ANALYSIS & DESIGN OF COMPRESSION MEMBERS

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.

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

λe = k1 + k2 λ2vv + k3λφ

2

k1 = 1.25, k 2 = 0.5, k3 = 0.60

λvv =

l / rvv

=2500 / 25.4

= 1.107ε π 2 E 1 π 2 x2x105

250 250

λ =b1 + b2 / 2t = 130 + 130/ 2(10) = 0.146φ π 2 E π 2 x2x105

ε 1250 250

λe = 1.25 + 0.5x1.107 2 + 60x0.1462 = 1.772

φ = 0.51 + α λ − 0.2 + λ2 = 0.51 + 0.491.772 − 0.2 + 1.7722 = 2.455

f cd =

f y

γ mo

= 250 1.1

− λ2 0.5 2.455 + 2.4552 − 1.7722 0.5φ + φ 2

fcd = 250 1.1 = 54.71N / mm 2 4.154

P = 54.7 x2506 = 137kNd

1000

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

k1 = 0.20, k2 = 0.35, k3 = 20

λvv =

l / rvv

= 1.107ε π 2 E

250

λ =b1 + b2 / 2t = 0.146φ π 2 Eε

250

λe= k + k

2λ 2 + k λ2 = 0.20 + 0.35x1.107 2 + 20x0.14062 = 1.102

1 vv 3 φ

f cd = f y

γ mo

= 250 1.1

− λ2 0.5 1.211 + 1.2112 − 1.0122 0.5φ + φ 2

φ = 0.51 + α λ − 0.2 + λ2 0.51 + 0.490.012 − 0.2 + 1.0122 = 1.211

f = fy γ mo = 227.27 = 137.45N / mm 2cd

φ + φ 2 − λ2 0.51.211 + 0.683

2506

Pd = 137.45x = 344.4kN

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

fcd = 107 − 2.6

x12.4 = 103.8N / mm 2 10

103.8x4312

Strength of the member = 447.6kN

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 le 0.85x3500 2975mm

rmin = 2.56cm P105le

=

2975

= 116.2r

min 25.6

Table 9c; P42; IS800

fcd = 94.6 − 6.2

x10.9 = 87.87 4 N / mm 2 10

87.84x4312

Safe load = 378.8kN

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

tf = 10.6mm Table 10, P44, IS800:2007.

Buckling class of cross section

h= 300

= 1.2 b 250

tf ≤ 40mm

Buckling about zz axis

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

About zz axis, α 0.21

γ Z = 129.5mm

γy = 54.1mm

4500 2 π 2 5 = 0.391= 250 x2x10

129.5

φ = 0.51 + α λ − 0.2 + λ2

= 0.51 + 0.210.391 − 0.2 + 0.3912 = 0.5965

f cd = f y γ mo φ + φ 2 − λ2 0.5 250

2 0.5= 0.5965+0.59652−0.391

1.10

f cd = 237.9N / mm 2

About y-y axis buckling class (b) α 0.34

λy = f y kL / r 2 / π 2

4500 2 2 5= 250 π x2x10 = 0.9366

54.1

φ = 0.51 + α λ − 0.2 + λ2

= 0.51 + 0.340.9366 − 0.2 + 0.93662 = 1.0638

fcd = 250

1.0638 + 1.06382 − 0.93662 0.5 1.10

f cd = 356.42N / mm2

Table 9(a) P40, IS800:2007.

kL =

4500 = 34.75

r 129.5

fcd = 220 − 4.75

x7 = 216.7 N / mm2 10

kL =

4500 = 83.2

r 54.1

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

fcd = 150 − 3.2

x16 = 144.48 10

Strength = 144.88x7485 = 1084.4kN

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.

Area = 262.93 + 250x1 = 225.86cm2

50x132

I zz = 215082.8 + 2 + 50x120 + 0.512

= 72198.9cm4

2 1x50I yy = 2504.8 + 62.9310 + 2.42 + 2

12

= 41257.6cm4

Imin 41257.6cm4

41257.6

rmin = = 13.5cm

kL = 5000 = 37r 135

fcd = 211 − 7 x13 = 201.9N / mm 2

10201.9x22586

Safe load = = 4560kN

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.

A= 253.66 = 107.32cm 2

I zz = 210008 = 20016cm 4

I yy = 2430.6 + 53.6617.5 − 2.442

= 25201.7cm 4

r=I min = 13.6cm

min A

kL =

700 = 51.2

r 13.6

Table 9c

fcd = 183 − 1.2

x15 = 181.2N / mm2 10

181.2x10732

Strength of the member = = 1944.6kN

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 = 9128mm2

From SP (6)

γ min = 11.66cm

l e = 0.65(6) = 3.9m = 390cm

kL =

390= 33.4

r 11.66 Table

9c class ‘c’

f cd = 211− 3.4

10 x13 = 206.6N / mm2

Safe load carrying capacity = 206.6x9128

= 1885.8kN 1000

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) = 60.2cm2

I x = 487.7 + 15.0518 − 2.342

= 15113.98cm4

γ =I

min = 15.85cmmin A

kL =

5000 = 31.5 Buckling class ' c'

r 158.5

fcd = 211 − 1.5

x13 = 209.05N / mm2 10

Safe load = 209.05x6020

= 1258.5kN1000

Ex.10 Determine the design strength of the column section ISHB 300 @ 58.8 kg/m. The effective length of the column is 3m.

fy = 250N / mm 2

h = 300

= 1.2b f 250

tf = 10.6 ≤ 40mm

z − z axis Buckling class ' c'kL

= le = 3000 = 23.17

r rzz129.5

kL

= 23.17

fcd = 224 − 3.17

x13 = 219.9N / mm2 10

y − y axis Buckling class 'b'

kL =

3000 = 55.45

r 54.1

fcd = 194 − 5.45

x13 = 186.9N / mm2 10

Design Strength = 186.9x7485

= 1398.9kN

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 f y = 100N / mm 2

Area required = 800mm2

Gross area = 800x1.25= 1000mm2 = 10cm2

Try ISA 90x90x8mm A = 13.79cm2, r = 1.95cmvv

λ = k + k λ2 + k λ2

e 1 2 vv 3 φ

λ = l rvv =0.85x275 / 1.95 =119.87

vv ε πE / 250π 2x2x10 5 0.1986

1250

λ = b1 + b2 / 2t =90 + 90/

2x8φ

ε π 2 E 1 π 2 x2x105

250 250

λe = k1 + k 2 λ2vv + k3 λφ

2 = 1.25 + 0.5(1.35 + 60(0.1267) 2

λe = 1.768

φ = 0.51 + 0.49(1.768 − 0.2 + 1.7682 = 2.25

f = f y / γ mo = 250 /1.1 = 38.57 N / mm 2cd

φ + φ 2 − λ2 0.5 2.25 + 2.252 − 1.7682 0.5

P =38.57 x1379 = 53.18kN < 80kNd 1000

Revise the section

Try ISA 100X100X10 Area = 1903mm2

λ = k + kλ2 + k3λ2

e 1 2 vv φ

k1 = 1.25; k2 = 0.5, k3 = 60

λ = l / rvv =0.85x275/1.94 = 120.49vv

ε π 2 E 1x π 2 x2x105 88.81250 250

λvv = 1.36

λ = 100 +100 /2x10 = 100 = 0.1126

φ π 2 x2x105 88.811x250

λe = 1.25 + 0.5(1.36)2 + 60x0.11262 = 1.713

φ = 0.51 + 0.49(1.713 − 0.2+ 1.7132 = 2.338

φ = 0.51 + 0.492.338 − 0.2 + 2.3382

φ = 3.756

f = 250 /1.1 = 32 N / mm 2cd

+ 3.7562 + 1.7132 0.53.75632 x1903

Pd = = 61kN < 80kN

Try ISA 130x130x10 Area = 25.06cm 2

λe k1 k2λ2vv k3λφ

2

k1 1.25; k2 0.5, k3 60

λ = l / rvv =0.85 x275/1.94 = 2.63vvπ 2 E π 2 x2x105

ε 1x250 250

λ =130 + 130/2x10 = 13 = 0.1463

φ π 2 x2x105 88.811x250

λe = 1.25 + 0.5(2.36) 2 + 60x0.14632 = 2.448

φ = 0.51 + 0.49(2.448 − 0.2+ 2.4482 = 4.047

f = 250 /1.1 = 31.26N / mm 2cd

+ 4.047 2 + 2.4482 0.54.0473126 x2506

Pd = = 78.33kN < 80kN

Try ISA 150X150X10

A=29.03cm2

rvv = 2.93cm

λe = k1 + k2λ2vv + k3λφ

2

k1 = 1.25; k2 = 0.5, k3 = 60

λ = l / rvv = 0.85x275/ 2.93 = 0.898vv π 2 E

π 2 x2x105

ε 1x250 250

λ =150 + 150/2x10 = 15 = 0.168

φ π 2 x2x105 88.811x250

λe = 1.25 + 0.5(0.898) 2 + 60x0.14632 = 1.83

φ = 0.51 + 0.49(1.83 − 0.2+ 1.832 = 2.574

f = 250 /1.1 = 51.84N / mm 2cd+ 2.5742 − 1.832 0.52.574

Strength = 150.5kN>80

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

λ = l / rvv =0.85x275/ 2.54 =92.03 = 1.036vv

π 2 E π2 x2x105 88.81ε 1x250 250

λ =130 +130/2x10 = 13 = 0.1463

φ π 2 x2x105 88.811x250

λe = 1.25 + 0.5(1.036)2 + 60x0.14632 = 1.752

φ = 0.51 + 0.49(1.752 − 0.2+ 1.7522 = 2.415

f 250 /1.1 43.89N /

mm2cd

2.4152 1.7522 0.52.415Safe 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 f y =100Nmm2

Area Required = 125000/100=1250mm2

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

Try two ISA 75x75x6 area = 17.32cm2

γ min = 2.3cm

Effective length kL = 0.85x380 = 323cm

kL 140.4

rmin

Table 9(c) f cd 66.2N / mm 2

66.2x1732

Safe strength = = 114.7kN < 125kN

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

Area = 24.42cm2

γ min = 2.44cm

kL=

323 = 132.4

rmin 2.442.4

Table 9(c) f cd = 74.3 − x8.1 = 72.4N / mm 2

Safe strength = 72.4x2242 /1000 162.3kN 125kN

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=800mm2

Area of 1 angle = 2000mm2 = 20cm2

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

Try 4 angle of ISA 130x130x12mm A = 29.82cm2

I x = I y = I min = 4473.8 + 29.8220 − 3.662 = 33742cm 4

γ min

= 33742= 16.82cm4x29.82

kL = 500 = 29.7

16.82r

min

f cd = 211.39N / mm 2

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

Try 4 angles of ISA 100x100x12 A = 22.59cm2

Imin = 4207 + 22.5920 − 2.922 = 27188.4cm4

r= 27188.4 = 17.35cmmin 4x22.59

kL = 500 = 28.8

17.35r

min

f cd = 212.56N / mm 2

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

Try 4 angles of ISA 90x90x10 A = 17.03cm2

Imin = 4126.7 + 17.0320 − 2.592 = 21154.5cm4

r= 21154.35 = 17.62cmmin 4x17.03

kL = 500 = 28.4

17.62r

min

fcd = 224 − 8.4

x13 = 213.08N / mm 2 10

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

Try 4 angles of ISA 80x80x10 A = 15.05cm2

Imin = 487.7 + 15.0520 − 2.342 = 19125.7cm 4

rmin = 17.82cm

kL = 500 = 28rmin 17.82

fcd = 224 − 8 x13 = 213.6N / mm 2

10Safe load = 213.6x4x1505/1000=1285.2>800kN

Try 4 angles of ISA 80x80x8 A = 12.21cm2

Imin = 472.5 + 12.2120 − 2.272 = 15642.99cm4

rmin = 17.89cmkL 500

27.95

fcd = 224 − 7.95

x13 = 213.67 N / mm 2

10

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

Try 4 angles of ISA 60x60x10 Area = 11cm2

Imin = 434.8 + 1120 − 1.852 = 14633.79cm 4

rmin = 18.24cmkL 500

= = 27.41

fcd = 224 − 7.41

x13 = 214.38N / mm 2 10

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

Try 4 angles of ISA 60x60x8mm A = 8.18cm2

Imin = 429 + 8.9620 − 1.772 = 12026.8cm 4

rmin = 19.17cm

kL = 500 = 26rmin 19.17

fcd = 224 − 6 x13 = 216.2N / mm 2

10

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

I zz 12545.2cm4

I yy = 2193.6cm 4

A = 74.85cm2

h = 300

= 1.2, t f = 10.6mm < 100mmb f250

For the buckling about zz axis – ‘b’12545.2

rzz = = 12.95cm

l e =450

= 34.75rzz 12.95

fcd = 216 − 4.75

x10 = 211.25N / mm 2 10

For the buckling about yy axis class ‘c’2193.6

ryy = = 5.41cm

le =

450 = 83.18

ryy 5.41

fcd = 136 − 3.2

x10 = 131.2N / mm 2

10

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 f y =125N/mm2 Area required = 1000x103/125=8000mm2 = 80cm2

Area of one channel = 45cm2

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

rzz = 13.66cm ryy = 15.21cm

About zz axis

le = 600 = 43.92r

zz 13.66

fcd = 198 − 3.9

x15 = 192.15N / mm 2

10

About yy axisle =

600= 39.34

ryy 15.21

fcd = 211 − 9.4

x3 = 208.18N / mm 2 10

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

Try ISLC 300 A = 84.22cm2

rzz = 11.98cm ryy = 15.32cm

le 600 50r

zz 11.98

fcd = 211 − 0.17

x13 = 210.78N / mm 2 10

le =

600= 39.17

ryy 15.32

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 = 400 ≤ θ ≤ 700 θ 450

Cl 7.6.3 Thickness of lacing bar =1

spacing + g + g = 1 250 + 60 + 60 = 6.17mm say 10mm60 60

Cl 7.6.5.1

Spacing of lacing ≤ 50rmin of one component of member

le

≤ 0.7rminwhole

250 + 60 + 60 = 12.89 ≤ 50 ≤ 0.739.17 = 27.42

28.7

Cl 7.6.6.3

kL≤ 145

r flat

λ of the lacing bar = 0.7l =0.7 x37 2 = 126.86 ≤ 145t 120 1 12

Check the bars for lacing in compression

Shear force = 2.5

x1000 = 25kN100

Force on the lacing bar = S cos ecθ = 25 cos ec45 = 8.84kN2n 2x2

For the flat angle, for λ 127

fcd = 83.7 − 7 x9.4 = 77.15N / mm 2

10

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

Check for the flat in Tension

= b − d tf y / γ m =

50 − 1810410 = 105kN

1.25

Or f y Ag / γ mo = 250 x50x10 /1.1 = 113.4kN > 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 kL of battened columns shall be 1.1 times ther

maximum actual slenderness ratio of the column.

kL = 1.1x39.17=43r

fcd = 198 − 3 x15 = 193.5N / mm2

10

Safe load = 193.5x 8422

1629.7kN 1000kN1000

Maximum spacing of the batten le≤ 0.7

rmin of one component of memberr

minwhole

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 Vb Vt

C

25000x85 28.72kN

NS 2x370

Vt = 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 M V

t c 25000x850 5312500Nmm2N 2x2

For end batten

Shear Stress = 28720/370x8=9.7N/mm2 <250

131.2N / mm2

3x1.1

6M 6x5312500 29.10N / mm 2 250

227 N / mm2

Bending stress =

td 2 8x3702 1.1

Hence safe

For Intermediate battens

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

Bending stress = 6x53/2500/8x3002=44.27N/mm2 < 250/1.1=227N/mm2

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

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

Adopt a pitch of 100mm

M r

Force due to moment = ∑r 2

= 5312500 x100 26.56kN1002 x1002

Resultant force = 9.572 26.562 28.23kN

45.3kN 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, 100b2c / 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 Pc is

f cu AP A 0.45 p ≥ Pcc g

p ycs

f cuP A 0.25 A p ycs g

p yc

Where Ac = gross sectional area of concrete. Casing in excess of 75mm from the steel section is neglected. Finish is neglected.

Ag =gross area of the steel section

fcu =characteristic strength of the concrete at 28 days. This should not exceed 40N/mm2.

pc =compressive strength of steel section determined using rx and rz in the determination of which p y ≤ 335N / mm 2

p y = 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 2502

A=6496mm

ryy =5.49cm

For the above cased column;

ry 0.2325 65mm≠ 0.2250 150 80mm

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

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

iii) 100 bc2 =100x325=32500mmd c

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

slenderness ratio = kL

3150

48.46r 65

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

fcd = 198 − 8.46

x15 = 185.3N / mm 2

10

The gross sectional area of concrete

Ac 325x325 105625mm 2

Compressive strength of concrete

25 185.3P = 6496 + 0.45x x105625 = 2084.5kN

c 250 1000

Short column strength

25x105625 250P 6496 0.25x 2284kN

cs 250 1000

Compressive strength of column = 2084.5kN