柱之設計Design of Columns

55

柱之設計 Design of Columns

- Column load transfer from beams and slabs

- Type of Columns

- Strength of Short Axially Loaded Columns

5-15-1

- Column Failure by Axial Load

柱之荷重 (由梁及版傳遞 )Column load transfer from beams and slabs

1) Tributary area method:

Half distance to adjacent columns

y

x

Load on column = area ดfloor load

Floor load = DL + LL

DL = slab thickness ดconc. unit wt.

Example: x = 5.0 m, y = 4.0 m, LL = 300 kg/m2, slab thickness = 10 cm

Floor load = 0.1(2,400) + 300 = 540 kg/m2

Load on column = (5.0)(4.0)(540) = 10,800 kg = 10.8 t

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.

Buckling Defined

Euler’s Formula

Effective Length Concept

Effective Length Concept

柱之種類 Type of Columns

橫箍筋 Tie

Longitudinalsteel

橫箍柱 Tied column

螺旋箍筋 Spiral

s = pitch

螺旋箍筋柱 Spirally reinforced column

承受軸向荷重短柱之強度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

Stres

s

P 0

f yf y

cf

F s = A s t f y

F c = ( A g - A s t ) cf

[ Fy = 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

Column_08

Pu

0Axial deformation

Initial failure

Axi

al l

oad Tied column

Lightspiral

ACI spiral

Heavy spiral

柱之破壞Column Failure by Axial Load

Pu

需要強度及強度折減係數

需要強度

U = 1.4D + 1.7L

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

U = 0.9D + 1.3W

強度折減係數 () :螺旋箍筋柱 = 0.75 橫箍筋柱 = 0.70

柱鋼筋之配置

橫箍筋間距 (s)

s ด 16 主筋直徑s ด 48 箍筋直徑s ด 柱之最小尺寸

箍筋之配置

xx

x ด15 cm

xx

x 15 cm

xx

x ด15 cm

x

x

xx

x 15 cm

x

x

螺旋箍筋柱

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

[ Fx = 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

ccore core

A A ff

A h s

3

f2

Define:2

4

( / 4)b core b

score core

A h A

h s h s

s 3

0.421gc

sy core

Af

f A

Rounding 0.42 to 0.45,

ACI Code:0.45

1gcs

y core

Af

f A

1) fy ด4,000 kg/cm2

2) Spiral diameter ด10 mm

3) 2.5 cm ดClear spacing ด7.5 cm

螺筋比

最小螺筋比