12.1 Introduction to column buckling
• Buckling: “Buckling can be defined as the
sudden large deformation of structure due to a
slight increase of an existing load under which
the structure had exhibited little, if any,
deformation before the load was increased.” No
failure implied!!!
Reinforced concrete
steel
Stability of equilibrium condition
• Easy to visualize for a ball on a surface
• Note that we use curvature to decide
stability, but the surface can curve up with
zero curvature
Euler formula
• For simply supported column
Pcr =
π 2 EI
L
2
or using I = Ar ,
L / r : Slenderness ratio
2
Pcr =
π 2 EA
(L / r)
2
Large displacements
• Slenderness ratio and yield stress govern type of
post-buckling response
Dashed lines represent behavior without yielding
Governing equation for beam column
d2y M
P
=
=−
y
2
dx
EI
EI
d2y P
+
y=0
2
dx
EI
Try : y ( x) = A cos(ω x) + B sin(ω x)
d2y
2
ω
=
−
( A cos(ω x) + B sin(ω x))
2
dx
P
Substitute into ODE :ω =
EI
⎛ P ⎞
⎛ P ⎞
⎟
⎜
y ( x) = A cos⎜
x ⎟ + B sin ⎜⎜
x ⎟⎟
⎝ EI ⎠
⎝ EI ⎠
Apply boundary conditions
y (0) = 0 and y ( L) = 0
⎛ P ⎞
P
⎜
⎟
B sin ⎜
L⎟ =0 →
L = nπ n =1,2,3,..
EI
EI
⎝
⎠
n 2π 2 EI
P=
n is the integer defining the buckling mode.
2
L
I
I = Ar 2 where r =
is the radius of gyration.
A
For a column with both ends pinned, n = 1 defines the critical buckling load
Pcr =
π 2 EI
or using I = Ar , Pcr =
L
L / r :Slenderness ratio
2
2
π 2 EI
(L / r)2
Compressive (normal) stress at critical buckling load : σ cr
Pcr
π 2E
=
=
A (L / r)2
Example 1
A 3m column with the cross section shown is constructed from two
pieces of timber, that act as a unit. If the modulus of elasticity of timber
is E=13 GPa, determine a) The slenderness ratio b) Critical buckling
load c) Axial stress in the column when the critical load is applied
• Propertiesof the cross section
A = 2(150)(50) =15,000 mm2
25(50)(150) + (50 + 75)(50)(150)
yc =
= 75mm from bottom
15,000
• Momentsof inertia about the centroid of the cross section
150(50)3
50(150)3
2
Ix =
+150(50)(50) +
+150(50)(50) 2 = 53.13 x 106 mm4
12
12
50(150)3 150(50)3
Iy =
+
=15.625 x 106 mm4
L 3000
12
12
a ) Slenderness ratio =
= 93
r
32
.
3
• Radii of inertia
Iy
Ix
rx =
= 59.51 mm and ry =
= 32.3 mm
A
A
⇒ r = rmin = 32.3 mm
b) Critical Buckling Load Pcr =
π 2 EA
(L / r)2
= 222.75 kN
Pcr
π 2E
c) Critical Buckling Stress σ cr =
=
=14.85 MPa
A (L / r)2
Example 2
C229 x 30 structural steel channels with E=200 GPa are used for a 12
m column. Determine the total compressive load required to buckle
the column if a) One channel is used b) Two channels are laced 150
mm back to back as shown
Section properties for one channel
A = 3795 mm 2 xc = 14.8mm
I xc = 25.3 x 106 mm 4
I yc =1.01 x 106 mm 4
Solution for single channel
.
Radii of inertia of a single channel
Ix
rx =
= 81.1 mm and ry =
A
⇒ r = rmin = 16.3 mm
Iy
A
= 16.3 mm
L 12000
Slenderness ratio :
=
= 736.2
r
16.3
π 2 EA
Critical Buckling load : Pcr =
= 13.82 kN
2
(L / r)
Two laced channels.
A = 2(3975) = 7590mm 3
Moments of inertia about the centroid of the cross section :
I x = 2 I x c = 50.6 x 106 mm 4
[
]
I y = 2 I yc + A(75 + 14.8) 2 = 63.23 x 106 mm 4
Radii of inertia :
Iy
Ix
= 81.7 mm and ry =
= 91.3 mm → r = rmin = 81.7 mm
A
A
L 12000
Slenderness ratio :
=
= 146.9
r
81.7
π 2 EA
Critical Buckling load : Pcr =
= 694.3 kN
2
(L / r)
rx =
Higher buckling modes
• With appropriate boundary
conditions can get higher
modes
P =4
σ cr ( 2 )
π 2 EI
= 4 Pcr
L
π 2E
P
= =4
2
(L / r)
A
2
Buckling modes : n = 1,2,3.
Imperfection approach
• Put in imperfection in the form of eccentricity
Eccentrically loaded pinned-end columns
• Moment equation
x
M ( x) = − Py − Pe
L
Solution of eccentric load
• Differential equation of equilibrium
2
d2y
k
ex
P
2
2
,
+
k
y
=
−
k
=
dx 2
L
EI
y = 0 for x = 0 , L
• General solution
ex
y = A sin k x + B cos k x −
L
• With boundary conditions
⎛ sin k x x ⎞
y = e ⎜⎜
− ⎟⎟
⎝ sin k L L ⎠
Reading assignment
Sections 12.3-4: Question: Why equation 12.26 does not really
guarantee buckling? What does it guarantee?
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