13 buckling
DESCRIPTION
ColumnsTRANSCRIPT
-
1! Critical Load! Ideal Column with Pin Supports! Columns Having Various Supports
BUCKLING OF COLUMNS
-
2Critical Load
Pcr
Pcr
P > Pcr
P > Pcr
-
3L/2
L/2
P
Ak
L/2
L/2
P
Ak=(L/2)
F
P
P
P tan
P tan
A
Fx = 0:+ FP =tan2
= kP tan2
)2
(2 LkP =
4kLPcr =
For small ,
-
4Unstableequilibrium
P
Neutralequilibrium
Bifurcation point
Stable equilibrium
O
4kLPcr =
-
5Ideal Column with Pin Supports
L
P Pcr
-
6x
L
P
P
x x
+ Mx = 0 ;
0=+ MP
PM =
Moment-curvature
PdxdEIM == 2
2
022
=+ PdxdEI
0)(22
=+ EIP
dxd
*0)( 222
=+ EIP
dxd
0'' 2 =+ c
)cos()sin( 21 xEIPCx
EIPC +=
P
P
M
N = 0
x
-
7 Boundary condition
0, == Lx
)sin(0)sin( nLEIP
==
0),1()0(0 221 =+= CCC
0,0 == x
)cos()sin( 21 xEIPCx
EIPC +=
0,0)sin( 11 = CLEIPC
,...3,2,1: == nnLEIP
x
L
P
P
x x
P
P
M
N = 0
x
-
8x
v
L
P
P
x
P
P
M
0
x
x
Critical Load Pcr
nLEIP
=
222 nLEIP
=
,...3,2,1: == nnLEIP
*222
=L
EInP
*22
=LEIPcr
-
9x
max
L/2L
P
P
xn = 1
P
P
L/2
L/2
n = 2
,...3,2,1,222
== nL
EInPcr
2
221L
EIPcr
= 2222
LEIPcr
=
-
10
Neutralequilibrium
Bifurcation point
Stable equilibrium
O
Pcr = L22EI
Unstableequilibrium
P
x
L
Pcr
Pcr
x
-
11
Critical Stress
2
2
)(KLEIPcr
=
ratiosslenderneseffectiverLK =
2
22
)()(
KLArE
=
2
2
)(rLK
EA
Pcr =
r = radius of gyration
2
2
)(r
KLE
cr =
K = effective-length factor, for pin-pin column K = 1
-
12
KL/r
89100125150175200225
cr (MPa)
25019712688644939
Structural steel A 36E = 200 GPa y = 250 MPa
Structural steelA 36R 40 (4000 kg/cm2)
400
200
0 KL/r
cr , MPa
100 200
100
50 150
300
2
32
2
2
)(
)10200(
)(r
KLMPa
rKL
Ecr
==
197 MPa
88 MPa49 MPa
2
2
)(r
KLE
cr =
89
y = 250
-
13
2
2
)(r
KLE
cr =
Aluminum E = 70 GPa y = 215 MPa
KL/r
5775100125150175200
cr (MPa)
215122.869.144.230.722.617.3
200
100
0 KL/r
cr , MPa
100 200
50
50 150
150
2
32
2
2
)(
)1070(
)(r
KLMPa
rKL
Ecr
==
57
215 MPa
69 MPa
30 MPa17 MPa
-
14
Example 1
A 7 m long A-36 steel tube having the cross section shown is to be used as apin-ended column. Determine the maximum allowable axial load the column cansupport so that it does not buckle or yield. Take the yield stress of 250 MPa
70 mm75 mm
Pcr
Pcr
7 m
-
15
Using Eq. 5 to obtain the critical load with Est = 200 GPa,
Pcr
Pcr
7 m
70 mm
75 mm
2
2
LEIPcr
=
2
4462
7
)07.04
075.04
)(10200[( =
= 241.4 kN
This force creates an average compressive stress in the column of
22 )070.0()075.0(43.241
==A
Pcrcr
= 106 MPa < Y = 250 MPa O.K
The maximum allowable axial load the column can support is 241.73 kN
-
16
Pcr
Pcr
7 m
70 mm
75 mm
= 106 MPa < Y = 250 MPa O.K
Pcr = cr A = (106 x 106) (.0752-.0702)
2
2
)(r
KLE
cr =
222
442 002631.
)070.075(.4/)070.075(. m
AIr =
==
5.136)00795.0(
)7)(1(==
rKL
2
32
2
2
)5.136()10200(
)(
MPa
rKL
Ecr
==
Alternate method:
= 241.7 kN
mmr 5.79=
-
17
Structural steelA 36R 40 (4000 kg/cm2)
400
200
0 KL/r
cr , MPa
100 200
100
50 150
300
197 MPa
88 MPa49 MPa
2
2
)(r
KLE
cr =
89
y = 250
136.5
cr = 106
-
18
Example 2
The A-36 steel W200x46 member show is to be used as a pin-connected column.Determine the largest axial load it can support before it either begins to buckle orthe steel yields.
x
x
y y4 m
-
19
Pinned - Pinned Column
2
2
LEIPcr
=
= 1887. 6 kN
APcr
cr =
AAP yallow ==
A-36 steel W200x46 A = 5890 mm2 , Ix = 45.5x106 mm4, and Iy = 15.3x106 mm4
x
x
y y
= 320.5 MPa > Y = 250 MPa
2
662
4)103.15)(10200(
=
610589056.1887
=
)105890)(10250( 266 mPa =
kN1472=
4 m
-
20
Columns Having Various Type of Supports
P
P
ML
P
x
u
( ) = PdxdEI 2
2
EIP
EIP
dxd
=+22
-----(7)
This equation is non-homogeneous because of thenonzero term on the right side. The solution consists of botha complementary and particular solution, namely,
++= )cos()sin( 21 xEIPCx
EIPC
The constants are determined from the boundary conditions. At x =0, = 0, so that C2 = -. Also,
)sin()cos( 21 xEIP
EIPCx
EIP
EIPC
dxd
=
At x = 0, d/dx = 0, so that C1 = 0. The deflection curve is therefore
)]cos(1[ xEIP
= -----(8)
-
21
Since the deflection at the top of the column is , that is, at x = L, = , we require
)2
cos()cos(0)cos( nLEIPorL
EIP
==
The smallest critical load occurs when n = 1, so that
2
2
4LEIPcr
= -----(9)
)]cos(1[ xEIP
= 0)cos( =LEIP-->
,0Since
-
22
L
P
Fixed - Pinned ends
L
P
Fixed fixed ends
K = 0.7
Le = L
K = 1
Le = 0.5L
K = 0.5
Le = 0.7L
Pinned -pinned ends
P
Note : K = effective-length factor
Effective Length (Le)
2
2
2
2
)/()( rKLEor
KLEIP crcr
==
K = 2
Fixed - free ends
L
P
Le = 2L
-
23
Example 3
A W 150x24 (A=3060 mm2, Ix = 13.4x106 mm4, Iy = 1.83x106 mm4) steel columnis 8 m long and is fixed at its ends as shown. Its load-carrying capacity isincreased by bracing it about the y-y (weak) axis using struts that are assumed tobe pin-connected to its mid-height. Determine the load it can support so that thecolumn does not buckle nor the material exceed the yield stress. Take E = 200GPa and y = 250 MPa
yy
x
x
3 m
5 m
P
-
24
W 150x24 A = 3060 mm2
rx = 66.2 mm
Fixed (top)Fixed (bottom)Kx = 0.5
Ix = 13.4x106 mm4Iy = 1.83x106 mm4
Pinned (top)Fixed (bottom)Ky = 0.7ry = 24.5 mm
Bucking x-x axis
kNmPaAP yY 765)103060)(10250(266 ===
kNm
mkPaKL
EIPx
xxcr 1653)85.0(
)104.13)(10200()(
)( 24662
2
2
=
==
Yield Stress (y)
kNm
mkPaKL
EIP
y
yycr 9.294)57.0(
)1083.1)(10200()(
)( 24662
2
2
=
==
Bucking y-y axis
E = 200 GPa , y = 414 MPa
yy
x
x
3 m
5 m
P
y-y axis buckling
5m
3 m
x-x axis buckling
8 m
-
25
NOTE
2
32
2
2
)(
)10200(
)(r
KLMPa
rKL
Ecr
==
KL/r
89100125150175200225
cr (MPa)
25019712688644939
Structural steel A 36E = 200 GPa
400
200
0 KL/r
, MPa
100 200
100
197 MPa
50 150
300
88 MPa49 MPa
89
y = 250
2
2
)(rLK
Ecr
=
42.602.66
)108)(5.0()(3
=
=xrKL
179
61.9 MPa
6.1785.24
)105)(7.0()(3
=
=yrKL
-
26
Example 4
The aluminum column is fixed at its bottom and is braced at its top by two rodsso as to prevent movement at the top along the x axis, If it is assumed to be fixedat its base, determine the largest allowable load P that can be applied. Use afactor of safety for buckling of F.S. = 3.0. Take Eal = 70 GPa, y = 215 MPa, A =7.5(10-3) m2, Ix = 61.3(10-6) m4, Iy = 23.2(10-6) m4.
x
z
y
5 m
P
Rod
-
27
5 m
y-y axis bucklingx-x axis buckling
5 m
Eal = 70 GPa, y = 215 MPa, A = 7.5(10-3) m2
rx = 90 mm
Free (top)Fixed (bottom)Kx = 2
Ix = 61.3(10-6) m4 Iy = 23.2(10-6) m4
Pinned (top)Fixed (bottom)Ky = 0.7ry = 50 mm
Bucking x-x axis
kNmPaAP yY 1612)105.7)(10215(236 ===
kNkNSF
PP crallow 43631308
.===
Yield Stress (y)
kNm
mkPaKL
EIP
y
yycr 1308)57.0(
)102.23)(1070()(
)( 24662
2
2
=
==
Bucking y-y axis
kNm
mkPaKL
EIPx
xxcr 425)52(
)103.61)(1070()(
)( 24662
2
2
=
==
kNkNSF
PP crallow 1413425
.===
x
z
y
5 m
P
Rod
-
28
NOTE
Aluminum E = 70 Gpa y = 215 MPa
2
32
2
2
)(
)1070(
)(r
KLMPa
rKL
Ecr
==
KL/r
5775100125150175200
cr (MPa)
215122.869.144.230.722.617.3
1.11190
1052)(3
=
=mm
mmr
KLx
7050
1057.0)(3
=
=mm
mmr
KLy
-
29
Example 5
Determine the maximum load P the column can support before it either begins tobuckle or the steel yields. Assume that member BC is pinned at its end for the x-xaxis and fixed for y-y axis buckling. Take E = 200 GPa, y = 250 MPa.
2 m
P
4 m
1 m35 mm
4
53
x
y35 mm
25 mm
AC
B
-
30
2 m
P
4 m
1 m35 mm
4
53
x
y35 mm
25 mm
AC
B
F
+ MA = 0: 0)2()4)(53( =+ PF
*65
= PF
)035.0)(025.0()495(
)10200(65
2
92 =
P
495]
)035.0)(025.0()035.0)(025.0(
121[
)5(1)(2/1
3 == xrKL
346]
)025.0)(035.0()025.0)(035.0(
121[
)5(5.0)(2/1
3 == yrKL
A
rKL
EAF cr2
2
)(
==
P = 8.46 kN < 262.5 kN
Ax
Ay
kNP
kNAPF
Y
YYY
5.262
8.218)035.025)(.10250(65 3
=
====