Unit 14 ( DESIGN OF SLENDER COLUMNS )

REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 14 DESIGN 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 ratios

l ey b

14.2

l ex h

and

are greater than ___________________________.

An unbraced slender column is one in which both ratios

l ey l ex and is h b

greater than _________________________. 14.3

The strength of a slender column is reduced by _____________________.

14.4

The deflection of slender column is significant compared with the ________________________ moment.

14.5

For slender columns, _________________ should be considered.

14.6

The additional moment of slender column is caused by the column ________________________.


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FEEDBACK 14a

Now check your answers.

14.1

15

14.2

10

14.3

transverse deflection

14.4

initial moment

14.5

additional moment

14.6

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

N Figure 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 load a u = column deflection at ultimate limit.

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

au = β a Kh

And βa is calculated using equation 34 as follows;

 1  l e  βa =   '   2000  b 

2

βa 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;

2

M add

Nh  l e  =   K 2000  b ' 

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

The deflection, au is calculated using equation ________________.

14.9

βa 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 , βa is equal to _____________________. b'

14.13 When

le = 60 , βa is equal to ______________________. b'

14.14 When h = 500 mm, K = 1.0, βa = ______________________. 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 .


14.17 Calculate N bal

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for the column section given below using fcu as 40

N/mm2.

400mm

550 mm

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


FEEDBACK 14b

Here are the answers. 14.7

M add = Na u

14.8

au = β a Kh

14.9

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 kN

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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 bal

=

3500 − 3100 3500 − 3000

=

400 500

= 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


And,

M i = 0.6 M 2 − 0.4 M 1

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≥ 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.

1 2

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

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

moment

of

____________________________.

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

__________________________.

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

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


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

350mm

14.29 Calculate the minimum moment of the column in Question 10, if N = 4000 kN.

14.30 From the following information, determine the maximum design moment of the column: M2 = 500 kNm Mi = 460 kNm Madd = 325 kNm Nemin = 100 kNm M1 = 800 kNm

FEEDBACK 14c


Your answers should be as follows;

14.19 Total moment, Mt 14.20 symmetrical bending 14.21 symmetrical 14.22 Assymmetrical bending 14.23 Nemin 14.24 0.05h 14.25 M2 , Mi + Madd , M 1 + 14.26

M add 2

M 1 = 100 kNm M 2 = 200 kNm M i = 0.4 M 1 + 0.6 M 2

14.27

= 0.4( −100 ) + 0.6( 200 ) = 80 kNm

14.28 emin = 0.05 h

and eminN

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= 0.05 × 350 = 17.5 mm

But, e min should not be less than 20 mm. Therefore, the minimum eccentricity is 20 mm.

14.29 M = Ne min = 4000 × 20 ×10 −3 = 80 kNm

14.30

Maximum design is taken as the greatest value derived from M 2, Mi, Madd , Nemin and M1 . Therefore, the maximum design moment is 800 kNm.


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INPUT 4

14.31 Design Example A braced slender column of dimensions 300 mm x 450 mm carry an axial load of 1700 kN and end moments of 70 kNm and 10 kNm at ultimate limit state. This load and moments induced a double curvature about x-axis as shown below. The effective heights are lex = 6.75m and ley = 8.0 m . The characteristic strength of materials are fcu = 30 N/mm2 and fy = 460 N/mm2.


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M2= 70 kNm

b = 450

d’= 60 d = 240

x

x h = 300

SECTION

M1=10 kNm 14.4:Initial End Moments

Solution


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Slenderness ratio: l ex 6.75 = = 22.5 > 15 h 0.3 l ey b

=

8.0 = 17 .8 > 15 0.45

This shows that the column is slender and will be designed as such: at the time when double curvature occurs we have, M1 = - 10 kNm and M i = 0.6 M 2 + 0.4 M 1

= 0.4(−10 ) +0.6(70 ) = 38 kNm

therefore M i is greater than 0.4 M 2

Additional moment produced by column deflection is, 2

M add

Nh  l e  =   K 2000  b '  2

1700 × 300  6750  3 =   ×1.0 ×10 2000  300 

= 129 kNm

With K = 1.0 as the initial trial value, the total moment is,


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M t = M i + M add

= 38 + 129 = 167 kNm

N 1700 ×10 3 = bh 450 × 300

=

12 6

= 2.0

M 167 × 10 6 = bh 2 450 × 300 2

= 4.12

From the Column Design Chart No.27, 100 Asc = 3.2 bh

and K = 0.65

Now, for the second trial we shall use K = 0.65, and recalculate Mt, we have:


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M t = M i + M add

= 38 + 83.91 = 121.91 kNm

From Chart No.27, Asc =

2.2bh 100

= 2.2 × 450 ×

300 10

= 2970 mm2

Now check the final value of K from the design chart, N bal = 0.25 f cu bd

= ( 0.25 × 30 × 450 × 240 ) x 10 -3 kN = 810 kN N uz = 0.45 f cu bh +0.87 f y Asc

= ( 0.45 × 30 × 450 × 300 + 0.87 × 460 × 2970 ) x 10 -3 kN = 3011 kN

Now, calculate the reduction factor as follows;


K=

=

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N uz − N ≤1 N uz − N bal 3011 −1700 3011 − 810

= 0.6

This is equal to the final value in column 5 tabulated below;

(1)

2)

(3)

(4)

(5)

K

Mt

M bh 2

100 Asc bh

K

1.0

167

4.12

3.2

0.65

0.65

122

3.0

0.6

Table 14.1: Iteration Process of K Value

The iteration stops when value in column (1) is approximately equal to value in column (5). Therefore the area of reinforcement required is 2970 mm2. Hence, provide 4 T 32 (Asc = 3218 mm2 )

Ties: Minimum size =

1 ×32 = 8 mm 4

Maximum spacing = 12 x 32 = 384 mm centres.


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Use R8 at 375 centres.

Details of the reinforcements are shown below;

2T32 R8 – 375

2T32

2T32 14.5: Rectangular Columns

SUMMARY


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This unit should have enabled you to design a reinforced concrete braced slender column according to BS 8110 requirements. The BS 8110`s design procedure for designing this column is summarised below: Step 1: Calculate the effective height, l e Step 2: Calculate the total moment about a minor axis for N and Mt Mt is taken as the greatest among the following (i) to (iv); (i) (ii)

M t = M i + M add Mt = M2

(iii)

M t = M1 +

(iv)

M t = Ne min

1 M add 2

Step 3: Calculate the total bending moment on a major axis for N and M t if the ratio of the length of the longer side to that length of the shorter side is less

than 3, and if

le ≤ 20 . h

Step 4: For biaxial bending calculate the following; a)

Mty (total moment about the minor axis )


b)

Mtx (total moment about the major axis )

c)

design for N, Mty and Mtx

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Step 5: Calculate the reduction factor using the equation given below ; K=


Or K can be read-off from BS 8110 Part 3 Column Design Charts.

Congratulations! You

have

now

completed Unit 14 Turn back to page 1 of this unit. Have you achieved these objectives successfully? If your answer is YES, do the SelfAssessment. If your answer is NO, go through Unit 14 again.

SELF-ASSESSMENT


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Read and answer this question. Design the longitudinal reinforcement for the braced slender column in Figure shown below for bending about the minor axis, if N = 2500 kN, M 1y = 100 kNm, M2y = 120 kNm, fcu = 40 N/mm2 and fy = 460 N/mm2. Assume that the cover, c = 50 mm and try T40 bars for estimating the effective depth.

N y 400

x

500

x

(6.5 m)

y Slender column

FEEDBACK ON SELF-ASSESSMENT


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Now check your answers.

For bending about a minor axis, calculate Mi and Madd : a)

M i = 0.4 M 1 + 0.6 M 2

= (0.4)(100) + 0.6(M2) = 112 kNm b)

le 6500 = = 16.25 b' 400

c)

βa = 0.13

d)

K = 1.0

Remember that, h is the depth in the plane of bending, i.e. 400 mm.

M add = ( 2500 )( 0.13 )(1)( 0.4)

= 130 kNm M t = M i + M add

= 112 + 130 = 242 kNm

Mt = M2 = 120 kNm M t = M1 +

1 M add 2


1 2

= 100 + (130 ) = 165 kNm

M t = Ne min where, emin = (0.05 )( 400 ) = 20 mm

= (2500)(0.02) = 50 kNm

Taking the greatest value; Mt = 242 kNm

Now, fcu = 40 N/mm2 fy = 460 N/mm2 and

d 500 − 50 − 20 = = 0.86 . h 500

Therefore use Design Chart No. 38 (taking d/h = 0.85). N ( 2500 )(10 3 ) = bh 500 × 400

= 12.5 N/mm2

Mt (242 )(10 6 ) = bh 2 500 × 400 2

= 3.03 N/mm2

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From the design chart, Asc = 1.0% (2000 mm2) Provide 4 size 32 bars (3216 mm2 )

Ties: You should calculate the minimum size and spacing of the ties as follows; Minimum size =

1 x 32 = 8 mm 4

Maximum spacing = 12 x 32 = 384 mm

Use R8 at 375 centres.

The reinforcement details are as follows;


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400

2T32

R8 - 375 500

2T32

YOU SHOULD SCORE 80% OR MORE TO PASS THIS UNIT. IF YOUR SCORE IS LESS THAN 80%, YOU SHOULD WORK THROUGH THIS UNIT OR PARTS OF THIS UNIT AGAIN. GOOD LUCK!

“The power of work and the power of creativity can be your salvation” NICOLE KIDMAN in Washington Post

END OF UNIT 14


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GLOSSARY

ENGLISH

MALAY

eccentrically loaded column

tiang dibebani sipi

uniaxial bending

lenturan satu paksi

biaxial bending

lenturan dwi-paksi

transverse reinforcement

tetulang membujur

rectangular section

keratan segiempat bujur

square column

tiang segiempat sama

ties

tetulang pemaut

enhance moment

momen tertambah

major axis

paksi utama

minor axis

paksi kedua

symmetry

simetri

braced

berembat

unbraced

tak berembat

slender column

tiang langsing

reduction factor

factor pengurangan

Unit 14 ( DESIGN OF SLENDER COLUMNS )

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