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§7. Double Modulus Theory (By Engesser, 1989)
(Reduced)
ref. Text pp. 37~42 , Salmon’s pp. 303~306
Inelastic Bklg. of straight Column
Assumptions
1. Small   material nonlinearity only
2. Plane sections remain plane
 Bernoullis’ Hypothesis
3. The relationship between stress-strain in any
longitudinal fiber is given by the stress strain diagram of the material.
(Comp. – Tension, Same Relationship)
4. The col. section is at least singly symmetric
and the plane of bending is a plane of
symmetry
5. The axial load remains constant as the member
moves from the straight to deformed position
※ This theory gives higher strength
than test results.(Salmon pp.304)
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
Et
 cr
unloading path
P=PE
proportional
limit
E
E

permanent set
h2
h1
εdx
d
R
ds≒dx
N.A
Z2
Z1
2
inside
concave
cr
S2
S1
composition
e
1
outside
concave
C.G.
 1  E 1
 2  Et 2
E t ; the slope of stress-strain curve at    cr
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1 d 2 y d
small  theory 


R dx 2 dx
Curvature 
 1  Z1 y ,  2  Z 2 y
 1  Eh1 y ,  2  Et h2 y
or S1  EZ1 y , S2  Et Z 2 y

h1

h1
0
S1 dA  
h2
0
0
S2 dA  0 -------- (pure bending only)
S1 Z1  e dA  
h2
0
S2 Z 2  e dA  py --------( y measured from
C.G.)
h1
from  Ey Z1 dA  Et y
0
h2
0
Z 2 dA  0 (C+T=0)
h1
let Q1   Z1dA
0
h2
Q2   Z 2 dA statical moment of area
0
about neutral axis
EQ1  E t Q2  0 -------- ( Z 1  Z 2 both taken as +)
then the 2nd equil. eq. gives



y E 0 Z12 dA  Et 0 Z 22 dA  ey E 0 Z1 dA  Et 0 Z 2 dA  py
h1
h2
h1
h2
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EI1  Et I 2
I
Reduced modulus
let E 
depends
on
the
stress-strain
relationship of the material & the shape of the
cross-section.
E Iy  py  0 ; I : modulus of inertia about the C.G.
reduced Pr 
 2 EI
l2
introducing  r 
or  r
cr

2E
l r 
2
E
,   E t E  1.0
E
EI r y  py  0
r 
Et I 2 I 1
I
I
  2  1
E I
I
I
I
 r  Pcr A 
 2 E r
l r 
2
 Procedures for determining  r
1) For    diagram  prepare    diagram.
2) From step 1) prepare  r   curve.
3) From step 2) prepare  r  l r curve.
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§8. Tangent Modulus Theory
Assumption ; same as D.M. theory except #5.
the axial load increases during the
transition
from
straight
to
slight
bent position such that the increase
in average stress in compression is
greater than the decrease in stress
due to bending at the extreme fiber
on the convex side, i.e., no strain
reversal
side.
takes
The
place
on
the
compressible
convex
stress
increases at all point ; the tangent
modulus
governs
the
entire
cross-
section.
P
bent

Et
P+ P
cr
E
1
r
P
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P  P 
As P  P
Et Iy  Py  0
Pt   2 Et I  l 2
 t  Pt A   2 E l r 
2
hence,
t l r
where   Et E
curve
is
not
affected
by
cross-
section shape.
 Procedure
1) From    diagram  Establish    curve
2) From step 1), prepare  t  l r curve
 This theory gives lower strength than tested
ultimate strength.
Note; Shanley Concept – True column behavior
Load P
increment of load
PS
Bilinear stress-strain
relation
Pf
 C > T
 Double Modulus
Action
stress-strain
relation
Deflection d
Neglect ps  pt , pt may be considered as pcr
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H.W#5) An axially loaded simply supported column is made of
structural steel with the following mechanical properties.
E  30  103 ksi,  p  28.0 ksi,  y  36 ksi, and
 (ksi )
  Et / E
28.0
1.00
29.0
0.98
30.0
0.96
31.0
0.93
32.0
0.88
33.0
0.77
34.0
0.55
35.0
0.31
35.5
0.16
36.0
0.00
Determine the following ;
(a) The value of l / r which divides the elastic buckling
range and the inelastic buckling range.
(b) The values of
34.0,
35.0,
theory
and
 r and l / r for p / A = 28.0, 30.0, 32.0,
and
35.5
assuming
ksi,
that
using
the
the
double-modulus
cross-section
of
the
column is square.
(c) The critical average stress p / A for l / r = 20, 40, 60,
80,
100,
120,
140,
160,
180,
and
200.
Using
the
tangent-modulus theory in the inelastic range.
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From the result of (a), (b), and (c), plot
(d) The p A   r curve for the double-modulus theory.
(e) The p A  l r curves, distinguishing the portion of the
curve derived by the tangent-modulus theory from that
derived
by
the
double-modulus
theory.
Present
short
discussion.
(f)
The
current
1.5.1.3.1-3)
AISC
that
Specifications
the
allowable
specifies
value
of
(Sect.
p / A for
axially loaded column shall not exceed the following ;
(1) Eq. (1.5-1)
(2) Eq. (1.5-2)
(3) Eq. (1.5-3)
Plot these curves and superimpose them on the graph in (e).
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※ Column Strength
A.Inelastic Column Buckling Strength
without Any Imperfection.
 Tangent Modulus Theory, Pt 
 2 Et I
L2
 Reduced (Double) Modulus Theory, Pr 
Er 
 2 Er I
L2
E I 1  Et I 2
I
 Post Buckling Behavior (by Shanley, 1947)
Buckling commences at P  Pt
 After buckling, increase in stiffness due to
elastic unloading of some fibers in the sections.
 results in increase in load.
 without further yielding, P  Pr
 Further yielding
⇒ further decrease in stiffness
 P  Pm ax , Pt  Pmax  Pr
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B. Factors affecting column strength.
 Out-of-straightness.
imperfections
 Eccentricity of axial load.
 Material nonlinearity.
 End restraints.
 Residual stress.
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C. Design of Metal Columns as related
to Strength theories.
1. Emperical formulas based on column tests.
 not rational because they are entirely
dependent on test results.
 not rational in considering end restraint
(Earliest Column strength, since 1840s)
2. Formulas based on the yield limit state.
 The strength of a column is defined as the load
which
will
give
an
elastic
stress
for
an
initially imperfect column, equal to the yield
stress. ( Large displ. theory )
 popular since 1850s upto almost present.
(e.g.,British use of Perry-Robertson formula)
AASHTO ( 12th. Ed )
σ
σ all  max 
F.S.
σ y / F.S.

l
1   0.25 sec
2r

(σ a ) F.S. 


E

 lack of inelastic basis of column strength
can not rationally consider end restraint.
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3. Formulas based on the tangent–modulus theory.
 Two methods of treating the imperfections.
a. reduce
Pmax to Pt
emperically justified reduced column strength.
 basis for cold-formed columns and aluminum columns.
b. represented as flexural effects in the
interaction equation (SSRC).
 CRC-Column Strength Curve.(1960)
(CRC–Column
Research Council, former name of SSRC)
 cr
2
 1
y
4
and
 cr
1
 2
y

where
 
for  
for  
2
2
KL 1  y
r  E
 can accurately account for end restraints.
4. Formulas based on max. strength.
 Formulas based on a numerical fit of curves obtained
from
max.
strength
analyses
considering
imperfections and residual stresses.
 Modern trend.
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 SSRC Curves 1, 2, and 3 (1976)
(multiple column curves)
 Euro code 3 (ECS, 1994)
 Canadian Standard (CSA, 1994)
D. Effect of Residual Stresses.
1. Hot rolled shapes.
residual
sections,
stress
distribution
rolling
depends
temperature,
on
cooling
types
of
conditions,
straightening procedures, and metal properties.
t f  4.91
t w  3.07
heaviest
A  215 in2
d  22.42
b f  17.89
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※ Because of
this distribution
pattern of
residual stresses,
Pmax . ) minor  Pmax . ) major
for same
slenderness ratio.
Note when the
inelastic bklg.
begins depending
on the mag. of
flange tip
residual
stresses.
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2. Welded Columns
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Poxigencut  Puniv. mill
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Pheavy  Plight
P stress  relieved  Poxygencut  Puniversalmill
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 Inelastic Behavior of Column.
 Column strength curves for H-shaped sections
having compressive residual stresses at
flange tips.
When
Max. comp.
residual stress
When
 0.3 Fy
F
 0.5 , it is
Fy
considered in most
design codes that some
parts of column are in
inelastic range
 inelastic buckling.
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․SSRC Multiple Column Curves
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Algebraic
representations
of
the
three
column
strength curves were obtained by curve fitting,
and the resulting equations are given as Eqs. 3.13
through 3.15.
SSRC Curve 1
1. For 0    0.15
2. For 0.15 
  1.2
u   y
 u   y (0.990  0.122  0.3672 )
3. For 1.2 
  1.8
 u   y (0.051 0.8012 )
4. For 1.8 
  2.8
 u   y (0.008 0.9422 )
5. For
  2.8
(3.13)
 u   y 2 (  Euler curve)
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SSRC Curve 2
1. For 0    0.15
2. For 0.15 
  1.0
u   y
 u   y (1.035  0.202  0.2222 )
3. For 1.0 
  2.0
 u   y ( 0.111 0.6361  0.0872 )
4. For 2.0 
  3.6
 u   y (0.009 0.8772 )
5. For
  3.6
(3.14)
 u   y 2 (  Euler curve)
SSRC Curve 3
1. For 0    0.15
2. For 0.15 
  0.8
u   y
 u   y (1.093  0.622 )
3. For 0.8 
  2.2
 u   y ( 0.128  0.7071  0.1022 )
4. For 2.2 
  5.0
 u   y (0.008 0.7922 )
5. For
  5.0
(3.15)
 u   y 2 (  Euler curve)
82