6 column design

6. Design of Reinforced Concrete Column

• This chapter will discuss the following topics:

• Definition of R.C. columns.

• Classification of short / slender (long) columns.

• Determination of short/slender columns by using simplified method.

• Design of axially loaded (subject to axial load mainly) columns by design formulae.

• Design of columns subject to uniaxial bending (axial• Design of columns subject to uniaxial bending (axial load + one bending moment) by using the column design charts.

• Design of columns subject to biaxial bending (axial load + two bending moments).

RC Design and Construction – HKC 2004 (2nd Edition) 6-1

6.1 Definition of R.C. Columns

• In accordance with HK Code, R.C. column isdefined as a vertical member whose the greatergcross-sectional dimension does not exceed fourtimes the smaller dimension (i.e. h ≤ 4b). The( )requirements for design of columns are treated inHK Code: Section 6.2.1 and Section 9.5.

x

by y

hx


6.2 Classification of Column• A column may be considered braced in a given plane if

lateral stability to the structure as a whole is provided bywalls or bracing or buttressing designed to resist all lateralwalls or bracing or buttressing designed to resist all lateralforces in that plane. It should otherwise be considered asunbraced.

• A braced column is considered as short if both thel d ti l /h d l /b l th 15 If ithslenderness ratio lex/h and ley/b are less than 15. If either

ratio is greater than 15, the column is considered as slender(long).( g)

• An unbraced column is considered short if both theslenderness ratio lex/h and ley/b are less than 10. If eitherratio is greater than 10, the column is considered as slender(long)


(long).

6.2 Classification of Column

Where,– h is the column depth perpendicular to the x-x axis.

– b is the column width perpendicular to the y-y axis.p p y y

– lex is the effective height of column when bending isex g gabout x-x axis.

– ley is the effective height of column when bending isabout y-y axis.


6.2.1 Effective Height of Column

• The effective height (l ) of a column depends on:The effective height (le) of a column depends on:– the actual height between floor beams;

– the column sectional dimensions h * b;the column sectional dimensions h b;

– the end conditions;

– whether the column is braced or unbraced– whether the column is braced or unbraced.


6.2.2 - Effective Height Estimation• Cl. 6.2.1.1 gives the following general equation

for obtaining effective heights:

le = β loe β o

• where lo is the clear height between end restraintswhere lo is the clear height between end restraintsand β is a coefficient from tables 6.11 and 6.12 ofthe Code for braced and unbraced columnsrespectively.

N ll l i NOT l t l i β d l• Normally lex is NOT equal to ley since β and lohave different values when column bends about

dRC Design and Construction – HKC 2004 (2nd Edition) 6-6

x-x and y-y axes.

6.2.2 - Effective Height Estimation

• Most of the R.C. structures are monolithicconstruction and therefore the end condition iseither condition 1 or condition 2.

• The diagram below provides a betterThe diagram below provides a betterunderstanding of the condition 1 & 2.

Fld1

Floor

Beam

col

d2

Floor

col


6.2.2 - Effective Height Estimation

• If (d1) ≥ (d2) ⇒ condition 1If (d1) ≥ (d2) ⇒ condition 1

• If (d1) < (d2) ⇒ condition 2


6.3 Design Provision

• Minimum % of Reinforcement

– The min. % of reinforcement for both grade 250 andgrade 460 reinforcement isg

100 Asc/Acc = 0.8

Asc - Area of steel reinforcement in compression.

A Area of concreteAcc - Area of concrete.


6.3 Design Provision

• Maximum Area of Reinforcement

– The max. % of reinforcement ≤ 6% of the gross cross-sectional area of a vertically casted column.y

– The max. % of reinforcement ≤ 8% of the gross cross-The max. % of reinforcement ≤ 8% of the gross crosssectional area of a horizontally casted column.

– The max. % of reinforcement at laps ≤ 10% of the gross cross-sectional area of column. (see cl.9.5.1)


6.3 Design Provision• Requirements for Links / Binders / Transverse

Reinforcement (see figure 9.5)

– The φlink of links ≥ 6mm or 1/4 φmax,M.B. of the largest longitudinalbarsbars.

– The max. spacing ≤ 12 φmin,M.B. of the smallest longitudinal bar.

– The links should pass through every corner bar and each alternatebar in an outer layer.

– The links passing round the bar shall have an included angle of notmore than 135o.more than 135 .

– No bar is to be further than 150mm from a restrained bar.


Fig. 6.1 - Requirements for Links /Binders

• Link size ≥ 1/4*32 = 8mm

• Link spacing ≤ 12*20 = 240 mmLink spacing ≤ 12 20 240 mm

• Use R8-links-240 c/cCorner BarCorner Bar

alt.bar

T32T20 T20 Corner BarCorner Bar bar

o

T32T32

T20 T20

alt.bar

alt.bar

< 135o

T32T32

T20T20

bar

T20

T20T20

T20

Corner BarCorner Bar alt.b

T32T32

T32

T20T20


bar

Fig. 6.2 Part Plan of a Braced Structure C1 C2 C3

B1-1 (250 X 450) B1-2 (250 X 450)

S1 S1 S1 S12

(250

X 4

50)

(250

X 4

50)

2 (2

50 X

450

)

(250

X 4

50)

250

X 4

50)

(125) (125) (125) (125)B

4-2

B5-

2

B6-

2

B7-

2 (

B8-

2 (2(125) (125) (125) (125)

C4C5 C6

B2-1 (300 X 500) B2-2 (300 X 500)

50)

50)

50)

50)

50)

S1S1S1S1

B4-

1 (2

50 X

4

B5-

1 (2

50 X

4

B6-

1 (2

50 X

45

B7-

1 (2

50 X

45

B8-

1 (2

50 X

4

(125)(125)(125)(125)

C7 B3-1 (250 X 450) B3-2 (250 X 450)


C7C8 C9

( ) B3 2 (250 X 450)

6.4 Short Braced Axially Loaded Columns

– The Code gives the following expression for theultimate load N that a short braced axially loadedcolumn can support. (i.e. col. C5 in Fig. 6.2 supporting

t i l t f fl b B B2 1a symmetrical arrangement of floor beams: Beam B2-1and B2-2 have equal span, and beam B6-1 and B6-2have equal span.)have equal span.)

N = 0 4f A + 0 75A f (1)N 0.4fcuAc + 0.75Ascfy (1)

– where A is the net cross-sectional area of conc inwhere Ac is the net cross-sectional area of conc. incolumn.

– Asc is the area of vertical reinforcement.


Asc is the area of vertical reinforcement.

6.4 Short Braced Axially Loaded Columns

– Design Formula (1) allows for eccentricity due to t ti t l d t i d d d tconstruction tolerances and moment induced due to

pattern load (i.e. one span of beam 2B-1 is loaded with max. load (1.4*D.L. + 1.6* L.L.) and the other span of max. load (1.4 D.L. 1.6 L.L.) and the other span of beam 2B-2 loaded with min. load (1.0*D.L.))


6.4 Short Braced Axially Loaded Columns• Short braced columns supporting an approximately

symmetrical arrangement of beams

– These beams must be designed for uniformlydistributed loads.

– The span must not differ by more than 15% of thellonger span.

– The ultimate load is given by the expression:

N = 0.35fcuAc + 0.67fyAsc (2)


Example 6.1A short braced axially loaded column supporting asymmetrical arrangement of beams is reinforced with 4T25b Th l 325 hbars. The column is 325 mm square in section as shown inFig. 6.3. Find the ultimate axial load that the column cancarry and the spacing and the diameter of the links requiredcarry and the spacing and the diameter of the links required.The materials are grade 35 concrete and grade 460reinforcement.f


Solution : Ex. 6.1Steel area Asc = 4*491 = 1964 mm2

Concrete area Ac = 3252 - 1964 = 103661 mm2

N = 0.4fcuAc + 0.75fyAsc

= 0.4*35*103661 + 0.75*460*1964 = 2129 kN

Links size shall not be less than 6mm

or 1/4 * largest longitudinal bars

= 1/4*25 = 6.25 mm.


Solution : Ex. 6.1The spacing of binders ≤ 12 times the diameter of the smallest longitudinal bars = 12*25 = 300 mm.

∴ PROVIDE R8-300 c/c links

325

532

5

4T25 R8 - 300c/c

Fi 6 3


Fig. 6.3

Example 6.2A short braced axially loaded column supporting asymmetrical arrangement of beams carries an ultimate axiall d f 2300 kN Th l 325 * 325 F d hload of 2300 kN. The column size is 325 * 325 mm. Find thearea of the longitudinal reinforcement required and selectsuitable bars The materials are grade 30 concrete andsuitable bars. The materials are grade 30 concrete andgrade 460 reinforcement.


Solution: Ex. 6.2N = 0.4fcuAc + 0.75fyAsc

2300*103 0 4*30*(3252 A ) 0 75*460*A2300*103 = 0.4*30*(3252 - Asc) + 0.75*460*Asc

2∴ Asc = 3101 mm2

Provide 4T32 bars (Asc = 4*804 = 3216 mm2)

0 8033216*100A100 sc

6.0<

0.8>0.3325*325

3216100

bh

A100 sc ==

∴ O.K.

6.0 <


6.5 Uniaxial Bending (N + Mx or + My)

Code Provisions

– In a monolithically constructed braced frame, the axialforce in column can be calculated by assuming thaty gbeams are simply supported (i.e. by tributary area).

– If the arrangement of beam is symmetrical, the columncan be designed for axial load only as discussed in 6.4.g y

– The column may also be designed for axial load and aThe column may also be designed for axial load and amoment due to the nominal eccentricity.


6.5 Uniaxial Bending (Nominal eccentricity)

• Nominal EccentricityNominal Eccentricity– The min. design moment of a column should not be

less than the (ultimate load) * (min. eccentricity emin).( ) ( y min)

– The min. eccentricity emin equals to 0.05 times theoverall dimension of the column in the plane ofbending, but not more than 20mm, e.g.

• When h ≥ 400 mm, emin = 20 mm.

• When h < 400 mm, emin = 0.05*h


6.5 Uniaxial Bending– The design calculation of short column subject to

uniaxial bending is normally done by using ColumnD i Ch f BS8110 P 3 (1985)Design Charts of BS8110: Part3 (1985).

– When using the column design charts, one has toobserve the values of fcu, fy and the ratio of d/h and

l t th i t d i h tselect the appropriate design chart.

A h d i i f l d i h b d– As the derviation of column design charts are based onsymmetrical arrangement of reinforcement, specialattention should be paid in detailing or consult relevantattention should be paid in detailing or consult relevanttextbook if you do not follow the arrangement ofrebars.


Modes of Failure• The relative magnitude of the moment (M) and axial load (N) govern

the modes of failure. With large effective eccentricity (e = M/N) atensile failure is likely, but with a small eccentricity a compressiontensile failure is likely, but with a small eccentricity a compressionfailure is more likely. Point b in Fig. 6.4 presents a “balanced section”.That is to say the concrete reaches the ultimate strain (strain = 0.0035for f ≤ 60 MPa) and at the same time the reinforcement yields (strainfor fcu ≤ 60 MPa) and at the same time the reinforcement yields (strain≥ 0.002).

No

CompressionN

Nb l b

r

CompressionFailure

0M

Nbal b

Mbal

TensionFailure


Fig. 6.4 - Bending plus axial load chart with modes of failure

Example 6.3A short R.C. Column is subject to an ultimate axial load of5000 kN and a major ultimate moment of 300 kNm. The

l 400*600 D h fcolumn section is 400*600. Determine the area oflongitudinal reinforcement required. Given that fcu = 35N/mm2 and f = 460 N/mm2; Nominal cover = 40 mm; the linkN/mm and fy 460 N/mm ; Nominal cover 40 mm; the linksize is assumed to be 10 mm.


Solution: Ex. 6.3fcu = 35 N/mm2, fy = 460 N/mm2

d = 600 - 40 - 10 - 40/2 = 530 mm

d/h = 0.88 ≈ 0.85

N/bh = 5000*103/(400*600) = 20.8

M/bh2 = 300*106/(400*6002) = 2.08( )

From column design chart 33g

100Asc/bh = 2.6

∴ Asc = 2.6*400*600/100 = 6240 mm2sc

Provide 8T32 (Asc = 6432 mm2)


Provide 8T32 (Asc 6432 mm )

Example 6.4 - Column Design Using Design Charts

Fig. 6.5a and 6.5b show a part plan and an elevation ofan industrial building respectively, for which the centrecolumn C2 is to be designed. The frames at 3.5 mcentre, are braced against lateral forces, and support thef ll l dfollowing loads:

Dead load gk = 15 kN/m2,

Imposed load qk= 25 kN/m2k

Characteristic material strengths are fcu = 35 N/mm2 forthe concrete and fy = 460 N/mm2 for the steel. Nominalcover to reinforcement = 40 mm.


Example 6.4 - Fig. 6.5a (Part Plan)

3500

3

C1

C2(400 x 400) C3

(300 x 700) (300 x 700)B1 B2

500

( )

35

6500 4500


Example 6.4 - Fig. 6.5b (Elevation)

00

3/F 3B1 (300 x 700 dp) 3B2 (300 x 700 dp)

35

2/F2B1 (300 x 700 dp) 2B2 (300 x 700 dp)

3500

2/F

1/F1B1 (300 x 700 dp) 1B2 (300 x 700 dp)

P

3500

C1 C2 (400x400) C36500 4500

G/F

Section through the frame

C1 C2 (400x400) C3


Fig. 6.5b - Columns in an industrial building

Solution: Ex. 6.4 (Design Load) Maximum ultimate load at each floor per meter length of thebeam.

= 3.5*(1.4gk + 1.6qk)

= 3.5*(1.4*15 + 1.6*25) = 213.5 kN/m

Minimum ultimate load at each floor per meter length of thebeam.

= 3.5*(1.0gk)

= 3.5*(1.0*15) = 52.5 kN/m


Solution: Ex. 6.4 (Design Axial Load for C2)

The critical arrangement of load which will cause themaximum moment in column is shown in fig. 6.6a.

Column loads

3/F & 2/F = 2*213.5*11/2 = 2349 kN

1/F = 213.5*6.5/2 + 52.5*4.5/2 = 812 kN

Col. Self-weight, 0.42*24*3.5*3 = 40 kN

Total N =3201 kN

– As the difference in spans of beam B1 and B2 is greater than 15%,i lifi d d i f l f 6 4 b d Th lsimplified design formulae of 6.4 cannot be used. The column

shall be designed for axial load plus bending (i.e. using columndesign charts)


Solution: Ex. 6.4 (Design Moment)A simplified approach is used to determine the bending moment of column C2 (G/F - 1/F). The column moment is b i d b i h b l d d h l iobtained by using the unbalanced moment and the relative

stiffness (4EI/L) of members meeting at a joint.

The loading arrangement and the sub-frame for determining the column moments at the first floor is shown in fig 6 6bthe column moments at the first floor is shown in fig. 6.6b. Member stiffnesses are:-

K bh

LB

B

1 13

1 1

34

2

1

2 12

1

2

0 3 0 7

12 6 56 596 10= = = −. * .

* .. *

B1 1 6 5.


Solution: Ex. 6.4 (Design Moment)

K bh

LB1 2

3 34

2

1

2 12

1

2

0 3 0 7

12 4 59 528 10= = = −. * .

*. *

L B1 22 2 12 2 12 4 5* .

Kcol = = −0 46 095 10

44.

. *

∴ ΣK = (6.596 + 9.528 + 6.095*2)10-4 = 28.31*10-4

col 12 3 5* .

( )

Distribution factor for the column

= = =K

Kcol

Σ6 095

28 310 215

.

..

KΣ 28 31.


Solution: Ex. 6.4 (Design Moment)

Fixed end moments at joint P are:

FEM1B1 = 213.5*6.52 / 12 = 752 kNmFEM1B1 213.5 6.5 / 12 752 kNm

FEM1B2 = 52.5*4.52 / 12 = 89 kNm

∴ the unbalanced moment

= (752 - 89) = 663 kNm

Column moment M

0 215*663 143 kN= 0.215*663 = 143 kNm.


Solution: Ex. 6.4 (Design Reinforcement)Design column C2 (G/F - 1/F) reinforcement by using designchart 33.

N

bh= =

3201 10

400 40020 0

3*

*.

M

bh2

6

2

143 10

400 4002 23= =

*

*.

fcu = 35 N/mm2, fy = 460 N/mm2

bh 400 400*

y

d = 400 - 40 -10 - 32/2 = 334

d/h = 334/400 = 0.835 ≈ 0.85


Solution: Ex. 6.4 (Design Reinforcement)From column design chart no. 33,

62100

=Asc

∴ Asc = 2.6*4002/100 = 4160 mm2

6.2=bh

sc

Provide 4T32 + 2T25 (Asc = 4198 mm2)( sc )

0.8 > 62.2400

4198*1001002

==bh

Asc

O.K. 6.0 < 4002

∴bh

* By adopting the similar procedures, students can design the rest of the columns.


Figure 6.6

3/F1.4G + 1.6Q k k

1/F

2/F1.4G + 1.6Q k k

1.4G + 1.6Q k k1.0G k

213.5*6.5=1388 kN52.5*4.5=236 kN

1/F

1/F

C1 C2 C3

1B1 1B2

C2G/FG/F

P

(a) Critical loading arrangement for col. C2 at 1/F (b) Sub-frame and load for

max. col. moment-752kN 2

1B1

kNm -752kNm

-89 kNm -89 kNm

P1B1 1B2

C2

(c) Fixed End Moment

P


(c) Fixed End Moment

Fig. 6.6

6.5 Biaxial Bending (N + Mx + My)

– When a column is subjected to two moments plus axialload, the method as described in HK Code can be used.Basically, the design method converts biaxial bending

t t i l b di t b i imoments to an unaxial bending moment by increasingmoment about one axis. Hence column charts can beused.used.

Which bending moment to be increased depends on the– Which bending moment to be increased depends on therelative values of the moments and the column sectiondimensions. The amount of increase depends on thepratio of the design axial load to the axial load capacity.


6.5 Biaxial Bending (N + Mx + My)

Mx -design ultimate moment about the x-x axis.

-effective uniaxial design moment about theMx ' effective uniaxial design moment about the

x-x axis.

M d i lti t t b t th i

Mx

My - design ultimate moment about the y-y axis.

-effective uniaxial design moment about theM y '

y-y axis.

h - overall depth perpendicular to the x-x axis.h overall depth perpendicular to the x x axis.

- effective depth perpendicular to the x-x axis.

b ll idth di l t th i

h'

b - overall width perpendicular to the y-y axis.

- effective width perpendicular to the y-yb'


axis.

6.5 Biaxial Bending (Increase Bending Moment about one-axis)

IfM

h

M

bx y ' '≥f

h b' '

M Mh

Mx x y''

= + βM Mb

Mx x y'+ β

IfM Mx yIfM

h bx y ' '<

M Mb

M''β

Th ffi i t β i t k f T bl 6 14 f th

M Mh

My y x''

= + β

• The coefficient β is taken from Table 6.14 of theCode and as shown below.


6.5 Biaxial Bending (Increase Bending Moment about one-axis)

Values of the coefficient βN

bhf0 0.1 0.2 0.3 0.4 0.5 ≥ 0.6

bhfcu

β 1.00 0.88 0.77 0.65 0.53 0.42 0.30

Y

b

b'

X X

Y

Mx

X X h

h'

YMy


Example 6.5

The column section shown below is to be designed to resist anultimate axial load of 1800 kN plus moments of Mxx = 85 kNm

d M 95 kN Th h l hand Myy = 95 kNm. The characteristic material strengths arefcu = 30 N/mm2 and fy = 460 N/mm2.

Y

350

X X Mx = 85 kNm

00 X

4033

0

YMy = 95 kNm

280


Solution: Ex. 6.5 (Increased Moment)

M

hx

'.= =

85

3300 258

M

by

'.= =

95

2800 339

M

h

M

bx y

' '< ∴ Increase the bending moment Myh b

M Mb

hMy y x'

'

'= + β

N/bhfcu = 1800*103/(350*400*30) = 0.43

From design table 6.14, β = 0.50g , β

M y ' . * *= + =95 0 5280

33085 131 kNm


Solution: Ex. 6.5 (Increased Moment)As bending is about y-y axis,

d = 280 and d/h = 280/350 = 0.8

N/bh = 1800*103 / (350*400) = 11.4

My’/bh2 = 131*106/(400*3502) = 2.7

From column design chart no. 27,

100Asc / bh = 1.7sc

∴ Asc = 1.7*350*400/100 = 2380 mm2

Provide 6T25 (Asc = 2946 mm2)


Table 2 Area of Steel Reinforcement

Steel Reinforcement

Bar Size Area of one Bar (mm2)

8 50.310 78 510 78.512 11316 20120 31420 31425 49132 80440 125740 1257


Table 6.11 & 6.12 - Effective Height of a Column

RC Design and Construction – HKC 2004 (2nd Edition) 6-47Reproduced from HK Code

Table 6.14 - Values of the Coefficient β (Biaxial Bending)

Reproduced from HK Code


Figure 9.5 : Column transverse reinforcement

Reproduced from HK Code


Chart No. 27

11.4

2.7

Reproduced from BS8110 (1985)

2.7


Chart No. 33

20.8

20.0

2.08 2.23

Reproduced from BS8110 (1985)


Maximum Area of Reinforcement in Members

9.5.1 Longitudinal reinforcement• The area of total longitudinal reinforcement based on the gross cross-

ti l f l h ld t b l th 0 8% S lsectional area of a column should not be less than 0.8%. See clause9.9.2 for ductility requirement.

• Bars should have a diameter of not less than 12 mm.• The minimum number of longitudinal bars in a column should be four

in rectangular columns and six in circular columns. In columns havinga polygonal cross-section, at least one bar should be placed at eachp yg pcorner.

• The longitudinal reinforcement should not exceed the followingamounts, calculated as percentages of the gross cross-sectional area ofamounts, calculated as percentages of the gross cross sectional area ofthe concrete:

♦ vertically-cast columns - 6%;♦ horizontally cast columns 8%; and♦ horizontally-cast columns - 8%; and♦ laps in vertically-or horizontally-cast columns - 10%.

At laps, the sum of the reinforcement sizes in a particular layer shouldt d 40% f th b dth f th ti t th t l ti


not exceed 40% of the breadth of the section at that location.

Self-Assessment QuestionsQ1. Give the definition of a short R.C. column in a braced

structure.

x

Answers:b y y

(a) Both lex/h and ley/b ≤ 15.

(b) Both lex/h and ley/b ≤ 10.h

x

y

(c) Either lex/h or ley/b ≤ 15

(d) Either lex/h or ley/b ≤ 10y


Self-Assessment QuestionsQ2. Give the definition of R.C. column in accordance with

BS8110.

Answers:

(a) h/b > 4

(b) h/b ≤ 4 b(c) h/b > 3

(d) h/b ≤ 3

b

h


Self-Assessment QuestionsQ3. A short braced column is supporting a symmetrical

arrangement of main beams subject to dead and i d l dimposed loads.

Which equation would be used for its ultimate axial l d l l ti ?load calculation?

AAnswers:

(a) N = 0.4fcuAc + 0.75Ascfy

(b) N 0 45f A 0 87f A(b) N = 0.45fcuAc + 0.87fyAsc

(c) N = 0.35fcuAc + 0.67fyAsc


Self-Assessment Questions Q4. A short braced column has a 300 mm square section

and is made from grade 40 concrete and is reinforced i h 6T32 b Th l i lwith 6T32 bars. The column supports a symmetrical

arrangement of beams carrying dead and imposed load. Calculate its ultimate axial load capacityCalculate its ultimate axial load capacity.

Answers:Answers:

(a) 2679 kN

(b) 3027 kN(b) 3027 kN

(c) 3463 kN


Self-Assessment QuestionsQ5. A short braced column has a 300 mm square section

and is made from grade 40 concrete and is reinforced i h 6T32 b Th lwith 6T32 bars. The column supports an

approximately symmetrical arrangement of beams (i.e. difference in spans is less than 15%) carrying dead anddifference in spans is less than 15%) carrying dead and imposed load. Calculate its ultimate axial load capacity.p y

Answers:

(a) 2679 kN( )

(b) 3027 kN

(c) 3463 kN( )


Assignment No. 6AQ1 A short column has a 300 x 500 mm section and is

subjected to an ultimate axial load of 4000 kN and ad i b di f 300 kN b i jdesign bending moment of 300 kNm about its majoraxis. Calculate the area of reinforcement required.Given that (i) Nominal cover = 35 mm; (ii) Links size =Given that (i) Nominal cover 35 mm; (ii) Links size10 mm; (iii) fcu = 35 N/mm2, fy = 460 N/mm2.

Answer: As = 8100 mm2


Assignment No. 6 AQ2 A short column has a 300 x 500 mm section, and is

reinforced with 6T32 bars as shown in below. Thel i bj d l i i l l d f 3250column is subjected to an ultimate axial load of 3250

kN and an ultimate moment about its major axis.Determine the ultimate moment capacity of the columnDetermine the ultimate moment capacity of the column.Given that (i) Nominal cover = 35 mm; (ii) Links size =10 mm; (iii) fcu = 35 N/mm2, fy = 460 N/mm2.( ) cu y

Answer: M = 203 kNm300

3T32

500

3T32

5


Assignment No. 6 AQ3 A short column has a 300 x 500 mm section, and is

reinforced with 6T32 bars as shown in below. Thel i bj d l i i l l d dcolumn is subjected to an ultimate axial load and an

ultimate moment of 350 kNm about its major axis.Determine the ultimate axial load capacity of theDetermine the ultimate axial load capacity of thecolumn. Given that (i) Nominal cover = 35 mm; (ii)Links size = 10 mm; (iii) fcu = 35 N/mm2, fy = 460( ) cu y

N/mm2.

Answer: N = 2250 kN300

3T32

00

3T32

50

M=350 kNm


Assignment No. 6 AQ4 A 350 x450 mm short column is subjected to an

ultimate axial load of 3500 kN, and ultimate momentsf 280 kN d 180 kN b i j d iof 280 kNm and 180 kNm about its major and minor

axes respectively. Determine the area of reinforcementrequired Given that (i) Nominal cover = 40 mm; (ii)required. Given that (i) Nominal cover 40 mm; (ii)Links size = 10 mm; (iii) fcu = 40 N/mm2, fy = 460N/mm2.

Answer: As = 7400 mm2s


Assignment No. 6 AQ5 A 350 x450 mm short column is subjected to an

ultimate axial load of 3500 kN, and ultimate momentsf 200 kN d 180 kN b i j d iof 200 kNm and 180 kNm about its major and minor

axes respectively. Determine the area of reinforcementrequired Given that (i) Nominal cover = 40 mm; (ii)required. Given that (i) Nominal cover 40 mm; (ii)Links size = 10 mm; (iii) fcu = 40 N/mm2, fy = 460N/mm2.

Answers: As = 6770 mm2s


Assignment No. 6 AQ6 Fig. AQ6 shows the general arrangement of a braced

column in a frame. Determine whether the columnh i h l d f h fl I i ishown is a short or slender for each floor. It is given

that the floor to floor height is 4.5 m.

AAnswers:

(i) (ii)

2/F 1/F 1/F B2/F - 1/F 1/F - Base

( ) Sh ( ) Sh(a) Short (a) Short


Assignment No. 6 AQ7 If the column shown in Fig. AQ6 is an unbraced

column in a frame. Determine whether the column is ah l d f h fl I i i h h flshort or slender for each floor. It is given that the floor

to floor height is 4.5 m.

AAnswers:

(i) (ii)

2/F 1/F 1/F B2/F - 1/F 1/F - Base

( ) Sl d ( ) Sl d(a) Slender (a) Slender


Assignment No. 6 500 250

x y

xBeam A

Beam ABeam A 200 x 350

yx

2/F

Beam A

Beam B

4500

Beam B 250 x 550Floor to floor height = 4.5m

Beam A

1/F

4500

Base

Figure AQ6

FixedSupport


Figure AQ6

6 column design

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