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7/29/2019 Short Column
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Reinforced Concrete Design
Introduction to Columns
7/29/2019 Short Column
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Introduction to Columns
Three categories
Short compression blocks or pedestals
Short reinforced concrete columnsLong or slender reinforced concrete columns
Short compression blocks or pedestals
Height is less than three times it least lateral dimension.
ACI 2.2 and 10.14: May be designed with plain concrete with a
design compressive strength if 0.85fc where = 0.7.
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Introduction to Columns
Short reinforced concrete columns
A stocky member with little flexibility. Would fail first due to
material failure.The load it supports is controlled by the size of the cross section and
by the strength of the materials.
Long or slender reinforced concrete columnsHave large bending deformations that
result in secondary moments that reduce
the axial load capacity of the columns.
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Introduction to Columns
Type of Columns
Tied Columns
Spirally-Reinforced ColumnsComposite Columns
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Introduction to Columns
Axial Load Capacity of Columns
For spiral columns ( = 0.75):
For tied columns ( = 0.7):
The previous expressions are only to be used when there is no
moment or when it is very small. Moment is very small when
e < 0.10h for tied columns and when e < 0.05h for spiral columns.
h = outside diameter for round column
= total depth of square or rectangular column
'0.8 [0.85 ( ) ]n C g st y st P f A A f A
'0.85 [0.85 ( ) ]n C g st y st P f A A f A
u
u
MeP
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Introduction to Columns
Code requirements longitudinal bars
ACI 7.6.3: Clear distance between longitudinal bars shall not be less
than 1.5db nor less than 4cm.
ACI 10.9.2: Minimum number of longitudinal bars: 4 for bars within
rectangular or circular ties, 3 bars within triangular shapes, 6 for bars
enclosed within spirals.
ACI 10.9.1:
min
max
0.01
0.08
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Introduction to Columns
Code requirements for ties
ACI 7.10.5.1: Minimum tie size: 10 for longitudinal bars 32 or
smaller; 12 for larger longitudinal bars or for bundled bars.
ACI 7.10.5.2: Maximum center-to-center spacing shall not be more
than
16 times the diameter of longitudinal bars ,
48 times the diameter of the ties, or
the least dimension of the column.
ACI 7.10.5.3: The ties must be arranged so that every corner and
alternate longitudinal bar will have support provided by the corner ofa tie having an included angle not greater than 135o.
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Introduction to Columns
Code requirements for ties
I d i C l
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Introduction to Columns
Code requirements for spirals
ACI 10.9.3: Minimum spiral percentage:
Pitch can then be determined with this expression:
ACI 7.10.4: Spacing may not be less than 2.5cm
and may not be larger than 7.5cm.
Cover
ACI 7.7.1(c): Minimum cover = 4cm
'
0.45 1gc
s
sy c
Aff A
'
4
0.45 1
s
g cc
c sy
aS
A fD
A f
I d i C l
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Introduction to Columns Example 1
Design a tied column that is subjected to the following axial compression
loads: PDL
= 100 tons and PLL
= 50 tons.
The material properties are as follows' 2 2
250 / , 4200 / C ykg cm f kg cm
1.2 1.6 1.2 100 1.6 50 200u DL LLP P P ton
Calculate the ultimate load
Step 1: Calculate column dimension
Calculate the cross section area
Assume = 1%
' '0.52 [0.85 ( 0.85 )]u g C g y C P A f f f
21415gA cm
3
200 10 0.52 [0.85 250 0.01(4200 0.85 250)]gA
Assume column width = 25cm
141556.6 60
25h cm h cm
'0.8 [0.85 ( ) ]n C g st y st P f A A f A
I t d ti t C l
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Introduction to Columns
20.01 25 60 15
S CA A cm
Step 2: Calculate longitudinal reinforcement area
Choose 8 16
Step 3: Calculate stirrups
Choose stirrup diameter of 8mm
The spacing between ties is the smallest of
16(1.6) = 25.6cm
48(0.8) = 38.4cm
25cm
Use ties 8mm spaced @ 25cm
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Introduction to Columns
' 2 2250 / , 4200 / C ykg cm f kg cm
22.5
184
0.04480 25
g
' '
0.52 [0.85 ( 0.85 )]u g C g y C P A f f f
403.5u
P tons
0.52 80 25 [0.85 250 0.044(4200 0.85 250)]/1000uP
I t d ti t C l
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Introduction to Columns Example 3
Design a spiral column that is subjected to the following axial compression
loads: PDL
= 60 tons and PLL
= 60 tons.
The material properties are as follows' 2 2
280 / , 4200 / C ykg cm f kg cm
1.2 1.6 1.2 60 1.6 60 168u DL LLP P P tons
Calculate the ultimate load
Step 1: Calculate column diameter
Calculate the cross section area
Assume = 1%
' '0.595 [0.85 ( 0.85 )]u g C g y C P A f f f
2949gA cm
3168 10 0.595 [0.85 280 0.01(4200 0.85 280)]
gA
Taken as 35cm4 949
34.7D cm
I t d ti t C l
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Introduction to Columns
2 2
0.01 / 4 35 9.62S CA A cm
Step 2: Calculate longitudinal reinforcement area
Choose 7 14
Step 3: Calculate spiral reinforcement
Choose stirrup diameter of 8mm
DC = 35 - 4 - 4 = 27cm
'
4
0.45 1
s
g cc
c sy
aS
A fDA f
2
2
4(0.50)3.63
/ 4 35 2800.45 27 1
4200/ 4 27
S cm
Taken as 3.5cm (center-to-center)
Sc = 3.50.8 = 2.7 cm O.K within ACI limits.
Introduction to Columns
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Introduction to Columns
' 35 2(4) 2(0.8) 1.4 24D cm
Step 4: Check spacing between longitudinal reinforcement bars
12
sin(64.3) 10.41sin(51.430S cm
10.41 1.4 9.01 1.5(1.4) 4Sc cm cm OK