Timber and Steel Design Timber and Steel Design

Urban and Civil

Nov 29, 2013 (4 years and 5 months ago)

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￿
Sections used for Columns
￿
￿
Allowable Stress
￿
Effective Length
Timber and Steel Design
Timber and Steel Design
Lecture
Lecture
5
5
-
-
Analysis of Compression Members
S U R A N A R E E
INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY
SCHOOL OF CIVIL ENGINEERING
Tall Buildings
Tall Buildings
New York
WTC on September 11, 2001
Frame Buckling Mechanism
Frame Buckling Mechanism
How can column fail?
How can column fail?
￿
￿
Compression
Compression
￿
￿
Flexural or Lateral Buckling
Flexural or Lateral Buckling
￿
￿
Local Buckling
Local Buckling
￿
￿
Torsional
Torsional
Buckling
Buckling
Buckling
Buckling
Types
Types
Sections used for Columns
Sections used for Columns
Double angle
Wide flange
Tube
Pipe
Built-up sections
P
L
B
A
P
L
B
A
Buckling of column due to
Buckling=Type of failure which occurs when column is slender
Short
Yielding
Long
Buckling
Intermediate
Yielding & Buckling
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

Columns with Pinned Ends
Columns with Pinned Ends
P
L
B
A
Ideal column:-perfectly straight
-linearly elastic material
-no imperfections
P
L/2
B
A
y
x
υ
υ
max
P
P
M=-P
υ
B
y
x
υ
P
2
2
d
EI
dx
υ
=
Bending-moment
equation
2
2
0
dP
EI
dx
υ
υ
+=
Differential equation
of deflected curve
π
=
2
2
cr
EI
P
L
Euler’s formula
π
=
2
2
From
cr
EI
P
L
I=leastmoment of inertia
Which axis give the least moment of inertia?
a
a
b
b
Critical Stress
Critical Stress
Dividing critical load by cross-sectional area
2
2
AL
EI
A
P
cr
cr
π
σ
==
L/r
σ
cr
Euler’s curve
σ
pl
Proportional limit
0
AIr/=
2
2
)/(rL
E
cr
π
σ
=
Critical stress
rL/
Slenderness ratio
￿￿￿￿￿￿￿￿￿￿￿ 5-1
(a) W250x175 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ 5.0 ￿￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿ F.S. =2 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ Proportional Limit =2,500 ￿.￿./￿￿.
2 ￿￿￿ (b) ￿￿￿￿￿￿
(a) ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ 2.5 ￿￿￿￿ ￿￿￿￿￿￿￿ (a)W250×175 (A= 56.24 ￿￿.
2, r
x
= 10.4 ￿￿.,ry
= 4.18 ￿￿.)
￿￿￿ r ￿￿￿￿￿￿￿￿￿￿￿￿￿ ry
= 4.18 ￿￿.
L/r= 5.0(100)/4.18 = 119.6
￿￿￿￿￿￿￿￿￿￿￿
(
)
()
26
2
2.110
1,448
119.6
P
A
π
×
==
￿.￿./￿￿.
2
< 2,500 ￿.￿./￿￿.
2
OK
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
(b)L/r= 2.5(100)/4.18 = 59.81
(
)
()
26
2
2.110
5,791
59.81
P
A
π
×
==
￿￿￿￿￿￿￿￿￿￿￿ ￿.￿./￿￿.
2
< 2,500 ￿.￿./￿￿.
2
NG
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

Pipe or tube
column
Base plate
welded to column
Welded or bolted
connection
Stiffening plates
welded to flange
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿￿ (Inflection Point) ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿ KL
KL=L
K =1.0(a)
K =0.5(b)
KL =0.5LL
KL =0.7L
K =0.7
(c)
L
￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿
￿￿￿
K ￿￿￿￿￿￿￿￿
0.5
0.7
1.0
1.0
2.0
2.0
￿￿￿
K ￿￿￿￿￿￿￿
0.65
0.8
1.2
1.0
2.1
2.0
￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿ ￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿￿ ￿￿￿ ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
Effective Length Factor
Effective Length Factor
(
(
K
K
)
)
Buckling Axis
Buckling Axis
P
yy
x
x
xx
y
y
Deflected
positions
Buckling
in the x-x plane
Buckling
in the y-y plane
Columns normally buckle laterally (the cross-section moves sideways).
This is usually the y-y axis, but both must be considered.
It occurs about the axis which has the highest value of
L / r
.
Usually,
rx
> ry
x-x = strong axis
y-y = weak axis
But . . .
But . . .
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

x -x = Strong axis
(￿￿￿￿￿￿￿)
y -y = Weak axis
(￿￿￿￿￿￿)
= Lateral support against
xx
y
y
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

2 m
2 m
KL= 2 m
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

2 m
2 m
KL= 4 m
Column Buckling Strength from Experiment
Column Buckling Strength from Experiment
Weight
ab
Euler’sformula
Fy
L/r
Pu/A
Short Column
fails by
Yielding,
Long Column
fails by
Buckling
So,What happen to
Intermediate Column
?Fails by
???
Buckling
Buckling
strength
strength
vs. slenderness
vs. slenderness
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿

MAX 200
MAX 200
A
A
KL
r
Elastic
buckling
Inelastic
buckling
Failure stress
Band of
test data
Euler’s
formula
σ
=
cr
cr
P
A
Where is section A
Where is section A
-
-
A
A
Cc
KL/r
P/A
π
=
2
2
(/)
cr
P
E
A
KLr
Euler’s formula:
Elastic
Elastic
vs.
vs.
Inelastic
Inelastic
Buckling
Buckling

=−


2
2
(/)
1
2
cr
y
c
P
KLr
F
A
C
Elastic Buckling = 0.50Fy
()
22
2
2
/
c
EE
C
Lr
ππ
==
Fy
0.50Fy
y
c
F
E
C
2
2
π
=
(
)
()
()





=
+−
2
2
3
3
/
1
2
3//
5
388
y
c
a
cc
KLr
F
C
F
KLrKLr
CC
<
c
KL
C
r
()
π
=
2
2
12
23/
a
E
F
KLr
>
c
KL
C
r
Inelastic Buckling:
Elastic Buckling:
π
=
2
2
c
y
E
C
F
0.60Fy
Fa
/2
23/12
y
F
AISC
AISC
Formulas
Formulas
Allowable Compressive Stress
Allowable Compressive Stress
F.S. Variable
Cc
KL/r
P/A
π
=
2
2
(/)
cr
P
E
A
KLr

=−


2
2
(/)
1
2
cr
y
c
P
KLr
F
A
C
Fy
F.S. = 23/12
￿￿￿￿￿￿￿￿￿￿￿ 5-2
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ P ￿￿￿￿￿￿ W300 ×94 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
P
P
5 m
W300x106
￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿ W300×94 ￿.￿./￿.(A= 119.8 ￿￿.
2, r
y
= 7.51 ￿￿.)
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿-￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿￿ 5.1 ￿￿￿ K= 0.8
26
2(2.110)
128.8
2,500
c
C
π
×
==
0.8(500)
53
7.51
KL
r
==
￿￿￿￿￿￿￿￿￿
KL/r< Cc ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ (5.8)
3
3
53(53)(53)
F.S.1.81
38(128.8)8(128.8)
=+−=
2
2
2
(53)
12,500
2(128.8)
1,264 kg/cm
1.81
a
F




==
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿P= FaAg
= (1.26)(119.8) = 151 ￿￿￿
￿
30 cm
PL1.2
×
50 cm
y
39.2 cm
MC380
×
100
67.3 kg/m
￿￿￿￿￿￿￿￿￿￿￿ 5-3 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿.1 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿
P ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿￿￿￿￿
KL= 6.0
￿￿￿￿
￿￿￿￿￿￿￿￿ =￿￿.
12.15
42
.
231
)2.20)(71.85(2)6.0)(50(2.1
=
+
y
￿￿￿￿￿￿￿ MC380
￿100 ￿￿￿￿ 67.3 ￿￿./￿.
(A=85.71 ￿￿.
2, Ix=17,600 ￿￿.
4,
Iy
= 671 ￿￿.
4
, cy
= 2.50 ￿￿.)
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ A= 1.2(50) + 2(85.71) = 231.42 ￿￿.
2
Ix
= 2(17,600) + 2(85.71)(20.2-15.12)
2
+ (1/12)(50)(1.2)
3
+ (1.2)(50)(15.12-0.6)
2
= 52,281
￿￿.
4
Iy
= 2(671) + 2(85.71)(15+2.5)
2
+ (1/12)(1.2)(50)
3
= 66,339
￿￿.
4
￿￿￿￿￿￿￿￿￿￿￿￿￿ ￿￿.
03.15
42.231
281,52
==
r
600
40
15.03
KL
r
==
￿￿￿￿￿￿￿￿￿￿￿ ￿.1
Fa
= 1,337
￿.￿./￿￿.
2
P=F
a
Ag
= 1.337(231.42) = 309.4
￿￿￿
￿￿￿￿￿￿￿￿￿￿￿ 5-4
￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿ W250
×
×××66.5 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿￿￿￿￿￿ 7.2 ￿￿￿￿￿￿￿￿￿￿￿￿ x-x ￿￿￿￿￿￿￿￿￿￿￿￿ 3.6 ￿￿￿￿￿￿￿￿￿￿￿￿ y-y ￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿ A36 ￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿￿
￿￿￿￿￿￿￿ ￿￿￿￿￿￿￿ W250
×
×××66.5(A= 84.7 ￿￿.
2, r
x
= 10.8 ￿￿.,ry
= 6.29 ￿￿.)
1(360)
57
6.29
y
y
KL
r
==
(￿￿￿￿￿￿￿￿￿)
(￿￿￿￿￿￿)
0.8(720)
53
10.8
x
x
KL
r
==
￿￿￿￿￿￿￿￿￿￿￿ ￿.1Fa
= 1,263 ￿.￿./￿￿.
2
P = (1.263)(84.7) = 107 ￿￿￿
￿