Dr. K.U - BookSpar · 2013. 4. 4. · Ref. CL 7.5.2.1, P48, IS800:2007 Effective length factor is...

Department of Civil Engineering, MSRIT. KUM 1

COMPRESSION MEMBERS

Dr. K.U.MUTHU*

A perfectly straight member of linear elastic material is shown if figure.

The above member has a friction less hinge at each end, its lower end being fixed

in position while its upper end is free to move vertically but prevented from

deflecting horizontally. It is assumed that the deflections of the member remain

small.

The elastic critical load EP at which a straight compression member buckles

laterally can be determined by finding a deflected position which is one of

equilibrium.

Basic Strut Theory

yPdx

ydEI −=

2

2

(1)

Eulers critical load is obtained as

2

2

l

EIP

y

E

π= (2)

*, Professor & Head, Department of Civil Engineering, MSRIT, Bangalore 54

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In terms of the stress equation is

( )2

2

/ rKL

EpE

π= (3)

Strut with initial curvature

In practice, columns are generally not straight and the effect out of straightness on

strength is studied. Consider a strut with an initial curvature bent in a half sine

curve as shown in Figure.

If the initial deflection, at x from A is yo and the strut deflects ‘y” further under

load, P, the equilibrium equation is

( )oyyPdx

ydEI +=

2

2

(4)

Where deflection

=

l

xy

πsin (5)

If oδ is the deflection at the centre and δ the additional deflection caused by P,

then

( ) 1/

0

−=

PPE

δδ (6)

The maximum stress at the centre of the strut is given by




( )

yI

hP

A

PP

δδ ++= 0

max (7)

Where h is shown in figure

i.e.

( )2

0

Ary

hPpp cy

δδ ++= (8)

i.e. ( )

2

0

ry

hppp ccy

δδ ++= (9)

( )

−++=

1

1102

cE

ccy

ppry

hppp δ (10)

Denoting the Perry factor

2

0

ry

hδη = (11)

( )( )

+=−

cE

c

ccypp

pppp 1η (12)

On simplification it gave

( )( )cEcycE pppppp η=−− (13)

The value of pc, the limiting strength at which the maximum stress equal the

design strength, can be found by solving this equation and η is the Perry factor.

The minimum value of pc after solving the quadratic equation is obtained as

( ) 5.02

yEc ppp −±= φφ (14)

which is of the form




( ) 5.02

yEc ppp −−= φφ (15)

and 2

0

yr

hδη = (16)

the initial deflection 0δ is taken as (1/1000)th

of length of the column and hence

η is given by

=

=

yyyr

h

r

l

r

hl

10001000 2η (17)

and hence

=

yr

lαη (18)

and

=

yr

lζ (19)

a lower value of α was suggested by Robertson as α = 0.003 for column designs.

This approach was suggested in British code. ξ is the slenderness ratio. The total

effect of the imperfections (initial curvature, end eccentricity and residual stresses

on strength). They are combined in to the Perry constant η and is modified as

( )0001.0 λξη −= a (20)

and

=

yf

E2

0 02.0π

λ (21)

the value of 0λ gives the limit to the plateau over which the design strength py

controls the strut load. The Robertson’s constant ‘ a ’ is assigned different values

to give the different design curves.

As per IS 800-2007;

mo

y

mo

y

cd

fff

γγχ ≤= (22)

And χ = stress reduction factor for different buckling class, slenderness ratio and

yield stress.

( )[ ]5.022

1

λφφχ

−+= (23)




and ( )[ ]22.015.0 λλαφ +−+= (24)

α = imperfection factor given in Table 7, in P35, IS800:207.

λ =non dimensional effective slenderness ratio.

cc

y

f

f=λ (25)

and ccf Euler’s buckling stress = ( )2

2

/ rKL

Eπ (26)

and

r

KL effective slenderness ratio (or) the effective length KL to appropriate

radius of gyration, r, moγ = partial safety factor for material strength. It is noted

that the stress reduction factor χ depends on buckling class, slenderness ratio and

yield stress (Table 8, P36- 39, IS800-2007).




Ex.1 A single angle discontinuous member ISA 130x130x10mm with single

bolted connection is 2.5m long. Calculate the safe load carrying capacity of

the section. If it is connected by one bolt at each end.

yf =250Mpa. Class 7, 5.1.2, P48, IS800:2007.

2

3

2

21 φλλλ kkk vve ++=

60.0,5.0,25.1 321 === kkk

107.1

250

1021

4.25/2500

250

/

522===

xxE

rl vvvv

ππε

λ

( ) ( )146.0

250

1021

)10(2/130130

250

2/

522

21 =+

=+

=xxE

tbb

ππε

λφ

772.1146.060107.15.025.1 22 =++= xxeλ

( )[ ]22.015.0 λλαφ +−+=

= ( )[ ] 455.2772.12.0772.149.015.0 2 =+−+

[ ] [ ] 5.0225.022772.1455.2455.2

1.1250

−+=

−+=

λφφ

γ moy

cd

ff

( ) 2/71.54154.4

1.1250mmNfcd ==

kNx

Pd 1371000

25067.54==




Ex.2 In the above problem, if the single angle discontinuous strut is connected

with 2 bolts at each end connection, determine the safe load carrying capacity

of the section.

Fixed condition, Cl 7.5.1.2, P48, IS800:2007

20,35.0,20.0 321 === kkk

107.1

250

/

2==

E

rl vv

vv

πε

λ

( )146.0

250

2/

2

21 =+

=E

tbb

πε

λφ

102.11406.020107.135.020.0 222

3

2

21 =++=++= xxkkk vve φλλλ

[ ] [ ] 5.0225.022012.1211.1211.1

1.1250

−+=

−+=

λφφ

γ moy

cd

ff

( )[ ]22.015.0 λλαφ +−+=

( )[ ] 211.1012.12.0012.049.015.0 2 =+−+

[ ]2

5.022/45.137

683.0211.1

27.227mmN

ff

moy

cd =+

=−+

=λφφ

γ

kNxPd 4.3441000

250645.137 ==




Ex.3 A double angle discontinuous strut ISA 150x75x10mm long leg back to

back is connected to either side by gusset plate of 10mm thick with 2 bolts.

The length of the strut between the intersection is 3.5m. Determine the safe

load carrying capacity of the section.

Ref. CL 7.5.2.1, P48, IS800:2007

Effective length factor is between 0.7 and 0.85 Assume k=0.85

Effective length of the member = 0.85x3500=2975mm

2/8.1034.1210

6.2107 mmNxfcd =−=

Strength of the member = kNx

6.4471000

43128.103=

Ex.4 In the above problem if double angles discontinuous strut is connected

to one side of the gusset plate determine the safe load.

Effective length mmxle 2975350085.0 ==

rmin = 2.56cm P105

2.1166.25

2975

min

==r

le




Table 9c; P42; IS800

24 /87.879.1010

2.66.94 mmNxf cd =−=

Safe load = kNx

8.3781000

431284.87=

Ex.5 A rolled steel beam ISHB 300 @ 58.8 kg/m is used as a column. The

column is fixed in position but not in direction at both ends. Determine the

safe load carrying capacity in the section if the length of the column is 4.5m

mmt f 6.10= Table 10, P44, IS800:2007.

Buckling class of cross section

2.1250

300==

b

h




mmt f 40≤

Buckling about zz axis

Buckling class a Table 7, P35, IS800:2007.

About zz axis, 21.0=α

mmZ 5.129=γ

mmy 1.54=γ

( )

( )[ ]

( )[ ]

( )[ ]( )

( )

2

5.022

5.022

2

52

2

/9.237

391.05965.05965.010.1

250

5965.03912.02.0391.021.015.0

2.015.0

391.01025.129

4500250

mmNf

ff

xx

cd

moycd

=

−+

=

−+=

=+−+=

+−+=

=

=

λφφγ

λλαφ

π

About y-y axis buckling class (b) 34.0=α




( )

( )[ ]

( )[ ]

( )( )

2

5.022

2

2

52

2

22

/42.356

9366.00638.10638.110.1

250

0638.19366.02.09366.034.015.0

2.015.0

9366.01021.54

4500250

//

mmNf

f

xx

rkLf

cd

cd

yy

=

−+=

=+−+=

+−+=

=

=

=

λλαφ

π

πλ

Table 9(a) P40, IS800:2007.

2.831.54

4500

/7.216710

75.4220

75.345.129

4500

2

==

=−=

==

r

kL

mmNxf

r

kL

cd

Table 9(b), P41, IS 800:2007.

kNxStrength

xfcd

4.1084748588.144

48.1441610

2.3150

==

=−=




Ex.6 A built up column consists of two ISMC 400 @ 49kg/m and two plates of

500mmx10mm. The clear distance between back to back of channel is

200mm. One plate is connected to each flange. Determine the safe load

carrying capacity of the built u column if the effective length of column is 5m.

( ) ( )

[ ] ( )

4

23

2

9.72198

5.02015012

15028.150822

86.225150293.622

cm

xx

I

cmxArea

zz

=

+++=

=+=

( )[ ]

4

2

6.41257

12

501242.21093.628.5042

cm

xI yy

=

+++=

4

min 6.41257 cmI =

cmr 5.1386.225

6.41257min ==

37135

5000==

r

kL

2/9.2011310

7211 mmNxfcd =−=

Safe load = kNx

45601000

225869.201=




Ex.7 Calculate the safe load of a bridge compression member of two channels

ISMC 350 @ 421.1 kg/m placed toe to toe. The effective length of member is

7m. The widths over the back of the channel is 350mm and the section is

properly connected by lacings.

( )

( )

( )[ ]

4

2

4

2

7.25201

44.25.1766.536.4302

20016100082

32.10766.532

cm

I

cmI

cmA

yy

zz

=

−+=

==

==

2.516.13

700

6.13minmin

==

==

r

kL

cmA

Ir

Table 9c

2/2.1811510

2.1183 mmNxfcd =−=

Strength of the member = kNx

6.19441000

107322.181=




Ex.8 A column 6m high has its ends firmly built in. The column is built up

with two channels. ISMC 300 placed back to back with 180mm gap between

them. The channels are effectively laced together. Using IS800, determine the

safe load carrying capacity of the column.

Area = 9128mm

2

From SP (6)

4.3366.11

390

3909.3)6(65.0

66.11min

==

===

=

r

kL

cmml

cm

e

γ

Table 9c class ‘c’

2/6.20613

104.3211 mmNxfcd =−=

Safe load carrying capacity = kNx

8.18851000

91286.206=




Ex.9 A column height 5m is hinged at the ends. It is square in cross section

(plan) of side 360mm and consists of 4 angles of ISA 80x80x10mm at each

corner suitably laced. Find the minimum load on the column.

A = 4(15.05) = 60.2cm

2

( )[ ]

kNx

loadSafe

mmNxf

cclassBucklingr

kL

cmA

I

cm

I

cd

x

5.12581000

602005.209

/05.2091310

5.1211

''5.315.158

5000

85.15

98.15113

34.21805.157.874

2

minmin

4

2

==

=−=

==

==

=

−+=

γ




Ex.10 Determine the design strength of the column section ISHB 300 @ 58.8

kg/m. The effective length of the column is 3m.

2/250 mmNf y =

''

406.10

2.1250

300

cclassBucklingaxiszz

mmt

b

h

f

f

−

≤=

==

kNx

StrengthDesign

mmNxf

r

kL

bclassBucklingaxisyy

mmNxf

r

kL

r

l

r

kL

cd

cd

zz

e

9.13981000

74859.186

/9.1861310

45.5194

45.551.54

3000

''

/9.2191310

17.3224

17.23

17.235.129

3000

2

2

==

=−=

==

−

=−=

=

===




DESIGN OF COMPRESSION MEMBER

Ex.11 Design a single angle section discontinuous strut to carry a load of

80kN. The length of the member between c/c intersection is 2.75m

Axial load = 80kN

Permissible stress = 0.4 2/100 mmNf y =

Area required = 800mm2

Gross area = 800x1.25= 1000mm2 = 10cm

2

Try ISA 90x90x8mm A = 13.79cm2, cmrvv 95.1=

( ) ( )1986.0

87.119

250

1021

95.1/27585.0

250/ 52

2

3

2

21

===

++=

xx

x

E

rl

kkk

vvvv

vve

ππελ

λλλ φ

( ) ( )

250

1021

82/9090

250

2/

522

21

xx

x

E

tbb

ππε

λφ

+=

+=

[ ] 25.2768.12.0768.1(49.015.0

768.1

)1267.0(6035.1(5.025.1

2

22

3

2

21

=+−+=

=

++=++=

φ

λ

λλλ φ

e

vve kkk

[ ] [ ]

kNkNx

P

mmNf

f

d

moy

cd

8018.531000

137957.38

/57.38768.125.225.2

1.1/250/2

5.0225.022

<==

=−+

=−+

=λφφ

γ

Revise the section

Try ISA 100X100X10 Area = 1903mm2




( )

( ) ( )

[ ] 338.2713.12.0713.1(49.015.0

713.11126.060)36.1(5.025.1

1126.081.88

100

250

1021

102/100100

36.1

81.88

49.120

250

1021

94.1/27585.0

250

/

60,5.0;25.1

2

22

52

522

321

2

3

2

21

=+−+=

=++=

==+

=

=

===

===

++=

φ

λ

πλ

λ

ππε

λ

λλλ

φ

φ

x

xxx

x

xxx

x

E

rvvl

kkk

kkk

e

vv

vv

vve

( )[ ]

( )[ ]

2

2

5.022

2

06.2510130130

80611000

190332

/32713.1756.3756.3

1.1/250

756.3

338.22.0338.249.015.0

cmAreaxxISATry

kNkNx

P

mmNf

d

cd

=

<==

=++

=

=

+−+=

φ

φ

2

3

2


60,5.0;25.1 321 === kkk

( )63.2

250

1021

94.1/27585.0

250

/

522===

xxx

x

E

rvvlvv

ππε

λ




( ) ( )1463.0

81.88

13

250

1021

102/130130

52==

+=

xxx

x

πλφ

[ ] 047.4448.22.0448.2(49.015.0

448.21463.060)36.2(5.025.1

2

22

=+−+=

=++=

φ

λ xe

( )[ ]

kNkNx

P

mmNf

d

cd

8033.781000

25063126

/26.31448.2047.4047.4

1.1/250 2

5.022

<==

=++

=

Try ISA 150X150X10

A=29.03cm2

rvv = 2.93cm

2

3

2


60,5.0;25.1 321 === kkk

( )898.0

250

1021

93.2/27585.0

250

/

522===

xxx

x

E

rvvlvv

ππε

λ

( ) ( )168.0

81.88

15

250

1021

102/150150

52==

+=

xxx

x

πλφ

[ ] 574.283.12.083.1(49.015.0

83.11463.060)898.0(5.025.1

2

22

=+−+=

=++=

φ

λ xe

( )[ ]

2

5.022/84.51

83.1574.2574.2

1.1/250mmNfcd =

−+=

Strength = 150.5kN>80

Try ISA 130x130x10 A+25.06cm, rvv=2.54cm




( )036.1

81.88

03.92

250

1021

54.2/27585.0

250

/

522====

xxx

x

E

rvvlvv

ππε

λ

( ) ( )1463.0

81.88

13

250

1021

102/130130

52==

+=

xxx

x

πλφ

[ ] 415.2752.12.0752.1(49.015.0

752.11463.060)036.1(5.025.1

2

22

=+−+=

=++=

φ

λ xe

( )[ ]

2

5.022/89.43

752.1415.2415.2

1.1/250mmNfcd =

++=

Safe strength = 43.89x2506/1000=110kN>80kN

Ex.12 Design a double angle discontinuous strut to carry a load of 125kN, the

length between the intersection is 3.8m

Axial load = 125kN

Permissible stress 0.4 yf =100Nmm2

Area Required = 125000/100=1250mm2

Gross area required = 1250x1.25=1562.5mm2 = 15.63mm

2

Try two ISA 75x75x6 area = 17.32cm2

cm3.2min =γ

Effective length cmxkL 32338085.0 ==

4.140min

=r

kL

Table 9(c) 2/2.66 mmNfcd =

Safe strength = kNkNx

1257.1141000

17322.66<=

Hence revise the section, Try two angle of ISA 80x80x8




Area = 24.42cm2

cm44.2min =γ

4.13244.2

323

min

==r

kL

Table 9(c) 2/4.721.810

4.23.74 mmNxfcd =−=

Safe strength = kNkNx 1253.1621000/22424.72 >=

Ex.13 A column connects four equal angles arranged in the form of a square

section of side 400mm. Design the section if the column is to carry an axial

load of 800kN. The length of the column is 5m. Both the ends of the column

are restrained in position but not in direction.

Axial load = 800kN

Allowable compressive stress = 0.4x250=100N/mm2

Area of 4 angles = 800x103/100=800mm

2

Area of 1 angle = 2000mm2 = 20cm

2




Increase this area by 25%, Gross area of l angle = 20x1.25=25cm2

Try 4 angle of ISA 130x130x12mm A = 29.82cm2

( )[ ] 42

min 3374266.32082.298.4734 cmIII yx =−+===

cmx

82.1682.294

33742min ==γ

7.2982.16

500

min

==r

kL

2/39.211 mmNfcd =

Strength of the member = 211.39x4x2982/1000=2521>800kN

Hence revise the section

Try 4 angles of ISA 100x100x12 A = 22.59cm2

( )[ ] 42

min 4.2718892.22059.222074 cmI =−+=

cmx

r 35.1759.224

4.27188min ==

8.2835.17

500

min

==r

kL

2/56.212 mmNfcd =

Safe load = 212.56x4x2259/1000=212.6>80kN


( )[ ] 42

min 5.2115459.22003.177.1264 cmI =−+=

cmx

r 62.1703.174

35.21154min ==

4.2862.17

500

min

==r

kL

2/08.2131310

4.8224 mmNxfcd =−=

Safe load = 213x4x1703/1000=1450>800kN





( )[ ] 42

min 7.1912534.22005.157.874 cmI =−+=

cmr 82.17min =

2882.17

500

min

==r

kL

2/6.2131310

8224 mmNxfcd =−=

Safe load = 213.6x4x1505/1000=1285.2>800kN


( )[ ] 42

min 99.1564227.22021.125.724 cmI =−+=

cmr 89.17min =

95.2789.17

500

min

==r

kL

2/67.2131310

95.7224 mmNxfcd =−=

Safe load = 213.67x4x1221/1000=1043.6kN>800kN

Try 4 angles of ISA 60x60x10 Area = 11cm2

( )[ ] 42

min 79.1463385.120118.344 cmI =−+=

cmr 24.18min =

41.2724.18

500

min

==r

kL

2/38.2141310

41.7224 mmNxfcd =−=

Safe load = 214.38x4x1100/1000=943.2kN>800kN

Try 4 angles of ISA 60x60x8mm A = 8.18cm2

( )[ ] 42

min 8.1202677.12096.8294 cmI =−+=




cmr 17.19min =

2617.19

500

min

==r

kL

2/2.2161310

6224 mmNxfcd =−=

Safe load = 216.2x4x818/1000=707kN<800kN

Hence revise the section.

Adopt 4 angles of ISA 60x60x8mm

Ex.14 A rolled steel beam ISHB 300@ 58.8kg/m is used as a column. The

column is fixed in position but not in direction at both ends. Determine the

safe load carrying capacity of the section if the length of column is 4.5m

42.12545 cmI zz =

46.2193 cmI yy =

A = 74.85cm2

mmmmtb

hf

f

1006.10,2.1250

300<===




For the buckling about zz axis – ‘b’

cmrzz 95.1285.74

2.12545==

75.3495.12

450==

zz

e

r

l

2/25.2111010

75.4216 mmNxfcd =−=

For the buckling about yy axis class ‘c’

cmryy 41.585.74

6.2193==

18.8341.5

450==

yy

e

r

l

2/2.1311010

2.3136 mmNxfcd =−=

Strength of the member = 131.2x7485/1000=982kN

Ex.15 Design a built up column consisting of two channel sections placed

back to back with a clear spacing of 250mm between them. The column

carries an axial load of 1000kN and is having an effective height of 6m.

Design the lacing for the column.

Axial load = 1000kN

Assume the permissible compressive stress = 0.5 yf =125N/mm2

Area required = 1000x103/125=8000mm

2 = 80cm

2

Area of one channel = 45cm2




Try 2 channels of ISMC 350; area = 2x53.65=107.3cm2

cmrzz 66.13= cmryy 21.15=

About zz axis

92.4366.13

600==

zz

e

r

l

2/15.1921510

9.3198 mmNxfcd =−=

About yy axis

34.3921.15

600==

yy

e

r

l

2/18.208310

4.9211 mmNxfcd =−=

Safe load = 192.15x10730/1000-2061.76>1000kN hence OK

Try ISLC 300 A = 84.22cm2

cmrzz 98.11= cmryy 32.15=

5098.11

600==

zz

e

r

l

2/78.2101310

17.0211 mmNxfcd =−=

17.3932.15

600==

yy

e

r

l

Safe load = 183x8422/1000=1541kN>1000kN

Hence adopt the section.




Design of lacing

Cl 7.6.2 Minimum width of lacing bar = 3x16 (dia of bolt) = 48 say 50mm

Cl 7.6.4 Angle of inclination = 000 457040 ==≤≤ θθ

Cl 7.6.3 Thickness of lacing bar =

[ ] [ ] 10mmsaymmggspacing 17.6606025060

1

60

1=++=++

Cl 7.6.5.1

whole

e

r

l

memberofcomponentoneofr

lacingofSpacing

≤

≤

min

min

7.0

50

( ) ( ) 42.2717.397.05089.127.28

6060250=≤≤=

++

Cl 7.6.6.3

145≤

flatr

kL

λ of the lacing bar = 14586.126121

2377.0

120

7.0≤==

x

t

l

Check the bars for lacing in compression

Shear force kNx 251000100

5.2==

Force on the lacing bar = kNecx

ecn

S84.845cos

22

25cos

2==θ

For the flat angle, for 127=λ

2/15.774.910

77.83 mmNxfcd =−=

Safe load = 77.12x50x10/1000=38.6kN>8.84kN

Check for the flat in Tension

( ) ( ) ( )kNtfdb my 105

25.1

410101850/ =

−=−= γ




Or >== kNxxAf mogy 4.1131.1/1050250/γ for in the lacing bar

Ex.16 Design a battened column for the column shown in figure. Assume that

the channels are kept back to back.

The effective slenderness ratio

r

kLof battened columns shall be 1.1 times the

maximum actual slenderness ratio of the column.

r

kL = 1.1x39.17=43

2/5.1931510

3198 mmNxfcd =−=

Safe load = kNkNx 10007.16291000

84225.193 >=

whole

e

r

l

memberofcomponentoneofr

battentheofspacingMaximum

≤

minmin

7.0

Maximum spacing of batten = 143.5cm = 0.7(2.87)43= 86.4cm

Provide the battens at a spacing of 850mm

Provide 20mm bolts. For rolled, machine flameout, P74, Cl 10.2.4.2 ⇒ 1.5xhole

diameter = 1.5x20=33mm

Effective depth of batten

= 250+2(23.5)=297mm>2(100)=200mm

Overall depth of batten = 297+2(33)=363mm=370mm

Required thickness of batten = 1/50 (distance between inner most bolts.)




1/50(250+2x60)=7.4 say 8mm

Length of the batten = 250+2(100) =450mm

Provide 450x370x8mm

Size of intermediate batten

Effective depth = 3/4x297=222.75mm>2x100=200mm

Hence an effective depth of 225mm

Overall depth = 225+2x33=291say 300mm

Provide 450x300x8mm intermediate battens

Design forces

Transverse shear = V = 2.5/100x1000=25kN=25000N

Longitudinal shear kNx

x

NS

CVV t

b 72.283702

8525000===

tV = transverse shear = 25000N

C = c/c of battens, longitudinally = 850mm

N = number of parallel planes = 2

S = minimum distance between the centroid of the bolt = 370mm

Moment Nmmx

x

N

cVM t 5312500

22

85025000

2===

For end batten

Shear Stress = 28720/370x8=9.7N/mm2 < 2/2.131

1.13

250mmN

x=

Bending stress = 22

22/227

1.1

250/10.29

3708

531250066mmNmmN

x

x

td

M=<==

Hence safe

For Intermediate battens

Shear stress = 28720/300x8=11.97N/mm2 < 131.2N/mm

2

Bending stress = 6x53/2500/8x3002=44.27N/mm

2 < 250/1.1=227N/mm

2




Connections

Strength of the bolt = 45.3kN

Required number of bolts = 28.72/45.3 < 1.0

As the bending moment is also present, provide 3 bolts

Check Force in each bolt due to shear = 28.72/3=9.57kN

Adopt a pitch of 100mm

Force due to moment = ∑ 2

r

rM

= kNx

x56.26

100100

100531250022

=

Resultant force = kNkN 3.4523.2856.2657.922 <=+

Hence safe.




CASED COLUMNS

Encased I sections or filled hollow sections carries more load. In cased columns,

the advantages derived from the properties of concrete and steel are used. The

concrete is strong in stronger in compression and it provides greater rigidity. The

solid concrete casing assists in carrying the load and the entire load is resisted by

concrete and steel. The design of the above columns is currently based on IS

11384-1985. As the above code is on working stress method the guide lines given

in BS5950, Part I is presented here. The role of concrete is that it acts as a fire

protection for the encased steel columns and also prevents the column from

buckling about the weak axis. As per the BS5950, Part I the column must satisfy

the following specifications.

(i) The steel section is either a single rolled or fabricated I or H section

with equal flanges, channels and compound sections can also be used.

(ii) The steel section should not exceed 1000mmx500mm. The dimension

100mm is in the direction of web.

(iii) Primary structural connections should be made in the steel section.

(iv) The steel section is unpainted and free from dirt, grease, rust, scale etc.

(v) The steel section is encased in concrete of at least Grade 20, to BS

8110.

(vi) The cover on the steel is to be not less than 50mm. The corners may be

chamfered.

(vii) The concrete extends the full length of the member and is thoroughly

compacted.

(viii) The casing is reinforced with bars not less than 5mm diameter at a

maximum spacing of 200mm to form a cage of closed links and

longitudinal bars. The reinforcement is to pass through the centre of

the cover.

(ix) The effective length is not to exceed 40bc, 100b2

c / dc or 250 r

whichever is the least, where

bc = minimum width of solid casing.

dc = minimum depth of solid casing.

r = minimum radius of gyration of steel section.




BS5950, Part I guidelines for estimating the compressive strength of column.

a) The radius of gyration about yy axis is shown in figure, ry should be taken

as 0.2bc but not more than 0.2 (B+150) where B = overall width of flange.

The radius of gyration for the zz axis should be taken as that of the steel

section.

b) The compression resistance cP is

csc

y

cu

gc Ppp

AfAP ≥

+= 45.0

yc

y

cu

gcs pAp

fAP

+= 25.0

Where cA = gross sectional area of concrete. Casing in excess of 75mm from the

steel section is neglected. Finish is neglected.

gA =gross area of the steel section

cuf =characteristic strength of the concrete at 28 days. This should not exceed

40N/mm2.

cp =compressive strength of steel section determined using xr and zr in the

determination of which 2/335 mmNp y ≤

yp = design strength of steel

Cased Column




CASED COLUMN WITH AXIAL LOAD

Ex.17 An internal column in a building has an actual length of 4.5m centre to

centre of floor beams. The steel section is ISHB250 @ 51kg/m. Calculate the

compression resistance of the column if it is cased in accordance with the

codal provision. M25 concrete grade has been use. The casing has been made

325mm square.

Properties of ISHB 250

A=6496mm2

zzr =10.91cm

yyr =5.49cm

For the above cased column;

( )

( ) mm

mmry

801502502.0

653252.0

=+≠

==

i) effective length = 0.7 (4500) = 3150mm of cased column

ii) 40 cb =40(325) = 13000mm

iii) 100c

c

d

b2

=100x325=32500mm

iv) 250 r =250x54.9=13725mm

slenderness ratio = 46.4865

3150==

r

kL

refer Table 9(c) in P42, IS800:2007.




2/3.1851510

46.8198 mmNxfcd =−=

The gross sectional area of concrete

2105625325325 mmxAc ==

Compressive strength of concrete

kNxxPc 5.20841000

3.185105625

250

2545.06496 =

+=

Short column strength

kNx

xPcs 22841000

250

250

1056252525.06496 =

+=

Compressive strength of column = 2084.5kN




Column with axial load and moment

Ex.18 A stanchion carries an factorial axial load 500kN and a factored

bending moment of 250kNm. Design the section if the length is 6m and one

end of the column is restrained in position and direction whereas, other end is

restrained only in position but not in direction.

Try section ISWB 600@ 133.7kg/m

43.915.52

60008.0==

x

r

kL

Yura suggested

b

M

d

MPp

yzeff 5.72 ++= for initializing the size of the column. If the BM is

predominant then the equivalent BM can be found out from

2

dPMM uzeq +=

In this case;

kNxd

MPP z

eff 33.13336.0

25025002 =+=+=




Check the above section

2.14.2250

600>==

b

h

403.21 <= mmtc

Buckling about yy axis, buckling class ‘b’

2/7.1311610

43.1134 mmNxfcd =−=

Compressive strength of the trial section=131.7x17038/1000=2244kN>1333.33kN

Section properties

A = 177.38cm2

( ) 451006.1 cmI zz =

( ) 43107.47 cmI yy =

( ) ( ) 4/22/2 fwtftpz tHttHtbZ −+−=

( ) ( ) ( ) 3266.39864/3.2126002.112/3.216003.212502 cmxZ pz =−+−=

(P138, IS800:2007)

( ) ( ) ( ) 323

32 1.6834

2.113.212600

4

2503.212424/2 cmxttHbtZ wfffpy =−+=−+=

Cross section classification




0.1250

25050.2===

yfε

Outstanding flanges

ε4.987.53.21

2/250<==

ft

b (Table 2, P18)

Hence the flange is plastic

Web

( ) mmrtHd f 4.521)18(23.21260022 1 =−−=−−=

ε846.462.11

4.521<==

wt

d

Hence the cross section is plastic

Refer d 9.3.1.1 for plastic and compact sections

0.1≤++dz

z

dy

y

d M

M

M

M

N

N

N = factored applied axial force = 500kN

dN = design strength in compression = mx

yg fA

γ

= kNx

xx3.3872

10001.1

2501038.170 2

=

kNmx

xxxM dz 05.906

101.1

2501066.398616

3

==

0.14.005.906

250

3.3872

500<=+∴

Member buckling resistance in compression

Effective length of member = 0.8x6000=4800mm

22.197.249

4800==

z

z

r

kL




mmtb

h

r

kL

f

y

y

403.21;4.2250

600

43.915.52

4800

<===

==

Buckling about zz axis (Buckling class ‘a’)

2/226110

22.9227 mmNxfcd =−=

Buckling about minor axis (Buckling class ‘b’)

2/7.1311610

43.1134 mmNxfcd =−=

Safe compressive strength = kNkNx

50023361000

177387.131>=

Hence the section is conservative.

Member buckling resistance in bending

tionplasticfor

fZM

b

bdpbd

sec0.1=

=

β

β

=LTα Imperfection parameter = 0.21 for rolled steel section P54, cl 8.2.2

bcrf , = extreme fibre bending compressive stress

( )

5.02

2

2

,/

/

20

11

1.1

+=

ff

yLT

yLT

bcrth

rL

rL

Ef

π

43.91=y

LT

r

L

5.0

2

2

52

,3.21/600

43.91

20

11

43.91

1021.1

+=

xxf bcr

π

[ ] 25.0/64.320527.0148.259 mmN=+=




TL,λ = Non dimensionless slenderness ratio

883.064.320

250

,

==bcr

y

f

f

( )[ ]22.015.0 LTLTLTLT λλαφ +−+=

Strength reduction factor = ( )[ ]2863.02.0883.021.015.0 +−+

96.0=LTφ

=LTχ bending stress reduction factor to account for lateral Torsional buckling

22

1

XLLTLT χφφ −+=

748.0853.096.096.0

1

22=

−+=LTχ

2/170

1.1

250748.0 mmNx

ff

mo

y

LTbd ===γ

χ

Elastic lateral buckling moment

bcrpbcr fZM ,β=

=crM Elastic lateral buckling moment

kNmkNmxx

M cr 2503.127810

64.3201066.39866

3

>==

Hence it is safe

Moment amplification factors

( )dz

ZP

Pk 2.011 −+= λ

146.12336

5002.0883.01 =

−+=

171.12336

5008.018.01 =+=+= X

P

Pk

dz

ZZ




01

2 ==M

Mzψ

( )[ ]4.0,4.06.0max ψ+=mzC

6.0=mzC

134.03.1278

2506.0146.121.01 <=+=+ xx

M

MCk

P

P

cr

mzz

dz

Hence the section is safe against bending moment and axial force.




Questions on Compression members, Column Splices, Slab & Gusset Bases and Connections between beams & columns

1. Design a single angle discontinuous strut (equal & unequal angle) to carry a

compressive force of 500kN. The c/c distance between the joints is 3m.

Design also the connections using

a) M24 bolts of property class 5.6

b) M24 HSFG bolts of property class 10.9

c) Equivalent welded connections

2. Repeat the above problem using double angles (on same side & on either side

of gusset plate) for a force of 1000kN.

3. A discontinuous double angle strut is placed back to back on the same side of

the gusset plate 8mm thick. The angles are ISA 125x95x8 with c/c distance

between the joints =3m. Calculate the safe load when:

a) connected by one bolt at each end

b) connected by two or more bolts at each end

What will the % change of load if the above angles are placed on either side of

the gusset plate?

4. A single angle discontinuous strut ISA 130x130x12 is 3m between centre to

centre of intersections. Calculate the safe load when:

i) connected by one bolt at each end

ii) connected by two or more bolts at each end

5. A truss member has a length of 3.5m between the centre of joints. The force in

the member is 150kN compression due to DL & IL; 200kN due to DL & WL.

Design the member and the connection to a 10mm thick gusset plate. Adopt

single equal angle; single unequal angle; double equal angles & double

unequal angles.




6. Compute the strength of the column shown in figure

7. Design a builtup column to carry an axial load of 1400kN with the length of

column being 8m. The column is effectively held in position at both ends, but

not restrained against rotation at both ends. The C/S of the column is:

a) Two channels back to back (heel to heel)

b) Two channels toe to toe separated & unseparated

c) Two I- sections - ISHB & ISMB

d) I- section with cover plates

e) Four angles – equal & unequal (arranged as square or rectangular

section)

In all the cases, design also the lacings and battens, if applicable. Also, check

other end conditions as specified in the code. (Connections can be bolted or

welded)




8. a) Design a builtup column carrying an axial load of 1300kN. The height of

column is 7m & is effectively held in position at both ends, but restrained

against rotation at one end only. Adopt two channels toe to toe with the

width over the back of the channels being 400mm. Also, design a suitable

lacing and battens. Connections can be bolted or welded.

b) Repeat the design in 7a) with two channels back to back with a clear

spacing of 300mm between them.

9. Design a suitable slab base and gusset base in problems (6) & (7) assuming

plain concrete pedestal of grade M15. Design the pedestal also. Adopt suitable

bolts. SBC of soil is 150kN/m2.

10. Design a column using an ISHB section with cover plates to carry a

compressive load of 3000kN. The effective length of the column is 6m. Also,

design a suitable gusset base & plain concrete pedestal of M15 grade. Adopt

suitable bolts. SBC of soil is 200kN/m2

11. An upper storey column ISHB300 @ 58.8 kg/m carries a load of 1000kN & a

BM of 40kNm. It is spliced with a lower storey column ISHB400 @ 82.2

kg/m. Ends of the columns are machined. (Milled) Design a suitable splice.

Adopt suitable bolts or welds.

12. a) A column section ISHB400 @ 82.2 kg/m carries an axial load of 1200kN &

BM of 50kNm. Design a suitable column splice. Adopt bolts or welds of

suitable size.

b) Design a suitable splice for a 5m effective length ISHB450 @ 87.2 kg/m

column carrying an axial load of 1000kN & a BM of 50kNm. Assume the

surfaces to be unmilled. Adopt bolts or welds of suitable size.




13. An ISMB600 @ 122.6 kg/m transfer a reaction of 300kN framing into the

flange of a column ISHB400 @ 82.2kg/m. Design a suitable

a) Stiffened seated connection;

b) Unstiffened seated connection (simple seated)

c) Framed connection.

Adopt bolts or welds of suitable size.

14. Two secondary beams ISMB300 @ 58.8kg/m are directly welded on either

side of the web of the girder ISMB600 @ 122.6 kg/m. Each secondary beam

transfer an end reaction of 250kN. Design fillet field welded connection.

15. Repeat the above problem as a framed connection adopting bolts or welds of

suitable size.

16. A secondary beam ISMB400 @ 62.6 kg/m transmit an reaction of 300kN to a

main beam ISMB550 @ 86.9 Kg/m. Design a suitable framed connection

using bolts or welds of suitable size.

17. A stanchion factorial axial load of 750kN and factored Bending moment of

300 kNm. The effective length of the column is 5 m. Design the stanchion as

per IS 800:2007

18. A column of effective length 6.5m shown in fig is subjected to the design data

as follows.

Factored axial load at the top = 1250kN

Factored axial load at the bottom = 600kN




Moment about the major axis at the top = 100kNm

Moment about the major axis at the bottom = 55kNm

Check the adequacy of the section.

19. A column between the floor is provided with ISHB 300 @ 58.8kg/m.

Investigate its adequacy if the ultimate design loads and moments are as

follows

Axial compression = 2500kN

Ultimate Moments at Top

About Major axis = 350kNm

About Minor axis = 50kNm

Ultimate Moments at Bottom

About Major axis = 175kNm

About Minor axis = -75kNm

Effective length of the column = 6.0m

Dr. K.U. Muthu

Sri H. Narendra



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Dr. K.U - BookSpar · 2013. 4. 4. · Ref. CL 7.5.2.1, P48, IS800:2007 Effective length factor is...

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