Short Column

7/29/2019 Short Column

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Reinforced Concrete Design

Introduction to Columns


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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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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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Type of Columns

Tied Columns

Spirally-Reinforced ColumnsComposite Columns


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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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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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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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Code requirements for ties

I d i C l


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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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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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' 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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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.



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' 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

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