This document provides examples of calculating the factor resistance of steel columns and angles under axial compression loading. It determines the effective area considering local and global buckling effects. It calculates the critical buckling stress and compares it to design tables. For a double angle, it finds the factor resistance is 427 kN. For a W360x134 column with KLx=12m and KLy=6m, it calculates the factor resistance as 2654.6 kN.

1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example: 1
Determine the factor resistance of a double angle with short legs back to
back (152 X 89 X 9.5)mm, If (KLx) = 3 m , (KLy) = 6 m
Area = 4400 mm2
rx = 25.1 mm , ry = 74.5 mm
Solution:
(a) Check local buckling
b/t = 152/9.5 = 16 > 0.45 FyE / = 12.7
Account for local buckling Q = Qs * Qa
There is no stiffened elements (Qa = 1.0 , Q = Qs)
0.91 FyE / = 25.74
0.45 FyE / < b/t = 16 < 0.91 FyE /
Qs = 1.34 – 0.76*(b/t)* EFy / = 0.91 (Q = Qs = 0.91)
(b) Check global buckling
KLx/rx = 3000/25.1 = 119.5 , KLy /ry = 6000/74.5 = 80.5
Buckling @ x-x axis is critical
λc =
E
FyrKL
*
/
p
= (119.5/3.14)*(250/200000)0.5
= 1.345
cQ l* = 1.345* 91.0 = 1.283 < 1.5
Fcr = Q*(0.658) * Fy
= 0.91*(0.658) * (250) = 114.2 Mpa
ØPn = 0.85*Ag*Fcr
= 0.85*4400*114.2/1000 = 427 kN
Check using LRFD Tables: for Lx = 3m ØPn = 427 kN
Q l 2
0.91 * (1.345) 2
b
t
h
tg
X X
Y
Y

2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example: 2
Use the LRFD tables to determine the factor resistance for double angle
back to back (89 x 89 x 9.5) if (KLx)=3 m (KLy) = 6 m
And determine the number of required connectors needed.
Solution:
From Tables: rx = 27.3 mm , ry = 40.8 mm
KLx/rx = 3000/27.3 = 109.89 , KLy/ry = 6000/40.8 = 147
Since KLy/ry > KLx/rx , then buckling @ y-y axis is critical
From LRFD Tables:
Ly = 6000 mm , ØPn = 202 kN
Number of connectors needed for x-x axis; rz = 17.40 mm
K a/rz £ KLx/rx = 109.89
{ 3000/(n+1) } / 17.40 £ 109.89 , n > 0.56 for x-x axis
From Tables n = 3 for y-y axis
use 3 connectors along the length of the member
b
t
h
tg
X X
Y
Y

3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example : 3
A W360X134 section is used as an axially loaded member with(KLx) =
12 m and (KLy) = 6 m . Find the column factor compression resistance ?
Given :
Area = 17100 mm2
, Fy = 250 MPa
rx = 156 mm , ry = 94 mm
Solution :
(a) Check local buckling
Flange b/t = (369/2)/18 = 10.25 < 0.56 FyE / = 15.8
Web h/tw = 286/11.2 = 25.5 < 1.49 FyE / = 42.2
Then no local buckling will occur, Q =1.0
(b) Check global buckling
KLx/rx = 12000/156 = 76.92 , KLy /ry = 6000/94= 63.83
Buckling @ x-x axis is critical
λc =
E
FyrKL
*
/
p
= (76.92/3.14)*(250/200000)0.5
= 0.866 < 1.5
ØPn = 0.85*Ag*(0.658) * Fy
= 0.85*17100*(0.658) * (250) / 1000 = 2654.6 kN
Check Using LRFD Tables
Since, KLx/rx > KLy /ry ,
get the equivalent (KLy) = (KLx)/ (rx/ry) = 12000/1.66 = 7228.9 mm
From Tables: at KLy = 7000 mm Þ ØPn = 2710 kN
at KLy = 7500 mm Þ ØPn = 2590 kN
at KLy = 7228.9 mm Þ ØPn = 2655 kN
bf = 369mm
tf = 18mm
hw = 286mm
tw = 11.2mm
l 2
(0.866) 2

4. 4 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example :4
A W530X66 section is used for a column section. Find the column
factored compression resistance for two cases :
Case 1 : (KLx) = 5 m and (KLy) = 5 m .
Case 2 : (KLx) = 2 m and (KLy) = 2 m .
Given :
Area = 8370 mm2
, Fy = 250 MPa
rx = 205 mm , ry = 32 mm
Solution :
Check local buckling
Flange, bf/2*tf = (165/2)/11.4 = 7.236 < 0.56 FyE / = 15.8
(no local buckling) , (Qs = 1)
Web, hw/tw = 464/8.9 = 52.1 > 1.49 FyE / = 42.2
Account for local buckling of web Q = Qs * Qa
There is stiffened elements (Qs = 1.0 , Q = Qa)
Case 1 : (KLx) = 5 m and (KLy) = 5 m .
First assume Qa = 1.0 KL/rmin = 5000/32 = 156.13
λc =
E
FyrKL
*
/
p
= (156.13/3.14)*(250/200000)0.5
= 1.757 > 1.5
Fcr = (0.877/ λc2
)*Fy = 71 MPa
be = 1.91*t* ÷
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Fcr
E
t
h
Fcr
E
*/34.01*
= 1.91*8.9* ÷
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-
71
200000
*
1.52
34.0
1*
71
200000
= 589.7 mm > hw
Then Qa = 1.0, and Fcr = 71 MPa
ØPn = 0.85*Ag*Fcr
= 0.85*8370*71/1000 = 505 kN
tf = 11.4mm
bf = 165mm
hw = 464mm
tw = 8.9mm

5. 5 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Case 2 : (KLx) = 2 m and (KLy) = 2 m .
Trial # 1
First assume Qa = 1.0 KL/rmin = 2000/32 = 62.5
λc =
E
FyrKL
*
/
p
= (62.5/3.14)*(250/200000)0.5
= 0.703 < 1.5
Fcr = (0.658) * Fy = 203.2 Mpa
be = 1.91 * t * ÷
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Fcr
E
t
h
Fcr
E
*/34.01*
= 1.91 * 8.9 * ÷
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2.203
200000
*
1.52
34.0
1*
2.203
200000
= 424.1 mm < hw
b* = h – be = 464 – 424.1 = 39.9 mm
Qa =
Ag
effectiveArea.S
=
[ ]
Ag
tbAg w**
-
=
[ ]
8370
9.8*9.398370 -
= 0.957
Q = Qa = 0.957
Qc *l = 0.688 < 1.5 Area effective for web
Fcr = Q*(0.658) * Fy
= 0.957*(0.658) * (250) = 196.3 MPa
Difference in computed stresses = 100*
2.203
3.1962.203 -
= 3.4 > 1
Then try another trial;
Trial # 2
be = 1.91*t* ÷
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Fcr
E
t
h
Fcr
E
*/34.01*
= 1.91*8.9* ÷
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3.196
200000
*
1.52
34.0
1*
3.196
200000
= 429.57 mm < hw
lc 2
be/2
be/2
b* hw = 464mm
Q lc 2
0.957* ( 0.703) 2

6. 6 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
b* = h – be = 464 – 429.57 = 34.43 mm
Qa =
Ag
effectiveArea.S
=
[ ]
Ag
tbAg w**
-
=
[ ]
8370
9.8*43.348370 -
= 0.963
Q = Qa = 0.963
Qc *l = 0.690 < 1.5
Fcr = Q*(0.658) * Fy
= 0.963*(0.658) * (250) = 197.26 MPa
Difference in compression stresses = 100*
3.196
25.1973.196 -
= 0.48 < 1,
Then its good enough;
Fcr = 197.26 MPa
ØPn = 0.85*Ag*Fcr
= 0.85*8370*197.26/1000 = 1403.4 kN
0.963 * (0.703)2
Q lc 2