11. column stability supports
Post on 17-Feb-2018
217 Views
Preview:
TRANSCRIPT
-
7/23/2019 11. Column Stability Supports
1/46
COLUMN & SUPPORT
COLUMN STABILITYAND
SUPPORT
Ahmad Shufni Othman
July 2012
-
7/23/2019 11. Column Stability Supports
2/46
COLUMN & SUPPORT
LEARNING OUTCOMES
Explaining the following :
Column and support
Type of column failure
Type of column ends and support
Understanding the Euler Theorem and buckling load
(crippling load) formulae.
Determining the effective length and buckling load for a
slender column with different types of supports.
-
7/23/2019 11. Column Stability Supports
3/46
COLUMN & SUPPORT
COLUMNS (or struts)
Columns are vertical member of a structure, which issubjected to axial compressive loads. Struts is a member of
a structure which is not vertical with one or both ends
hinged or pinned jointed.
The cross-sectional dimensions are relatively small
compared with their length in the direction of thecompressive force.
They are also referred to as pillars, posts, and stanchions.
-
7/23/2019 11. Column Stability Supports
4/46
COLUMN & SUPPORT
Columns fail in compression. In civil
engineering they are often made of brittlematerial which is strong in compression such
as cast iron, stone and concrete. These
materials are weak in tension so it is important
to ensure that bending does not produce
tensile stresses in them. If the compressivestress is too big, they fail by crumbling and
cracking.
COLUMNS (or struts)
-
7/23/2019 11. Column Stability Supports
5/46
COLUMN & SUPPORT
TYPE OF COLUMNS FAILURE
(crushing)
All short columnsfail due to crushing
whereas long
columns fail due to
buckling and
crushing.
http://www.world-housing.net/uploads/101123_111_25.jpg -
7/23/2019 11. Column Stability Supports
6/46
COLUMN & SUPPORT
FORMATIONS OF SHORT COLUMNS
-
7/23/2019 11. Column Stability Supports
7/46
COLUMN & SUPPORT
Euler's buckling equation
What is buckling?
In engineering, buckling is a
failure mode characterized by a
sudden failure of a structuralmember subjected to high
compressive stresses, where
the actual compressive stress at
the point of failure is less than
the ultimate compressivestresses that the material is
capable of withstanding.A column under a concentric
axial load exhibiting the
characteristic deformation of
buckling
http://en.wikipedia.org/wiki/File:Buckled_column.svg -
7/23/2019 11. Column Stability Supports
8/46
COLUMN & SUPPORT
Buckling in columns
If the load on a column is applied through thecenter of gravity of its cross section, it is called
an axial load. A load at any other point in the
cross section is known as an eccentric load.
A short column under the action of an axial
load will fail by direct compression before itbuckles, but a long column loaded in the same
manner will fail by buckling (bending), the
buckling effect being so large that the effect of
the direct load may be neglected.
The intermediate-length column will fail by acombination of direct compressive stress and
bending.
eccentric load
axial load
The load at which the column just buckles is
known as buckling load or critical load or
crippling load.
-
7/23/2019 11. Column Stability Supports
9/46
COLUMN & SUPPORT
1757, mathematician Leonhard Euler (pronounced Oiler) derived
a formula that gives the maximum axial load that a long, slender,
ideal column can carry without buckling.
An ideal column is one that is perfectly straight, homogeneous,
and free from initial stress.The maximum load, sometimes called the critical load, causes
the column to be in a state of unstable equilibrium; that is, any
increase in the load, or the introduction of the slightest lateral
force, will cause the column to fail by buckling.
The formula derived by Euler for columns with no considerationfor lateral forces is given below. However, if lateral forces are
taken into consideration the value of critical load remains
approximately same.
Euler's Buckling Equation
-
7/23/2019 11. Column Stability Supports
10/46
COLUMN & SUPPORT
Euler's Buckling Equation
The following assumptions are made in the Eulers column theory;
1. Initially the column is perfectly straight, and the load applied is
truly axial.
2. The cross-section of the column is uniform throughout its length.
3. The column material is perfectly elastic, homogeneous and
isotropic, and thus obeys Hookes law.
4. The length of column is very large as compare to its cross-
sectional dimensions.
5. The shortening of column, due to direct compression (being
very small) is neglected.
6. The failure of column occurs due to buckling alone.
-
7/23/2019 11. Column Stability Supports
11/46
COLUMN & SUPPORT
Types of end cond i t ions o f columns
Euler's Buckling Equation
In actual practice, there are number of and conditions, for
column. For the Eulers column theory, there are four types of
end condition.
Both end
FixedBoth end
Pinned
(Hinged)
One end Fixed,
one end Pinned
One end Fixed,
one end Free
-
7/23/2019 11. Column Stability Supports
12/46
COLUMN & SUPPORT
http://localhost/var/www/apps/conversion/tmp/scratch_1//upload.wikimedia.org/wikipedia/commons/6/69/Buckledmodel.JPG -
7/23/2019 11. Column Stability Supports
13/46
COLUMN & SUPPORT
Sign convent ions
Euler's Buckling Equation
1. A moment, which tends to
bend the column with
convexity towards its initial
central line, as shown in figure(a), is taken as positive.
2. A moment, which tends to
bend the column with itsconcavity towards its initial
central line, as shown in figure
(b), is taken as negative.
Figure (a)
Figure (b)
-
7/23/2019 11. Column Stability Supports
14/46
COLUMN & SUPPORT
Derivation of Eulers Equation
Eulers Buckling Load for Columns or Struts
In 1757, Leonhard Euler developed a
relationship for the critical column
load which would produce buckling.
A very brief derivation of Euler's
equation goes as follows:
M
x
y
A loaded pinned-pinned column is
shown in the diagram. A top section of
the diagram is shown with the bending
moment indicated. In terms of theload P, and the lateral deflection y, we
can write an expression for the
bending moment M:
)1.......()( EqxPyM
-
7/23/2019 11. Column Stability Supports
15/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
We can also state that for beams and columns, the bending moment isproportional to the curvature of the beam, which, for small deflection
can be expressed as:
Where E = Young's modulus and I = moment of Inertia. Substituting
Equation 1 into Equation 2, we obtain the following differential
equation:
or
This is a second order differential equation which has the general
solution of:
)2.......(2
2
Eqdx
yd
EI
M
yEI
P
dx
yd
2
2
)3.......(02
2
Eqy
EI
P
dx
yd
)4.......(cossin
)sin()cos(
EqxEI
PBx
EI
PAy
cxBcxAy
-
7/23/2019 11. Column Stability Supports
16/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
We next apply boundary conditions: at x = 0, y = 0 and at x = L, y = 0.That is, the deflection of the column must be zero at each end since it
is pinned.
Applying the first boundary condition, it is noted that B must be zero
since cos (0) = 1. The second boundary condition implies that either A
must be zero (which leaves us with no equation at all) or that:
Noting that sin ( ) = 0, we can solve for P:
where Pcrstands for the critical load (or buckling load) at which the
column is predicted to buckle.
)5.......(0sin EqLEI
P
LEI
P
)6......(2
2
EqL
EIPcr
-
7/23/2019 11. Column Stability Supports
17/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
Therefore, Eulers Buckling Load is;
2
2
Le
EIPe
Where; E = Modulus of Elasticity of Column Material.
I = Minimum Moment of Inertia of Column Section.
Le = Equivalent length of Column or Effective Length
-
7/23/2019 11. Column Stability Supports
18/46
COLUMN & SUPPORT
Le= Equivalent length of Column or effective length(is defined as the distance between successive inflection
points or points of zero moment)
Eulers Buckling Load for Columns or Struts
-
7/23/2019 11. Column Stability Supports
19/46
COLUMN & SUPPORT
Moment of Inertia
OR
2nd. Moment Area
Eulers Buckling Load for Columns or Struts
axis
-
7/23/2019 11. Column Stability Supports
20/46
COLUMN & SUPPORT
Let us consider a column with both end hinged.
It is subjected to axial compressive load P.
Eulers Buckling Load for Columns or Struts
CASE 1 - Both end Hinged (pinned).
Let, d= diameter of the column
L = length of the column
-
7/23/2019 11. Column Stability Supports
21/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
CASE 1 - Both end Pinned (Hinged).
Let, d= diameter of the column
L = length of the column
Since Both Ends are Pinned, Le = L
Column material has Modulus of Elasticity = E
Moment of Inertia of Circular Section,
Due to axial compressive load P, the column will buckle
Now, at a distance of x from Point B.
Let us take a section x-x
Let the lateral deflection at section x-x be y.
4
64dI
Now, Eulers Formula will be,2
2
L
EIPe
-
7/23/2019 11. Column Stability Supports
22/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
CASE 2 With both end Fixed.
Since Both Ends are Fixed,
42
2
2
2
2
2
2
L
EI
L
EI
Le
EIPe
2
LLe
2
24
L
EIPe
&
-
7/23/2019 11. Column Stability Supports
23/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
CASE 3 With one end Fixed & one end Hinged (pinned).
When one end Fixed and one end Hinged,2
LLe
22
2
2
2
2
2
2
L
EI
L
EI
Le
EIPe
2
22
L
EIPe
CO & S O
-
7/23/2019 11. Column Stability Supports
24/46
COLUMN & SUPPORT
Eulers Buckling Load for Columns or Struts
CASE 4 With one end Fixed & one end Free.
When one end Fixed and one end Free, LLe 2
2
2
2
2
2
2
42 L
EI
L
EI
Le
EIPe
2
2
4L
EIPe
A column with one fixed and one free
end (a), will behave as the upper-half of
a pin-connected (hinged) column (b).
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
25/46
COLUMN & SUPPORT
Summary
Eulers Buckling Load for Columns or Struts
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
26/46
COLUMN & SUPPORT
The Buckling Stress, cr, is the Euler Buckling Load divided bythe columns cross-sectional area:
2
2
ALe
EI
A
Pecr
Buckling Stress (Crippling Stress)
Eulers Buckling Load for Columns or Struts
Radius of Gyration
A
Ir
If all of the cross-sectional area of a column were massed a distance r away
from the neutral axis, the lumped-area cross-section would have the same
Moment of Inertia as the real cross-section. ris termed the RADIUS OF
GYRATION and is given by:
The Moment of Inertia I, can be expressed in terms of Radius of Gyration, r
as:2ArI
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
27/46
COLUMN & SUPPORT
2
ArI
Buckling or Crippling Stress in terms of Effective Length and Radius ofGyration.
Eulers Buckling Load for Columns or Struts
2
2
ALe
EI
A
Pecr
2
2
2
22
2
22 )(
rL
E
L
Er
AL
ArE
)( ratiosslendernessr
Lwhere
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
28/46
COLUMN & SUPPORT
Slenderness Ratio
r
Ls
SLENDERNESS RATIO is a measure of how long the column is
compared to its cross-section's effective width (resistance to bending
or buckling). The Slenderness Ratio, s, is simply the column's Length
divided by the Radius of Gyration.
2
22
2
2
2
s
E
Le
rE
Le
EIPe
Eulers Buckling Load for Columns or Struts
Applying the Slenderness Ratio and the
Radius of Gyration reduces the Euler BucklingFormula to:
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
29/46
COLUMN & SUPPORT
EXAMPLE 1
A steel rod 5 m long and 4 cm d iameter is us ed as a colum n, wi th one end
fixed and the other free. Determine the crippling load by Eulers formula.Take E as 2.0 x 106 kg/cm2
Solution
Given. Length, L = 5 m = 500 cm
Dia. Of column, d= 4 cm
Moment of Inertia,
444
446464 cmdI
Crippling load (critical load or buckling load),
Since the column is fixed at one end and free at the other, therefore
equivalent length of the column, Le= 2L = 2 x 500 = 1000 cm
kgLe
EIPe 1.248
1000
4100.22
62
2
2
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
30/46
COLUMN & SUPPORT
EXAMPLE 2
A hol low al loy tube 5 m long w ith diameters 4 cm and 2.5 cm respect ive ly
was fo und to extend 6.4 mm under a tens i le load of 60 kN. Find th ebuc kl ing load for the tube, when used as a strut wi th both ends pinn ed.
Als o f ind th e safe load on the tube, taking factor o f safety as 4.
Solution
Given. Length, L = 5 m
Outer diameter, D= 4 cm = 0.04 m
Moment of Inertia,474444
1006.1)025.004.0(64
)(64
mdDI
Therefore the Area,
Inner diameter, d= 2.5 cm = 0.025 m
242222
1066.7)025.004.0(4)(4 mdDA
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
31/46
COLUMN & SUPPORT
EXAMPLE 2 : Solution (cont.)
Deflection, = 6.4 mm = 6.4 X 10-3
m
Buckling Load (Crippling Load or Critical Load),
Since the column is pinned at its both end, therefore equivalent length of
the column, Le= L = 5 m
kNLe
EIPe 56.2
5
)1006.1()1012.6(2
772
2
2
Tensile load, P = 60 kN
And Youngs Modulus,
l
27
34 /1012.6)104.6()1066.7(
560mkN
lA
PL
Ll
APE
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
32/46
COLUMN & SUPPORT
EXAMPLE 2 : Solution (cont.)
We know that the safe load on the tube,
kN64.04
56.2
SafetyofFactor
LoadBucklingLoadSafe
Safe Load
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
33/46
COLUMN & SUPPORT
EXAMPLE 3
An I sect ion jo ist 40 cm x 20 cm x 2 cm and 6 m long is u sed as a strut
with both ends fixed. What is Eulers crippling load for the column?. TakeYoungs modulus for the joist as 2.0 x 106 N/cm2
Solution
Given; Length, L = 6 m
Outer width, b= 20 cm
4333
11
3c67.36682)36184020(
12
1)(
12
1mdbbdI
XX
Therefore, Moment of Inertia about X-X;
Inner depth, d1= 40 2x2 = 36 cm
Inner width, b1= 20 2 = 18 cm
Outer depth, d= 40 cm 40 cm
20 cm
2 cm
2 cm
2 cm
xx
y
y
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
34/46
COLUMN & SUPPORT
EXAMPLE 3 : Solution (cont.)
433c67.26902467.2666)236(
12
1)202(
12
12 mI
YY
And ;
Since IYY is less than IXX, therefore the joist will tend to buckle along Y-Y
axis. Thus we shall take the value of Ias IYY= 2690.67 cm4
Since the column is fixed at its both ends, therefore equivalent length of
the column,cm300
2
600
2
LLe
Crippling Load,
kNNLe
EIPe 5909.590129
300
)67.2690()102(2
62
2
2
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
35/46
COLUMN & SUPPORT
EXAMPLE 4
A T-sect ion 150 mm x 120 mm x 20 mm is used as a strut of 4 m lon g w ith
hinged at its both ends. Calculate the crippling load, if Youngs modulusfor t he material be 2.0 x 106 N/cm2
Solution
Given; Length, L = 4 m
First, determine the c.g. of the section; 20 mm
20 mm
xx
y
y
120 mm
150 mm
Flange; a1= 150 X 20 = 3000 mm2
y1= 20 / 2 = 10 mm
Web; a2= 100 X 20 = 2000 mm2
y2
= 20 + 100 / 2 = 70 mm
y26/102 cmNE
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
36/46
COLUMN & SUPPORT
2sec
23
1sec
23
1212 tiontionXX ay
bday
bdI
EXAMPLE 4 : Solution (cont.)
And,
We know that the moment of inertia the section about X-X axis,
mmaa
yaya
y 3420003000
)70(2000)10(3000
21
2211
442
3
2
3
1060936200012
10020243000
12
20150mmI
XX
44
33
1056912
20100
12
15020mmI
YY
Since IYYis less than IXX, therefore the column will buckle along Y-Y axis,
Thus we shall take the value ofIasIYY
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
37/46
COLUMN & SUPPORT
EXAMPLE 4 : Solution (cont.)
The column is pinned at its both ends, therefore the equivalent length ofthe column, Le = L = 4 m = 4000 mm
24
2
6
26/102
1010
1020/102 mmN
mm
NcmNE
Youngs Modulus,
Crippling Load (or Buckling Load),
kNNLe
EIPe 2.700197.67
4000
10569)102(2
442
2
2
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
38/46
COLUMN & SUPPORT
EXAMPLE 5
If an al lowable strain for a column are 0.00035 and the min imum radius o f
gyrat ion are 50 mm before the co lumn beg in to buck le . Determine themaximum length for that colum n, that are f ixed at one end and free on th e
other.
Since the column is fixed at one end and free at the other, therefore
equivalent length of the column, Le= 2L
Solution
Buckling Load (or Crippling Load),2
2
2
2
2
2
4)2( L
EI
L
EI
Le
EIPe
We know that,
2
22
2
22
2
2
44.
)(
4.
L
Er
LA
ArE
LA
EI
A
Pecr
2
where,
ArI
A
Ir
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
39/46
COLUMN & SUPPORT
And we know that, Youngs Modulus,
cr
crE
2
22
4,previously
L
Ercr
2
22
2
22
4
4.
L
r
LE
Er
E
cr
cr
mmL
L
L
4196
00035.04
50
45000035.0
22
2
2
22
EXAMPLE 5 : Solution (cont.)
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
40/46
COLUMN & SUPPORT
EXAMPLE 6
A squ are steel sect ion , 38 mm X 50 mm, f ixed at both end and loaded
axially. If the buc klin g streng th = 23.9 N/mm2and E = 21.7 kN/mm2 .Determine the minim um length for that sect ion.
45
33
1029.212
3850
12 mm
bd
I
Solution
Secon d m oment area,
Column is fixed at both ends, therefore,2
LLe
And, the Bu ckl ing Lo ad,
22
2
2
2
2 4
2 L
EI
L
EI
Le
EIPe
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
41/46
COLUMN & SUPPORT
Buc kl ing Strength,
2
532
)3850(
)1029.2)(107.21(49.23
L
EXAMPLE 6 : Solution (cont.)
2
24
AL
EI
A
Pecr
mmL
L
2077
9.23)3850(
)1029.2)(107.21(4 5322
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
42/46
COLUMN & SUPPORT
CLASS ASSIGNMENT
QUESTION 1
A colum n of t imb er sect ion, 15 cm X 20 cm is 6 m lo ng w ith both ends
being fixed. If the Youngs modulus for timber = 17.5 kN/mm2, determ ine;
i . Buck ling load .
i i . Safe load for the co lumn i f fac tor of safety = 3.
A sol id ro und bar 3 m lon g and 5 cm in diameter is u sed as a strut . If the
Youngs modulus for the strut, 2.0 x 105 N/mm2 , determine the cr ippl in g
(or bu ckl ing) load, when the given s trut is used w ith the fo l low ing
cond i t ions ;
i . B oth of th e en ds h in ged .
i i . One end of the stru t is f ixed and the other end is f ree.i i i. Both the ends of stru t are f ixed.
iv . One end is f ixed and the other end is h inged.
QUESTION 2
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
43/46
COLUMN & SUPPORT
CLASS ASSIGNMENT (cont.)
QUESTION 3a) A ho l low cy l indr ical cast i ron co lumn wi th both ends f ixed has to
carry a safe load of 250 kN with a factor of safety of 5. Determine the
m inim um diameter of the co lumn. Take the internal diameter as 0.8
t imes the external diameter and the maximum crus hin g stress as 550
N/mm2.
QUESTION 4
A simply sup por ted beam of length 4 m is sub jected to a uni form lydistr ib uted load of 30 kN/m over the wh ole span and def lects 15 mm at the
centre. Determine the buc kl ing loads w hen th is b eam is us ed as a column
wi th the fo l lowing cond i t ions;
i . One end i s fi xed and the o ther end i s h inged .
i i. Bo th the ends pin join ted .
b) If an al lowable stra in are 0.00055. Determine the maximum length for
the co lumn .
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
44/46
COLUMN & SUPPORT
QUESTION 5Determine the rat io of buck l ing strengths of two c olum ns, one hol low and
the other so l id. Both are made of th e same mater ial and have the same
length, cros s-sect ional area and end con dit ion s. The internal diameter of
ho l low c olumn is half of i ts external diameter.
CLASS ASSIGNMENT (cont.)
A sol id round bar 4 m long and 5 cm in diameter was found to extend
4.6 mm under a tensi le load of 50 kN. This bar is used as a strut wi th
both ends pinned. Determine the buc kl ing load for the bar and also the
safe load taking a factor of safety of 4.
QUESTION 6
COLUMN & SUPPORT
-
7/23/2019 11. Column Stability Supports
45/46
COLUMN & SUPPORT
CLASS ASSIGNMENT (cont.)
QUESTION 7Determine Eulers crippling load for an I-
sect ion jo ist 40 cm X 20 cm X 1 cm and 5
m long wh ich is used as a st ru t wi th both
ends fixed. Take Youngs modulus for the
jo is t as 2.1 X 105N/mm2.
20 cm
1 cm
1 cm
1 cm
40 cm
Calculate the Eulers critical load for a
st rut of T-sect ion , the f lange width
being 10 cm , overall depth 8 cm and
both f lange and stem 1 cm thick. The
strut is 3 m lo ng and is bui l t -in at both
ends. Take Youngs modulus for the
strut as 2 x 105 N/mm2 .
QUESTION 8
1 cm
1 cm
8 cm
10 cm
COLUMN & SUPPORT
http://www.zazzle.com/words_of_wisdom_poster_learning_collage_print-228074748364917573?print_width=23&print_height=28.4762&rf=238145465983783136 -
7/23/2019 11. Column Stability Supports
46/46
COLUMN & SUPPORT
It is the resu lt of,
preparat ion,
learn ing from fai lure.
and
http://www.zazzle.com/words_of_wisdom_poster_learning_collage_print-228074748364917573?print_width=23&print_height=28.4762&rf=238145465983783136
top related