Unit 14 ( DESIGN OF SLENDER COLUMNS )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT14/

UNIT 14

DESIGN OF SLENDER COLUMNS

GENERAL OBJECTIVE

To be able to identify the braced slender columns design principles according to

BS 8110 requirements.

At the end of this unit you will be able to calculate:

1. the minimum eccentricity, .

2. the deflection , 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.

1

OBJECTIVES

SPECIFIC OBJECTIVES


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, 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 is similar.

2

INPUT 1


Now do the following.

14.1 A braced slender column is a column in when both the ratios and

are greater than ___________________________.

14.2 An unbraced slender column is one in which both ratios and is

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

________________________.

3

ACTIVITY 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.

4

FEEDBACK 14a


14.7 Additional Moment ,

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

The additional moment, is calculated as follows;

5

N

au

INPUT 2

N

Figure 14.1: Deflection of a rectangular column.


Where, N = the ultimate axial load

= column deflection at ultimate limit.

is calculated using equation 32 of the code given below ;

And βa is calculated using equation 34 as follows;

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

ratio where is the effective height of column in the plane considered and

, is the dimension of the smaller column.

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

are present in both axes, x and y, the value is equal to and .

6

Kha au


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:

Where, and

(used for symmetrical reinforced concrete columns).

Therefore, can be rewritten as;

depends on but 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.

7

ACTIVITY 14b


Now do the following exercise. Fill in the blanks.

14.7 The additional moment, is calculated using equation

_________________________.

14.8 The deflection, au is calculated using equation ________________.

14.9 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 , is equal to _____________________.

14.13 When , is equal to ______________________.

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

14.15 When = 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 .

14.17 Calculate for the column section given below using fcu as 40 N/mm2.

8


14.18 Calculate the reduction factor, K if = 3500kN, = 3000 kN and

N = 3100 kN.

9

550 mm

400mm

FEEDBACK 14b


Here are the answers.

14.7

14.8 Kha au

14.9 Table 3.23

14.10 Reduction Factor, K

14.11

14.12 0.07

14.13 1.80

14.14

14.15 600 kNm

14.16

=

= 3447 kN

14.17

=

10


= 2200 kN

14.18

=

=

= 0.8

11

INPUT 3


14.19 Slender Braced Columns

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

below;

The total moment is . The initial moment Mi to be used is simply

the initial moment at an end of the column. BS8110 recommends that M i be taken

as:

(M2 being the larger) for symmetrical bending

And, but not less than for assymmetrical

bending i.e. bending in double curvature. This is shown in Fig 3.20, BS

8110.

12

N

e au

Figure14.2: Deflection of Slender Braced Column


The two equations can be combined as follows:

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

A column in any case is designed for a moment of at least Nemin. Hence we have

further condition that

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

whichever is lesser.

Now do the following exercise.

13

ACTIVITY 14c


Fill in the blanks.

14.19 ___________________ is equal to the sum of initial moment, and the

additional moment

14.20 For symmetrical bending, the initial moment is equal to

____________________________.

14.21 For _________________ bending,

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 ;

a)

b)

c)

14


d)

e)

f)

g)

h)

14.26 For the slender column experiencing initial moments at both ends (shown

below), determine the values of and assuming double curvature.

14.27 Calculate of the column in Question 8 assuming single curvature.

14.28 Calculate the design minimum eccentricity for the column section shown

below;

15

100 kNm

200kNm

Figure 14.3: End Moments in Slender Column


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

16

350mm

200mm

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 , and eminN

14.26

14.27

=

= 80 kNm

14.28

=

= 17.5 mm

17


But, should not be less than 20 mm.

Therefore, the minimum eccentricity is 20 mm.

14.29

=

= 80 kNm

14.30 Maximum design is taken as the greatest value derived from M2, Mi,

Madd , Nemin and M1 . Therefore, the maximum design moment is 800

kNm.

18

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.

Solution

Slenderness ratio:

19

SECTION

14.4:Initial End Moments

M1=10 kNm

M2 = 70 kNmb = 450

d = 240

h = 300

d’= 60

xx


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

=

= 38 kNm

therefore is greater than

Additional moment produced by column deflection is,

=

= 129 kNm

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

20


= 38 + 129

= 167 kNm

=

= 2.0

= 4.12

From the Column Design Chart No.27,

and

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

21


= 38 + 83.91

= 121.91 kNm

From Chart No.27,

=

= 2970 mm 2

Now check the final value of K from the design chart,

= ( ) x 10 -3 kN

= 810 kN

= ( ) x 10 -3 kN

= 3011 kN

Now, calculate the reduction factor as follows;

22


=

= 0.6

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

(1) 2) (3) (4) (5)

K Mt K

1.0 167 4.12 3.2 0.65

0.65 122 3.0 0.6

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 mm 2 )

Ties:

Minimum size = = 8 mm

Maximum spacing = 12 x 32 = 384 mm centres.

23

Table 14.1: Iteration Process of K Value


Use R8 at 375 centres.

Details of the reinforcements are shown below;

24

2T32

2T32

2T32

R8 – 375

SUMMARY

14.5: Rectangular Columns


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,

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)

(iii)

(iv)

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 .

Step 4:

For biaxial bending calculate the following;

a) Mty (total moment about the minor axis )

b) Mtx (total moment about the major axis )

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c) design for N, Mty and Mtx

Step 5:

Calculate the reduction factor using the equation given below ;

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 Self-

Assessment. If your answer is NO, go through Unit 14 again.

26

SELF-ASSESSMENT


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, M1y = 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.

27

(6.5 m)

N

400

500xx

y

y

Slender column

FEEDBACK ON SELF-ASSESSMENT


Now check your answers.

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

a)

= (0.4)(100) + 0.6(M2)

= 112 kNm

b)

c)

d) K = 1.0

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

= 130 kNm

= 112 + 130

= 242 kNm

Mt = M2 = 120 kNm

28


=

= 165 kNm

where,

= (2500)(0.02)

= 50 kNm

Taking the greatest value;

Mt = 242 kNm

Now,

fcu = 40 N/mm2

fy = 460 N/mm2

and .

Therefore use Design Chart No. 38 (taking d/h = 0.85).

= 12.5 N/mm 2

= 3.03 N/mm 2

29


From the design chart,

Asc = 1.0% (2000 mm2)

Provide 4 size 32 bars (3216 mm 2 )

Ties:

You should calculate the minimum size and spacing of the ties as follows;

Minimum size = x 32 = 8 mm

Maximum spacing = 12 x 32 = 384 mm

Use R8 at 375 centres.

The reinforcement details are as follows;

30


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

31

R8 - 375

2T32

2T32

400

500


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

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Documents

Unit 14 ( DESIGN OF SLENDER COLUMNS )