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
• Study the stability of elastic columns
• Determine the critical load Pcr
• The effective length
• 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
Concerns before
Stable? Unstable?
New concern
(10.1)
(10.2)
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Since
(10.2)
The system is stable, if The system is unstable
if
A new equilibrium state may be established
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The new equilibrium position is
(10.3)
or
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After the load P is applied, there are three
possibilities
1. P lt Pcr equilibrium ? 0 -- stable
2. P gt Pcr equilibrium ? ? -- stable
3. P gt Pcr unstable the structure collapses,
? 90o
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10.3 Eulers Formula for Pin-Ended Columns
Determination of Pcr for the configuration in
Fig. 10.1 ceases to be stable
Assume it is a beam subjected to bending moment
(10.4)
(10.5)
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Defining
(10.6)
(10.7)
The general solution to this harmonic function is
(10.8)
B.C.s
_at_ x 0, y 0 ? B 0 _at_ x L, y 0
Eq. (10.8) reduces to
(10.9)
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(10.9)
Therefore,
1. A 0 ? y 0 ? the column is straight!
2. sin pL 0 ? pL n? ? p n? /L ?
(10.6)
Since
We have
(10.10)
For n 1
-- Eulers formula
(10.11)
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Substituting Eq. (10.11) into Eq. (10.6),
(10.11)
(10.6)
Therefore,
Hence
Equation (10.8) becomes
(10.12)
This is the elastic curve after the beam is
buckled.
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(10.9)
1. A 0 ? y 0 ? the column is straight!
2. sin pL 0 ? pL n? ?
If P lt Pcr ? sin pL ? 0
Hence, A 0 and y 0 ? straight configuration
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Critical Stress
Introducing
Where r radius of gyration
Where r radius of gyration
(10.13)
L/r Slenderness ratio
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10.4 Extension of Eulers Formula to columns with
Other End Conditions
Case A One Fixed End, One Free End
(10.11')
(10.13')
Le 2L
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Case B Both Ends Fixed
At Point C
RCx 0 Q 0 ?
Point D inflection point ? M 0 ? AD and
DC are symmetric
Hence, Le L/2
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Case C One Fixed End, One Pinned End
M -Py - Vx
Since
Therefore,
The general solution
The particular solution
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Substituting
into the particular solution, it follows
As a consequence, the complete solution is
(10.16)
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(10.16)
B.C.s
_at_ x 0, y 0 ? B 0 _at_ x L, y 0 ?
(10.17)
Eq. (10.16) now takes the new form
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Taking derivative of the question,
B.C.s
_at_ x L, dy/dx ? 0
(10.18)
(10.17)
(10.19)
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Solving Eq. (10.19) by trial and error,
Since
Therefore,
Solving for Le
Case C
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)
If Le/r ltlt 1,
Eq. (10.36) reduces to
(10.37)
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(No Transcript)
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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 Load
Assumptions in the preceding sections -- A
column is straight -- Load is applied at the
center of the column -- ? lt ?y
Reality may violate these assumptions -- use
empirical equations and rely lab data
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Test Data
Facts 1. Long Columns obey Eulers
Equation 2. Short Columns dominated by ?y
3. Intermediate Columns mixed behavior
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Empirical Formulas
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Real Case Design using Empirical Equations
1. Allowable Stress Design
Two Approaches
2. Load Resistance Factor Design
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Structural Steel Allowable Stress Design
Approach I -- w/o Considering F.S.
1. For L/r ? Cc long columns
Eulers eq.
2. For L/r ? Cc short interm. columns
where
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Approach II -- Considering F.S.
1. L/r ? Cc
(10.43)
2. L/r ? Cc
(10.45)
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10.7 Design of Columns under an Eccentric Load
(10.56)
1. The section is far from the ends 2. ? lt ?y
(10.57)
Two Approaches
(I) Allowable Stress Method (II) Interaction
Method
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I. Allowable-Stress Method
(10.58)
-- ?all is obtained from Section 10.6. -- The
results may be too conservative.
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II. Interaction Method
Case A If P is applied in a plane of
symmetry
(10.59)
(10.60)
(Interaction Formula)
-- determined using the largest Le
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Case B If P is NOT Applied in a Plane of
Symmetry
(10.61)