Unit 13 ( Design Of Short Braced Columns ) 5s5r2

REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 13 DESIGN OF SHORT BRACED COLUMNS

OBJECTIVES

GENERAL OBJECTIVE To understand how to design short braced reinforced concrete columns according to BS 8110 requirements. SPECIFIC OBJECTIVES

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.

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INPUT 1

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; N uz = 0.45 f cu Ac +0.87 f y Asc

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

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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;

N = 0.4 f cu Ac + 0.75 f y Asc

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, N = 0.4 f cu Ac + 0.75 f y Asc

Asc =

=

N − 0.4 f cu Ac 0.75 f y 1700 ×10 3 − 0.4(30 )( 300 ) 2 0.75 × 460

= 1797 mm2 Use 4T25 bars (Asc = 1960 mm2) Ties

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Minimum size =

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1 × 25 4

= 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.

T2 5

R8-300

T25

R8

4T25

T25

T25

Section

Elevation Figure 13.1: Detail of reinforcement for column.

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 s approximately symmetrical arrangement of beams. The ultimate axial load is calculated using equation 39 of the code and is reproduced below;

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N = 0.35 f cu Ac +0.67 f y Asc

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.

8m

8m

8m

C1

8m

Figure 13.2: Column layout plan

Given the data; i)

N = 1900 kN (from uniformly distributed load)

REINFORCED CONCRETE STRUCTURAL DESIGN

ii)

b= 300 mm

iii)

h= 300 mm

iv)

fcu = 30 N/mm2

v)

fy = 460 N/mm2

vi)

fyv=250N/mm2

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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. N = 0.35 f cu Ac +0.67 f y Asc

Asc =

=

N − 0.35 f cu Ac 0.67 f y 1900 ×10 3 − 0.35 (30 )( 300 × 300 ) 0.67 × 460

= 3099 mm2

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

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Minimum size of ties =

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1 × 25 4

= 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;

T25

T20

T25

R8 - 225

R8 -225

T20

T20

4T20 + 4T25

T25

T20

T25

SECTION

Figure 13.2: Details Of The Reinforcement

ACTIVITY 13

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Complete these statements. 13.1

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

13

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Now, check your answers. 13.1

deflection

13.2

slenderness

13.3

axial

13.4

eccentricity

13.5

eccentricity

13.6

N = 0.4 f cu Ac + 0.75 f y Asc

13.7

N = 0.35 f cu Ac +0.67 f y Asc N = 0.35 f cu Ac +0.67 f y Asc

Asc =

=

N − 0.35 f cu Ac 0.67 f y 2500 ×10 3 − 0.35 ( 40 )( 300 )( 500 ) 0.67 ( 460 )

= 1298 mm2 Use 4T25 (Asc = 1964 mm2)

“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”

INPUT 2

REINFORCED CONCRETE STRUCTURAL DESIGN

13.4

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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

M N and as follows; bh bh 2

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N 3050 ×10 3 = bh 300 × 400

= 25.4 N/mm2

M 61 × 10 6 = bh 2 300 × 400 2

= 1.27 N/mm2

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

d = 400 − 30 − 13

= 357 mm

d 357 = h 400

= 0.89

For fcu = 35 N/mm2 and fy = 460 N/mm2, we should use Chart No. 34 (Using d = 0.90 ) h

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From the chart, read the

100 Asc bh

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value.

100 Asc = 3.2 (Minimum value = 0.4, maximum value = 6.0) bh

Therefore, Asc =

3.2 × 300 × 400 100

= 3840 mm2

Asc 3840 = 2 2

= 1920 mm2

Provide this area in each opposite face. (4T25 on each opposite face ) .

Ties Minimum diameter =

1 × 25 4

= 6.25 mm Maximum spacing = 12 X 25 = 300 mm Provide R8 at 250 centres. The details are shown below;

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x 2T25

2T25

300

2T25 2T25

400

x Figure 13.3:Details of longitudinal reinforcement

* 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 d 257 = h 300

= 0.86

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Using Chart No. 33 (for

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d = 0.85 ) , h

N 3050 ×10 3 = bh 400 × 300

= 25.4 N/mm2

M 45.8 × 10 6 = bh 2 400 × 300 2

= 1.27 N/mm2

100 Asc = 3.3 bh

Asc =

3.3 × 400 × 300 100

= 3960 mm2

Provide 6T32 (Asc = 4827 mm2) Ties Minimum diameter =

1 ×32 4

= 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.

300 y

y

400 Figure 13.4: Details of reinforcement

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SUMMARY

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1.

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The general method in deg 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.

SELF-ASSESMENT

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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 ing 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

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fy = 460 N/mm2 d = 0.85 h

(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

d = 0.85 h

(8 marks)

6. Design the ties for the column in Question 5. (3 marks)

ON SELF-ASSESSMENT

Check your score 1

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1.

N uz = 0.45 f cu Ac +0.87 f y Asc

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…………………………………

1 = 0.45 (35 )( 500 ×300 ) + 0.87 ( 460 )( 3218 ) N…………….. 1

= 3650 kN. …………………………………………….

(3 marks) 2.

N = 0.4 f cu Ac + 0.75 f y Asc

Asc =

1 N − 0.4 f cu Ac ………………………………………..... 0.75 f y

=

2500 ×10 3 − 0.4( 40 )( 400 × 300 ) 1 ………………….... 0.75 × 460

= 1681 mm2……………………………………………

Provide 4T25 ( Asc = 1964 mm2 )…………………………

1

1 (4 marks)

3.

N = 0.35 f cu Ac +0.67 f y Asc

Asc =

N − 0.35 f cu Ac 1 ……………………………………… 0.67 f y

=

3000 ×10 3 − 0.35 (30 )( 450 × 450 ) ………………….. 1 0.67 × 460

= 2835 mm2……………………………………………..

Provide 6T25 ( Asc = 2946 mm2 )………………………….

1

1 (4 marks)

4.

N 2300 ×10 3 = ……………………………………….. bh 300 × 500

= 15.33 N/mm2………………………………………..

1 1

1

1

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Mx 300 × 10 6 = bh 300 × 500 2

1 2

= 4.00 N/mm …………………………………….. (3 marks) Use chart no. 38 for

1

d = 0.85 …………………………. h

1

100 Asc = 2.3 ……………………………………………… bh

1

2.3 × 300 × 500 Asc = …………………………………….. 100

1

= 3450 mm2……………………………………………

1

Provide 6T32 ( Asc = 4825 mm2 )……………………….

(5 marks) 5.

In this case, b = 500 mm and h = 300 mm. 1

N 2300 ×10 3 = …………………………………… bh 500 × 300

1

= 15.33 N/mm2……………………………….. My

120 × 10 6 …………………………………. = bh 2 500 × 300 2

= 2.67 N/mm2 ………………………………

1

1

1 From chart no. 38………………………………....

1

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100 Asc = 1.3 ……………………………………… bh

Asc =

1.3 × 500 × 300 100

1

2

= 1950 mm …………………………………

Provide 4T25 ( Asc = 1963 mm2 )…………………

1 (8 marks)

6. Minimum diameter =

1 × 25 4

= 6.25 mm……………………..

1

1 Maximum spacing = 12 x 25 = 300 mm centres……… 1 Use R8 at 275mm centres……………………..

(3 marks)

END OF UNIT 13 “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.”