## 17 Reinforced Concrete Design Design of Column 1

Column load transfer from beams and slabs 1) Tributary area method: Half distance to adjacent columns y x Load on column =area ×floor load y x
17

Reinforced Concrete Design

Design of Column 1  Column load transfer from beams and slabs  Type of Columns  Strength of Short Axially Loaded Columns  Column Failure by Axial Load  Lateral Ties and Spirals Mongkol JIRAVACHARADET

SURANAREE UNIVERSITY OF TECHNOLOGY

INSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERING

Tributary Area When loads are evenly distributed over a surface, it is often possible to assign portions of the load to the various structural elements supporting that surface by subdividing the total area into tributary areas corresponding to each member. Half the load of the table goes to each lifter.

6m

Half the 100 kg/m2 snow load on the cantilevered roof goes to each column. The tributary area for each column is 3 m x 3 m. So the load on each column is 100 (3 x 3) = 900 kg

3m

100 kg/m2

Column load transfer from beams and slabs 1) Tributary area method: Half distance to adjacent columns Load on column = area × floor load

y y

x

x

6m

6m

4.5 m

9m

12 m

9m

All area must be tributed to columns

C3

C1

C3

C1 6m

C2

C4

C4

C2

6m C4

C2

C4

C2

4.5 m C1

C3

C3

9m

12 m

C1

9m

C1 : Corner column

C3 : Exterior column

C2 : Exterior column

C4 : Interior column

2) Beams reaction method: Collect loads from adjacent beam ends B2

B1

B4 RB1

RB2

RB1 RB2

B1

C1 B3

B2

Load summation on column section for design ROOF

Design section

Load on 2nd floor column = Roof floor + Column wt.

Design section

Load on 1st floor column = load on 2nd floor column + 2nd floor + Column wt.

2nd FLOOR

1st FLOOR

Ground level

Design section Footing

Load on pier column = load on 1st floor column + 1st floor + Column wt.

     

  C1 (A-6)

 

RB2

3.50 m

T1

RB4

 0.3 x 0.3 m

RB19



B5

3.50 m

B4

B4

0.3 x 0.3 m

B5

 

B5 B4

B4 B5

 

1.50 m

 0.4 x 0.4 m

RB2 RB4 RB19 T1 Col.Wt. Floor load

= 5280 kg = 4800 kg = 4416 kg = 960 kg = 756 kg = 16212 kg

2B5 = 10764 kg 2B4 = 14736 kg Col.Wt. = 756 kg Floor load = 26256 kg Cum. load = 42468 kg 2B5 = 10764 kg 2B4 = 14736 kg Col.Wt. = 576 kg Floor load = 26076 kg Cum. load = 68544 kg

Type of Columns

Tie

Spiral

Longitudinal steel s = pitch

Tied column

Spirally reinforced column

Strength of Short Axially Loaded Columns Short columns are typical in most building columns. P0

Steel

fy Stress

f c′ Section A-A

A

A

Concrete

.001

.002 Strain

.003

P0

[ ΣFy = 0 ] P0 = Ast fy + fc′( Ag − Ast ) From experiment:

f c′ fy

fy

Fs = Ast fy Fc = (Ag - Ast) f c′

P0 = Ast fy + 0.85fc′( Ag − Ast ) where Ag = Gross area of column section Ast = Longitudinal steel area

Column Failure by Axial Load

Pu ∆

Axial load

Pu

0

Initial failure

Tied column

Axial deformation ∆

Heavy spiral ACI spiral Light spiral

       : U = 1.4D + 1.7L

  : U = 0.75(1.4D + 1.7L + 1.7W)



U = 0.9D + 1.3W

"#  (φ) :     

φ = 0.75

   

φ = 0.70

  & '( )  **+& , & '(  ACI 1)   !"#   2) &'    ∅ 9 ".".   ) ≤ DB32 *  &'    ∅ 12 ".".   ) DB36 *  DB40 3) !! (s) s ≤ 16 ∅

  

s ≤ 48 ∅    s≤



,"  

4) "") *    ."! / 135o *  ."!"   "! / 15 0". x

x

x

x ≤ 15 cm x

x > 15 cm

x

x

x

x

x

x x ≤ 15 cm

x

x

x > 15 cm

  & ' Initial shape

Pu Spiral

Final shape

f2

f2

Increase of compressive strength due to lateral pressure:

ff = fc′ + 4.1f2 Good design: Strength lost by spalling = Strength gain from f2

( Ag − Acore )(0.85fc′) = Acore (4.1f2 )

1

hcore Core

s

s

Ab fy Spiral Ab fy [ ΣFx = 0 ]

hcore s f2 = 2 Ab fy

f2 = 2

2 Ab fy

2

hcore s

1

4.1(2 Ab fy )  Ag  − 1 (0.85fc′) =  hcore s  Acore 

3

Define:

ρs

ρs =

Abπ hcore 4 Ab = 2 (π hcore / 4)s hcore s

3

 0.42fc′  Ag ρs = − 1  fy  Acore  Rounding 0.42 to 0.45,

 0.45fc′  Ag ACI Code: ρ s = − 1  fy  Acore 

)  **+& , & ' 1) Minimum width or diameter: hmin ≥ 20 cm 2) Reinforcement ratio: 1% ≤ ρg ≤ 8% (usually ≤ 5%) 3) Can use bundled bars in corners (≤ 4) 4) 2.5 cm ≤ Clear stirrup spacing ≤ 8 cm 5) Spiral diameter: db ≥ 9 mm 6) Lap splices: Lsp ≤ min { 48 db , 30 cm }

Minimum Cover for Column Reinforcement Min. Cover

Condition

Reinforcement

Cast against earth

all sizes

7 cm

Exposed to weather or earth

DB20 - DB60 DB16 and smaller

5 cm 4 cm

No exposure

main reinforcement, ties, and spirals

4 cm

Limits on percentage of reinforcement

0.01 ≤  ρ g = Ast / Ag  ≤ 0.08 Lower limit:

To prevent failure mode of plain concrete

Upper limit:

To maintain proper clearances between bars

ACI Strength Provision: Pu  φ Pn Spirally reinforced column:

Pn = 0.85[0.85fc′( Ag − Ast ) + fy Ast ], φ = 0.75 Tied column:

Pn = 0.80[0.85fc′( Ag − Ast ) + fy Ast ], φ = 0.70

Working Stress Design (WSD) of Short Column Spirally reinforced column:

P = Ag (0.25fc′ + fs ρ g ),

ρ g = Ast / Ag

Tied column:

P = 0.85 Ag (0.25fc′ + fs ρ g ),

ρ g = Ast / Ag

where fs = 0.40fy but not exceed 2,100 kg/cm2

Length Effects ACI permits neglect of length effect when

kLu M ≤ 34 − 12 1 r M2

M1

for braced system

M1

where (34 - 12M1/M2) may not exceed 40 M1 = The smaller bending moment

+

-

M2 = The larger bending moment M1/M2 is positive for single curvature and negative for double curvature

kLu ≤ 22 r

for unbraced system

M2

M2

Example 11.1 Design for Pure Compression Design a concentrically loaded square column with ties providing lateral reinforcement. Service dead and live loads are 180 and 90 tons, respectively The column has an unsupported height of 3.0 m and is braced against sidesway. Use f’c = 240 kg/cm2 and fy = 4,000 kg/cm2. 1) Determine required strength Pu = 1.4D + 1.7L = 1.4(180) + 1.7(90) = 405 tons 2) Check column slenderness. Assume an 50-cm square column k = 1.0 for braced compression member 1 4 2 r = 0.3(50) = 15 cm I/A = h /h = 12

(

)

1/12 h

kLu 1.0 × 3.0 × 100 = = 20 < 34 − 12(M1 / M2 ) = 22 r 15

Neglect length effects

3) Design for column reinforcement Required Pn = Pu/φ = 405/0.70 = 578.6 ton for tied column:

Pn = 0.80[0.85fc′( Ag − Ast ) + fy Ast ]

578.6 × 1,000 = 0.80(0.85×240(50×50 - Ast) + 4,000Ast) Ast = 56.2 cm2

50 cm

USE 12DB25 (Ast = 58.9 cm2, ρg = 2.36%) 4) Select lateral reinforcement

12DB25 [email protected]

USE RB9 ties with DB25 longitudinal bars Spacing not greater than:

16 (2.5) = 40 cm 48 (0.9) = 43.2 cm column size = 50 cm

USE RB9 @ 40 cm

50 cm