Column Design (4.11-4.13)
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Column Design(4.11-4.13)
MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering
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Real world examples
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Continue…
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Jack
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Column failure
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Quebec Bridge designed by Theodore Cooper. It collapsed in 1907.
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Introduction
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“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
2crEIP
l
P
P
“Short”AP crcr
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Buckling of Columns (4.12)
For a simply supported beam (pin-pin)
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y, v
x
F.B.D
y, v
v(x)
0
)(
)()(
2
2
2
2
Pvdx
vdEI
PvxMdx
vdEI
xPvxM
The solution to the above O.D.E is:
xEIPBx
EIPAxv cossin)(
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Buckling of Columns (4.12)
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.
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xEIPBx
EIPAxv cossin)(
0 & sin 0PB A lEI
sin 0P Pl l nEI EI
where n = 1, 2, …
22( )cr
EIP nl
2
2crEIP
l
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Buckling of Columns (4.12)
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, butnon-linear analysis is required to proceedfurther
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( ) sin xv x Al
Pcr
A
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Buckling of Columns (4.12)
Define the radius of gyration as
Pcr can be written as
If l/k is large, the beam is long and slender. If l/k is small, the beam is short an stumpy.
To account for different boundary conditions (other than pin-pin), use the end-condition constant, C.
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2 IkA
2
2( / )crP EA l k
2
2( / )cr
crP C EA l k
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Buckling of Columns (4.12)
Theoretical end condition constants for various boundary conditions (Figure 4-18 in textbook) (a) pin-pin (b) fixed-fixed (c) fixed-free (d) fixed-pin
Values of C for design are given in Table 4-2 in the textbook
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Buckling of Columns (4.12)
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1.Stability
(3) Neutral equilibrium
(1) Unstable equilibrium
(2) Stable equilibrium
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Buckling of Columns (4.12)
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crP P crP P crP P
FF
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Example 4-16
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Example
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Column AB carries a centric load P of magnitude 15 kips. Cables BC and BD are taut and prevent motion of point B in the xz-plane. Using Euler’s formula and a factor of safety of 2.2, and neglecting the tension in the cables, determine the maximum allowable length L.Use E = 29 x 106 psi.
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How to prevent buckling? 1) Choose proper cross-sections
2) End conditions 3) Change compression to tension
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(cable stayed bridges)
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Critical Stress (4.12)
Plot σcr=Pcr/A vs. l/k 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
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?
No failureFailure
Sy
Johnson Curve
Note:Sy = yield strength = σy
l/k
l/k
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Critical Stress (4.13)
Johnson equation
Where
Notice as l/k 0, Pcr/A Sy
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2crP la bA k
21
2y
y
Sa S and b
CE
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Critical Stress (4.13)
To find the transition point (l/k)1 between the Johnson and Euler regions
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2 22
2
2
1
12
2
yy
y
Euler Johnson
SC E lSCE kl
k
l CEk S
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Design of Columns (4.13)
Procedure Calculate (l/k)1 and l/k If l/k ≥ (l/k)1 use Euler’s equation
If l/k ≤ (l/k)1 use Johnson’s equation
Notice we haven’t included factor of safety yet…Column Design20
2
2( / )crP C EA l k
2crP la bA k
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Example 4-19
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ExampleFind the required outer diameter, with F.S.=3, for the column shown below. 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.
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t
do
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ExampleFind the maximum allowable force, Fmax, that can be applied without causing pipe to buckle. Assume L = 12 ft, b = 5 ft, do = 2 in, t = 0.5 in, and the material is low carbon steel.
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t
do
F
L
b
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Struts or short compression members
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Strut - short member loaded in compression
If eccentricity exists, maximum stress is at B with axial compression and bending.
Note that it is not a function of length
If bending deflection is limited to 1 percent of e, then from Eq. (4-44), the limiting slenderness ratio for strut is