12-Examples on Compression Members (Steel Structural Design & Prof. Shehab Mourad)

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