Transcript Chapter 10 Columns
Chapter 10 Columns
10.1 Introduction Column = vertical prismatic members subjected to compressive forces Goals of this chapter: 1. Study the stability of elastic columns 2. Determine the critical load P cr 3. The effective length 4. Secant formula
Previous chapters: -- concerning about (1) the strength and (2) excessive deformation (e.g. yielding) This chapter: -- concerning about (1) stability of the structure (e.g. bucking)
10.2 Stability of Structures Concerns before:
P A
allow
PL AE
cr
New concern:
P cr
(
L
2
)sin sin
K
(
2
) Stable?
Unstable?
(10.1) (10.2)
Since sin
P cr
(
L
2
)
P cr
4
K
(
2
)
(10.2) The system is stable, if
P cr
4
The system is unstable if
P cr
4
A new equilibrium state may be established
P
(
L
2
)sin
M
K
(
2
)
The new equilibrium position is:
PL
4
K
sin
(10.3) or
sin
PL
4
K
After the load P is applied, there are three possibilities: 1. P < P cr – equilibrium &
= 0 - stable 2. P > P cr – equilibrium &
=
- stable 3. P > P cr – unstable collapses,
– the structure = 90 o
10.3 Euler’s Formula for Pin-Ended Columns Determination of P cr for the configuration in Fig. 10.1 ceases to be stable Assume it is a beam subjected to bending moment:
2
d y
dx
2
M EI
P EI y
(10.4)
dx
2
P EI y
0
(10.5)
Defining:
p
2
P EI
2
d y
dx
2 2
p y
0
(10.6) (10.7) The general solution to this harmonic function is:
y
A
sin
px
B
cos
px
(10.8) B.C.s:
@
x = 0, y = 0
B = 0
@
x = L, y = 0 Eq. (10.8) reduces to
A
sin
pL
0
(10.9)
A
sin
pL
0
(10.9) Therefore, Since We have For n = 1 1. A = 0
y = 0
the column is straight!
2. sin pL = 0
pL = n
p = n
/L
p
2
n
2
2
L
2
p
2
P EI P
n
2 2
EI L
2
P cr
2
EI L
2
(10.6) (10.10) -- Euler’s formula (10.11)
Substituting Eq. (10.11) into Eq. (10.6),
P cr
2
EI L
2
(10.11)
p
2
P EI
(10.6) Therefore, Hence
p p
2
P cr
EI L
2
EI
2
L EI
Equation (10.8) becomes
2
L
2
y
A
sin
L x
This is the elastic curve after the beam is buckled.
(10.12)
A
sin
pL
0
(10.9) 1. A = 0
y = 0
2. sin pL = 0
the column is straight!
pL = n
P cr
2
EI L
2
If P < P cr
sin pL
0 Hence, A = 0 and y = 0
straight configuration
Critical Stress:
cr
P cr A
2
EI
2
L A
Introducing
I
Ar
2
Where r = radius of gyration
r x
cr I A x
,
2
E
(
L r
)
2
r y
I y A
,
r z
I z A
Where r = radius of gyration L/r = Slenderness ratio (10.13)
10.4 Extension of Euler’s Formula to columns with Other End Conditions Case A: One Fixed End, One Free End
P cr
2
EI L
2
e
cr
2
E
(
L e r
)
2
L e = 2L (10.11
' ) (10.13')
Case B: Both Ends Fixed At Point C R Cx = 0 Q = 0
VQ It
0
Point D = inflection point
M = 0
AD and DC are symmetric Hence, L e = L/2
Case C: One Fixed End, One Pinned End M = -Py - Vx
2
d y dx
2
M EI
P EI y
V EI x
Since Therefore,
p
2
P EI
2
d y
dx
2 2
p y
V EI x
The general solution:
y
A
sin
px
B
cos
px
The particular solution:
y
V x
Substituting
p
2
P EI
into the particular solution, it follows
y
V P x
As a consequence, the complete solution is
y
A
sin
px
B
cos
px
V P x
(10.16)
y
A
sin
px
B
cos
px
V P x
B.C.s: (10.16)
@
x = 0, y = 0
B = 0
@
x = L, y = 0
A
sin
pL
V P L
Eq. (10.16) now takes the new form (10.17)
y
A
sin
px
V P x
Taking derivative of the question,
dy dx
Ap
cos
px
V P
B.C.s:
@ x = L, dy/dx = = 0
( .
( .
) )
Ap
cos
pL
V P A
sin
pL
V P L
tan pL
pL
(10.18) (10.17) (10.19)
Solving Eq. (10.19) by trial and error,
pL
p
/
L
p
2
Since
p
2
P EI
P
2
p EI
Therefore,
P cr
L
2
EI
2
EI L
2
e
Solving for L e L e = 0.699L
0.7 L
L
2
EI
Case C /
L
2
Summary
10.5
* Eccentric Loading; the Scant Formula
Secant Formula:
P A
1
max
ec r
2
sec(
1 2
P L e
)
EA r
If L e /r << 1, sec(
1 2
P L e
)
EA r
1
Eq. (10.36) reduces to
P A
1
max
ec r
2
(10.36) (10.37)
10.6 Design of Columns under a Centric Load
10.6 Design of Columns under a Centric Load Assumptions in the preceding sections: -- A column is straight -- Load is applied at the center of the column --
<
y Reality: may violate these assumptions -- use empirical equations and rely lab data
Test Data:
Facts: 1. Long Columns: obey Euler’s Equation 2. Short Columns: dominated by
y 3. Intermediate Columns: mixed behavior
Empirical Formulas:
Real Case Design using Empirical Equations: 1. Allowable Stress Design Two Approaches: 2. Load & Resistance Factor Design
Structural Steel – Allowable Stress Design Approach I -- w/o Considering F.S.
1. For L/r
C c [long columns]
:
cr
2
E
2
[Euler’s eq.] 2. For L/r
C c [short & interm. columns]
:
where
cr
Y
[
1 2
C c
2 2
]
C c
2 2
Y
2
E
Approach II -- Considering F.S.
1. L/r
C c :
all
cr
2
E
1 92
L r
2
2. L/r
C c :
all
cr
Y
[
1 1 2
(
C c
2
) ] (10.43) (10.45)
10.7 Design of Columns under an Eccentric Load
centric
bending
(10.56) 1. The section is far from the ends 2.
<
y
max
P A
Mc I
(10.57) Two Approaches: (I) Allowable Stress Method (II) Interaction Method
I. Allowable-Stress Method
P A
Mc I
all
(10.58) --
all is obtained from Section 10.6.
-- The results may be too conservative.
II. Interaction Method Case A: If P is applied in a plane of symmetry:
all
all
1
(10.59) (
(
1
(Interaction Formula) (10.60)
all
centric
-- determined using the largest L e
Case B: If P is NOT Applied in a Plane of Symmetry:
(
)
all centric
M z x
max
(
/ )
all bending I x
M x x
max
(
/ )
all bending I x
1
(10.61)