg. 0.0 168 7.75 --- 2.72 14-No. 4Positive 59.3 168 7.75 1.91 2.72 14-No. 4

iddleStrip

Int. Neg. 48.0 168 7.75 1.55 2.72 14-No. 4Interior SpanColumnStrip

Positive 59.3 120 7.75 1.91 1.94 10-No. 4

MiddleStrip

Positive 39.5 168 7.75 1.27 2.72 14-No. 4

*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




Two-Way Slabs


P a g e 5 o f 7

b1 = 20 + (7.75/2) = 23.875 in.

b2 = 20 + 7.75 = 27.75 in.

b1 /b2 = 0.86

From Fig. 5, γf = 0.62*

γfMu = 0.62 x 73.4 = 45.5 ft-kips

Required As = 45.5/(4 x 7.75) = 1.47 in.2

Number of No. 4 bars = 1.47/0.2 = 7.4,say 8 bars

Must provide 8-No. 4 bars within an

effective slab width = 3h + c2 = (3 x 9) +20 = 47 in.

Provide the required 8-No. 4 bars byconcentrating 8 of the column strip bars(12-No. 4) within the 47 in. slab width overthe column.

Check bar spacing:

For 8-No. 4 within 47 in. width: 47/8 =5.9 in. < 18 in. O.K.

For 4-No. 4 within 120 – 47 = 73 in. width:73/4 = 18.25 in. > 18 in.

Add 1 additional bar on each side of the47 in. strip; the spacing becomes 73/6 =12.2 in. < 18 in. O.K.

Reinforcement details at this location areshown in the figure on the next page (seeFig. 6).

∗

∗The provisions of Sect. 13.5.3.3 may be utilized; however, they are not in this example.




Two-Way Slabs


P a g e 6 o f 7

1 ′ - 8 ″

3 ′ - 1 1 ″

C o l u m n s t r i p

–

1 0 ′ - 0 ″

5′-6″

3 - N o .

4

8 - N o .

4

3 - N o .

4

Check the combined shear stress at theinside face of the critical transfer section.

c

M

A

Vv u

c

uu

/

γ+=

Factored shear force at edge column:

Vu = 0.285[(24 x 10.83) – (1.99 x 2.31)]Vu = 72.8 kips

When the end span moments aredetermined from the Direct DesignMethod, the fraction of unbalanced

moment transferred by eccentricity ofshear must be 0.3Mo = 0.3 x 282.2 =84.7 ft-kips (Sect. 13.6.3.6).

γv = 1 – γf = 1 – 0.62 = 0.38

c2 /c1 = 1.0

c1 /d = 20/7.75 = 2.58

Interpolating from Table 7, f1 = 9.74 andf2 = 5.53

Ac = f1 d2 = 9.74 x 7.752 = 585.0 in.2




Two-Way Slabs


P a g e 7 o f 7

J/c = 2f2d3 = 2 x 5.53 x 7.753 = 5,148 in.3

psi99524

148

000248

085

8002v

u

u

...

,

,..

.

,

=+=

××+=

Determine allowable shear stress φvc from

Fig. 4b:

bo /d = (2b1 + b2)/d

bo /d = [(2 x 23.875) + 27.75]/7.75 = 9.74βc = 1

φvc = 215 psi > vu = 199.4 psi OK

Reinforcement Details

The figures below show the reinforcementdetails for the column and middle strips.The bar lengths are determined fromFig. 13.3.8 of ACI 318-99.

1′-8″ 2′-0″

20′-0″

5′-6″

14-No. 4 13-No. 4

2-No. 4 13-No. 4

Standardhook (typ.)

Class A tension splice

5′-6″

3′-8″

5′-6″

3′-8″

12-No. 4

6″

Column strip

1′-8″ 2′-0″

20′-0″

4′-0″

14-No. 4 14-No. 4

7-No. 4 7-No. 4

Standardhook (typ.)

4′-0″ 4′-0″

6″

Middle strip

6″

3′-0″ 7-No. 4

7-No. 4

time saving design

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