柱之設計 design of columns 5. 柱之設計 design of columns - column load transfer from beams...
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柱之設計Design of Columns
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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