Chapter 10 Columns. 10.1 Introduction Column = vertical prismatic members subjected to compressive...
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Transcript of Chapter 10 Columns. 10.1 Introduction Column = vertical prismatic members subjected to compressive...
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 Pcr
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
allow
PA
cr
PLAE
22( )sin ( )crLP K (10.1)
sin (10.2)
Concerns before:
New concern: Stable?
Unstable?
22( ) ( )crLP K
sin
4 /crP K L (10.2)
The system is stable, if
The system is unstable if
Since
4 /crP K L
4 /crP K L
A new equilibrium state may be established
After the load P is applied, there are three possibilities:
1. P < Pcr – equilibrium & = 0 -- stable
2. P > Pcr – equilibrium & = -- stable
3. P > Pcr – unstable – the structure
collapses, = 90o
10.3 Euler’s Formula for Pin-Ended Columns
Determination of Pcr for the configuration in Fig. 10.1 ceases to be stable
2
2
d y M Py
dx EI EI
2
2 0d y P
ydx EI
Assume it is a beam subjected to bending moment:
(10.5)
(10.4)
2 Pp
EI
22
2 0d y
p ydx
sin cosy A px B px
Defining:
The general solution to this harmonic function is:
(10.7)
(10.6)
(10.8)
B.C.s:
@ x = 0, y = 0 B = 0
@ x = L, y = 0
Eq. (10.8) reduces to
0sinA pL (10.9)
0sinA pL
2 22
2
np
L
2 Pp
EI
(10.9)
Therefore,
1. A = 0 y = 0 the column is straight!
2. sin pL = 0 pL = n p = n /L
Since (10.6)
We have2 2
2
n EIP
L
For n = 12
2cr
EIP
L
(10.10)
-- Euler’s formula (10.11)
Substituting Eq. (10.11) into Eq. (10.6),
2
2cr
EIP
L
2 Pp
EI
2 22
2 2crP EI
pEI L EI L
(10.6)
(10.11)
Therefore,
Hence pL
Equation (10.8) becomes sinx
y AL
(10.12)
This is the elastic curve after the beam is buckled.
If P < Pcr sin pL 0
0sinA pL
1. A = 0 y = 0 the column is straight!
2. sin pL = 0 pL = n 2
2cr
EIP
L
(10.9)
Hence, A = 0 and y = 0 straight configuration
Critical Stress:
Introducing
2
2( )cr
ELr
2
2cr
cr
P EIA L A
2I Ar
Where r = radius of gyration
, ,y zxx y z
I IIr r r
A A A
Where r = radius of gyration (10.13)
L/r = Slenderness ratio
10.4 Extension of Euler’s Formula to columns with
Other End Conditions
2
2cre
EIP
L
2
2( )cr
e
ELr
(10.11')
(10.13')
Case A: One Fixed End, One Free End
Le = 2L
Case B: Both Ends Fixed
At Point C
RCx = 0
Q = 0 0VQIt
Point D = inflection point M = 0 AD and DC are symmetric
Hence, Le = L/2
2
2
d y M P Vy x
dx EI EI EI
2 Pp
EI
22
2
d y Vp y x
dx EI
Case C: One Fixed End, One Pinned End
M = -Py - Vx
Since
Therefore,
The general solution:
The particular solution:
sin cosy A px B px
2
Vy x
p EI
2 Pp
EI
Vy x
P
Substituting into the particular solution, it follows
As a consequence, the complete solution is
sin cosV
y A px B px xP
(10.16)
sinV
A pL LP
sinV
y A px xP
(10.16)
B.C.s:
@ x = 0, y = 0 B = 0
@ x = L, y = 0
(10.17)
Eq. (10.16) now takes the new form
sin cosV
y A px B px xP
Taking derivative of the question,
cosdy V
Ap pxdx P
cosV
Ap pLP
sinV
A pL LP
B.C.s: @ x = L, dy/dx = = 0
(10.18)
(10.17)
10 1710 18
( . )( . ) tan pL pL (10.19)
Solving Eq. (10.19) by trial and error,
2 24 4934 4 4934 20 19064. . / . /pL p L p L
2 22
20 19.P EIp P p EI
EI L
2
2 2
20 19.cr
e
EI EIP
L L
Since
Therefore,
Case C
Solving for Le
Le = 0.699L 0.7 L
Secant Formula:
(10.36)
2
11
2
max
sec( )e
PA ec P L
r EA r
11
2sec( )e
P LEA r
If Le/r << 1,
Eq. (10.36) reduces to
21
maxPecAr
(10.37)
10.6 Design of Columns under a Centric LoadAssumptions 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
Facts:
1. Long Columns: obey Euler’s Equation
2. Short Columns: dominated by y
3. Intermediate Columns: mixed behavior
Test Data:
Real Case Design using Empirical Equations:
Two Approaches:
1. Allowable Stress Design
2. Load & Resistance Factor Design
1. For L/r Cc [long columns]:
[Euler’s eq.]
Structural Steel – Allowable Stress Design
Approach I -- w/o Considering F.S.
2
2( / )cr
EL r
2
212
( / )[ ]cr Y
c
L rC
2. For L/r Cc [short & interm. columns]:
where2
2 2c
Y
EC
1. L/r Cc :
(10.43)
Approach II -- Considering F.S.
2
21 92. . ( / )cr
all
EF S L r
2. L/r Cc :
211
2/
[ ( ) ]. . .cr Y
allc
L cF S F S C (10.45)
10.7 Design of Columns under an Eccentric Load
centric bending
max
P McA I
(10.56)
(10.57)
(I) Allowable Stress Method
(II) Interaction Method
Two Approaches:
1. The section is far from the ends
2. < y
I. Allowable-Stress Method
all
P McA I
(10.58)
-- all is obtained from Section 10.6.
-- The results may be too conservative.
II. Interaction Method
1/ /
all all
P A Mc I
1/ /
( ) ( )all centric all bending
P A Mc I
(10.59)
(10.60)(Interaction Formula)
all centric -- determined using the largest Le
Case A: If P is applied in a plane of symmetry: