Q factor calculation according to IS 801:1975 for columns

Name of the member: 200x80x25x4

Calculated Sectional Properties of the Member as per IS 811:1987

A = 1520.96W = 200 mmB = 80 mm

L = 25 mmT = 4 mm

R = 6 mmr = 8 mm

Ixx = 9182920.34Iyy = 1262661.26

28.81

Radius of curve = 8 mm

MI @ centroidal axis

Elements LengthWeb 180 0 0

Flange 120 98 1152480Lips 30 82.5 204187.50

Corners 50.24 94.90 452500.09TOTAL = 380.24 mm 1809167.59

TOTAL = 2295730.09

918.29

Distance of the C.G. from centre of arc =

mm2

mm4mm4

ryy =

for 900 corners =

d' from XX Ld2

MI1 = x104 mm4

Mid-Line Dimensions

Area calculations neglacting the curve effect:Centroid of the mid line section

Elements Length LWeb 196 76

Flange 76 76Lips 23 23

TOTAL = 394.00 mm 23Therefore, Area = 1576

973.39136.06

Computation of effective widths as per IS:

1. For Unstiffened Element, (Lips) 2. For Stiffened Elements, (Flange)

3.75

theirfore fc = 144Uneff length =

if b > w, then b = wif b < w, then b =

3. For Stiffened Elements, (Web)

mm2

MI1 = x 104 mm4MI2 = x 104 mm4

w/t lip =

if w/tlip

w/t = 45.00

b/t = 40.11

160.44 mm

uneff length = 39.56 mm

5.Determination of safe load,

Length = 1000 mm

l/r = 34.71if (l/r) < (l/r)min then, fa =

97.04

therefore, permissible load = 147.59

befft =

N/mm2

Q factor calculation according to IS 801:1975 for columns

Calculated Sectional Properties of the Member as per IS 811:1987 Sectional Properties of the Member as per Mid Line

fy = 240 A =E = 204646 W =

B =180 mm L =

t = 4 mm T =R =

60 mm r =t = 4 mm Ixx =

Iyy =15 mm

t = 4 mm

1.57 r*4 50.24 mm Radius of curv

MI @ centroidal axis MI @ YY axis, axis is assumed at leftmost surface of web

MI Elements Length Ld486000.00 Web 180 2 360

Flange 120 40 4800562.50 Lips 30 78 2340

Nr Corner 25.12 4.91 123.26486562.50 Far Corner 25.12 75.09 1886.34

TOTAL = 380.24 mm 9509.60TOTAL = 527004.95

25.00947 mm

MI of fange @ their CG = 360005.096

Linear Iyy = 315.67

N/mm2N/mm2

wweb =

wflange =

wlip= ryy =

for 900 corners =

d' from YY

mm3

Xcg =

126.27

Centroid of the mid line sectionXi LiXi38 288838 2888 X = 23.53299 mm76 1748 Y = 98 mm76 1748

2. For Stiffened Elements, (Flange)

w/t = 15

b/t = 10.66

42.65 mm

Uneff length = 37.35 mm

if b > w, then b = wif b < w, then b = 42.65 mm

4. Determination of factor Q,

1213.31

MI2 = x104 mm4

befft =

befft=

Aeff = mm2

Q = 0.80

5. Determination of Cc and (l/r)min

Cc = 129.79

(l/r)min = 145.31

5.Determination of safe load,

kN

Sectional Properties of the Member as per Mid Line

1576.00 fy = 240200 mm E = 204646

80 mm25 mm 196 mm

4 mm t = 4 mm6 mm8 mm 76 mm

9733922.67 t = 4 mm1360594.95

21 mm29.38 t = 4 mm

8 mm 1.57 r*4 50.24

MI @ YY axis, axis is assumed at leftmost surface of web

720.00192000.00182520.00

604.87141650.48517495.35

mm2 N/mm2N/mm2

wweb =

wflange =mm4mm4

wlip=

for 900 corners =

Ld2

mm3

mm3

x 103 mm3

Computation of effective widths as per Mid Line:

1. For Unstiffened Element, (Lips) 2. For Stiffened Elements, (Flange)

5.25 w/t = 19

b/t = 19.96

49.00theirfore fc = 144

Uneff length = 31.00

if b > w, then b = wif b < w, then b =

4. Determination of factor Q,3. For Stiffened Elements, (Web)

1328.98

w/t lip =

if w/tlip

w/t = 49.00Q = 0.84

b/t = 41.31

165.25 mm 5. Determination of Cc and (l/r)min

uneff length = 30.75 mm Cc =

(l/r)min =

5.Determination of safe load,

Length = 1000 mm

l/r = 34.03if (l/r) < (l/r)min then, fa =

102.53

therefore, permissible load = 161.59 kN

befft =

N/mm2

Sectional Properties of the Member as per given in IS 811:1987

A =W =B =L =T =R =r =

Ixx =Iyy =

mm Radius of curve =

Computation of effective widths as per IS:

1. For Unstiffened Element, (Lips)

theirfore fc =

ryy =

w/t lip =

if w/tlip

3. For Stiffened Elements, (Web)

w/t =

b/t =

uneff length =

5.Determination of safe load,

Length =2. For Stiffened Elements, (Flange)

l/r =if (l/r) < (l/r)min then, fa =

mm therefore, permissible load =

mm

49.00 mm

4. Determination of factor Q,

befft =

mm2

5. Determination of Cc and (l/r)min

129.79

141.34

5.Determination of safe load,

Sectional Properties of the Member as per given in IS 811:1987

1500.00 fy = 240200 mm E = 204646

80 mm25 mm 180 mm

4 mm t = 4 mm6 mm8 mm 60 mm

9030000.00 t = 4 mm1240000.00

15 mm28.75 t = 4 mm

8 mm 1.57 r*4 50.24 mm

Computation of effective widths as per IS:

1. For Unstiffened Element, (Lips) 2. For Stiffened Elements, (Flange)

3.75 w/t = 15

b/t = 10.66

42.65 mm144

Uneff length = 37.35 mm

if b > w, then b = wif b < w, then b = 42.65 mm

mm2 N/mm2N/mm2

wweb =

wflange =mm4mm4

wlip=

for 900 corners =

if w/tlip

4. Determination of factor Q,3. For Stiffened Elements, (Web)

1272.3545.00

Q = 0.8540.11

160.44 mm 5. Determination of Cc and (l/r)min

uneff length = 19.56 mm Cc = 129.79

(l/r)min = 140.92

5.Determination of safe load,

1000 mm

34.78if (l/r) < (l/r)min then, fa =

102.98

therefore, permissible load = 154.47 kN

Aeff = mm2

N/mm2

Sectional Properties of the Member as per given in IS 811:1987

Q factor calculation according to BS5950 for columns

Name of the member: 200x80x25x4Sectional Properties of the Member

A = 1500W = 200 mm 180B = 80 mm 60L = 25 mm 15T = 4 mmR = 6 mm

Ixx = 9030000.00 fy = 240Iyy = 1240000.00 E = 204646

1. Effctive breadth of web 2. Effctive width of flanges

h = 0.33

K1 = 5.71 K2 = 0.634

therefore, K1 = 5.71 therefore, K2 =

Pcr = 521.26 Pcr = 3288.89

Fcr/(Pcr/Lm) = 0.40 Fcr/(Pcr/Lm) =

beff/b = 0.983 beff/b = 1.000

beff = 176.94 mm beff = 60.00

Gross Area as pe Mid Line Dimension = 1576 mm2

Lips 184Flanges 608

web 784Total = 1576 mm2

Reduced Area of the section = 1490.16 mm2

mm2Webeff B1 =

Fleff B2 =Lipeff =

mm4mm4

Lips 120Corners 182.4 Table 11 page 29 IS 801:1975Flanges 480

Web 707.76Total 1490.16 mm2

therefore, Q = 0.95

The Compressive Strength of the Member =

Q factor calculation according to BS5950 for columns

mmmmmm

2. Effctive width of flanges 3. Effctive width of lips

K = 0.425 conservtive for unstiffened elements

4 therefore, K = 0.425

Pcr = 5591.11 N/mm2

0.0635 Fcr/(Pcr/Lm) = 0.037

beff/b = 1.000

mm beff = 15.00 mm

N/mm2N/mm2

310.99 kN

Q factor calculation according to BS5950 for columns

conservtive for unstiffened elements

Sheet1Sheet2