Mechanics of Materials chp11
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Transcript of Mechanics of Materials chp11
MECHANICS OF
MATERIALS
Dr. Eddy El Tabach
Assistant professor
NDU University
Barsa campus
CHAPTER
11 Buckling of
Columns
BAU
MECHANICS OF MATERIALS
11 - 2
Stability of Structures
• In the design of columns, cross-sectional area is
selected such that
- allowable stress is not exceeded
allA
P
- deformation falls within specifications
specAE
PL
• After these design calculations, may discover
that the column is unstable under loading and
that it suddenly becomes sharply curved or
buckles (lateral deflection).
MECHANICS OF MATERIALS
11 - 3
Stability of Structures
The maximum axial load that a column can support when it is on
the verge of buckling is called the critical load, Pcr.
Any additional loading will cause the column to buckle and
therefore deflect laterally.
MECHANICS OF MATERIALS
11 - 4
Euler’s Formula for Pin-Ended Beams
• Consider an axially loaded beam.
After a small perturbation, the system
reaches an equilibrium configuration
such that
02
2
2
2
yEI
P
dx
yd
yEI
P
EI
M
dx
yd
• Solution with assumed configuration
can only be obtained if
2
2
2
22
2
2
rL
E
AL
ArE
A
P
L
EIPP
cr
cr
r is the radius of gyration
MECHANICS OF MATERIALS
11 - 5
Euler’s Formula for Pin-Ended Beams
s ratioslendernesr
L
tresscritical srL
E
AL
ArE
A
P
A
P
L
EIPP
cr
crcr
cr
2
2
2
22
2
2
• The value of stress corresponding to
the critical load,
• Preceding analysis is limited to
centric loadings.
MECHANICS OF MATERIALS
11 - 6
Extension of Euler’s Formula
• A column with one fixed and one free
end, will behave as the upper-half of a
pin-connected column.
• The critical loading is calculated from
Euler’s formula,
length equivalent 2
2
2
2
2
LL
rL
E
L
EIP
e
e
cr
ecr
MECHANICS OF MATERIALS
11 - 7
Extension of Euler’s Formula
MECHANICS OF MATERIALS
11- 8
Sample Problem 10.1
An aluminum column of length L and
rectangular cross-section has a fixed end at B
and supports a centric load at A. Two smooth
and rounded fixed plates restrain end A from
moving in one of the vertical planes of
symmetry but allow it to move in the other
plane.
a) Determine the ratio a/b of the two sides of
the cross-section corresponding to the most
efficient design against buckling.
b) Design the most efficient cross-section for
the column. L = 20 in.
E = 10.1 x 106 psi
P = 5 kips
FS = 2.5
MECHANICS OF MATERIALS
11 - 9
Sample Problem 10.1
• Buckling in xy Plane:
12
7.0
1212
,
23121
2
a
L
r
L
ar
a
ab
ba
A
Ir
z
ze
zz
z
• Buckling in xz Plane:
12/
2
1212
,
23121
2
b
L
r
L
br
b
ab
ab
A
Ir
y
ye
yy
y
• Most efficient design:
2
7.0
12/
2
12
7.0
,,
b
a
b
L
a
L
r
L
r
L
y
ye
z
ze
35.0b
a
SOLUTION:
The most efficient design occurs when the
resistance to buckling is equal in both planes of
symmetry. This occurs when the slenderness
ratios are equal.
MECHANICS OF MATERIALS
11- 10
Sample Problem 10.1
L = 20 in.
E = 10.1 x 106 psi
P = 5 kips
FS = 2.5
a/b = 0.35
• Design:
2
62
2
62
2
2
cr
cr
,
6.138
psi101.10
0.35
lbs 12500
6.138
psi101.10
0.35
lbs 12500
kips 5.12kips 55.2
6.138
12
in 202
12
2
bbb
brL
E
bbA
P
PFSP
bbb
L
r
L
e
cr
cr
y
ye
in. 567.035.0
in. 620.1
ba
b