Unit 13 ( DESIGN OF SHORT BRACED COLUMNS )

REINFORCED CONCRETE STRUCTURAL DESIGN

C4301/UNIT13/ 1

UNIT 13 DESIGN OF SHORT BRACED COLUMNS

OBJECTIVES

GENERAL OBJECTIVE To unders understan tand d how to design design short short braced braced reinfo reinforced rced concret concretee column columnss according to BS 8110 requirements. SPECIFIC OBJECTIVES

At the end of this unit you will be able to; 1.

calcu calculat latee the the area area of long longit itud udin inal al reinf reinfor orcem cemen entt of of shor shortt brace braced d axia axiall lly y loaded columns.

2.

calc alculate ate the area of longitu itudinal rei reinfor forcem cement of short brace aced column columnss carryi carrying ng an approx approxima imately tely symmet symmetric rical al arrang arrangeme ement nt of beams.

3.

calculate the distance of ties.

4.

calculate the diameter of ties.

5.

calc alculate ate the area of longitu itudinal rei reinfor forcem cement of short brace aced columns carrying axial load and moment.

6.

use use BS8 BS811 110 0 col colum umn n des desig ign n cha chart rtss to to get get the the bar bar stee steell are areaa

7.

sketc etch the details of rei reinfor forcem cement ent.


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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 N uz, 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 A s

where, Nuz = ultimate axial load Ac = net area of column Asc = area of longitudinal reinforcement f cu cu = characteristic strength of concrete f y = 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 curre ntly acting on the axis of a column. But this perfect condit condition ion rarely rarely occurs occurs becaus becausee there there will will always always be some some inaccu inaccuracy racy 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

f cu Ac

= 0.4

+ 0.75

f y A sc sc

The area of the main reinforcement, i.e. the longitudinal reinforcement can be calculated using this equation if f cu cu, f y 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 2 the column is 300 mm square. f cu cu and f y are 30 and 460 N/mm respectively. 2 Use f yv yv = 250 N/mm for ties. What is the area of main reinforcement?

Solution

Using equation 38 of BS 8110, N

f cu Ac

= 0.4

A sc

=

=

+ 0.75

f y A sc sc

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 (A sc = 1960 mm 2) Ties

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Minimum size =

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

Brac Br aced ed

Shor Sh ortt

Colu Column mn

carr carryi ying ng

an

Appr Approx oxim imat atel ely y

Symm Symmet etri rica call

Arrangement of Beams

Bending moment in this type of column is very small. This is because the column column suppor supports ts approx approxima imatel tely y symmet symmetrica ricall arrange arrangemen mentt of beams. beams. The ulti ultima mate te axia axiall load load is calcu calculat lated ed usin using g equa equati tion on 39 of the the code code and and is reproduced below;

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N

= 0.35

f cu Ac

+0.67

f y A sc sc

This equation can be used when the following requirements are fulfilled; a) The beam beam spans spans does not differ differ by by 15% 15% of the longes longest. t. b) The beams beams are design designed ed for uniformly uniformly distrib distributed uted loads. 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 = 190 1900 kN (fro (from m unif unifo ormly rmly distr istrib ibu uted ted load load))

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ii)

b= 300 mm

iii)

h= 300 mm

iv)

fcu = 30 N/mm2

v)

f y = 460 N/mm2

vi)

2 fyv=250N/mm

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) colu column mn is is shor shortt and and brac braced ed b) loads loads are are unifo uniforml rmly y distr distribu ibuted ted c) span spanss are are equ equal

Therefore, equation 39 of BS 8110 can be used. N

= 0.35

Asc =

=

f cu Ac

+0.67

f y A sc sc

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 (A sc = 1257 + 1964 = 3221 mm 2 ) Ties

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Minimum size of ties =

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 13.1

The The term shor shortt colum column n indica indicate tess that no corr correct ectio ion n is needed needed to take take account for additional bending due to _______________________ _______________________..

13.2 13.2

In practi practice, ce, a colum column n is never never cons constru tructe cted d absolu absolutel tely y plumb plumb nor nor is the the load applied truly _____________________. _____________________.

13.3 13.3

Some Some _______ __________ ______ ______ ______ _____ _____ ___ of of the load load exists exists which which induc induces es a degree of bending.

13.4 13.4

The The degr degree ee of ____ ______ ____ ____ ____ ____ ____ ____ ____ ____ ____ __ acce accept pted ed as perm permis issi sibl blee within these limits is 0.05h

13.5 13.5

The expr express ession ion for the the capac capacity ity of of the colu column mn to resist resist load load is is given given by by the equation _______________________ _____________________________. ______.

13.6 13.6

The expr express ession ion for for the the capaci capacity ty of colu column mn to resi resist st load load and and allowi allowing ng for the effect of small moments is ______________________ ____________________________. ______.

13.7 13.7

Desi Desig gn a colu colum mn for for N = 250 2500 kN with within in a sys system tem of beam eams of approximately equal spans, if b = 300 mm and h = 500 mm. Use 2 2 f cu cu = 40 N/mm and f y = 460 N/mm .

FEEDBACK 13


C4301/UNIT13/ 9

Now, check your answers. 13.1

deflection

13.2

slenderness

13.3

axial

13.4

eccentricity

13.5

eccentricity

13.6

N

= 0.4

13.7

N

= 0.35

f cu Ac

+0.67

f y Asc

N

= 0.35

f cu Ac

+0.67

f y A sc sc

f cu Ac

Asc =

=

+ 0.75

f y 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 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”

INPUT 2


13.4 13.4

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Short Short Brac Braced ed Colu Columns mns Carryi Carrying ng Axia Axiall Load Load and and Mome Moment nt

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 load loadin ing. g. When When deali dealing ng with with mome moment nt,, h relat relates es to the the dime dimens nsio ion n in the 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, M x = 30.6 kNm but the minimum moment, i.e. 0.05Nh is 61kNm about x-axis. The following data are known; 2 b = 300 mm, h = 400 mm, f cu 35N/mm2, f y = 460N/mm2, f yv cu = 35N/mm yv = 250 N/mm ,

cover to main reinforcement, c = 30 mm.

Solution

First calculate the ratios

N bh

and

M bh 2

as follows;


N

=

bh

3050 300

10

×

×

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3

400

= 25.4 N/mm2

M bh

2

=

61× 10

6

300 300 × 400 400

2

= 1.27 N/mm2

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

d = 400 400 − 30 − 13

= 357 mm

d h

=

357 357 400

= 0.89

2 2 For f cu cu = 35 N/mm and fy = 460 N/mm , we should use Chart No. 34 (Using

d h

=

0.90 )


100 A sc

From the chart, read the 100 A sc bh

=

bh

value.

3.2 (Minimum value = 0.4, maximum value = 6.0)

Therefore, A sc

=

3.2 × 300

×

400

100

= 3840 mm2

A sc 2

3840 =

C4301/UNIT13/ 12

2

= 1920 mm2

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

Ties

Minimum diameter =

1 4

×

25

= 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 Note that that the the long longit itud udin inal al reinf reinfor orcem cemen entt is arran arrange ged d so that that they they are are symmetrical about the axis of bending, i.e. the x-axis.

13.6 13.6

Design Design Exampl Examplee (Be (Bendi nding ng About About The Minor Minor Axis) Axis)

The term minor axis refers to the y-axis. The bending moment in this direction is denoted as M y. 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 h

=

257 300

= 0.86


Using Chart No. 33 (for N

=

bh

3050 400

10

× ×

d h

=

C4301/UNIT13/ 14

0.85 ) ,

3

300

= 25.4 N/mm2

M bh

2

=

45.8 × 10

6

400 400 × 300 300

2

= 1.27 N/mm2

100 100 A sc

bh

A sc

=

=

3.3

3.3 × 400 × 300 100

= 3960 mm2

Provide 6T32 (A sc = 4827 mm 2) 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


SUMMARY

C4301/ C4301/UNI UNIT13 T13// 15


1.

C4301/UNIT13/ 16

The The gen gener eral al meth method od in desi design gnin ing g bra braced ced and and unb unbrac raced ed shor shortt col colum umns ns is to obtain the axial loads and moments from analysis.

2.

The The mome moment nt from from anal analys ysis is is is com compa pare red d with with the the mini minimu mum m mome moment nt of of Nemin and the larger value is taken.

3.

If cert certai ain n crit criter eria ia are are met met,, it it is is not not neces necessa sary ry to calc calcul ulat atee the the mome moment nts. s.

4.

If a colu column mn is not not sub subje ject cted ed to to a sig signi nifi fica cant nt mom momen entt (les (lesss tha than n Ne Nemin), equation 38 of BS 8110 can be used.

5.

Equa Equati tion on 39 can can be be use used d for for bra brace ced d sho short rt col colum umn n whe when n the the fol follo lowi wing ng criteria are met; a)

The The bea beams ms carr carry y a unif unifor orml mly y dis distr trib ibut uted ed impo impose sed d load loadss

b) b)

The The bea beam m spa spans ns do not not dif diffe ferr by by mor moree tha than n 15% 15% of the the lon longe gest st beam.

2.

Design Charts of Par Part 3, BS 811 8110 can be used for 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 The maxi maximu mum m spac spacin ing g of of the the tie tiess in in a col colum umn n is is 12 12 tim times es the the smal smalle lest st diameter of the main bars.

SELF-ASSESMENT


C4301/UNIT13/ 17

Answer all the questions.

1.

A 500 500mm mm by 300 300 mm mm sho short rt colu column mn is rein reinfo forc rced ed with with 4T32 4T32 bars bars.. If If 2 f cu cu = 35 N/mm , calculate the design ultimate capacity, design if N uz of

the section if it is subjected to axial load only. (3 marks)

2.

A sho short bra braced ced axi axially loa loaded col column 400 400mm by 30 300 mm is is to be be designed for an axial load of 2500 k. Calculate the area of longitudinal reinforcement allowing some eccentricity. Use f cu cu and f y as 40 and 460 N/mm2 respectively. (4 marks)

3.

Desi Design gn the the long longit itud udin inal al rei reinf nfor orce ceme ment nt for for a 450m 450mm m by by 450 450 mm shor shortt braced column supporting an approximately symmetrical arrangement of beams. The axial load, N is 3000 kN. Use f cu and f y as 30 and 460 N/mm2 respectively. (4 marks)

4.

Design the longitudinal reinforcement for a 500mm by 300 mm column column section if N = 2300 kN and M x = 300 kNm. M x 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 ; 2 f cu cu = 40 N/mm

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f y = 460 N/mm2 d h

=

0.85

(8 marks)

5.

Desi Design gn the the long longit itud udin inal al rei reinf nfor orce ceme ment nt for for a 500m 500mm m by by 300 300 mm shor shortt braced column given that N = 2300 kN , My = 120 kNm, f cu cu = 40

N/mm2 , f y = 460 N/mm 2 and

d h

=

0.85

(8 marks)

6. Design Design the the ties ties for for the colum column n in Quest Question ion 5. (3 marks)

FEEDBACK ON SELF-ASSESSMENT

Check your score

1

C4301/UNIT13/ 19


1.

N uz

= 0.45

f cu Ac

+0.87

= 0.45 (35 )(500

×

f y A sc …………………………………

1

300 ) + 0.87 ( 460 )(3218 ) N……………..

1

= 3650 kN. …………………………………………….

(3 marks) 2.

N

f cu Ac

= 0.4

A sc

=

=

+ 0.75

N − 0.4 f cu Ac 0.75 f y

2500

10

×

3

−

f y A sc

1 ……………………………………….....

0.4( 40 )(400

×

300 )

0.75 × 460

…………………....

= 1681 mm2……………………………………………

1 1

1 2

Provide 4T25 ( A sc = 1964 mm )……………… ……… ………………… ………… (4 marks) 3.

N

= 0.35

Asc =

=

f cu Ac

+ 0.67

N − 0.35 f cu Ac

0.67 f y 3000

10

×

3

−

f y A sc sc

………………………………………

0.35 (30 )(450

0.67 × 460

×

450 )

1

………………….. 1

= 2835 mm2……………………………………………..

Provide 6T25 ( A sc = 2946 mm 2 )………………… ……… ………………… ………..

1

1 (4 marks)

4.

N bh

2300 =

300 300

×10

3

500 × 500

………………………………………..

= 15.33 N/mm2………………………………………..

1 1

1

1


M x

300 300 × 10

=

bh

C4301/UNIT13/ 20

6

300 × 500

2

1

= 4.00 N/mm2…………………………………….. (3 marks) Use chart no. 38 for

100 A sc

bh

d h

=

1

0.85 ………………………….

1 =

2.3 ………………………………………………

1 A sc

=

2.3 × 300

×

500

100

…………………………………….. 1

= 3450 mm2……………………………………………

1 2

Provide 6T32 ( A sc = 4825 mm )………………………. (5 marks) 5.

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

2300 =

bh

500 500

×10

1

3

……………………………………

× 300 300

1

= 15.33 N/mm2……………………………….. M y bh

2

=

120 120 × 10

1

6

500 × 300 300

2

………………………………….

= 2.67 N/mm2 ………………………………

1

1 From chart no. 38………………………………....

1


100 A sc

bh

A sc

=

= 1 .3

C4301/UNIT13/ 21

………………………………………

1.3 × 500

×

300

100

1 = 1950 mm2…………………………………

Provide 4T25 ( A sc = 1963 mm 2 )…………… ……… ………… ……

1 (8 marks)

6. Mini Minimu mum m dia diame mete terr =

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.”

Unit 13 ( DESIGN OF SHORT BRACED COLUMNS )

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