lecture
twenty
steel construction:
columns & tension members
Cor-Ten Steel Sculpture By Richard Serra Museum of Modern Art Fort Worth, TX (AISC - Steel Structures of the Everyday)
A
RCHITECTURAL
S
TRUCTURES
:
F
ORM,
B
EHAVIOR, AND
D
ESIGN
A
RCH 331
HÜDAVERDİ TOZAN
Structural Steel
• standard rolled shapes
(W, C, L, T)
• tubing
• pipe
Design Methods (revisited)
• know
– loads or lengths
• select
– section or load
– adequate for
strength and
no buckling
Allowable Stress Design (ASD)
• AICS 9
th
ed
• slenderness ratio
– for kl/r
C
c
= 126.1 with F
y
= 36 ksi
= 107.0 with F
y
= 50 ksi
2
2
23
12
.
.
r
K l
E
S
F
f
F
a
cr i t i ca l
r
Kl
C
c
and Euler’s Formula
• Kl/r < C
c
– short and stubby
– parabolic transition
• Kl/r > C
c
– Euler’s relationship
– < 200 preferred
E
C
2
2
Short / Intermediate
• L
e
/r < C
c
– where
.
.
2
1
2
2
S
F
F
C
r
K l
F
y
c
a
3
3
8
8
3
3
5
.
.
c
c
C
r
K l
C
r
K l
S
F
Unified Design
• limit states for failure
1. yielding
2. buckling
F
e
– elastic buckling stress (Euler)
g
c r
n
c
0
.
9 0
P
F
A
n
a
P
P
n
c
u
P
P
y
e
y
F
.
F
or
F
E
.
r
K L
44
0
71
4
y
e
y
F
.
F
or
F
E
.
r
K L
44
0
71
4
Unified Design
• P
n
= F
cr
A
g
– for
– for
– where
y
F
F
cr
y
F
.
F
F
E
.
r
KL
e
y
4
71
0
658
e
cr
y
F
.
F
F
E
.
r
K L
877
0
71
4
2
2
r
KL
E
F
e
Procedure for Analysis
1. calculate KL/r
•
biggest of KL/r with respect to x axes and y axis
2. find F
a
or F
cr
from appropriate equation
•
tables are available
3. compute P
allowable
= F
a
A or P
n
= F
cr
A
g
•
or find f
actual
= P/A
4. is P
P
allowable
(P
a
P
n
/
)? or is P
u
P
n
?
•
yes: ok
1. guess a size (pick a section)
2. calculate KL/r
•
biggest of KL/r with respect to x axes and y axis
3. find F
a
or F
cr
from appropriate equations
•
or find a chart
4. compute P
allowable
= F
a
A (P
n
/
= F
cr
A
g
)
or P
n
= F
cr
A
g
•
or find f
actual
= P/A
Procedure for Design (cont’d)
5. is P
P
allowable
(P
a
P
n
/
)? or is P
u
P
n
?
•
yes: ok
•
no: pick a bigger section and go back to step 2.
6. check design efficiency
•
percentage of stress =
•
if between 90-100%: good
•
if < 90%: pick a smaller section and
go back to step 2.
%
P
P
c
r
100
Beam-Column Design
• moment magnification (P-
)
C
m
– modification factor for end conditions
= 0.6 – 0.4(M
1
/M
2
) or
0.85 restrained, 1.00 unrestrained
P
e1
– Euler buckling strength
)
P
/
P
(
C
B
e
u
m
1
1
1
f a c t o r e d
m a x
u
B
M
M
1
2
2
1
r
Kl
EA
P
e
Beam-Column Design
• LRFD (Unified) Steel
– for
– for
P
r
is required, P
c
is capacity
c
- resistance factor for compression = 0.9
- resistance factor for bending = 0.9
0
1
9
8
2
0
.
M
M
M
M
P
P
:
.
P
P
n x
b
u x
n x
b
u x
n
c
u
c
r
0
1
2
2
0
.
M
M
M
M
P
P
:
.
P
P
n x
b
u x
n x
b
u x
n
c
u
c
r
Design Steps Knowing Loads
(revisited)
1. assume limiting stress
•
buckling, axial stress,
combined stress
2. solve for r, A or S
3. pick trial section
4. analyze stresses
5. section ok?
Rigid Frame Design
(revisited)
• columns in frames
– ends can be “flexible”
– stiffness affected by beams
and column = EI/L
– for the joint
• l
c
is the column length of each column
• l
b
is the beam length of each beam
b
c
l
EI
l
EI
G
Rigid Frame Design
(revisited)
• column effective length, k
A
• steel members can
have holes
• reduced area
• increased stress
Tension Members
(AISC - Steel Structures of the Everyday)
