1717

Mongkol JIRAVACHARADET

Reinforced Concrete DesignReinforced Concrete Design

S U R A N A R E E INSTITUTE OF ENGINEERING

UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING

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

Tributary AreaTributary 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.

100 kg/m23 m

6 m

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

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

9 m 12 m 9 m

4.5 m

6 m

6 m

All area must be tributed to columns

C1

C1 : Corner column

C2

C2 : Exterior column

C3

C3 : Exterior column

C4

C4 : Interior column

9 m 12 m 9 m

4.5 m

6 m

6 m

C1

C1 C1

C2C2

C2

C3

C3C3

C4 C4

C4

2) Beams reaction method:

B1 B2

RB1

RB1 RB2

RB2

Collect loads from adjacent beam ends

C1B1 B2

B3

B4

Load summation on column section for design

Design section

Design section

Design section

ROOF

2nd FLOOR

1st FLOOR

Footing

Ground level

Load on pier column

= load on 1st floor column

+ 1st floor + Column wt.

Load on 1st floor column

= load on 2nd floor column

+ 2nd floor + Column wt.

Load on 2nd floor column

= Roof floor + Column wt.

�������������� �������������

���������

������3.50 m

0.3 x 0.3 m

C1 (A-6)

������

���� ����3.50 m

0.3 x 0.3 m

���� ����

������1.50 m

0.4 x 0.4 m

RB2 = 5280 kg

RB4 = 4800 kg

RB19 = 4416 kg

T1 = 960 kg

Col.Wt. = 756 kg

Floor load = 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

T1

RB2

RB4

RB19

B4

B5

B4

B5

B4

B5

B4

B5

Type of Columns

Tie

Longitudinal

steel

Tied column

Spiral

s = pitch

Spirally reinforced column

Strength of Short Axially Loaded Columns

Short columns are typical in most building columns.

P0

A A

∆

Section A-A

.001 .002 .003

fy

cf ′

Steel

Concrete

Strain

Stress

P0

fyfy

cf ′

Fs = Ast fyFc = (Ag - Ast) cf ′

[ ΣFy = 0 ]

0 ( )st y c g stP A f f A A′= + −

From experiment:

0 0.85 ( )st y c g stP A f f A A′= + −

where

Ag = Gross area of column section

Ast = Longitudinal steel area

Pu

0Axial deformation ∆

Initial failure

Axial load Tied column

Light

spiral

ACI spiral

Heavy spiral

Column Failure by Axial Load

∆

Pu

������������� ���������������������������

������������������������������������� :

U = 1.4D + 1.7L

���������������������������� :

U = 0.75(1.4D + 1.7L + 1.7W)

���� U = 0.9D + 1.3W

����"#������(φφφφ) :

��������� ���� ��� 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".

xx

x ≤ 15 cm

xx

x > 15 cm

xx

x ≤ 15 cm

x

x

xx

x > 15 cm

x

x

�����������&�'��

PuInitial shape

Final shapef2

Spiralf2

Increase of compressive strength due to lateral pressure:

24.1f cf f f′= +

Good design: Strength lost by spalling = Strength gain from f2

2( )(0.85 ) (4.1 )g core c coreA A f A f′− = 1

Core

Spiral

s

hcore

Ab fy

Ab fy

s

[ ΣFx = 0 ] hcore s f2 = 2 Ab fy

2

2 b y

core

A ff

h s= 2

12

4.1(2 )1 (0.85 )

g b y

c

core core

A A ff

A h s

′− =

3

Define:2

4

( / 4)b core b

s

core core

A h A

h s h s

πρ

π= =

ρs 3

0.421

gcs

y core

Af

f Aρ

′= −

Rounding 0.42 to 0.45,

ACI Code:0.45

1gc

s

y core

Af

f Aρ

′= −

)��������*����*+�&�,���&�'��

1) Minimum width or diameter: hmin ≥ 20 cm

4) 2.5 cm ≤ Clear stirrup spacing ≤ 8 cm

2) Reinforcement ratio: 1% ≤ ρg ≤ 8% (usually ≤ 5%)

3) Can use bundled bars in corners (≤ 4)

5) Spiral diameter: db ≥ 9 mm

6) Lap splices: Lsp ≤min { 48 db , 30 cm }

Minimum Cover for Column Reinforcement

Condition

Cast against earth

Exposed to weather

or earth

No exposure

Reinforcement

all sizes

DB20 - DB60

DB16 and smaller

main reinforcement,

ties, and spirals

Min. Cover

7 cm

5 cm

4 cm

4 cm

Limits on percentage of reinforcement

0.01 / 0.08g st gA Aρ ≤ = ≤

Lower limit: To prevent failure mode of plain concrete

Upper limit: To maintain proper clearances between bars

ACI Strength Provision: Pu � φφφφ P

n

Spirally reinforced column:

0.85[0.85 ( ) ], 0.75n c g st y stP f A A f A φ′= − + =

Tied column:

0.80[0.85 ( ) ], 0.70n c g st y stP f A A f A φ′= − + =

Working Stress Design (WSD) of Short Column

Spirally reinforced column:

ρ ρ′= + =(0.25 ), /g c s g g st gP A f f A A

Tied column:

ρ ρ′= + =0.85 (0.25 ), /g c s g g st gP A f f A A

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

Length Effects

ACI permits neglect of length effect when

1

2

34 12 for braced systemukL M

r M≤ −

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

22 for unbraced systemukL

r≤

M1

M2

-

M1

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/cm

2.

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

r = 0.3(50) = 15 cm ( )4 21/ / 1/12

12I A h h h= =

1 2

1.0 3.0 10020 34 12( / ) 22

15ukL

M Mr

× ×= = < − =

Neglect length

effects

3) Design for column reinforcement

Required Pn = Pu/φ = 405/0.70 = 578.6 ton

for tied column: 0.80[0.85 ( ) ]n c g st y stP f A A f A′= − +

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

Ast = 56.2 cm2

USE 12DB25 (Ast = 58.9 cm2, ρg = 2.36%)

4) Select lateral reinforcement

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

50cm 12DB25

RB9@0.40