buckling and eccentric load
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A. J. Clark School of Engineering Department of Civil and Environmental Engineering A. J. Clark School of Engineering Department of Civil and Environmental Engineering A. J. Clark School of Engineering Department of Civil and Environmental Engineering A. J. Clark School of Engineering Department of Civil and Environmental Engineering A. J. Clark School of Engineering Department of Civil and Environmental Engineering
Third EditionLECTURE
2810.4Chapter
ADVANCED TOPICS IN
MECHANICS-I
byDr. Ibrahim A. Assakkaf
SPRING 2003ENES 220 Mechanics of Materials
Department of Civil and Environmental EngineeringUniversity of Maryland, College Park
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 2ENES220 Assakkaf
ADVANCED TOPICS INMECHANICS-I
1. Buckling: Eccentric Loading References:
Beer and Johnston, 1992. Mechanics ofMaterials , McGraw-Hill, Inc.
Byars and Snyder, 1975. EngineeringMechanics of Deformable Bodies , ThomasY. Crowell Company Inc.
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 3ENES220 Assakkaf
ADVANCED TOPICS INMECHANICS-I
2. Torsion of Noncircular Members andThin-Walled Hollow Shafts
References Beer and Johnston, 1992. Mechanics of
Materials, McGraw-Hill , Inc. Byars and Snyder, 1975. Engineering
Mechanics of Deformable Bodies , ThomasY. Crowell Company Inc.
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 4ENES220 Assakkaf
ADVANCED TOPICS INMECHANICS-I
3. Introduction to Plastic Moment References:
Salmon, C. G. and Johnson, J. E., 1990.Steel Structures Design and Behavior ,Chapter 10, HarperCollins Publishers Inc.
McCormac, J. C., 1989. Structural SteelDesign , Ch. 8,9, Harper & Row, Publishers,Inc.
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 5ENES220 Assakkaf
Buckling: Eccentric Loading
Introduction The Euler formula that was developed
earlier was based on the assumption thatthe concentrated compressive load P onthe column acts though the centroid of thecross section of the column (Fig. 1).
In many realistic situations, however, this isnot the case. The load P applied to acolumn is never perfectly centric.
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 6ENES220 Assakkaf
Buckling: Eccentric Loading
Introduction
Figure 1. Centric Loading
P
P
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 7ENES220 Assakkaf
Buckling: Eccentric Loading
Introduction
Figure. 2. Eccentric Loading
P
P
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 8ENES220 Assakkaf
Buckling: Eccentric Loading
Introduction In many structures and buildings,
compressive members that used ascolumns are subjected to eccentricconcentrated compressive loads (Fig. 2) aswell as moments.
The difference between a column and abeam is that in a column, the magnitude ofP is much much greater than that of abeam.
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 9ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Denoting by e the eccentricity of the load,that is, the distance between the line ofaction of P and the axis of the column, asshown in Fig. 3a, the given eccentric loadcan be replaced by a centric force P and acouple M A = Pe (Fig. 3b).
It is clear that no matter how small P and e ,
the couple M A will cause bending in thecolumn, as shown in Fig. 3c.
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 10ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula
L
P
P
L
P
P
e M A = Pe
M A = Pe
=
P M A = Pe
P
M A = Pe
ymax
A
B
A A
B B
Figure 3
(a) (b) (c)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 11ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula As the eccentric load is increased, both the
couple M A and the axial force P increase,and both cause the column to bend further.
Viewed in this way, the problem of bucklingis not a question of determining how longthe column can remain straight and stableunder an increasing load, but rather how
much it can be permitted to bend under the
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 12ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula The increasing load, if the allowable stress
is not to be exceeded and if the deflection
y max is not to become excessive. The deflection equation (elastic curve) for
this column can written and solved in amanner similar to column subjected tocentric loading P .
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 13ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula:
Drawing the free-body diagram of a portion AQof the column of Figure 3 and choosing thecoordinate axes as shown in Fig. 4, thebending moment at Q is given by
Pe Py M Py x M A ==)( (1)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 14ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula
P M A = Pe
P
M A = Pe
ymax
A
B
P M A = Pe y A
Q
M ( x)
x
x
Figure 4(a) (b)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 15ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
Recalling that the relationship between thecurvature and the moment along the column isgiven by
Therefore, combining Eqs. 1 and 2, yields
( ) x M dx
yd =2
2
EI Pe y
EI P
dx yd =2
2
(2)
(3)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 16ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
Moving the term containing y in Eq. 3 to the leftand setting
as done earlier, Eq. 3 can be written as
EI P p =2
e p y pdx
yd 222
2
=+
(4)
(5)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 17ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
The general solution of the differential equation(Eq. 5) is
Using the boundary condition y = 0, at x = 0,Eq. 6 gives
e px B px A y += cossin (6)
e B = (7)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 18ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
Using the other boundary condition at the otherend: y = 0, at x = L, Eq. 6 gives
Recalling that( ) pLe pL A cos1sin =+ (8)
2sin2cos1
and2
cos2
sin2sin
2 pL pL
pL pL pL
=
= (9)
(10)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 19ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
Substituting Eqs. 9 and 10 into Eq. 8, we obtain
And substituting for A and B into Eq. 6, theelastic curve can be obtained as
2tan
pLe A =
+= 1cossin2tan px px pL
e y
(11)
(12)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 20ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
The maximum deflection is obtained by setting x = L/2 in Eq. 12. The result is
y max becomes infinite when the argument of thesecant function in Eq. 13 equals /2. While thedeflection does not become infinite, itnevertheless becomes unacceptably large, andP would reach the critical value P cr .
= 12
secmax L
EI P
e y (13)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 21ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd): Based on that, when the argument of the
secant function equals /2, that is
Therefore, the critical load is obtained as22
= L EI P
(14)
2
2
L
EI P cr
= (15)
which is the same for column under centric loading.
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 22ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
Solving Eq. 15 for E I and substituting into Eq.13, the maximum deflection can be expressedin an alternative form as
The maximum stress occur at midspan of thecolumn (at x = L/2), and can computed from
= 12
secmaxcr P
P e y
I c M
A P max
max +=
(16)
(17)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 23ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
The moment at the midspan of the column ismaximum ( M max ) and is given by
Substituting this value of M max and theexpression for y max of Eq. 13 into Eq. 17 andnoting that I = Ar 2, the result is
( )e y P M Py M A +=+= maxmaxmax (18)
+=
2sec1 2max
L EI P
r ec
A P
(19)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 24ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Derivation of the formula (contd):
An alternative form of Eq. 19 is obtained bysubstituting for y
maxfrom Eq. 16 into Eq. 18.
Thus
Where P cr is Euler buckling load for centricloading as given by Eq. 15.
+=cr P
P r ec
A P
2sec1 2max
(20)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 27ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant FormulaThe secant formula for a column subjectedto eccentric compressive load P is given by
+
=
r L
EA P
r ec A
P
21
sec1 2
max
e = eccentricity A = area of cross section
E = modulus of elasticity r = min radius of gyrationc = distance from N . A. to extreme fiber.
= effective length for column
L`
P
P
e
L
(21)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 28ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula An alternative form for the secant formulais give by
+
=
r L
EA P
r ec
A P
21
sec1 2
max
e = eccentricity A = area of cross section E = modulus of elasticity r = min radius of gyrationc = distance from N . A. to extreme fiber.
= effective length for column
L`
P
P
e
L
(22)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 29ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula General Notes on the Secant Formula:
The formula of Eq. 21 is referred to as thesecant formula; it defines the force per unitarea, P / A, which causes a specified maximumstress max in a column of given effectiveslenderness ratio, / r , for a given value of theratio ec /r 2, where e is the eccentricity of theapplied load.
Since P or P/A appears in both sides of Eqs. 21or 22, it necessary to solve these equations by
L
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 30ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formulaby an iterative procedure (trial and error) toobtain the value of P or P/A corresponding to agiven column and loading condition.
The curves shown in Figs. 5 and 6 areconstructed for steel column using Eq. 21.
These curves make it possible to determine theload per unit area P/A , which causes thecolumn to yield for given values of the ratio / rand ec /r 2.
L
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 31ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant FormulaFigure 5.Load per unit area, P/A, causing yieldin column (USCustomary units)(Beer & Johnston 1992)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 32ENES220 Assakkaf
Buckling: Eccentric Loading
Figure 6.Load per unit area, P/A, causing yieldin column (SI)(Beer & Johnston 1992)
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 33ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant FormulaIt should be noted that for small values of
/r , the secant is almost equal to unity inEqs. 21 and 22, and P/A (or P ) may beassumed equal to
L
2
max
2
max
1 or
1
r
ec A
P
r
ec A P
+=
+= (23)
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 34ENES220 Assakkaf
Buckling: Eccentric Loading
The Secant Formula Notes on Figures 5 and 6:
For large value of / r , the curvescorresponding to the various values of the ratioec/r 2 get very close to Eulers curve, and thusthat the effect of the eccentricity of the loadingon the value of P/A becomes negligible.
The secant formula is mainly useful forintermediate values of / r .
L
L
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 35ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1The axial load P is applied at a pointlocated on the x axis at a distance e fromthe geometric axis of the W 250 58rolled-steel column AB (Fig. 7). When P =350 kN, it is observed that the horizontaldeflection of the top of the column is 5 mm.Using E = 200 GPa, determine (a) the
eccentricity e of the load, (b) the maximumstress in the column.
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 36ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd)
e
y
x
z
3.2 m
W 250 58
P
A
BFigure 7
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 37ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd) For W 250 58 (see Handout or Fig. 8):
One-end fixed, one-end free column,
Therefore
m0502.0 m105.184
m1085.0 m1073.18
m1042.7 m100.87
36
26
2326
==
==
==
y y
x y
x
r S
r I
A I
( ) m4.62.32 == L
( )( )( ) kN6.9024.6 1073.1810200 266
2
2
=
==
L
EI P ycr
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 38ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd)Figure 8
Beer andJohnston
1992
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 39ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd) (a) From Eq. 16,
( )[ ]
mm6.33m00633.0 1)9782.0cos(
1005.0
19782.0sec16.902
3502
sec005.0
12
secmax
===
==
=
ee
ee
P P
e ycr
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 40ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd) (b) From Eq. 20,
( )
MPa68.62 N/m044,615,68
6.902350
2sec
0503.0
102
203)33.6(
11042.710350
2sec1
2
2
3
3
3
2max
==
+=
=+=
cr P P
r ec
A P
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 41ENES220 Assakkaf
Buckling: Eccentric Loading
Example 1 (contd) An alternate solution for Part (b):
P
e
ym
P M
[ ]
MPa67.68105.184
966.3
1042.7
350
m-kN966.300633.0005.0350)(
10350
mm33.6
mm5
63
3
=
+
=+=
=+=+==
==
y
m
m
m
S
M
A
P
e y P M
N P
e
y
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 42ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 An axial load P is applied at a point locatedon the x axis at a distance e = 0.60 in. fromthe geometric axis of the W8 28 rolled-
steel column AC (Fig. 9). Knowing that thecolumn is free at its top B and fixed at itsbase A and that y = 36 ksi and E = 29 10 6 psi, determine the allowable load P if afactor of safety of 2.5 with respect to yieldis required.
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 43ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd)e
y
x
z
6 ft
W 8 28
P
A
BFigure 9
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 44ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) For W 8 28 (see Handout or Fig. 10):
One-end fixed, one-end free column,
Therefore
in62.1 3.2675in2535.6
in25.8 in7.21 22
===
==
y
y
r c
A I
( ) ft1262 == L
( )( )
747.062.1
2675.36.0 and 89.88
62.11212
22 ====
r ec
r L
y
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 45ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd)Figure 10
Beer andJohnston1992
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 46ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) Since y = 36 ksi, and E = 29 10 6 ksi, Fig. 5
can be used:We read
Thus
kips75.123)25.8(1515e, therefor ;15 === A P A P
kips505.495.275.123
FSallowable === P P
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 47ENES220 Assakkaf
Figure 5.Load per unit area, P/A, causing yieldin column (USCustomary units)(Beer & Johnston 1992)15
Buckling: Eccentric Loading
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 48ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) General Iterative Procedure:
Suppose that we do not have the curves provided in Fig. 5, or we do have thecurves but our problem consists of acolumn that has different material (e.g.,
y = 50 ksi), how can we evaluate theeccentric load P for Example 2?
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 49ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) A general trial and error (iterative)procedure can be used as follows:
Using Eqs. 20 or 22, we assume an initial(guess) value for P in the right-hand side of theequation; let it be 20 kips, hence
( )
( )( )
kips80.163
89.8825.81029
2021
cos
1747.01
25.836
21
sec1
3
2
max =
+
=
+
=
r L
EA P
r ec
A P
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 50ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd)The revised value P = 163.80 kips can nowbe substituted in the right-hand side of thesame equation to produce yet anotherrevised value as follows:
( )
( )( )
kips01.103
89.8825.81029
80.16321
cos
1747.01
25.836
21
sec1
3
2
max =
+
=
+
=
r L
EA P
r ec
A P
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 51ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) A third iteration using a revised value for P = 103.01 kips, gives
( )
( )( )
kips79.132
89.8825.81029
01.10321
cos
1747.01
25.836
21
sec1
3
2
max =
+
=
+
=
r L
EA P
r ec
A P
LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 52ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd) The iterative procedure is continued until
the value of the eccentric load P convergesto the exact solution of 123.53 kips, asshown in the spreadsheet (Excel) result ofTable 1.
Therefore,
kips504.495.253.123
FSallowable === P P
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LECTURE 28. ADVANCED TOPICS IN MECHANICS-I (BUCKLING-ECCENTRIC LOADING) Slide No. 53ENES220 Assakkaf
Buckling: Eccentric Loading
Example 2 (contd)
P (kip) 20.00 123.63
163.79 123.48103.01 123.55132.79 123.52119.10 123.53125.59 123.53
122.56 123.53123.98 123.53123.31 123.53
+
=
r
L
EA
P
r
ec
A P
2
1sec1 2
max
Initial Value of P
Table 1. Spreadsheet Result
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