Buckling and Eccentric Load

8/10/2019 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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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.



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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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.



Introduction

Figure 1. Centric Loading

P

P


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Introduction

Figure. 2. Eccentric Loading

P

P



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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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.



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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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



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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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)



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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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)




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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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)




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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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)




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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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.




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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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)




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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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)



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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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



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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The Secant FormulaFigure 5.Load per unit area, P/A, causing yieldin column (USCustomary units)(Beer & Johnston 1992)



Figure 6.Load per unit area, P/A, causing yieldin column (SI)(Beer & Johnston 1992)


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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)



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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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.



Example 1 (contd)

e

y

x

z

3.2 m

W 250 58

P

A

BFigure 7


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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



Example 1 (contd)Figure 8

Beer andJohnston

1992


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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



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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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



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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Example 2 (contd)e

y

x

z

6 ft

W 8 28

P

A

BFigure 9



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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Example 2 (contd)Figure 10

Beer andJohnston1992



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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Figure 5.Load per unit area, P/A, causing yieldin column (USCustomary units)(Beer & Johnston 1992)15




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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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



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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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



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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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

Buckling and Eccentric Load

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