Unit 14 ( DESIGN OF SLENDER COLUMNS )

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REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 14DESIGN OF SLENDER COLUMNS

OBJECTIVES

GENERAL OBJECTIVE To be able to identify the braced slender columns design principles according to BS 8110 requirements.

SPECIFIC OBJECTIVES At the end of this unit you will be able to calculate: 1. the minimum eccentricity, emin . 2. the deflection , a u using equation 32 3. the reduction factor, K using equation 33 4. the additional moment using equation 35 5. the initial end moments using equation 36 6. the area of longitudinal reinforcement. 7. the size and spacing of lateral reinforcement (ties). 8. sketch the reinforcements details in slender columns.

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INPUT 1

14.1

Slender Columns

A braced slender column is defined as a column in which the effective height/depth ratio is greater than 15. The strength of slender columns is significantly reduced by transverse deflections. The slenderness effect reduces its load-carrying capacity. If the column is short, the deflection is small and hence the additional moment is negligible, compared with the initial moment. If the column is slender, the deflection is no longer small, then the additional moment becomes significant compared to the initial moment. The additional moment should then be considered if the effective height/depth ratio is greater than 15. If the column is very slender, the column will quickly collapse and such a failure is called instability failure.

The additional moment, M add is caused by the deflection of slender column. Hence the design moment will be greater than the initial moment obtained from the structural analysis. The design moments for braced and unbraced slender columns are different. Thus, their design moments will also be different however the calculation for M add is similar.

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ACTIVITY 14a

Now do the following. 14.1 A braced slender column is a column in when both the ratiosl ex h

and

l ey b

are greater than ___________________________.l ey l ex and is h b

14.2

An unbraced slender column is one in which both ratios greater than _________________________.

14.3 14.4

The strength of a slender column is reduced by _____________________. The deflection of slender column is significant compared with the ________________________ moment.

14.5 14.6

For slender columns, _________________ should be considered. The additional moment of slender column is caused by the column ________________________.

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FEEDBACK 14a

Now check your answers.

14.1 14.2 14.3 14.4 14.5 14.6

15 10 transverse deflection initial moment additional moment deflection.

If all your answers are correct, please proceed. Otherwise please go back to the INPUT in this section and do the activity.

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INPUT 2

14.7

Additional Moment,

M add

Consider a column acted by axial load N and an end moment Mi as shown below; N

au

NFigure 14.1: Deflection of a rectangular column.

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The additional moment, M add is calculated as follows;

M add = Na u

Where, N = the ultimate axial loada u = column deflection at ultimate limit.

a u is calculated using equation 32 of the code given below ;

au = a Kh

And a is calculated using equation 34 as follows;

1 l e a = ' 2000 b

2

a can also be obtained from Table 3.23 of BS 8110, which in dependent on the

ratio

le where l e is the effective height of column in the plane considered and b'

b ' = b , is the dimension of the smaller column.

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However, according to clause 3.8.3.6 for columns in biaxial bending, where

M add are present in both axes, x and y, the value

le is equal to b'

l ex and h

l ey b

.

K is the reduction factor to correct the deflection and to take into consideration the effect of axial load. K can be calculated using equation 33 of the code. This is shown below:

K=

N uz N 1 N uz N bal

Where, N uz = 0.45 f cu Ac +0.87 f y Asc and N bal = 0.25 f cu bd (used for symmetrical reinforced concrete columns).

Therefore, M add can be rewritten as;

M add

Nh l e = K 2000 b '

2

N uz depends on

Asc but

Asc is not yet known, K is then taken as 1.0. The

iteration process is to be continued until the K value obtained is equal to or approximately equal to the value assumed earlier.

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ACTIVITY 14b

Now do the following exercise. Fill in the blanks. 14.7 The additional moment, M add is calculated using equation

_________________________. 14.8 14.9 The deflection, au is calculated using equation ________________.a can be obtained from Table __________________of the code.

14.10 __________________ is the reduction factor to correct deflection. 14.11 The reduction factor can be calculated using equation

________________________. 14.12 When

le = 12 , a is equal to _____________________. b' le = 60 , a is equal to ______________________. b'

14.13 When

14.14 When h = 500 mm, K = 1.0, a = ______________________. 14.15 When a u = 200mm and N = 3000 kN, the additional moment is equal to _________ kN. 14.16 A column of dimension 300mm x 400mm is reinforced with 4T32 bars. If fcu = 40 N/mm2 and fy = 460 N/mm2, calculate N uz .

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14.17 Calculate N bal N/mm2.

for the column section given below using fcu as 40

400mm

550 mm

14.18 Calculate the reduction factor, K if N uz = 3500kN, N bal = 3000 kN and N = 3100 kN.

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FEEDBACK 14b

Here are the answers. 14.7 14.8 14.9M add = Na u

au = a Kh Table 3.23

14.10 Reduction Factor, K 14.11 K = 14.12 0.07 14.13 1.80 14.14 0.25 1.0 500 = 225 mm 14.15 600 kNm 14.16 N uz = 0.45 f cu Ac +0.87 f y Asc = 0.45 40 300 400 + 0.87 460 3216 10 3 = 3447 kNN uz N 1 N uz N bal

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14.17

N bal = 0.25 f cu bd

= 0.25 40 400 550 10 3 = 2200 kN 14.18 K =N uz N N uz N bal3500 3100 3500 3000400 500

= =

= 0.8 1

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INPUT 3

14.19 Slender Braced Columns

Consider a column experiencing an earlier eccentricity of axial load as shown below;

N

e

au

Figure14.2: Deflection of Slender Braced Column

The total moment is M t = M i + M add . The initial moment Mi to be used is simply the initial moment at an end of the column. BS8110 recommends that Mi be taken as:M i = 0.6 M 2 + 0.4 M 1 (M2 being the larger) for symmetrical bending

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And,

M i = 0.6 M 2 0.4 M 1

0.4 M 2 but not less than

0 .4 M 2

for

assymmetrical bending i.e. bending in double curvature. This is shown in Fig 3.20, BS 8110. The two equations can be combined as follows:

M i = 0.4 M 1 + 0.6 M 2 0.4 M 2

Where, Mi is the smaller initial end moment (taken as negative if the column is bent in double curvature) and M2 is the larger initial end moment, which is always taken as positive.

BS8110 imposes a further condition that M t M 1 + M add

1 2

A column in any case is designed for a moment of at least Ne min. Hence we have further condition that M t Ne min

emin is the design minimum eccentricity and is taken as 0.05h or 20 mm, whichever is lesser.

ACTIVITY 14c

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Now do the following exercise.

Fill in the blanks.

14.19 ___________________ is equal to the sum of initial moment, M i and the additional moment M add

14.20 For

symmetrical

bending,

the

initial

moment

is

equal

to

____________________________.

14.21 For _________________ bending, M i = 0.6 M 2 0.4 M 1

14.22 ______________________ is bending in double curvature.

14.23 A

slender

column

is

designed

for

a

minimum

moment

of

__________________________.

14.24 The design minimum eccentricity is taken as ___________ or 20 mm whichever is less. 14.25 Pick four (4) from the following criteria in determining the maximum design moment ;

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a) M 2 b) 0.4 M 1 c) 0.6 M 2 d) M i + M add e) M 1 +M add 2

f) 0.4 M 2 g) 0.4 M 1 + 0.6 M 2 h) emin N 14.26 For the slender column experiencing initial moments at both ends (shown below), determine the values of M 1 and M 2 assuming double curvature.

100 kNm

200kNmFigure 14.3: End Moments in Slender Column 14.27 Calculate M i of the column in Question 8 assuming single curvature.

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14.28 Calculate the design minimum eccentricity for the column section shown below; 200mm

350mm

14.29 Calculate the min

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