Transcript Normal Stress (1.1-1.5) - North Carolina State University

Buckling of Slender Columns
(10.1-10.4)
MAE 314 – Solid Mechanics
Yun Jing
Buckling of Slender Columns
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Introduction

Up to this point, we have designed structures with constraints based on
material failure.



Find the state of stress and strain due to applied loads for certain simple structures.
Compare the stress state to a maximum stress criterion or the strain state to a
maximum strain criterion and determine at what load the structure will fail.
In this chapter, we learn how to design structures with constraints based on
structural failure.




No point in the material reaches the maximum stress or strain criterion.
The structure at equilibrium becomes unstable.
Any small change in the loading or any imperfection in the structure will not
remain at equilibrium.
This is generally due to compressive loading.
Buckling of Slender Columns
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Example: Pinned Column
Column in Equilibrium
roller
supports
deformable
column

Equilibrium state for P<PCR
Equilibrium state for P=PCR
Will a structure experience material failure or structural failure first?


What happens as P increases?
1. The point A moves downward and the
beam shortens.
2. At a certain value, PCR, the beam
“buckles” and suddenly changes shape.
3. This new shape may not support the
full load.
Depends on the material, geometry, etc.
In general:


Long slender columns typically fail first due instability.
Short wide columns typically fail first due to the material.
Buckling of Slender Columns
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Example: Pinned Column



Why does this occur?
To find out, let’s consider a simpler system.
When is the deformed equilibrium position stable?



Move point C a small amount to the right.
If the system moves back to the vertical
position it is stable.
If the system moves away from the vertical
position it is unstable.

Spring moment: M = K (2Δθ)
Draw a free body diagram of the top bar.
Sum moments about point C.

 L
 L
M

P
sin



M

P


  sin   2K
 C 2
2
Assume sin   


rigid bars
torsional spring
angle of rotation
of spring
M  K
4K 
 PL

 L 
M


2
K



P







 C 2
2
L




Buckling of Slender Columns
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Example: Pinned Column

What happens for different values of P?

For P < 4K / L, ΣMC < 0



For P = 4K / L, ΣMC = 0



Point C does not move.
Neutrally Stable Equilibrium
For P > 4K / L, ΣMC > 0



Point C moves back to the left.
Stable Equilibrium
Point C moves to the right.
Unstable Equilibrium
4K 
 L 
M

P




 C  2  L 
We define PCR to be the critical load P at which the system moves from
a stable equilibrium to an unstable equilibrium.
Buckling of Slender Columns
4K
PCR 
L
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Euler’s Formula




This chapter uses slender pinned column as its basis because it is a commonly
used member.
The same concepts can be applied to other structures.
Next step is to derive Euler’s formula for pin-ended columns.
Solve for deflection curve of buckled beam.
d2y M

dx2 EI
Moment at some
distance x: M  Py
d 2 y Py

dx2 EI
d2y P

y0
dx2 EI
Define p2 = P / EI
d2y
2

p
y0
2
dx
Solution:
Buckling of Slender Columns
y  A sin px  B cos px
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Critical Load

Boundary conditions

y(0) = 0: y(0)  0  B  0

y(L) = 0: y( L)  0  A sin pL  0  pL  n
n 2 2
p L  n   P  2 EI
L
2 2


2
The lowest value of P corresponds to n = 1.
PCR 

2
2
2
L
EI
y ( x )  A sin

L
x
PCR n1 
We cannot solve for the amplitude (A)
using a stability analysis.
Other values of n correspond to higher
order modes.
n=1
n=2
Buckling of Slender Columns
n=3
n=4
2
L2
EI
4 2
 PCR n  2  2 EI
L
9 2
 PCR n 3  2 EI
L
16 2
 PCR n  4  7 2 EI
L
Critical Stress

Knowing the critical load, we can now calculate the critical stress.
 CR


Evaluate at section of beam with smallest I.
For a rectangular beam, I = bh3/12.
 CR 


PCR  2 EI


A
AL2
 2 Ebh3
12bhL2

 2 Eh2
12L2

 2E 
1 


2 
12  L / h  
Typical plot of σCR vs. L / r for steel (r is the
radius of gyration)
Critical stress depends on the slenderness
yield stress determines
failure
ratio L /r .
Buckling of Slender Columns
buckling determines
failure
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Rectangular Cross-Section

What happens when the column cross-section is rectangular?

i.e. In which direction does it buckle?
I11  I 22 
PCR 1 1  PCR 2  2
So the beam buckles about the 2-2 axis first.
PCR 
2
L2
EI
Buckling of Slender Columns
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Other End Conditions



We can extend Euler’s formula to columns with other end conditions.
Replace the length L with an “equivalent” or “effective” length Le.
L is the actual length of the beam & Le is the length for use in PCR.
PCR 
2
L2e
EI
Buckling of Slender Columns
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Example Problem
Knowing that the torsional spring at B is of constant K and
that the bar AB is rigid, determine the critical load Pcr.
Buckling of Slender Columns continued
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Example Problem
Determine the critical load of an aluminum tube that is 1.5 m
long and has a 16-mm outer diameter and a 1.25-mm wall
thickness. Use E = 70 GPa.
1.25 mm
16 mm
Buckling of Slender Columns continued
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Example Problem
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.
Buckling of Slender Columns continued
W10 x 22
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