Chapter 10 Columns
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Transcript of Chapter 10 Columns
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Chapter 10 Columns
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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
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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)
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10.2 Stability of Structures
allowPA
crPLAE
22( )sin ( )crLP K (10.1)
sin (10.2)
Concerns before:
New concern: Stable?
Unstable?
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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
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22( )sin ( )LP M K
4 sinPLK
The new equilibrium position is:
4sinPLK
(10.3)
or
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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
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10.3 Euler’s Formula for Pin-Ended ColumnsDetermination of Pcr for the configuration in Fig. 10.1
ceases to be stable
2
2d y M P ydx EI EI
2
2 0d y P ydx EI
Assume it is a beam subjected to bending moment:
(10.5)
(10.4)
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2 PpEI
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)
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0sinA pL
2 22
2npL
2 PpEI
(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
2n EIPL
For n = 12
2crEIPL
(10.10)
-- Euler’s formula (10.11)
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Substituting Eq. (10.11) into Eq. (10.6),
2
2crEIPL
2 PpEI
2 22
2 2crP EIpEI L EI L
(10.6)
(10.11)
Therefore,
Hence pL
Equation (10.8) becomes sin xy AL
(10.12)
This is the elastic curve after the beam is buckled.
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If P < Pcr sin pL 0
0sinA pL
1. A = 0 y = 0 the column is straight!
2. sin pL = 0 pL = n 2
2crEIPL
(10.9)
Hence, A = 0 and y = 0 straight configuration
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Critical Stress:
Introducing
2
2( )crE
Lr
2
2cr
crP EIA L A
2I Ar
Where r = radius of gyration
, ,y zxx y z
I IIr r rA A A
Where r = radius of gyration (10.13)
L/r = Slenderness ratio
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10.4 Extension of Euler’s Formula to columns with
Other End Conditions
2
2cre
EIPL
2
2( )cr
e
ELr
(10.11')
(10.13')
Case A: One Fixed End, One Free End
Le = 2L
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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
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2
2d y M P Vy xdx EI EI EI
2 PpEI
22
2
d y Vp y xdx 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
2Vy xp EI
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2 PpEI
Vy xP
Substituting into the particular solution, it follows
As a consequence, the complete solution is
sin cos Vy A px B px xP
(10.16)
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sin VA pL LP
sin Vy 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 cos Vy A px B px xP
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Taking derivative of the question,
cosdy VAp pxdx P
cos VAp pLP
sin VA pL LP
B.C.s: @ x = L, dy/dx = = 0
(10.18)
(10.17)
10 1710 18
( . )( . ) tan pL pL (10.19)
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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 EIEI L
2
2 220 19.
cre
EI EIPL L
Since
Therefore,
Case C
Solving for Le
Le = 0.699L 0.7 L
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Summary
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10.5* Eccentric Loading; the Scant Formula
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Secant Formula:
(10.36)
2112
max
sec( )e
PA ec P L
r EA r
1 12
sec( )eP LEA r
If Le/r << 1,
Eq. (10.36) reduces to
21
maxPecAr
(10.37)
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10.6 Design of Columns under a Centric
Load
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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
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Facts:
1. Long Columns: obey Euler’s Equation
2. Short Columns: dominated by y
3. Intermediate Columns: mixed behavior
Test Data:
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Empirical Formulas:
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Real Case Design using Empirical Equations:
Two Approaches:1. Allowable Stress Design
2. Load & Resistance Factor Design
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1. For L/r Cc [long columns]:
[Euler’s eq.]
Structural Steel – Allowable Stress Design
Approach I -- w/o Considering F.S.
2
2( / )crE
L r
2
212
( / )[ ]cr Yc
L rC
2. For L/r Cc [short & interm. columns]:
where2
2 2c
Y
EC
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1. L/r Cc :
(10.43)
Approach II -- Considering F.S.
2
21 92. . ( / )cr
allE
F S L r
2. L/r Cc :
2112
/[ ( ) ]. . .cr Y
allc
L cF S F S C (10.45)
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10.7 Design of Columns under an Eccentric Load
centric bending
maxP McA I
(10.56)
(10.57)
(I) Allowable Stress Method
(II) Interaction Method
Two Approaches:
1. The section is far from the ends2. < y
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I. Allowable-Stress Method
allP McA I
(10.58)
-- all is obtained from Section 10.6.
-- The results may be too conservative.
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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:
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1max max/ //( ) ( ) ( )
x x x x
all centric all bending all bending
M z I M x IP A
(10.61)
Case B: If P is NOT Applied in a Plane of Symmetry: