9789810682627 ppt 08 [compatibility mode]

8/6/2019 9789810682627 Ppt 08 [Compatibility Mode]

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Cha ter 8:Cha ter 8:

Structural Analysis 7Structural Analysis 7thth Edition in SI UnitsEdition in SI UnitsRussell C. HibbelerRussell C. Hibbeler

DeflectionsDeflections

Deflection diagrams & the elastic curveDeflection diagrams & the elastic curve

Deflections of structures can come from loads,Deflections of structures can come from loads,temperature, fabrication errors or settlementtemperature, fabrication errors or settlement

In designs, deflections must be limited in order toIn designs, deflections must be limited in order toprevent cracking of attached brittle materialsprevent cracking of attached brittle materials

A structure must not vibrate or deflect severely forA structure must not vibrate or deflect severely forthe comfort of occupantsthe comfort of occupants

e ec ons a spec e po n s mus e e erm nee ec ons a spec e po n s mus e e erm neif one is to analyze statically indeterminateif one is to analyze statically indeterminatestructuresstructures


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In this topic, only linear elastic material responseIn this topic, only linear elastic material responseis consideredis considered

This means a structure subjected to load willThis means a structure subjected to load willreturn to its original undeformed position after thereturn to its original undeformed position after theload is removedload is removed

It is useful to sketch the shape of the structureIt is useful to sketch the shape of the structure

computed results & to partially check the resultscomputed results & to partially check the results


This deflection diagram rep the elasticThis deflection diagram rep the elasticcurve for the points at the centroidscurve for the points at the centroidsof the crossof the cross--sectional areas along eachsectional areas along eachof the membersof the members

If the elastic curve seems difficult toIf the elastic curve seems difficult toestablish, it is suggested that theestablish, it is suggested that themoment dia ram be drawn firstmoment dia ram be drawn first

From there, the curve can be constructedFrom there, the curve can be constructed


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Due to pinDue to pin--andand--roller support, the disp at A & Droller support, the disp at A & Dmust be zeromust be zero

Within the region ofWithin the region ofve moment,ve moment,

the elastic curve isthe elastic curve isconcave downwardconcave downward

Within the region of +ve moment,Within the region of +ve moment,

There must be an inflection pointThere must be an inflection pointwhere the curve changes fromwhere the curve changes fromconcave down to concave upconcave down to concave up

Draw the deflected shape of each of the beams.

Example 8.1Example 8.1


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In (a), the roller at A allows free rotation with no deflection whilethe fixed wall at B prevents both rotation & deflection. The

SolutionSolution

.

In (b), no rotation or deflection occur at A & B

In (c), the couple moment will rotate end A. This will causedeflections at both ends of the beam since no deflection isossible at B & C. Notice that se ment CD remains un-deformed

since no internal load acts within.

In (d), the pin at B allows rotation, so the slope of the deflectioncurve will suddenly change at this point while the beam is

SolutionSolution

.

In (e), the compound beam deflects as shown. The slope changesabruptly on each side of B.

In (f), span BC will deflect concave upwards due to load. Since

the beam is continuous, the end spans will deflect concavedownwards.


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Elastic Beam theoryElastic Beam theory

To derive the DE, we look at an initially straight beamTo derive the DE, we look at an initially straight beamthat is elastically deformed by loads appliedthat is elastically deformed by loads appliedperpendicular to beamperpendicular to beams xs x--axis & lying in xaxis & lying in x--v plane ofv plane ofsymmetrysymmetry

Due to loading, the beamDue to loading, the beamdeforms under shear & bendingdeforms under shear & bending

If beam L >> d, greatestIf beam L >> d, greatestdeformation will be caused by bendingdeformation will be caused by bending

When M deforms, the angleWhen M deforms, the anglebetween the cross sections becomes dbetween the cross sections becomes d


The arc dx that rep a portion of the elastic curveThe arc dx that rep a portion of the elastic curveintersects the neutral axisintersects the neutral axis

The radius of curvature for this arc is defined asThe radius of curvature for this arc is defined asthe distance,the distance, , which is measured from ctr of, which is measured from ctr ofcurvature Ocurvature O to dxto dx

Any arc on the element other than dx is subjectedAny arc on the element other than dx is subjected

The strain in arc ds located at position y from theThe strain in arc ds located at position y from theneutral axis isneutral axis is

dsdsds /)'(


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dydsddxds )('and

If the material is homogeneous & behaves in aIf the material is homogeneous & behaves in alinear manner, then Hookelinear manner, then Hookes law appliess law applies

E/

yd

y

The flexure formula also appliesThe flexure formula also applies

IMy/


Combining these eqns, we have:Combining these eqns, we have:

elasticityofmodulussmaterial'the

determinedbetoisepoint wherat thebeamin themomentinternal

curveelasticon thepointspecificaatcurvatureofradiusthe

1

E

M

EI

M

axisneutralabout thecomputedinertiaofmomentsbeam'theI


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rigidity;flexural ddxEI

2/32

22

/1

/1,veasaxisv

dxdv

dxvd

dxEI

Md

2/32

22

])/(1[/Therefore,

dxdvdxvd

EIM


This eqn rep a nonThis eqn rep a non--linear second DElinear second DE

==

The slope of the elastic curve for most structuresThe slope of the elastic curve for most structuresis very smallis very small

Using small deflection theory, we assume dv/dx ~Using small deflection theory, we assume dv/dx ~00

EI

M

dx

vd

2

2


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By assuming dv/By assuming dv/dxdx ~ 0~ 0 dsds, it will approximately, it will approximatelyequal toequal to dxdx

This implies that points on the elastic curve willThis implies that points on the elastic curve willonly be displaced vertically & not horizontallyonly be displaced vertically & not horizontally

2 2 2 21 ( / )ds dx dv dv dx dx dx

The double integration methodThe double integration method

M = f(x), successive integration ofM = f(x), successive integration of eqneqn 8.4 will8.4 willyield the beamyield the beams slopes slope

tantan = dv/= dv/dxdx == M/EIM/EI dxdx

EqnEqn of elastic curveof elastic curve

V = f(x) =V = f(x) = M/EIM/EI dxdx

The internal moment in regions AB, BC & CD mustThe internal moment in regions AB, BC & CD must

e wr en n erms o xe wr en n erms o x11, x, x22 an xan x33


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Once these functions are integrated & theOnce these functions are integrated & theconstants determined, the functions will give theconstants determined, the functions will give theslope & deflection for each region of the beamslope & deflection for each region of the beam

It is important to use the proper sign for M asIt is important to use the proper sign for M asestablished by the sign convention used inestablished by the sign convention used inderivationderivation

, , ,, , ,will be measured counterclockwise from the xwill be measured counterclockwise from the x--axisaxis


The constants of integration are determined byThe constants of integration are determined byevaluating the functions for slope or displacementevaluating the functions for slope or displacementat a particular point on the beam where the valueat a particular point on the beam where the valueof the function is knownof the function is known

These values are called boundary conditionsThese values are called boundary conditions

Here xHere x11 and xand x22 coordinates are valid within thecoordinates are valid within the


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Once the functions for the slope & deflections areOnce the functions for the slope & deflections areobtained, they must give the same values for slopeobtained, they must give the same values for slope& deflection at point B& deflection at point B

This is so as for the beam to be physicallyThis is so as for the beam to be physicallycontinuouscontinuous

The cantilevered beam is subjected to a couple moment Mo at itsend. Determine the eqn of the elastic curve. EI is constant.

Example 8.1Example 8.1


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By inspection, the internal moment can be representedthroughout the beam using a single x coordinate. From the free-

SolutionSolution

, ,

Integrating twice yields:

o

2

2

Mdx

vdEI o

21

2

1

2 CxC

xMvEI

CxMdx

vEI

o

o

Using boundary conditions, dv/dx = 0 at x = 0 & v = 0 at x = 0then C1 = C2 =0.

SolutionSolution

Substituting these values into earlier eqns, we get:

EI

xMv

EI

xMdxdv

oo

2;

/with2

Max slope & disp occur at A (x = L) for which

EI

LMv

EI

LM oA

oA

2;

2


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The +ve result for A indicates counterclockwise rotation & the+ve result for vA indicates that it is upwards.

SolutionSolution

In order to obtain some idea to the actual magnitude of the slope.

Consider the beam to:

Have a length of 3.6m

Be made of steel having Est = 200GPa

If this beam were designed w/o a FOS by assuming the allowablenormal stress = yield stress = 250kNm/2.

SolutionSolution

A W6 x 9 would be found to be adequate

radmmkN

mkNmA 0529.0

)10)(10(8.6][/)10(200[

)6.3(204126

2

6

mmmmkN

mkNmvA 3.95

)10)(10(8.6][/)10(200[2

)6.3(204126

2

6

2


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Since 2A= 0.00280(10-6)


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Moment-Area theorems

The change dThe change d in the slope of the tangent onin the slope of the tangent oneither side of the elementeither side of the element dxdx = the lighter shade= the lighter shadearea under the M/EI diagramarea under the M/EI diagram

Integrating from point A on the elastic curve toIntegrating from point A on the elastic curve topoint B, we havepoint B, we have

dxMB

AB /

ThisThis eqneqn forms the basis for the first momentforms the basis for the first moment--areaareatheoremtheorem


Theorem 1Theorem 1

The chan e in slo e between an 2 oints on theThe chan e in slo e between an 2 oints on theelastic curve equals the area of the M/EI diagramelastic curve equals the area of the M/EI diagrambetween the 2 pointsbetween the 2 points

The second momentThe second moment--area theorem is based on thearea theorem is based on therelative deviation of tangents to the elastic curverelative deviation of tangents to the elastic curve

..of the vertical deviation dt of the tangents on eachof the vertical deviation dt of the tangents on eachside of the differential element,side of the differential element, dxdx


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Since slope of elastic curve & its deflection areSince slope of elastic curve & its deflection areassumed to be very small, it is satisfactory toassumed to be very small, it is satisfactory toapproximate the length of each tangent line by x &approximate the length of each tangent line by x &the arcthe arc dsds by dtby dt

Using s = rUsing s = r dt =dt = xdxd

UsingUsing eqneqn 8.2, d8.2, d = (M/EI)= (M/EI) dxdx

e ver ca ev a on o e angen ae ver ca ev a on o e angen a wrwr ee

tangent at B can be found by integrationtangent at B can be found by integrationdx

EI

Mxt

B

ABA /


Centroid of an areaCentroid of an area

dAdA

/

dxEIM

xt

x

B

ABA

B.&AbetweenareatheofcentroidthetoA

throughaxisverticalthefromdistancex


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Theorem 2Theorem 2

The vertical deviation of the tan ent at a oint AThe vertical deviation of the tan ent at a oint Aon the elastic curveon the elastic curve wrtwrt the tangent extended fromthe tangent extended fromanother point (B) equals theanother point (B) equals the momentmoment of the areaof the areaunder the M/EI diagram between the 2 points (A &under the M/EI diagram between the 2 points (A &B)B)

This moment is computed about point A where theThis moment is computed about point A where thedeviation is to be determineddeviation is to be determined


Provided the moment of a +ve M/EI area from AProvided the moment of a +ve M/EI area from Ato B is computed, it indicates that the tangent atto B is computed, it indicates that the tangent atpoint A is above the tangent to the curve extendedpoint A is above the tangent to the curve extendedfrom point Bfrom point B

--ve areas indicate that the tangent at A is belowve areas indicate that the tangent at A is belowthe tangent extended from Bthe tangent extended from B


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It is easier to solve the problem in terms of EI & substitute thenumerical data as a last step.

SolutionSolution

T e 10 N oa causes t e eam to e ect.

Here the tangent at A is always horizontal.

The tangents at B & C are also indicated.

By construction, the angle between tan A and tan B (B/A) isequivalent to B.

SolutionSolution

ACCABB // ;

Applying Theorem 1, is equal to the area under the M/EIdiagram between points A & B.

mEI

kNm

EI

kNmm

EI

kNmABB / )5(

50100

2

1)5(

50

/B A

EI

kNm2375


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Substituting numerical data for E & I

SolutionSolution

kNm375 2

The ve sign indicates that the angle is measured clockwise from

A.

In a similar manner, the area under the M/EI diagram between

rammkN

00521.0])10)(10(360][/)10(200[ 412626

.

SolutionSolution

2

radkNm

EIm

EIACC

00694.0500

:haveweEI,ofvaluesnumericalngSubstituti

)10(2

2

/

mmkN ])10)(10(360][/)10(200[


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ConjugateConjugate--Beam methodBeam method

The basis for the method comes from similarityThe basis for the method comes from similarityequationsequations

To show this similarity, we can write these eqn asTo show this similarity, we can write these eqn asshownshown

2

2d Mww

dx

dV

2

2d v M

dx EI EIM

dx

d


Or integrating,Or integrating,

dxdxM

v

wdxV dxwdxM

dxM


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Here the shear V compares with the slopeHere the shear V compares with the slope , the, themoment M compares with the disp v & themoment M compares with the disp v & theexternal load w compares with the M/EI diagramexternal load w compares with the M/EI diagram

To make use of this comparison we will nowTo make use of this comparison we will nowconsider a beam having the same length as theconsider a beam having the same length as thereal beam but referred to as thereal beam but referred to as the conjugate beamconjugate beam,,


The conjugate beam is loaded with the M/EIThe conjugate beam is loaded with the M/EIdiagram derived from the load w on the real beamdiagram derived from the load w on the real beam

From the above comparisons, we can state 2From the above comparisons, we can state 2theorems related to the conjugate beamtheorems related to the conjugate beam

Theorem 1Theorem 1

The slope at a point in the real beam is numericallyThe slope at a point in the real beam is numerically

conjugate beamconjugate beam


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Theorem 2Theorem 2

The dis . of a oint in the real beam is numericallThe dis . of a oint in the real beam is numericallequal to the moment at the corresponding point inequal to the moment at the corresponding point inthe conjugate beamthe conjugate beam

When drawing the conjugate beam, it is importantWhen drawing the conjugate beam, it is importantthat the shear & moment developed at thethat the shear & moment developed at thesu orts of the con u ate beam account for thesu orts of the con u ate beam account for thecorresponding slope & disp of the real beam at itscorresponding slope & disp of the real beam at its

supportssupports


Consequently from Theorem 1 & 2, the conjugateConsequently from Theorem 1 & 2, the conjugatebeam must be supported by a pin or roller sincebeam must be supported by a pin or roller sincethis support has zero moment but has a shear orthis support has zero moment but has a shear orend reactionend reaction

When the real beam is fixed supported, both beamWhen the real beam is fixed supported, both beamhas a free end since at this end there is zero shearhas a free end since at this end there is zero shear& moment& moment


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Determine the max deflection of the steel beam. The reactionshave been computed. Take E = 200GPa, I = 60(106)mm4

Example 8 .13Example 8 .13


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The conjugate beam loaded with the M/EI diagram is shown.Since M/EI diagram is +ve, the distributed load acts upward.

SolutionSolution

T e externa reactions on t e conjugate eam are eterminefirst and are indicated on the free-body diagram.

Max deflection of the real beam occurs at the point where theslope of the beam is zero.

Assuming this point acts within the region 0x9m from A we canisolate the section.

Note that the peak of the distributed loading was determined fromproportional triangles,

SolutionSolution

/ (18 / ) / 9

' 0

0y

w x EI

V

F

45 1 2 02

6.71 (0 9 )

x x EI EI

m x m OK


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Using this value for x, the max deflection in the real beamcorresponds to the moment M.

SolutionSolution

Hence,

0')71.6(3

171.6

)71.6(2

2

1)71.6(

45

0ve,asmomentsiseanticlockwWith

MEIEI

M

The ve sign indicates the deflection is downward .

SolutionSolution

mmm

mmmmmmkN

kNmEI

kNmM

8.160168.0

)])10/(1()10(60][/)10(200[

2.201

2.201'

44344626

3

max

9789810682627 ppt 08 [compatibility mode]

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