Unit 13 ( DESIGN OF SHORT BRACED COLUMNS )

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/

UNIT 13

DESIGN OF SHORT BRACED COLUMNS

GENERAL OBJECTIVE

To understand how to design short braced reinforced concrete columns

according to BS 8110 requirements.

At the end of this unit you will be able to;

1. calculate the area of longitudinal reinforcement of short braced axially

loaded columns.

2. calculate the area of longitudinal reinforcement of short braced

columns carrying an approximately symmetrical arrangement of

beams.

3. calculate the distance of ties.

4. calculate the diameter of ties.

5. calculate the area of longitudinal reinforcement of short braced

columns carrying axial load and moment.

6. use BS8110 column design charts to get the bar steel area

7. sketch the details of reinforcement.

1

OBJECTIVES

SPECIFIC OBJECTIVES


13.1 Short columns

The effect of deflection or bending to a short column is minimal compared to a

slender column. Therefore, the short column normally fails in compression

due to the crushing of the concrete. Short column is usually designed to resist

maximum bending moment about the critical axis only. This is stated in clause

3.8.4.3 of the code. The maximum axial load that can withstand a column is

denoted by Nuz, which is calculated based on the ultimate capacity of concrete

and reinforcements. Nuz is calculated using the equation given below;

where,

Nuz = ultimate axial load

Ac = net area of column

Asc = area of longitudinal reinforcement

fcu = characteristic strength of concrete

fy = characteristic strength of reinforcement

13.2 Short Braced Axially Loaded Columns

A column can be designed as axially loaded when there is no significant

moment in the column. An example of this is in precast column where there is

no continuity among the structural elements. The equation given can be used

2

INPUT 1


when the axial load is currently acting on the axis of a column. But this perfect

condition rarely occurs because there will always be some inaccuracy in

alignment of the reinforcement and the formworks. To allow some eccentricity

in the column, the concrete and steel stress is reduced. After the reduction, the

equation becomes;

The area of the main reinforcement, i.e. the longitudinal reinforcement can be

calculated using this equation if fcu, fy and Ac are known. This is shown in the

following example.

13.2.1 Design Example

A short braced column is to carry an axial load of 1700kN. The dimension of

the column is 300 mm square. fcu and fy are 30 and 460 N/mm2 respectively.

Use fyv = 250 N/mm2 for ties. What is the area of main reinforcement?

Solution

Using equation 38 of BS 8110,

=

= 1797 mm 2

Use 4T25 bars (Asc = 1960 mm2)

Ties

3


Minimum size =

= 6.25 mm

Maximum size = 12 x 25 = 300 mm

Hence, use R8 at 300 centres.

Details of the reinforcements are shown below in Figure 13.1.

13.3 Braced Short Column carrying an Approximately Symmetrical

Arrangement of Beams

Bending moment in this type of column is very small. This is because the

column supports approximately symmetrical arrangement of beams. The

ultimate axial load is calculated using equation 39 of the code and is

reproduced below;

4

R8-3004T25

Section

R8

T25

T25

T25 T25

Elevation

Figure 13.1: Detail of reinforcement for column.


This equation can be used when the following requirements are fulfilled;

a) The beam spans does not differ by 15% of the longest.

b) The beams are designed for uniformly distributed loads.

The application of this equation is shown in the following example.

13.3.1 Example

C1 is a short braced column carrying 4 beams on its sides. The beams are of

equal span of 8.0 m. Refer figure 13.2.

Given the data;

i) N = 1900 kN (from uniformly distributed load)

5

Figure 13.2: Column layout plan

C1

8m

8m

8m 8m


ii) b= 300 mm

iii) h= 300 mm

iv) fcu = 30 N/mm2

v) fy = 460 N/mm2

vi) fyv=250N/mm2

Calculate the area of longitudinal reinforcement required to carry the load and

sketch details of the reinforcement.

Solution

The following BS 8110`s requirements are met;

a) column is short and braced

b) loads are uniformly distributed

c) spans are equal

Therefore, equation 39 of BS 8110 can be used.

Asc =

=

= 3099 mm 2

Use 4T20 and 4T25 (Asc = 1257 + 1964 = 3221 mm2 )

Ties

6


Minimum size of ties =

= 6.25 mm

Maximum distance of ties = 12 x 20

= 240 mm

Use R8 at 225 centres.

Details of the reinforcement are as shown below;

7

R8 -225

4T20 + 4T25

SECTION

T25 T20 T25

T20 T20

T20 T25T25

R8 - 225

ACTIVITY 13

Figure 13.2: Details Of The Reinforcement


Complete these statements.

13.1 The term short column indicates that no correction is needed to take

account for additional bending due to _______________________.

13.2 In practice, a column is never constructed absolutely plumb nor is the

load applied truly _____________________.

13.3 Some ________________________ of the load exists which induces a

degree of bending.

13.4 The degree of ________________________ accepted as permissible

within these limits is 0.05h

13.5 The expression for the capacity of the column to resist load is given by

the equation _____________________________.

13.6 The expression for the capacity of column to resist load and allowing

for the effect of small moments is ____________________________.

13.7 Design a column for N = 2500 kN within a system of beams of

approximately equal spans, if b = 300 mm and h = 500 mm. Use

fcu = 40 N/mm2 and fy = 460 N/mm2.

8

FEEDBACK 13


Now, check your answers.

13.1 deflection

13.2 slenderness

13.3 axial

13.4 eccentricity

13.5 eccentricity

13.6

13.7

Asc =

=

= 1298 mm 2

Use 4T25 (Asc = 1964 mm 2 )

“Are your answers the same as the printed answers here? Never mind if your

answers are not correct. Please do them again until you get the right answer”

9

INPUT 2


13.4 Short Braced Columns Carrying Axial Load and Moment

In many cases bending moment on a column may be large in relation to the

load or may be applied to a column, which is already carrying a substantial

load. For these cases, design charts are used. The bending moment to be

considered is the largest out-of-balance moment resulting from conditions of

loading. When dealing with moment, h relates to the dimension in the

direction of bending, which is not necessarily the larger dimension.

Column design charts are given in Part 3 of BS 8110. They are for columns

with equal steel in opposite faces. The area of reinforcement obtained from the

chart, Asc is the total area required, half of which is to be placed in each face.

13.5 Design Example (Bending About The Major Axis )

Design a short braced column for N = 3050kN, Mx = 30.6 kNm but the

minimum moment, i.e. 0.05Nh is 61kNm about x-axis. The following data are

known;

b = 300 mm, h = 400 mm, fcu = 35N/mm2, fy = 460N/mm2, fyv = 250 N/mm2,

cover to main reinforcement, c = 30 mm.

Solution

First calculate the ratios and as follows;

10


= 25.4 N/mm 2

= 1.27 N/mm 2

To calculate the effective depth, we have to assume a size of bar, say 25 mm

hence,

= 357 mm

= 0.89

For fcu = 35 N/mm2 and fy = 460 N/mm2, we should use Chart No. 34 (Using

)

11


From the chart, read the value.

(Minimum value = 0.4, maximum value = 6.0)

Therefore,

= 3840 mm 2

= 1920 mm 2

Provide this area in each opposite face.

(4T25 on each opposite face ) .

Ties

Minimum diameter =

= 6.25 mm

Maximum spacing = 12 X 25 = 300 mm

Provide R8 at 250 centres.

The details are shown below;

12


* Note that the longitudinal reinforcement is arranged so that they are

symmetrical about the axis of bending, i.e. the x-axis.

13.6 Design Example (Bending About The Minor Axis)

The term minor axis refers to the y-axis. The bending moment in this direction

is denoted as My. The column in the previous example is to be designed for

My = 5.8kNm and N = 3050 kN.

Now, for bending about the minor axis, (bending in x-x direction);

h = 300 mm, b = 400 mm

d = 300 -30 – 13 (assuming T25 bars are used)

= 257 mm

= 0.86

13

x

x

400

300

2T25

2T25

2T25

2T25

Figure 13.3:Details of longitudinal reinforcement


Using Chart No. 33 (for ,

= 25.4 N/mm 2

= 1.27 N/mm 2

= 3960 mm 2

Provide 6T32 (Asc = 4827 mm2)

Ties

Minimum diameter =

= 8 mm

Maximum spacing = 12 x 32 = 384 mm

Hence, use R10 at 350 centres.

Details of the reinforcement are shown in Figure 13.4.

14

y y

300

400

Figure 13.4: Details of reinforcement

REINFORCED CONCRETE STRUCTURAL DESIGN C4301/UNIT13/ 15

SUMMARY


1. The general method in designing braced and unbraced short columns is

to obtain the axial loads and moments from analysis.

2. The moment from analysis is compared with the minimum moment of

Nemin and the larger value is taken.

3. If certain criteria are met, it is not necessary to calculate the moments.

4. If a column is not subjected to a significant moment (less than Nemin),

equation 38 of BS 8110 can be used.

5. Equation 39 can be used for braced short column when the following

criteria are met;

a) The beams carry a uniformly distributed imposed loads

b) The beam spans do not differ by more than 15% of the longest

beam.

2. Design Charts of Part 3, BS 8110 can be used for a short braced

column carrying axial load and bending moment.

3. When moment is considered, h is taken as the dimension in the

direction of bending.

4. The minimum diameter of ties in a column is one quarter of the

smallest diameter of the main bars.

5. The maximum spacing of the ties in a column is 12 times the smallest

diameter of the main bars.

16

SELF-ASSESMENT


Answer all the questions.

1. A 500mm by 300 mm short column is reinforced with 4T32 bars. If

fcu = 35 N/mm2, calculate the design ultimate capacity, design if Nuz of

the section if it is subjected to axial load only.

(3 marks)

2. A short braced axially loaded column 400mm by 300 mm is to be

designed for an axial load of 2500 k. Calculate the area of longitudinal

reinforcement allowing some eccentricity. Use fcu and fy as 40 and 460

N/mm2 respectively.

(4 marks)

3. Design the longitudinal reinforcement for a 450mm by 450 mm short

braced column supporting an approximately symmetrical arrangement

of beams. The axial load, N is 3000 kN. Use fcu and fy as 30 and 460

N/mm2 respectively.

(4 marks)

4. Design the longitudinal reinforcement for a 500mm by 300 mm

column section if N = 2300 kN and Mx = 300 kNm. Mx is the bending

moment about the major axis and bending is in the y – axis. Assume

the column is short and braced. The following information is known ;

fcu = 40 N/mm2

17


fy = 460 N/mm2

(8 marks)

5. Design the longitudinal reinforcement for a 500mm by 300 mm short

braced column given that N = 2300 kN , My = 120 kNm, fcu = 40

N/mm2 , fy = 460 N/mm2 and

(8 marks)

6. Design the ties for the column in Question 5.

(3 marks)

Check your score

18

FEEDBACK ON SELF-ASSESSMENT

1


1. …………………………………

= N……………..

= 3650 kN. …………………………………………….

(3 marks)

2.

……………………………………….....

= …………………....

= 1681 mm 2 ……………………………………………

Provide 4T25 ( Asc = 1964 mm 2 ) …………………………

(4 marks)

3.

Asc = ………………………………………

= …………………..

= 2835 mm 2 ……………………………………………..

Provide 6T25 ( Asc = 2946 mm 2 ) ………………………….

(4 marks)

4. ………………………………………..

= 15.33 N/mm 2 ………………………………………..

19

1

1

1

1

1

1

11

1

1

11

1

1

1

1


= 4.00 N/mm 2 ……………………………………..

(3 marks)

Use chart no. 38 for ………………………….

………………………………………………

……………………………………..

= 3450 mm 2 ……………………………………………

Provide 6T32 ( Asc = 4825 mm 2 )……………………….

(5 marks)

5. In this case, b = 500 mm and h = 300 mm.

……………………………………

= 15.33 N/mm 2 ………………………………..

………………………………….

= 2.67 N/mm 2 ………………………………

From chart no. 38………………………………....

20

1

1

1

1

1

1

1

1

1

1

1

1


………………………………………

= 1950 mm 2 …………………………………

Provide 4T25 ( Asc = 1963 mm 2 ) …………………

(8 marks)

6. Minimum diameter =

= 6.25 mm……………………..

Maximum spacing = 12 x 25 = 300 mm centres………

Use R8 at 275mm centres……………………..

(3 marks)

21

1

1

1

1

1

“The power to live with joy and victory,”

says Norman Vincent Peale “is available

to you and me. This power can lead you

to a solution to your problem, help you to

meet your difficulties successfully and fill

your heart with peace and contentment.”

END OF UNIT 13

Unit 13 ( DESIGN OF SHORT BRACED COLUMNS )

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