Buckling
general
Up to this point, the stress and deflections have been proportional to an applied load:
y
bending stress proportional to moment
e.g. σ x = M⋅
I
maximum deflection of a simply supported beam subject to uniform load per unit length q =>
y max =
5
⋅
q⋅ L
4
deflection proportional to the uniform load.
384 E⋅ I
This is not always the case, such as when compressive loads with/without lateral loads act on a column (beam).
Moments, stresses and deflections will NOT be proportional to axial loads, but will be dependent (not proportional)
to deflections, thus sensitive to slight initial deflections and/or eccentricities in the application of the load.
Euler buckling (derived from general case of beam-columns including lateral load q(x))
consider:
q
y P
P
P


ΣFy_dir = 0 =>  V +
q
M
x
V
dx
V+dV/dx*dx
dy
P
M+dM/dx*dx
d

d
V⋅ dx − V + q⋅ dx = 0 => q( x) := − V( x)
dx
d
x

and ΣM A = 0 => M +
(1)
 d M  ⋅dx − M −  V +  d V ⋅dx ⋅dx − q( x)⋅dx⋅ dx − P⋅ d y⋅ dx = 0 =>




2
dx
dx 

dx  
d
d
M − P⋅ y = V
dx
dx
as in previous bending: M ( x) := −E⋅I⋅
(2)
2
d
2
dx
y ( x) =>
−
2
d
2
M ( x) =
E⋅ I⋅
dx
d  d
d 
using (1) and (-) the derivative of (2) wrt x; −  M − P⋅ y  =
dx 
dxdx
4
d
4
y ( x)
dx
 d V = q(x) =>
dx 
−


2
P
d
=>
y ( x) + P⋅
y ( x) = q(x) or setting k :=
4
2
E⋅ I
dx 
dx

 q( x)
4 
2 d2
d 
euler buckling uses q = 0 which we will do now
y ( x) + k ⋅
y ( x) =
4
2
E
⋅
I
dx 
dx

E⋅ I⋅
1 of 33
4
d
notes_27_buckling_notes.mcd
y ( x) := A⋅sin(k⋅x) + B⋅cos (k⋅x) + C⋅ x + D
general solution is:
check =>

 0
2 d2
y ( x) + k ⋅
y ( x) → 
4
2
dx 
dx
 0
4
d
now apply to column with pinned ends:
x
y P
P
boundary conditions are: y(0) = y(L) = 0
and
2
d
2
y ( 0) =
dx
2
d
2
y ( L) = 0 (no bending moment at the ends)
dx
y(0) = 0 => B + D = 0
y(L) = 0 => A⋅ sin( k⋅ L) + B⋅ cos( k⋅ L) + C⋅ L + D = 0
2
d
2
2
2
2
y ( 0) = 0 => −k ⋅ A⋅ sin( k⋅ 0) − k ⋅ B⋅ cos( k⋅ 0) = - k ⋅ B = 0 => B and D = 0
dx
2
d
2
2
y ( L) = 0 => −k ⋅ A⋅ sin( k⋅ L) = 0 => A⋅ sin( k⋅ L) = 0 => C = 0 from the y(L) = 0 relation above
dx
this leaves A⋅ sin( k⋅ L) = 0 which has a non trivial solution only when sin( k⋅ L) = 0 => k*L = n*π
2
 n⋅ π  = P/(E*I) or .... solution defining that force P as P
recall that k^2 = P/(E*I) => k^2 = 
cr
 L 
2 E⋅ I
when Pcr := ( n⋅ π ) ⋅
the displacement is then: y ( x) := A⋅ sin( k⋅ x) where A can be any value. i.e.
2
L
with P < Pcr the trival solution applies y = 0, but at Pcr y(x) can be >0 and arbitrary.
2 E⋅ I
minimum P occurs when n = 1 => Pcr := π ⋅
cr
2 of 33
L
2
sometimes labeled P
E
notes_27_buckling_notes.mcd
let's look at a few other sets of boundary conditions and determine the Pcr by inspection:
A. clamped - free
x
Pcr
y
P
B. clamped - clamped
x
y
Pcr
P
C. clamped - clamped, free to
translate
x
Pcr
y
D. clamped - pinned, not free to
translate
x
Pcr
y
P
this one is not obvious (at least to me!!) let's apply boundary conditions to the general solution:
y ( x) := A⋅sin(k⋅x) + B⋅cos (k⋅x) + C⋅ x + D
3 of 33
notes_27_buckling_notes.mcd
boundary conditions are: y(0) =
2
d
d
y ( 0) = 0 ; clamped at 0and y ( L) =
y ( L) = 0 (no displacement or bending
2
dx
dx
moment at L)
y(0) = 0 => B + D = 0
d
y ( 0) = 0 => A*k + C = 0
dx
y(L) = 0 => A⋅ sin( k⋅ L) + B⋅ cos( k⋅ L) + C⋅ L + D = 0
2
d
2
2
y ( L) = 0 => −k ⋅ ( A⋅ sin( k⋅ L) + B⋅ cos( k⋅ L)) = 0
dx
=> C⋅ L + D = 0
(3)
(1)
(2)
(3)
=> A⋅ sin( k⋅ L) + B⋅ cos( k⋅ L) = 0
(4)
has only trivial solution A = B = C = D = 0
solve for A in terms of B using (1), (2) and (3)
(3) =>
C = -D/L
(2) =>
A = -C/k = -B/(k*l)
(1) => D = -B
=> C = B/L
−sin( k⋅ L)
−sin( k⋅ L)
+ cos( k⋅ L)  = 0 =>
+ cos( k⋅ L) = 0 =>
k
⋅
L
k⋅ L


k*L = sin(k*L)/cos(k*L) = tan(k*L) a transcendental equation - solve graphically: k_L := 0 , 0.01 .. 10
π
intersection approaches k*L = n*π/2, with first occurrence at ~ 3⋅ = 4.7 is between 4.49 and 4.5 by trial and error
2
B⋅ 
(4) =>
kl := 4.4934 , 4.49341 .. 4.49342
10
4.4938
tan ( kl)4.4936
tan(k_L)
0
k_L
kl
4.4934
4.4932
4.493395 4.4934 4.4934054.493414.493415
10
0
5
kl
10
k_L
value found by successive iteration k_L := 4.49341 tan(k_L) = 4.49342
2 E⋅ I
i.e. k^2 = 4.49342^2/L^2 or ... Pcr := 4.49341 ⋅
2
L
to see in general form multiply and divide by π^2
2
Pcr := π ⋅
E⋅ I
 π ⋅L

 4.49341 
4 of 33
2
and
π
4.49341
= 0.6992 ~ 0.7 =>
or ...
k_L1 := 4.5
root(tan(k_L1) − k_L1, k_L1) = 4.4934
E⋅ I
2
Pcr := π ⋅
(0.7⋅L)
notes_27_buckling_notes.mcd
2
we stated at the introduction to this segment that buckling was a situation where deflection was not proportional
to applied force (i.e. general definition of force includes moment). In Euler buckling the deflection is proportional to
axial force up to Pcr - note the proportionality is strain in the axial direction. Now let's look at a problem where
the axial force is combined with a tranverse force Q (a point force). For simplicity we will locate it at the center of
a beam-column so we can use symmetry. see Timoshenko & Gere section 1-3 for an arbitrary placement. (figure
later)
x
Q
y P
P
Q
In this case the equations are as follows: M ( x) :=
using: M ( x) := −E⋅I⋅
2
d
2
2
d
y ( x) =>
dx
 y( x) +
2

dx
P
E⋅ I
2
specific solution:
⋅x + P⋅ y
y ( x) := c⋅x
c
⋅ y ( x)  = −

Q⋅ x
P
2⋅ E⋅I
E⋅ I
⋅ y ( x) →
and as above let k^2 = P/(E*I). leads to solution:
y ( x) := A⋅cos(k⋅x) + B⋅sin(k⋅x) −
c :=
Q⋅ x
 0 =

c
−Q
y ( x) :=
2⋅ P
2⋅ P
the boundary conditions are y(0) = 0 => A := 0
and
d  L
= 0 =>
y
dx  2 
B :=
Q
2⋅ P⋅k
⋅
1
L
cos k⋅ 
 2
=>
sin(k⋅x)
Q
Q
y ( x) :=
⋅
−
⋅x
P⋅k
2⋅ P
consider2⋅deflection
(maximum):
 midpoint
 atL the
L

sin k⋅
L
L
L
Q
Q   L
 2 − Q ⋅L

=> y   =
=
y
⋅
⋅  tan k⋅
− k⋅ 
2
2⋅ P⋅k   2 
 2  2⋅ P⋅k cos k⋅ L  2⋅ P 2
 2

 2
L
following Timoshenko: let u := k⋅ and using some algebra and
2
substitutions; k :=
2⋅ u
L
2
2
; P := E⋅I⋅k ; P := E⋅I⋅4⋅
u
L
3
2
=>
Q
2⋅ P⋅k
=
Q

u


L
2⋅  E⋅I⋅4⋅
3
2
L
1 Q⋅ L 3
1 Q⋅ L  3
=> y   =
⋅
⋅
⋅
⋅
⋅(tan( u) − u)


3
3
2
48 E⋅ I
48
E
⋅
I
 
u
u

now why did we (Timoshenko) go to all that trouble?
which finally is =
5 of 33
notes_27_buckling_notes.mcd
2

=
⋅
2⋅ u
L
1
16
⋅
Q⋅ L
3
3
E⋅ I⋅u
−
Q⋅ x
2⋅ E⋅I
−Q
2⋅ P
⋅x
3
The deflection at L/2 is now in a form
1 Q⋅ L
⋅
48 E⋅ I
: the deflection due to the force Q times a multiplier
P
with the following properties: (again with some substitutions) k :=
3
so when P is small ~ 0 ;
3
⋅(tan( u) − u) -> 1
E⋅ I
L
u :=
;
2
⋅k ; u :=
L
2
⋅
P
E⋅ I
which can be seen by expanding tan(u) in a
u
3


u

+ ..... ;
tan( u) = u +
⋅(tan( u) − u) =
⋅ u+
− u =~ 1 or plotting: u := 0.001, 0.011 .. 1
3
3
3
3

u
u 
3
3
u
series:
3
also as u => π/2 tan(u) => ∞
2
at this value u :=
3
u
3
2 E⋅ I
⋅ ( tan ( u) − u) 1.5
P := π ⋅
L
1
2
2 E⋅ I
0
0.5
1
L
Q⋅ L
2
16⋅E⋅I
Q
2⋅ P⋅k
⋅
sin(k⋅x)
L
cos k⋅
 2
−
Q
2⋅ P
6 of 33
for u , P
2
⋅
E⋅ I
or
2
⋅
P
Pcr
2⋅(1 − cos( u)) 
Q⋅ L  tan( u) 
 2
 and the maximum moment is Mmax := 4 ⋅  u 
 u ⋅cos( u) 
small ( => 0) and => ∞ as u => π/2 or P => Pcr
notes_27_buckling_notes.mcd
π
2
=
L
2
⋅
P
E⋅ I
in the definition for u =>
⋅x and using this same approach: the slope at y = 0
these are both in the form of the effect of Q * a multiplier that
=> 1
π
P
cr
using Pcr := π ⋅
u :=
d
y ( 0) =
dx
2
⋅
which is the value for P above.
u
recalling that y ( x) :=
L
before leaving this approach to buckling, let's consider when q(x) is not 0 or a point but is uniform per unit length:
for a pinned column: see this link
Reference:C:\Documents and Settings\Dave Burke\My Documents\structures\overall_technical\buckling\eqn_11_3_1.mcd(R)
the result is: (for wmax) as stated in Hughes equation (11.3.1)
wmax( ξ ) :=
M max :=
2
 24 
ξ 
L
P


(
)
with ξ := ⋅
⋅
⋅ sec ξ − 1 −
2 
2 E⋅ I
384⋅E⋅I  4 
 5⋅ξ

5⋅ q⋅L
4
 2⋅ ( 1 − sec( ξ ))

2
8 
ξ


q⋅ L
2
⋅
note that ξ is the u above and w is y. The section is titled "use of the magnification factor".
a few more things to deveop for euler and general elastic (and other) buckling:
I
2
recall the definition of the radius of gyration = ρ defined such that I := ρ ⋅A or ρ :=
A
and
the definition of stress is force (P) / area (A).
2 E⋅ I
π ⋅
σ cr :=
Pcr
σ E :=
A
Pcr
2 E⋅ I
Pcr := π ⋅
A
L
σ E :=
2
L
2
E
2
σ E := π ⋅
A
 L

ρ
2
a typical plot of euler stress vs L/ρ (slenderness ratio):
Le_over_ρ := 80, 81 .. 300
σ Y := 30000
6
E := 30⋅10
σ E( Le_over_ρ) :=
2
π ⋅E
2
Le_over_ρ
4
5 .10
4
4 .10
4
.
σ E(Le_over_ρ)3 10
σY
as can be seen, the euler stress is yield when
2
2
π ⋅E
4
2 .10
2
= σ y or Le_over_ρ :=
Le_over_ρ
Le_over_ρ = 99.3
π ⋅E
σY
i.e.
4
1 .10
0
to understand the slenderness ratio better and the
difference between slender and "squat" (short fat)
columns consider the following example.
0
100
200
300
Le_over_ρ
7 of 33
notes_27_buckling_notes.mcd
Residual stresses from rolling or welding lead to reductions in modulus:
4
3 .10
d
σ approximates a parabolic
dε
shape above a value of σav defined as the
the decrease in E =
4
2 .10
σY
structural proportional limit typically σ spl :=
σ av
4
1 .10
2
e.g.
above the proportional limit:
0
5 .10
0
4
( )
Ets σ av :=
0.001 0.0015 0.002
ε σ av
( )
(
)
⋅E
σ spl⋅ ( σ Y − σ spl)
σ av⋅ σ Y − σ av
redefining because the modulus is reduced only above the proportional limit

( )
Ets σ av := if σ av > σ spl ,

(
) 
⋅E, E
σ spl⋅ ( σ Y − σ spl)

σ av⋅ σ Y − σ av
4
3 .10
4
.
σ av 2 10
σ spl
4
1 .10
0
7
1 .10
0
7
2 .10
Ets σ av
( )
7
3 .10
with some algebra and designating σult as σav an expression for σult vs. Le_over_ρ results:
Le_over_ρ := 50 .. 180

σ spl

σY
σ ult( Le_over_ρ) :=  1 −

⋅1 −

σ spl 

⋅σ Y

σ Y  σ E( Le_over_ρ) 
4
σ E( Le_over_ρ) :=
σY
0
2
π ⋅E
2
Le_over_ρ
2
σY
50
100
150
200
Le_over_ρ
8 of 33
σY
recall from above:
σ ult (Le_over_ρ)
σ E(Le_over_ρ)
⋅
notes_27_buckling_notes.mcd
if we define a ratio λ :=
σY
defined as the column slenderness parameter which
σE
=>
σ E( λ) :=

σ spl

σY
σ ult( λ ) :=  1 −
σ ult becomes

⋅1 −

σ spl  2
⋅λ  ⋅σ Y
σY  
applying when σspl < σult < σY, and limiting σE to σY =>
  σY  

σ E( λ) := min  σ Y   =>
 2
 λ 




σ spl

σY
σ ult( λ) := ifλ ≤ 2,  1 −

⋅ 1 −

σ spl  2
σ Y

⋅λ  ⋅σ Y,
2
σY  
λ 
λ := 0.0, 0.01 .. 2
1−
reduces to:
λ
2
4
σ ult ( λ ) 0.8
σY
when σspl/σY = 0.5 as
shown here.
0.6
σY
0.4
0.2
0
0.5
1
1.5
2
λ
2 E⋅ I
PE = π ⋅
L
9 of 33
2
2
2
λ
appears above (as λ^2)
σ E( λ )
σY
ρ =
I
PE
A
A
2 E⋅ I 1
= σE = π ⋅
⋅
L
2 A
2
E⋅
=π ⋅
L
I
A
2
notes_27_buckling_notes.mcd
2
2 E⋅ρ
=π ⋅
L
2
2
=π ⋅
E
 L

ρ
2
= σE
other factors that affect column behavior are not being perfectly straight and application of the load off center
these are termed eccentricity in geometry and load application. Consider first geometry: for a column with an
initial deflection δ(x) (ref: Hughes pp 394 ff):
wt
2
w(x)
d
2
w+
dx
P
P
E⋅ I
⋅ ( δ + w) = 0
P
x
δ(x)
assuming a sinusoidal (Fourier series) deflection results in a deflection wT :=
PE
PE − P
⋅δ
this has the property we saw earlier, a result with a magnification factor φ as
wT := φ⋅ δδ.
When P is small the deflection matches δ. When P approaches the euler value, the
deflection is very large. The deflection is continuous, not proportional to P. The load
deflection curves for eccentric columns are shown in fig 11.8 andbelowδ 
PE := 10
wT := 0.01 , 0.015 .. 0.3
P wT , δ := PE⋅  1 −
wT 

(
)
0.4
wT
10
(
)
P ( w T , 0.02)
P w T , .01
0.2
0
0
(
5
)
5
0
10
P w T , 0.01
0
0.1
0.2
wT
total displacement vs. applied force P
figure 11.8
eccentricity in load application is derived in Timoshenko with the same form with different magnification factor:
π P 
φ:= sec  ⋅
M := P⋅ e⋅ φ :
 2 PE 
wt
w(x)
P
P
e
10 of 33
notes_27_buckling_notes.mcd
x
This factor approximates the factor for geometry, as can be seen from the following plots so it is simpler to use the
geometry magnification factor for both effects
1
π

P_over_Pe := 0, 0.01 .. 0.6 φ1(P_over_Pe) :=
φ2(P_over_Pe) := sec ⋅ P_over_Pe
1 − P_over_Pe
2

3
1.15
φ 1(P_over_Pe)
φ 2(P_over_Pe) 1.1
2
φ 2(P_over_Pe)
φ 1(P_over_Pe) 1.05
1
1
0
0.2
0.4
0.95
0.6
0
0.2
P_over_Pe , P_over_Pe
0.4
0.6
P_over_Pe
this allows us to combine the two eccentricities δ (geometry) and e (off center load) such that ∆ := δ + e
the "magnified" deflection becomes
the moment from the applied force P then becomes M := P⋅ ∆⋅
∆φ
∆*φ
the total stress in a column accounting for both compression and bending is then:
w
w(x)
P
P
σ max :=
P
A
+
x
δ(x)
e
P⋅ ∆⋅
∆φ
Z
where Z is the modulus in the direction (extreme fiber) that undergoes compression. this
rc :=
expression can be rearranged as follows: defining
Z
A
=
I 1
=
⋅
c A
ρ
2
c
where c is the distance of the
extreme fiber (compression side) to the neutral axis.
σ max :=
P
A

∆
∆⋅ φ


Z
⋅1 +
A


=
σ max :=

A 
P
⋅1 +
∆
rc
⋅φ


where
∆
rc
=
eccentricity ratio.
estimates for eccentricity ratio have been accepted based on experimental evidence as proportional to slenderness
Le
Le 
Le
∆
P 
and σ max becomes σ max := ⋅  1 + α⋅ ⋅φ
ratio
i.e.
= α⋅
rc
A 
ρ
ρ
ρ 
11 of 33
notes_27_buckling_notes.mcd
if we now replace φ by φ :=
PE
PE − P
; designate P as P
ult
and declare "failure" when σ max =
σY
Le
Le 



α⋅ ⋅PE
α⋅


Pult
Pult
ρ
ρ 
eqn 11.2.1 rearranged => σ Y :=
we obtain σ Y :=
⋅1 +
⋅1 +
PE − Pult 
Pult 
A 
A 
1−

PE 

Le 

α⋅

ρ 
eqn 11.2.2
or in terms of stress σ Y := σ ult⋅  1 +
σ ult is the applied stress that will result in yield
σ ult 

1−

σE 

after accounting for the magnification factor as developed above.
if we define R :=
λ :=
L
π⋅ρ
⋅
σY
E
(
2
σ ult
σY
; η :=
α⋅L
ρ
;
and use the column slenderness parameter λ :=
to solve eqn 11.2.2 above for R :=
)
(1 − R)⋅ 1 − λ ⋅R =
η⋅R
σ ult
σY
σY
σE
or
we obtain a quadratic equation for R
with solution (taking the negative sign in the quadratic term) and a lot of algebra:
the Perry Robertson column formula results:
λ := 0.01 , 0.02 .. 1.5
R( α , λ ) :=
α 1 := 0.003
2
1

 ⋅  1 + 1 + η (α , λ )  − 1 ⋅  1 + 1 + η (α , λ )  − 1 
2 
2
2
2
4 
λ
λ
λ 
 



  σY  

as above euler = σ E( λ) := min  σ Y   and tangent modulus
 2
 λ 




σ spl

σY
σ ult( λ) := ifλ ≤ 2,  1 −
12 of 33

⋅ 1 −

σ spl  2
σ Y

⋅λ  ⋅σ Y,
2
σY  
λ 
notes_27_buckling_notes.mcd

η ( α , λ ) :=  α ⋅ π⋅

E

σY 
⋅λ
comparison of these different approaches =>
1
(
R α 1, λ
)
0.8
σ E( λ )
σY
0.6
σ ult ( λ )
σY
0.4
0.2
0
0.2
0.4
0.6
0.8
Perry Robertson
Euler
tangent modulus
this is CCB taking R( α , λ ) :=
σ ult( α , λ )
σY
 Lcol 
 ρ 
calculating λ and η := α⋅ 
1
13 of 33
σ a :=
P
A
1.4
1.6
with α := .002 from Table 11.1 and assuming Le := .7⋅L
Lcol in
σ ult is the applied stress that will result in yield after accounting for the
magnification factor
σ ult = R⋅σ Y
1.2
λ
 σa 
γRCCB := γ C⋅ 
 σ ult 
notes_27_buckling_notes.mcd
Page 322 S&J: The civil engineers use a curve similar to this for column design Table in manual of steel
construction:
the loads include a resistance factor (1/PSF) of 0.85. From LRFD spec E-2. Curve 2 fit to out-of-straightness
=1/1500.



2
σ c( λ ) := if  λ ≤ 1.5, 0.658 ⋅σ Y ,
SSRC guide S&J pg 322
λ
0.877
λ
2
⋅σ Y


1
critical stress/yield stress
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0
0.2
0.4
0.6
civ eng practice
ultimate stress
euler
0.8
length ratio =
1
1.2
1.4
1.6
Check using column table pg 2-35 Manual of Steel Construction (MSC)
KL=10ft, nominal diameter = 10 in, extra strong. φ = resistance factor. Load = 465,000 lbs
redefining values for MSC
le := 10⋅12 A := 16.1
ρ := 3.63
φ := 0.85
σ Y := 36000 E := 29000000
λ :=
le
ρ
⋅
σY
λ = 0.3707
2
π ⋅E
SSRC guide S&J pg 322



λ
14 of 33
6
E := 30⋅10
0.877
λ
reset yield, modulus and curve to general values
σ Y := 30000
2
σ c( λ ) := if  λ ≤ 1.5, 0.658 ⋅σ Y ,
P( λ ) = 465117
2
⋅σ Y


P( λ ) := σ c( λ ) ⋅ A⋅ φ
compares to 465 ksi in table



2
σ c( λ ) := if  λ ≤ 1.5, 0.658 ⋅σ Y ,
notes_27_buckling_notes.mcd
λ
0.877
λ
2
⋅σ Y


Accurate design curves, figures 11.11, 11.12, 11.13
define E to SI units per text
E := 200000
choose α based on column shape, α = 0.002 for circular
α := 0.002
Le_over_ρ := 1 .. 150
(
)
λ Le_over_ρ , σ Y :=
(
Le_over_ρ
)
π

(
⋅
σY
E

)
η α , Le_over_ρ, σ Y := if  λ Le_over_ρ , σ Y ≤ 0.2, 0,  α⋅π⋅


 ⋅ λ Le_over_ρ , σ − 0.2 
( (
)
Y)
σY 

E
2


1 + η ( α , Le_over_ρ , σ Y) 
1 + η ( α , Le_over_ρ, σ Y) 
1 
1 
1


σ u( α , Le_over_ρ, σ Y) :=
⋅1 +
−
⋅1 +
−
2 
2
2
2
4 
λ ( Le_over_ρ , σ Y)
λ ( Le_over_ρ , σ Y)
λ ( Le_over_ρ , σ Y) 
 



400
σ u( α , 90 , 200) = 147.5465
350
text has ~ 150 fig 11.11
300
σ u(α , Le_over_ρ , 400)
250
σ u(α , Le_over_ρ , 300)
σ u(α , Le_over_ρ , 200) 200
150
100
50
0
50
100
Le_over_ρ
15 of 33
notes_27_buckling_notes.mcd
150
revisiting the Perry Robertson relationship with λ offset
λ := 0.01 , 0.02 .. 1.5
α 1 := 0.003




η ( α , λ ) := if  λ ≤ 0.2, 0,  α⋅π⋅

1 
σ u( α , λ ) :=  ⋅ 1 +

2
 
2
6
α 2 := 0.002
E := 30⋅10
 ⋅ ( λ − 0.2)

σY 

E
2

1 + η (α , λ ) 
1 

−
⋅ 1+
−
⋅σ Y
2
2
4 
λ 
λ



1 + η (α , λ ) 
λ
redefine E to Emglish units
1
1
(
σu α 1 , λ
)
0.9
)
0.8
σY
(
σu α 2 , λ
σY
σ E( λ )
0.7
σY
σ ult ( λ )
0.6
σY
σ c( λ )
0.5
σY
0.4
0.3
0
0.2
0.4
0.6
0.8
λ
16 of 33
notes_27_buckling_notes.mcd
1
1.2
1.4
1.6
Timoshenko Th of Elas Stab sect 1.7
1
can be used as approximation of all amplification factors, χ(u), η(u) and λ(u) good for
P
P/Pe<0.6
1−
Pcr
u :=
k⋅ l
k_sq :=
2
Pcr :=
E⋅ I
π ⋅E⋅I
l
u :=
1
φ( u) :=
1−
 2⋅ u 

 π 
2
2
2
P
π
2
⋅
P
P_over_Pcr :=
3⋅(tan( u) − u)
3
 2⋅uu 

 π 
P_over_Pcr :=
2
P_over_Pcr( u) :=
or using u=k*l/2 and k=sqrt(P/EI) and EI=Pcr*l2 /π2
amplification factor for:simply supported,
Q at center 1-14
uniform distribution q 1-21
χ ( u) :=
2
π ⋅E⋅I
2
P_over_Pcr :=
Pcr
k ⋅E⋅I 2
⋅l
η ( u) :=
(
couple at ends 1-33
)
2
12⋅ 2⋅sec( u) − 2 − u
λ ( u) :=
u := 0.001, 0.002 ..
π
2
u :=
π
2
⋅
2
2
P
Pcr
moment for couple
2⋅(1 − cos( u))
sec( u)
2
u ⋅cos( u)
4
5⋅ u
u
 2⋅ u 

 π 
 k⋅ l 

π
χ ( u) η ( u) λ ( u) sec ( u)
,
,
,
φ( u) φ( u) φ( u) φ( u)
− 0.1
notice scale
10
8
χ ( u)
φ ( u)
φ ( u)
χ ( u)
η ( u)
6
λ ( u)
1.1
φ ( u)
4
sec( u)
sec( u)
1.05
φ ( u)
2
0
1.15
φ ( u)
η ( u)
λ ( u)
1.2
1
0
0.5
1
1.5
0.95
17 of 33
0
0.5
1
u
u
notes_27_buckling_notes.mcd
1.5
P_over_Pcr := 0, 0.01 .. 0.6
φ1(P_over_Pcr) :=
fig 1-9 T&G
π

φ2(P_over_Pcr) := sec ⋅ P_over_Pcr
2

1
1 − P_over_Pcr
2.5
φ 1(P_over_Pcr)
φ 2(P_over_Pcr)
2
φ 1(P_over_Pcr)
φ 2(P_over_Pcr)
1.1
1.12
1.5
1
1
0
0.2
0.4
a1 := 1
0.4
0.6
< 12% variation in text for P_over_Pcr < 0.5
watch curve change shape as α -> 1. This is result if second term dominates
initial shape. i.e. y1 has two terms.
α := 1.1
α⋅a1
1− α
 π⋅x  + α⋅a2 ⋅sin 2⋅π⋅x 

 l  22 − α  l 
⋅sin
0
5
y1(x, α )
10
15
0
1
2
x
18 of 33
0.6
a2 := 10
x := 0, 0.1 .. l
y 1( x, α ) :=
0.2
P_over_Pcr
P_over_Pcr
l := 4
0
notes_27_buckling_notes.mcd
3
4
Beam column notes
(at x= l/2)
u :=
wmax( u) :=
l
2
E⋅ I
2
 24 
u 
⋅  sec( u) − 1 −
2 
384⋅E⋅I  2 
 5⋅ u

5⋅ q⋅l
4
⋅
(
M max := M 0 + P⋅φ⋅ wmax( u) + ∆
)
M max( u) :=
q⋅ l
2
8
⋅


2⋅(1 − sec( u)) 
(11.3.2) see
 note below

2
u
re:sign
∆ = total eccentricity as in column
comparison of magnification factors;
using u=k*l/2 and k=sqrt(P/EI) and EI=Pcr*l2 /π2
P_over_Pcr := 0.0001 , 0.01 .. 0.6
φ1(P_over_Pcr) :=
q := 1 P := 1 I := 1
M 0 := 1
φ := 1
∆ := 0
P
⋅
u(P_over_Pcr) :=
1
1 − P_over_Pcr
π
2
⋅ P_over_Pcr

u(P_over_Pcr)

2
24⋅  sec(u(P_over_Pcr)) − 1 −
φ2(P_over_Pcr) :=
5⋅u(P_over_Pcr)
2

4
1.003
φ 2(P_over_Pcr)
1.002
φ 1(P_over_Pcr)
1.001
1
0
0.1
0.2
0.3
0.4
0.5
P_over_Pcr
M max1(P_over_Pcr) := M 0⋅ 


2⋅(sec(u(P_over_Pcr)) − 1) 
u(P_over_Pcr)
(
2


M max2(P_over_Pcr) := M 0 + P⋅φ⋅ wmax(u(P_over_Pcr)) + ∆
sign is reversed in text. see P&G eqn 1-23
)
which after using wmax and P = P_over_Pcr * Pcr becomes with ∆ = 0
M max2(P_over_Pcr) := M 0⋅  1 + P_over_Pcr⋅φ1(P_over_Pcr)⋅π ⋅
2 5

P_over_Pcr := 0.5

48 
φ1(P_over_Pcr) = 2
using φ estimate
analytical
M max1(P_over_Pcr) = 2.0299
M max2(P_over_Pcr) = 2.0281
M max2(P_over_Pcr = 0.5) = M 0⋅  1 +
comparison of analytical vs. M due to bending plus P * displacement (magnified)
P_over_Pcr := 0.0001 , 0.01 .. 0.99
19 of 33
notes_27_buckling_notes.mcd

5
48
⋅π
2

1.004
M max1(P_over_Pcr)
M max2(P_over_Pcr)
1.002
1
0
0.2
0.4
0.6
0.8
P_over_Pcr
This assumption regarding magnification factor φ allows using a relationship similar to the Perry-Robertson above
with some additional terms to account for the bending moment and displacement (magnified) due to the transverse
loading. This is combined with the eccentricity due to the column eccentricity
(
M max := M 0 + P⋅φ⋅ δ 0 + ∆
=>
σY =
Pult
A
+
M0
Z
+
)
σ max :=
(
Pult⋅ δ 0 + ∆
with
5⋅ q⋅l
4
384⋅E⋅I
M 0 :=
Pult 

 1 − P ⋅Z
E

M max
Z
using
φ1(P_over_Pcr) :=
σ u( λ , η , µ ) :=
σY
π
rc :=
σE
⋅
σY
E
σ max := σ Y
2
1

 ⋅  1 − µ + 1 + η  − 1 ⋅  1 − µ + 1 + η  − 1 − µ  ⋅σ
Y
2 
2
2
2 
4 
λ 
λ 
λ 
 

8
Le_over_ρ
1
1 − P_over_Pcr
which after rearranging and defining some non-dimensional factors
becomes:
2
λ ( Le_over_ρ) :=
λ ( Le_over_ρ) :=
20 of 33
q⋅ l
A
+
)
values due to uniform distributed loading
(pressure*breadth).
δ 0 :=
P
η :=
ρ
2
rc :=
c
(δ0 + ∆ )
Z
rc = core radius
⋅A
notes_27_buckling_notes.mcd
I
 M0 

 Z  = M0
µ=
A⋅ c
η := α⋅Le_over_ρ
Le_over_ρ +
σY
δ0
rc
Z⋅σ Y
figure 11.14 is parameterized by η and µ
λ := 0.01 , 0.02 .. 2
η := 0.2
µ 1 := 0.2
µ 2 := 0.6
1
(
σu λ , η , µ 1
)
σY
0.8
(
σu λ , η , µ 2
)
σY
σ E( λ )
0.6
σY
σ ult ( λ )
σY
0.4
0.2
0
0.5
1
1.5
λ
a few values from text to check. To avoid singularity in mathcad, λ set close to 0
λ 0 := 0.0001
(
)
σ u λ 0 , 0.2, 0
σY
(
)
σ u λ 0 , 0.4, 0.2
σY
21 of 33
= 0.8333
= 0.5714
text has 0.83, 0.57, 0.38, 0.29
(
)
σ u λ 0 , 0.6, 0.4
σY
(
= 0.375
)
σ u λ 0 , 0.4, 0.6
σY
= 0.2857
notes_27_buckling_notes.mcd
2
Beam Column
(page 401 of text)
this was copied from notes 28
beam column and added here
clamped clamped beam column
q
P
P
reset:
 x2
M ( x) := −q⋅ 
2
M ( 0) →
M 
L
 2
−1
12
→
M ( L) →
⋅ q⋅L
−
L⋅ x
L
2
+
L
2
y ( x) :=
12 
2
y 
L
 2
1
24
−1
12
⋅ q⋅L
→
1
q
⋅L
384 E⋅ I
E⋅ I
4
⋅
3
 24
−
δ o := y 
L⋅ x
L
12
+
2 2
L ⋅x
24
L

δo →
 2
1
⋅
q
384 E⋅ I
⋅L
4
2
1
M center :=
⋅ q⋅L
⋅
 x4
q
2
M end :=
24
−1
12
⋅q⋅L
L
⋅q⋅L
L
2
moments at center and ends
2
==============================================================================
find locations where M(x) = 0
Given M ( x) = 0
1

 1 1 2
Find( x) →   + ⋅3 ⋅L
 2 6 
1 

1 1 2 
 − ⋅3 ⋅L  x :=
2 6   1
1
distance between M(x) = 0
x1 − x2 simplify →
δ o := y 
max deflection (at center)
L
 2
deflection at M(X) = 0
22 of 33
y ( x1)
δo
expand →
1
3
⋅L⋅3
 1 + 1 ⋅ 3 ⋅LL

2 6 
x2 :=
 1 − 1 ⋅ 3 ⋅LL

2 6 
= 0.5774
ν := 0.577
2
ν :=
1
3
δo →
4
4
9
9
⋅L
L
1
q 4
⋅
⋅L
384 E⋅ I
= 0.4444
notes_27_buckling_notes.mcd
1
3
δ o :=
1
⋅
q
384 E⋅ I
⋅L
L
4
γ := 0.444
observations of clamped/clamped beam due to uniform load
fraction of length between M(x) = 0
moment at center
ν := 0.577
moment at ends
fraction of δo associated with end section
deflection due to q
δ o :=
γ := 0.444
1
⋅
q
384 E⋅ I
⋅L
L
4
0
0.5
1
make into two "simply supported" (based on M(x) = 0) problems
P
23 of 33
q
P
notes_27_buckling_notes.mcd
M center :=
M end :=
1
24
−1
12
⋅q⋅L
L
⋅q⋅L
L
2
2
q
P
P
P
P
q
P
P
ends
center
Lcent := ν⋅L
L
Lend := ( 1 − ν ) ⋅L
L
δ cent := ( 1 − γ ) ⋅δδ o
δ cent := γ ⋅δδ o
M center →
1
24
⋅q⋅L
M end →
2
−1
12
⋅ q⋅L
2
apply beam column pinned pinned relationships to each segment
(
M max := M 0 + P⋅φ⋅
φ δ0 + ∆
using
)
σ max :=
P
A
+
M max
Z
φ1(P_over_Pcr) :=
σY =
Pult
A
+
M0
Z
+
(
Pult⋅ δ 0 + ∆
)
Pult 

 1 − P ⋅Z
E

=>
σ max := σ Y
1
1 − P_over_Pcr
which after rearranging and defining some non-dimensional factors
becomes:
σ u( λ , η , µ ) :=
24 of 33
2

1
 ⋅  1 − µ + 1 + η  − 1 ⋅  1 − µ + 1 + η  − 1 − µ  ⋅σ
Y
2 
2
2
2 
4 
λ 
λ 
λ 
 

notes_27_buckling_notes.mcd
see beam column summary at
this point