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Chap 4. Beam – Column
§4.1 Beam-Column with Conc. Loading.
EIy IV  w

EIy IV  0
Static behavior
EI u  0
y  a1 x 3  a 2 x 2  a 3 x  a4
static shape function
EIy IV  Py  0
y  a1 cos kx  a2 sin kx  a3 x  a4
buckling shape function
※
If
we
function,
use
we
static
will
displ.
have
a
function
little
as
error
shape
because
buckling shape is different from static
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  ymax
l
x
2

Ql 3 3tan u  u
kl
, u
3
48 EI
2
u
    0  ( u)   0
3(tan u  u)
u3
a m p l i f iicoant
3t a nu  u

1
m a g n i f iicoant
u3
fa ctor
u3 2 5 17 7
tan u  u   u 
u  
3 15
315


2
5
   0  1  u2 
where u2 
17 4

u  
105

k 2 l 2 Pl 2  2
P


2
.
46
4
4 EI  2
PE
2


 P 
P

  L L 
then    0 1  0.984
 0.998


PE
 PE 


2


P  P 
  L L 
  0 1 
 


PE  PE 


1
 0
Amplificat ion Factor
P
1
Pcr
Q
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○ The presence of compression force weakens the
bending stiffness
○
The
presence
of
axial
tension
increase
the
bending stiffness.
P
The max. bending moment
Ql
Ql PQl 3
1
M m ax 
 P 

4
4 48EI 1  P / PE
 Ql 

Ql 
Pl 2
1
P
1

1

1  0.82



4  12EI 1  P / PE  4 
PE 1  P / PE 
P

 1  0.18
Ql 
PE
or M m ax 
4  1  P / PE








※
manual
0.18
M0
amplification factor
current
P
PE
AISC
involve
term
moment
magnification
factor)
for
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§4.2 Beam-Column with Distributed Loading.

5 wl4
1
0 
,    0 
384EI
 1  P / PE
M0 
wl 2
8



 1  0.3 P / PE
M max  M 0 
 1  P / PE



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§4.3
Effects
of
Axial
load
on
Bending
Stiffness
- Slope deflection Equations with Axial Force
The moment at a distance x from the origin is
M ( x )   EIy"  M ab  Py   M ab  M ba  P (  a   b )
x
l
Mx
EI
EIy  Py  M x
y  
y IV  k 2 y  0 , y  A sin kx  B cos kx  Cx  D
y( 0)   a , y( 0)   a 

B.C.
y( l )   b , y( l )   b 
M ab
EI
M ba
y( l ) 
EI
y( 0)  
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After eliminating the const. A, B, C and D in y and
solving
for
M ab & M ba
,
yields
the
slope
deflection equations
EI
 b
 S1 a  S2 b  S1  S2  a

l
l
EI
 b
M ba 
 S2 a  S1 b  S1  S2  a

l
l
M ab 
where S1 
1   cot 
 csc   1
, S2 

2 tan 2
2 tan  2
1
1

  kl 

Pl 2
(coupling of axial and bending together
EI
⇒ till now nobody use it)
Special case when P  0
M AB 
2 EI
2 a   b  
l
S1  4 , S2  2
Bleich P.212, Value of S1 , S2 tabulated
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○ Derivation of Slope-Deflection Equation
with Axial Tension
Note : Although a general model may be used, the
complexity
of
computation
is
reduced
by
using
superposition of simplified model.
The moment at a distance x from the origin is
M x  M ab  Py  M ab  M ba 
x
l
EIy   M x
x
x

 EIy  Py  M ab   1   M ba
l
l

Let k 2 
y  k 2 y 
P
EA
M ab  x
 M k
  1   ba
EI  l
 EI l
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the general solution is
y  A sinh kx  B coshkx 
M ab  x
 M x
  1   ba
P l
P l

M ab
P
y(0)  0
B
y( l )  0
0  A sinh kl 
M ab
M
coshkl  ba
P
P
let kl  
A
M ab cosh  M ba 1

P sinh 
P sinh 
 y
y 
M ab  cosh 
x
 M  sinh kx x 
sinh kx  coshkx   1   ba 
 

P  sinh 
l
P  sinh 
l

M ab  1
cosh  coshkx  sinh  sinh kx  M ba  1 k coshkx 
  k

 

P  l
sinh 
P
sinh  

 l
Recall identity
cosh  x   cosh cosh x  sinh  sinh x
 y 
M ab   coshk ( l  x )  M ba   coshkx 
 1 
 1

Pl 
sinh 
 Pl  sinh 

 a  y(0) 
M ab
 coth   1  M ba  csch  1
Pl
Pl
Multiplying numerator & denominator by l & P  k 2 EI
M ab l
ba l
 coth   1  M
 csch  1
2 2
2 2
k l EI
k l EI
M
M
or  a  ab  n  ba  f
--(a)
a 

where  

EI
l
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1
n 
2
f 
1
2
 coth   1
1   csch 
Likewise,  b  y (l )
M ab
 csch  1  M ba  coth   1
Pl
Pl
from which
b 
M ab l
ba l
 csch  1  M
 coth   1
2 2
2 2
k l EI
k l EI
M l
M l
or  b   ab  f  ba  n --(b)
b 


solving Eqs. (a) & (b) for M ab and M ba gives
M ab 
  a n   b f 
 n2   f2
M ba 
  b n   a f 
 n2   f2
Let S  S1 
C  S2 
 n2 
 f2 
1

4
1

4
(c)
(d)
n
   f2
2
n
f
 n2   f2
1  
2
coth2   2  coth  
1  
2
csch2   2  csch 
 n2   f2 
1

4

2
coth2   2  coth    2 csch2   2  csch 
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Recall identity
coth2 u  csch2 u  1
 n2   f2 
1

4

2

 2  coth   csch 

1 
1
 cosh 
  2

3 
 
 sinh  sinh 




1 
2
cosh   1

3 
sinh 
 

  coth   1
 S1  S 

2
cosh  1
sinh 
1  tanh 2  2  1  tanh 2  2
cosh   1
1  tanh 2  2

2 tanh  2
sinh 
1  tanh 2  2
2 tanh 2  2

 tanh  2

2 tanh 2
 S1  S 
  coth   1  coth   1

2 tanh  2
  2 tanh  2
1

Likewise
S2  C 
1   csch
2 tanh  2
1

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If the beam-column undergoes translation at both
ends without end rotation
The moment at a distance x from the origin is
M x  M  Py  2 M  P 
y  
x
l
Mx
EI
x
 2x

 EIy   Py  M 
 1   P
l
 l

or y  k 2 y 
M  2x
x

 1  k 2

EI  l
l

y  A sinh kx  B coshkx 
y0  0 : 0  B 
M
P
M  2x
x

 1  

P  l
l

 B
yl    :   A sinh  
M
P
M
M
cosh  

P
P
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A sinh  
M
cosh  1
P
 A
M cosh   1
P sinh 
 y
M  sinh kx
cosh  1  coshkx  2 x  1   x

P  sinh 
l
l

The relationship between
M &  is obtained from
the condition
dy
0
dx x  l
y 
0

l

M  k coshkx
2 


cosh


1

k
sinh
kx


P  sinh 
l  l
M  k cosh
cosh  1  k sinh   2   

P  sinh 
l l

M  k cosh2   sinh 2   k cosh  2 

 
P 
sinh 
sinh 
l
M
 csch   coth  - 2
l
lP
2
2
M
M n   f
   n   f   

 n   f



 M
M
M
 
 2  2
   Pl  EI Pl 


EI l
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 M 
  n   f  






S

C
 n2   f2 l
l
  S  C  a   b  / l
(e)
superimposing (e) on to Eqs (c) & (d) yields
M ab 
a b 
EI 


S


C


S

C
a
b
l 
l 
M ba 
a b 
EI 


C


S


S

C
a
b
l 
l 
compare
  b
EI
 S 1 a  S 2 b  S 1  S 2  a

l
l
  b
EI
M ba 
 S 2 a  S 1 b  S 1  S 2  a

l
l
M ab 
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○ Slope Deflection Equations with Axial
Tension
The moment at a distance x from the origin is
M x  M ab  Py  M ab  M ba  P
y  
x
l
Mx
EI
 EIy  Py   M ab  M ab  M ba  P
let k 2 
x
l
P
EI
Then y  k 2 y  f ( x )
where f ( x )  linear function of x
 y IV  k 2 y  0
y  A sinh kx  B coshkx  Cx  D
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B.C’s
1. y(0)  0
2. y( l )  
3. y(0)   a
, 4. y( l )   b
,
or y(0)  
M ab
EI
y( l ) 
,
M ba
EI
After elimination the consts A, B, C and D using the
B.C’s in 4 and solving for M ab and M ba gives
0  B  D --(a)
  A sinh kl  B coshkl  Cl  D --(b)
y  Ak coshkx  Bk sinh kx  C
 a  Ak  C --(c)
 b  Ak cosh kl  Bk sinh kl  C --(d)
0
1
0 1  A   0 

 sinh kl
coshkl l 1  B    

    
k
0
1 0  C   a 



 k coshkl k sinh kl 1 0  D   b 
By Cramer’s Rule
A
0
1
0 1

a
b
cosh kl
0
l
1
 Da
1 0
k sinh kl 1 0
0
1
0 1
sinh kl
cosh kl
k
0
l
1
1 0
 Dd
k cosh kl k sin kl 1 0
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1
0 1 0
Da   a coshkl l 1  
k sinh kl 1 0  b
1
1
coshkl 1
k sinh kl 0
let   kl
0 1
1

Da   a  k sinh 

l 1 cosh 

1
1
   b
1
cosh 
1
0 1
 k sinh 
1
 1
  a   sinh   1  cosh     b   b cosh   k cosh 
  a cosh    sinh   1   b 1  cosh    k sinh 
sinh 
l
1
sinh 
Dd  
k
k cosh 
1 0
k
1 0 k cosh 

k
k cosh 
1
cosh 
k
1
k sinh 
cosh 
l
0
k sinh 
1
1
l
sinh 

1 k cosh 
cosh 
k sinh 
  k  k cosh   cosh   k sinh   k sinh 2   k cosh2 

k
 2k  2k cosh   k sinh 
A
Da
Dd
0
B
sinh 
k
k cosh 
0
0 1
 l 1
a 1 0
b 1 0
 Db
Dd
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sinh 
Db  
k
k cosh 
 l
a 1
b 1
  a sinh   b  k cosh    a  cosh    b sinh   k 
  a  cosh   sinh     b sinh       k  k cosh  
y  Ak coshkx  Bk sinh kx  C
y  Ak 2 sinh kx  Bk 2 coshkx
M ab   EIy ( 0 )
k 2  cosh   sinh   a  sinh     b  k  k cosh   / l 
  EI
 2  2 cosh    sinh  


k  cosh   sinh   a  sinh     b  k  k cosh   
EI 
l

l
 2  2 cosh    sinh  
let S1  S 
  cosh   sinh  
2  2 cosh    sinh 
recall identities
sinh   2 sinh  2 cosh 2
cosh   cosh2  2  sinh 2  2  1  2 sinh 2  2
Dividing
numerator
&
denominator
of
S
by
sinh 
gives
S
  coth   1
2
 2 coth   
sinh 
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I
21  cosh  
sinh 


2 1  1  2 sinh 2  2 

2 sinh  2 cosh  2
 2 tanh  2
 S

  coth   1
  2 tanh  2
 coth   1
1
2 tanh  2

“Same result”
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Example
Let
b  0 , a  b  0
EI
S1 a
l
then M ab 
S1  M ab
P
l
: S1  measure of bending moment
EI a
kl      l
P
Pl 2
P



EI
 2 EI
PE
2.05 2 EI
Pcr 
, kl  2.05  4.49
l2

4.49
interpolation
value ;P212
if
kl  4.49,
we
have
negative
stiffness
which
that
means
the
moment and the rotation
are oppositely directed. That is , kl  4.49 ,  A is
induced
by
the
adjacent
member,
kl  4.49,  A
is
resisted by the adjacent member.
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○Stress Amplification in Columns
Concentric Load on a column
P
for perfectly homogeneous & straight Column.
A
Bending effect
f 
f 
P M

y , M  P  y0
A I
y0 : initial displ.
preamplified displ.
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○ Rectangular Moment Curve
l l
Py0  
2 4
Py0 l 2

8
○ Triangle Moment Curve
l 1 2 l
Py0    
2 2 3 2

Py0 l 2
12
○ Moment – Area Theorem No.2
Py0 l 2
, Rect
8
Py0 l 2
EIy1 
, Triangle
12
EIy1 
 y2 
Py0 l 2
y1 
10 EI
Py1 l 2
10 EI
Py2 l 2
10 EI
.............
y3 
< reach of Equil. position >
 y  y0  y1  ........
 y0 
Pl 2
Pl 2
y0 
y1  ..........
10 EI
10 EI
  Pl 2   Pl 2  2  Pl 2  3

  
  
  ...........
 y 0 1  
  10 EI   10 EI   10 EI 

y0

 Pl 2 

1  
 10 EI 
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Pl 2
 1 ->
infinite series if
10EI
Diverge
Pl 2
10EI
 1.0  Pcr  Pe  2
10EI
l
PE  Pcr exact  9.872EI
l
f 
P Py0 c 
1


A
I  1  P / PE

M
c 
1
 
c   y0 
I
I   1  P / PE

 
 P 
 
Secant Formula
※ initial imperfectness
l


col. 
 y0 
800

 in general


l
beam
 y0 
400


slenderness ratio
y0 c y0 l c

l rr
r2
initial crookedness
AISC SPEC. : y0 
1
2
shape factor
for 50
l / 200
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Stress Amplification in Columns
The
behavior
increasing
of
load
calculating
the
a
can
compression
be
bending
seen
member
most
stresses
under
clearly
and
by
lateral
deflections that occur as the end load is applied.
Consider
a
normal
straight,
slender
member
supporting a nominal axial load P. The loads of
the member are assumed free to rotate in this case.
If the member were perfectly straight and the load
was perfectly centered, the stress in the column
at any section would be simply f a  P / A .
No actual
member ever would be perfectly straight nor would
the load be perfectly centered. Even when great
efforts are made to achieve such perfection in
laboratory tests, it is not completely attained.
Therefore, the actual case is best represented by
assuming a slight initial imperfection of loading
or
an
initial
crookedness
represented
by
a
deflection y0 at mid height of the member (Fig.1).
When a load P is acting on the column the stresses
at the mid-height sections are
P Mc

A
I
where compressive bending stresses are taken as
f 
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positive.
At
any
section
of
the
column,
the
bending moment is the load times the eccentricity,
and the bending moment diagram has the same shape
as the curve of the deflected member (Fig.1). This
bending moment produces a further deflection at
the mid-height.
y1 
1 Py0 l 2
10 EI
the constant 1 / 10 is taken as a mean value for a
deflected curve of more or less uniform curvature
as shown.

l 1 2 l Py0 l 2 
Triangular
Moment
curve
:
Py

   
0

2 2 3 2
12 


Py0 l 2 
l l


: Py0  

 Rectangula r
2 4
8 
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For
other
cases
of
initial
curvature
or
eccentricity, the moment may vary between limits
of 1 / 8 to 1 / 12 . Because of the added deflection y1 ,
there will be an increased bending moment Py 1 , and
additional deflection y2 , and so on.
y2 
1 Py1 l 2
10 EI
Continuing
this
process,
the
total
deflection
becomes
y  y0  y1  y2  
Pl 2
Pl 2
 y0 
y0 
y1  
10EI
10EI
2
 Pl 2 
 Pl 2 
 y0  
 y0  
 y0  
 10EI 
 10EI 
2
3


Pl 2  Pl 2   Pl 2 
 y0 1 
 
  
  
10EI  10EI   10EI 


The series in bracket is the multiplier by which
the initial deflection y0 is increased under load
P to give the final deflection y
For
values
of
P
less
than
10 EI / l
at that load.
2
(i.e.
Euler
buckling load) the terms in the series are less
than unity and the series is convergent, having
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the limit
1
Pl 2
1
10 EI
The final deflection is y  y0
1
Pl 2
1
10 EI
This requires a slight modification that will be
discussed later if the curve of initial deflection
differs greatly from the uniform curvature assumed.
For
any
value
Pl 2
P
1
10 EI
of
the
series
is
divergent. This indicates that any small initial
deflection will be indefinitely magnified at the
load Pc  PE 
10EI
or greater.
l2
Let’s denote PE 
 2 EI
l
2

10EI
l2
Then the total stress of at mid-height of a column
is
f 
P Py0 c
1

A
I 1  P / PE
then PE 
 2 EI
l2

let
 2 EAr2
l2
P
 fa , since I  Ar 2
A
or fc 
 2E
l
 
r
2
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further
1
1

1  P PE 1  fa f c
2
Py0 c Py0 c
yc
c y
and

 f a 02  fa   0
2
I
Ar
r
r c
Then the total stress
2
1
c y
f  fa  fa   0
 r  c 1  fa fc
the magnitude of the bending stress
2
1
c y
fa   0
 r  c 1  fa fc
depends upon P
represented in f a ; the shape of the cross section
y 
c
  ; the initial curvature  0 
r
 c 
Since f c 
the
column
 2E
l
 
r
2
, the stiffness of the material of
and
the
slenderness
ratio
also
are
involved. It is convenient to make the expression
for stress dimensionless by dividing by f c , then
2
f
f c y
f
1
 a  a  0
fc
fc fc  r  c 1  fa fc
c
the value of shape factor  
r
2
range from 1.0 for a
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section
in
which
all
the
area
is
assumed
concentrated in the figs. to 3.0 for a rectangular
section and 4.0 for a circular section.
Rolled shapes of the proportions generally used
for Column having
c r 2
in the vicinity of 1.4.
about the strong axis and 3.8 about the weak axis.
I – beam(S shape) run to values of 5.0 and over
about the weak axis.
Reasonable values of y0 c are more difficult to
determine since the initial crookedness may be the
result
of
either
lack
of
straightness
of
the
member itself of imperfection of the lignment of
loading
through
the
connections.
Pending
the
better establishment of the values, the combined
constant
c r 2
y0
c
has been assumed to range from
0.01 to 1.0
Since it is usually more convenient to express
the initial crookedness y0 in terms of the length
of the member, the quantity
y0 c
may be rewritten
r2
as
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slenderness ratio
y0 c y0 l c

l rr
r2
lack of straightness
shape factor
The tolerances for straightness of Rolled shapes
are expressed in terms of length, as for example,
y0 
l
, as a max. figure in some specifications
1000
[Current
AISC
(7th
edition
P1-125)
50′or
less
0.5″max]
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Homework
1. Compute the critical load, Pcr , of the column using
a) Matrix Method
b) C 2 -method
c) F D M
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2. Compute Pcr of following truss
30°
3. Find Pcr using slope-deflection method
Pcr
152