2-Lectures LEC 13

8/12/2019 2-Lectures LEC 13

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Chapter 5

Deflection and StiffnessDeflection and Stiffness

A. Aziz Bazoune

23-Oct-07 Chapter 5 1


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OutlineOutline

1. Spring Rates2. Deflection in Tension, Compression & Torsion

3. Deflection due to Bending

4. Strain Energy5. Castiglianos Theorem

6. Statically Indeterminate Problems

7. Compression MembersLong Columns with Central Loading

Intermediate Length Columns with Central Loading



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Long Columns with Central LoadingLong Columns with Central Loading IntermediateIntermediate--Len th Columns with CentralLen th Columns with Central

LecLec. 13. 13

LoadingLoading



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55--11 Long Columns with Central Loading11 Long Columns with Central Loading

Figure 5-18

(a) Both ends rounded or pivoted


(b) Both end fixed(c) One end free(d) One end rounded and pivoted

and one end fixed


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Referring to the Figure shown,assume a bar of length loaded

by a force acting along the

centroidal axis on rounded orpinned ends.

P

L


The bar is bent in the positivedirection. This requires a negativemoment, and hence

Remember Eq. (5-12)

Py=

2

2

d yM EI

dx=

y


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Equating between the two above

equations gives

2

2 0

d y Py

dx EI

+ =


The previous equation represents asecond order ordinary differentialequation generally known forsimple harmonic motion . It can be

written as2

2

2 0

d yy

dx+ =


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where is known as the frequency of oscillations.

The general solution of the previous differential equation is

2P EI =

( ) ( )sin cosy A x B x = +


B.Cs

The first B.C. yields B=0, and the second leads to

( )

( )

0 0

0

y

y L

=

=


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The foregoing is satisfied if either or . The first of these corresponds to a condition of no buckling and yields a trivial

solution. The second case is satisfied if

or

( )sin 0A L =

0A =

( ) ( )1,2,3,L n n = =

( )sin 0L =


Solving for gives

( )1,2,3,L n nEI= =

P

( )

2 2

2 1,2,3,n EIP nL

= =


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The value of has a physical significance, as it determines thesmallest value of for which a buckled shape or mode can occur under

static loading. Therefore, the critical load for a column with pinned/rolled end is

P

( )2

2 1,2,3,

cr

EIP n

L

= =

Euler Column FormulaEuler Column Formula( )1n =


The preceding result, after L. Euler (1707-1783), is known as EulersFormula; the corresponding load is called the Euler Buckling Load.

https://ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=me&chap_sec=09.1&page=theory


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The First Three Mode ShapesThe First Three Mode Shapes



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The radius of gyration is a distance, (from a reference axis or point)

at which the entire area of a section may be considered to bedistributed for the purpose of calculating the moment of inertia. Thus

or

The critical unit load is obtained by substituting the above relation intoEuler column formula,

Ik =

2I Ak=

2P E

k


(5-43)

where is called as the slenderness ratio, that is used in classifyingcolumns according to their length. The quantity is the critical

unit load. It is the load per unit area necessary to place the column in acondition ofunstable equilibrium.

( )2A L k=

( )L k( )crP A


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The previous equations are valid only for round-round end conditions.

For different end conditions

1. Critical Load

Effect of End ConditionsEffect of End Conditions

2

crEI

P C=


(5-47)2. Critical Unit Load

where is end condition constant (Fig. 5-18 and table 5-2, userecommended values)

( )

2

2( / )

crP E

L k

C =

C


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23-Oct-07 Chapter 5 1423-Oct-07 Chapter 5 14

Figure 5-18(a) both ends rounded or pivoted; (b) both ends fixed; (c) one end free, one end fixed;(d) one end rounded and pivoted and one end fixed


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End-Condition ConstantColumn End Conditions Theoretical

ValueConservativeValue

RecommendedValue

Table 5-2 End-Condition Constants for Euler Columns(To be used with Eq. (5-47)

C


F xe -Free 1/4 1/4 1/4

Rounded-rounded 1 1 1

Fixed-rounded 2 1 1.2

Fixed-Fixed 4 1 1.2


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Eq. (5-47) can be written as

1. Critical Load

Effective LengthEffective Length

2 2

2 2( )

cr

e

EI I

C

EP

L L

= =

: is the effective

length

el


2. Critical Unit Load

( )

2 2 2

2 22

2( ) 1( / )

cr

e

P E E E

LA Lk

C

LC

k k

= = =


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Effective Lengths of Columns for Various End ConditionsEffective Lengths of Columns for Various End Conditions



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55--14 Intermediate14 Intermediate--Length Columns withLength Columns with

Central LoadingCentral Loading

Many columns formulas have been proposed over the years. The parabolicor J. B. Johnson Formula seems to be the most preferred one amongdesi ners.


The general form of the parabolic formula is

2

1

12

cr yy

SP L L LSA k E kC k

=


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When Eq. 5-47 is solved for various values of the unit load in terms of the slenderness ratio , we obtain the curvePQR shown in Figure 5-19.

crP Al k



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A column is pinned at both ends has a length of 4.50 ft and a solid circularcross-section of 1.25 in diameter. If it is made of steel AISI 1020 CD whatwould be a safe column loading?

Material AISI 1020 CD Steel = Sy = 57 ksi (Table A-20)

E = 30 Mpsi (Table A-5)

ProblemProblem

SolutionSolution


Ends: Pinned = Rounded-rounded case = C=1 (Table 5-2) Eq. (5-48)

Calculate the slenderness ratio

1/2 1/22 2 6

3

1

2 2 1 30 10101.9

57 10y

l CE

k S

= = =

l

k


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Calculate the slenderness ratio

ProblemProblem

( )l k

( )( )

( ) ( )

4 2

2

64 4

4.5 12172.8

4 1.25 / 416

ft in

in

l ll k

I A d d

l l

dd

= =

= = = =


Since == Eulers Column

Therefore

( ) ( )1l k l k >

2 2 6

2 2

30 109906

( / ) (172.8

1

)

crP EC

A L k

= = =

2 2

1.259906 9906 9906 12150

4 4lbcr

DP A = = = =


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Determine the critical load on a steel column having a rectangular cross-section, 12 mm X 18 mm and a length of 280 mm? It is proposed to use AISI1040 HR Steel. The lower end of the column is inserted into a close-fittingsocket and welded securely. The upper end is pinned.

Material AISI 1020 HR Steel

ProblemProblem

SolutionSolution


y

E = 207 GPa (Table A-5)

Fixed-rounded end Condition

C=1.2 (Table 5-2)

Radii of gyration

( )1 12xxxx

bIk

A

= =

3h

b

185.196

12 12mm

h

h= = =


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It is clear that and

On the other hand we have from Eq. 5-48

ProblemProblem

( )1 12yyyy

I hk

A

= =

3b

b h

123.464

12 12mm

b= = =

yy xxk k< ( ) ( )/ / 280 / 3.464 80.83yyl k l k

= = =


Since

Apply the formula for Johnsons column to find the critical load on thesteel column.

( )2 2 9

612 2 1.2 207 10/ 130

290 10y

CEl kS

= = =

( ) ( )1

/ / Johnson's Columnl k l k <


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ProblemProblem

2

266 6

9

1

2

290 10 1290 10 80.8 234 10

2 21 07 1.2 0

2N/m

cr y

y

SP LS

k ECA

=

= =


Therefore

( )6 6 6234 10 234 10 18 12 10 50,544 NcrP A = = =


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Leonhard Euler

Portrait by Johann Georg Brucker

BornApril 15, 1707(1707-04-15)Basel, Switzerland

DiedSeptember 18 [O.S. September 7] 1783St Petersburg, Russia

Leonhard Paul Euler (pronounced Oiler; IPA [l]) (April

15, 1707 September 18 [O.S. September 7] 1783) was apioneering Swiss mathematician and physicist, who spentmost of his life in Russia andGermany. He published more papers than any othermathematician of his time.[2]

Euler made important discoveries in fields as diverse as

calculus and graph theory. He also introduced much of themodern mathematical terminology and notation, particularlyfor mathematical analysis, such as the notion of a mathematical function.[3] He is alsorenowned for his work in mechanics, optics, and astronomy.

http://en.wikipedia.org/wiki/Leonhard_Euler


ResidencePrussiaRussiaSwitzerland

Nationality Swiss

Field Mathematics and physics

InstitutionsImperial Russian Academy of Sciences

Berlin AcademyAlma mater University of Basel

Religion Calvinist[1]

u er s cons ere o e e preem nen ma ema c an o

the 18th century and one of the greatest of all time. He isalsoone of the most prolific; his collected works fill 6080 quartovolumes.[4] A statement attributed to Pierre-Simon Laplaceexpresses Euler's influence on mathematics: "Read Euler,read Euler, he is the teacher (master) of us all".[5]

Euler was featured on the sixth series of the Swiss 10-francbanknote[6] and on numerous Swiss, German, and Russianpostage stamps. The asteroid 2002 Euler was named in hishonor. He is also commemorated by the Lutheran Church ontheir Calendar of Saints on May 24.


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