Ce 579: Structral Stability and Design
Ce 579: Structral Stability and Design
Ce 579: Structral Stability and Design
Amit H. Varma
Assistant Professor
School of Civil Engineering
Purdue University
Ph. No. (765) 496 3419
Email: ahvarma@purdue.edu
Office hours: M-T-Th 10:30-11:30 a.m.
Chapter 3. Structural Columns
1 E I x v P v P x0 0
2 E I y u P u P y0 0
3 E I w ( P r02 G KT ) u ( P y0 ) v ( P x0 ) 0
Where
Ix I y
r0 x y
2 2
0
2
0
A
Column buckling – doubly symmetric section
For a doubly symmetric section, the shear center is located at the
centroid xo= y0 = 0. Therefore, the three equations become uncoupled
1 E I x v P v 0
2 E I y u P u 0
3 E I w ( P r02 G KT ) 0
Take two derivatives of the first two equations and one more derivative
of the third equation.
1 E I x viv P v 0
2 E I y u iv P u 0
3 E I w iv ( P r02 G KT ) 0
P P P r 2
G KT
Let , Fv2 Fu2 F2 0
E Ix E Iy E Iw
Column buckling – doubly symmetric section
3
iv F2 0
All three equations are similar and of the fourth order. The
solution will be of the form C1 sin lz + C2 cos lz + C3 z + C4
Need four boundary conditions to evaluate the constant C1..C4
For the simply supported case, the boundary conditions are:
u= u”=0; v= v”=0; = ”=0
Lets solve one differential equation – the solution will be valid for
all three.
Column buckling – doubly symmetric section
v iv Fv2 v 0
Solution is
v C1 sin Fv z C2 cos Fv z C3 z C4
v C1 Fv2 sin Fv z C2 Fv2 cos Fv z
The coefficient matrix 0
Boundary conditions :
v(0) v(0) v( L) v( L) 0 Fv2 sin Fv L 0
sin Fv L 0
C2 C4 0 v(0) 0 Fv L n
C2 0 v(0) 0 P n
Fv
C1 sin Fv L C2 cos Fv L C3 L C4 v( L) 0 E Ix L
C1 Fv2 sin Fv L C2 Fv2 cos Fv L v( L) 0 n2 2
Px 2 E I x
L
0 1 0 1 C1 0 Smallest value of n 1:
0 1 0 0 C2 0
2 E Ix
sin Fv L cos Fv L L 1 C3 0 Px
Fv2 sin Fv L Fv2 cos Fv L 0 0 C4 0 L2
Column buckling – doubly symmetric section
Similarly, Similarly,
sin Fu L 0 sin F L 0
Fu L n F L n
P n P r02 G KT n
Fu F
E Iy L E Iw L
n2 2 n2 2 1
Py 2 E I y P 2 E I w G KT 2
L L r0
2 E Iy Smallest value of n 1:
Smallest value of n 1: Py 2
L
n2 2 1
P 2 E I w G KT 2
2 E Ix L r0
Px 1
L2
2 E Iy
Summary Py 2
2
L
2 E Iw 1
P 2
G KT 2 3
L r0
Column buckling – doubly symmetric section
These are, flexural buckling about the x and y axes and torsional
buckling about the z axis.
As you can see, the three buckling modes are uncoupled. You must
compute all three buckling load values.
The smallest of three buckling loads will govern the buckling of the
column.
Column buckling – boundary conditions
Consider the case of fix-fix boundary conditions:
viv Fv2 v 0
Solution is
v C1 sin Fv z C2 cos Fv z C3 z C4 The coefficient matrix 0
v C1 Fv cos Fv z C2 Fv sin Fv z C3 Fv L sin Fv L 2 cos Fv L 2 0
Boundary conditions : Fv L Fv L Fv L
2 sin F L cos 2sin 0
2 2
v
v(0) v(0) v( L) v( L) 0 2
C2 C4 0 v(0) 0 Fv L
n
2
C1 Fv C3 0 v(0) 0
2n
C1 sin Fv L C2 cos Fv L C3 L C4 v( L) 0 Fv
L
C1 Fv cos Fv L C2 Fv sin Fv L C3 v( L) 0 4 n2 2
Px E Ix
L2
0 1 0 1 C1 0 Smallest value of n 1:
Fv 0 1 0 C2 0
2 E Ix 2 E Ix
sin Fv L cos Fv L L 1 C3 Px
0 0.5 L
2
K L
2
Fv cos Fv L Fv sin Fv L 1 0 C4 0
Column Boundary Conditions
2 E Iy
Py 2
K L
2
y
2 E I 1
P w
G KT 2 3
z
2
K L r0
Ky Y K y
rx rx rx
P 2 E I w 2 A 1
G K r
x x y
2 T x r2 I I A Y
PY
K L
z rx
P 2 E I w 2 1
G K r
x x y Y
T x 2
r I I
2
PY
K L
z rx
P 578.26
2
0.2333
PY L
rx
Column buckling – example.
2
1.6
1.4
1.2
Yield load PY
Cannot be exceeded
1
0.8
The first equation for flexural buckling about the x-axis (axis of
symmetry) becomes uncoupled.
E I x v P v 0 (1)
Equations (2) and (3) are still
E I x v P v 0
iv
coupled in terms of u and .
viv Fv 2 v 0
P 2 E I y u P u P y0 0
where, Fv 2
E Ix 3 E I w ( P r02 G KT ) u ( P y0 ) 0
v C1 sin Fv z C2 cos Fv z C3 z C4
Boundary conditions
These equations will be satisfied by
sin Fv L 0
the solutions of the form
2 E Ix u=C2 sin (z/L) and =C3 sin (z/L)
Px
( K x Lx ) 2
Buckling mod v C1 sin Fv z
Column Buckling – Singly Symmetric Columns
E I y u P u P y0 0 (2)
E I w ( P r02 G KT ) u ( P y0 ) 0 (3)
E I y u iv P u P y0 0
E I w iv ( P r02 G KT ) u ( P y0 ) 0
z z
Let , u C2 sin ; C3 sin
L L
Therefore, substituting these in equations 2 and 3
z z z
4 2 2
2 E Iy 2 E Iw 1
Let , Py 2
and P 2
G K T 2
L L r0
Py P C2 P y0 C3 0
P P r02C3 P y0 C2 0
Py P P y0 C2
2 0
P y0 ( P P ) r0 3
C
Py P P y0
0
P y0 ( P P) r02
Column Buckling – Singly Symmetric Columns
y02 2 (1 2 )
( Py P ) ( Py P ) 4 Py P (1 2 )
2
r0
r0
P
y02
2 (1 2 ) Thus, there are two roots for P
r0
Smaller value will govern
y02
4 Py P (1 r 2 ) y 2
( Py P ) ( Py P ) 2 1 0 4 Py P (1 2 )
0
2 ( P P )
( Py P ) P P y 1 1 r0
y02 ( P P ) 2
2 (1 2 ) y
P
y 2 r0
2 (1 02 )
r0
Column Buckling – Singly Symmetric Columns
The critical buckling load will the lowest of Px and the two roots
shown on the previous slide.
If the flexural torsional buckling load govern, then the buckling
mode will be C2 sin (z/L) x C3 sin (z/L)
This buckling mode will include both flexural and torsional
deformations – hence flexural-torsional buckling mode.