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