Transcript Normal Stress (1.1-1.5)

Column Design
(4.11-4.13)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
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Column Design
Real world examples
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Column Design
Continue…
Jack
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Column Design
Column failure
Quebec Bridge designed by Theodore Cooper. It collapsed in 1907.
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Column Design
Introduction
“Long” columns fail structurally (buckling failure)
“Short” columns fail materially (yielding failure)
We will examine buckling failure first, and then transition
P
to yielding failure.



“Long”
Pcr 
P
 2 EI
l
“Short”
2
Pcr   cr A
P
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P
Column Design
Buckling of Columns (4.12)
For a simply supported beam (pin-pin)

M ( x)   Pv( x)
y, v
v(x)
y, v
F.B.D 
d 2v
EI 2  M ( x)   Pv
dx
d 2v
EI 2  Pv  0
dx
The solution to the above O.D.E is:
v( x)  A sin
P
P
x  B cos
x
EI
EI
x
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Column Design
Buckling of Columns (4.12)
v( x)  A sin
P
P
x  B cos
x
EI
EI
Apply boundary conditions: v(0)=0 & v(l)=0

B  0 & A sin
P
l 0
EI
Either A = 0 (trivial solution, no displacement) or

sin
P
P
l 0
l  n
EI
EI
where n = 1, 2, …

The critical load is then

The lowest critical load (called the Euler buckling load) occurs
when n=1.
 2 EI
Pcr  (n ) 2
Pcr 
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EI
l2
l2
Column Design
Buckling of Columns (4.12)
The corresponding deflection at the critical load is

v( x)  A sin
x
l
The beam deflects in a half sine wave.

Pcr
The amplitude A of the buckled beam cannot
be determined by this analysis. It is finite, but
non-linear analysis is required to proceed
further

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Column Design
A
Buckling of Columns (4.12)
Define the radius of gyration as

Pcr can be written as

k2 
I
A
Pcr
 2E

A (l / k )2


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 pinpin), use the end-condition constant, C.

Pcr C 2 E

  cr
2
A (l / k )
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Column Design
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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Column Design
Buckling of Columns (4.12)
1.Stability
(1) Unstable equilibrium
(2) Stable equilibrium
(3) Neutral equilibrium
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Column Design
Buckling of Columns (4.12)
F
F
P  Pcr
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P  Pcr
P  Pcr
Column Design
Example 4-16
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Column Design
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Column Design
Example
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 xzplane. 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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Column Design
How to prevent buckling?

1) Choose proper cross-sections

2) End conditions
3) Change compression to tension

(cable stayed bridges)
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Column Design
Note:
Sy = yield strength = σy
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
Johnson Curve
Sy
l/k
Failure
No failure
?
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l/k
Column Design
Critical Stress (4.13)

Johnson equation
Pcr
l
 a b 
A
k

2
Where
2
 Sy  1
a  S y and b  

2


 CE

Notice as l/k  0, Pcr/A  Sy
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Column Design
Critical Stress (4.13)

To find the transition point (l/k)1 between the Johnson and
Euler regions
Euler  Johnson
2
2
S


C E
1 l
y

S



y
 
2
2

CE
k
l


k
2
 
2 2CE
l
  
Sy
 k 1
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Column Design
Design of Columns (4.13)

Procedure


Calculate (l/k)1 and l/k
If l/k ≥ (l/k)1 use Euler’s equation
Pcr C 2 E

A (l / k )2

If l/k ≤ (l/k)1 use Johnson’s equation
Pcr
l
 a b 
A
k

2
Notice we haven’t included factor of safety yet…
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Column Design
Example 4-19
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Column Design
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Column Design
Example
Find 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.
t
do
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Column Design
Example
Find 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.
F
t
L
do
b
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Column Design
Struts or short compression members




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
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Column Design