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7-Column Design Shig
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Column Design(4.11-13)
MAE 316 Strength of Mechanical ComponentsY. Zhu
Column Design1
Introduction
Column Design2
Long columns fail structurally (buckling failure) Short columns fail materially (yielding failure) We will examine buckling failure first, and then transition
to yielding failure. P
P
Long
2
2
LEIPcr
P
P
ShortAP crcr
Buckling of Columns (6.2) For a simply supported beam (pin-pin)
Column Design3
y, v
x
F.B.D y, v
v(x)
0
)(
)()(
2
2
2
2
PvdxvdEI
PvxMdxvdEI
xPvxM
The solution to the above O.D.E is:
xEIPBx
EIPAxv cossin)(
Buckling of Columns (6.2)
Apply boundary conditions: v(0)=0 & v(L)=0
Either A = 0 (trivial solution, no displacement) or
The critical load is then
The lowest critical load (called the Euler buckling load) occurs when n=1.
Column Design4
xEIPBx
EIPAxv cossin)(
0sin&0 LEIPAB
nLEIPL
EIP 0sin where n = 1, 2,
22)(LEInPcr
2
2
LEIPcr
Buckling of Columns (6.2) The corresponding deflection at the critical load is
The beam deflects in a half sine wave.
The amplitude A of the buckled beam cannotbe determined by this analysis. It is finite, however. Non-linear analysis is required to proceed further.
Column Design5
LxAxv sin)(
Pcr
A
Buckling of Columns (6.2) Define the radius of gyration as
Pcr can be written as
If L/rg is large, the beam is long and slender. If L/rg is small, the beam is short and stumpy.
To account for different boundary conditions (other than pin-pin), use effective length, Le.
Column Design6
AIrg 2
2
2
)/( gcr
rLE
AP
crge
cr
rLE
AP 2
2
)/(
Buckling of Columns (6.2) Effective lengths for various boundary conditions
Design values (AISI)
Column Design7
(a) fixed-free, (b) pin-pin, (c) fixed-pin, (d) fixed-fixed, (e) fixed-non-rotating
(a) Le = 2.1L, (b) Le = L, (c) Le = 0.8L, (d) Le = 0.65L, (e) Le = 1.2L
Critical Stress (6.3) Plot cr=Pcr/A vs. L/rg
Long column: buckling occurs elastically before the yield stress is reached. Short column: material failure occurs inelastically beyond the yield stress. A Johnson Curve can be used to determine the stress for an intermediate
column
Column Design8
?
No failure
Failure
Sy
Johnson Curve
Note:Sy = yield strength = ySu = ultimate strength = uSp = proportional limit
Critical Stress (6.3) Johnson equation
As Le/rg 0, Pcr/A Sy To find the transition point (Le/rg)c between the Johnson and
Euler regions
Column Design9
2
2
2
4
g
eyy
cr
rL
ES
SAP
ycg
e
g
eyy
SE
rL
rL
ES
S
rgLe
E
JohnsonEuler
2
2
2
2
2
2
2
4
Design of Columns (6.6) Procedure
Calculate (Le/rg)c and Le/rg If Le/rg (Le/rg)c use Eulers equation
If Le/rg (Le/rg)c use Johnsons equation
Notice we havent included factor of safety yet
Column Design10
crge
cr
rLE
AP 2
2
)/(
2
2
2
4
g
eyy
cr
rL
ES
SAP
ExampleFind the required outer diameter, with F.S.=3, for the column shownbelow. Assume P = 15 kN, L = 50 mm, t = 5 mm, Sy = 372 MPa, E =207,000 MPa, and the material is 1018 cold-drawn steel.
Column Design11
t
do
ExampleFind the maximum allowable force, Fmax, that can be applied withoutcausing pipe to buckle. Assume L = 12 ft, b = 5 ft, do = 2 in, t = 0.5 in,and the material is low carbon steel.
Column Design12
t
do
F
L
b