internal forces.ppt

8/9/2019 INTERNAL FORCES.ppt

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Engineering Mechanics: Statics

Chapter 6:

Internal Forces


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Chapter Objectives

To show how to use the method osections or determining the internal

loadings in a member! To generali"e this procedure b#

ormulating e$uations that can be

plotted so that the# describe theinternal shear and momentthroughout a member!


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Chapter Outline

Internal Forces %eveloped inStructural Members

Shear and Moment E$uationsand %iagrams

&elations between %istributed

'oad( Shear and Moment


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)!* Internal Forces%eveloped in Structural

Members The design o an# structural or

mechanical member re$uires thematerial to be used to be able to

resist the loading acting on themember

These internal loadings can bedetermined b# the method o sections


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MembersConsider the +simpl# supported, beam

To determine the internal loadings acting on thecross section at C( an imaginar# section ispassed through the beam( cutting it into two

-# doing so( the internal loadings becomee.ternal on the F-%


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MembersSince both segments /0C and C-1 were in

e$uilibrium beore the sectioning(e$uilibrium o the segment is maintainedb# rectangular orce components and aresultant couple moment

Magnitude o the loadings is determined

b# the e$uilibrium e$uations


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MembersForce component N( acting normal to the

beam at the cut session and V( acting t

angent to the session are 2nown asnormal

or a.ial orce

and the shear orceCouple moment M is

reerred as the bending

moment


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MembersFor 3%( a general internal orce and couplemoment resultant will act at the section

4# is the normal orce( and 5. and 5" are

the shear componentsM# is the torisonal or

twisting moment( and

M. and M" are thebending moment

components


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MembersFor most applications( these

resultant loadings will act at thegeometric center or centroid /C1 othe sections cross sectional area

0lthough the magnitude o eachloading di7ers at di7erent points

along the a.is o the member( themethod o section can be used todetermine the values


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MembersFree -od# %iagrams Since rames and machines are composed

o multi8orce members( each o these

members will generall# be subjected tointernal shear( normal and bending loadings

Consider the rame with the blue

section passed through todetermine the internal loadings

at points 9( and F


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MembersFree -od# %iagrams F-% o the sectioned rame

0t each sectioned member( there is anun2nown normal orce( shear orce and bendingmoment

3 e$uilibrium e$uations cannot be used

to ;nd < un2nowns( thus dismemberthe rame and determine

reactions at each connection

l


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MembersFree -od# %iagramsOnce done( each member ma# be sectioned at

its appropriate point and appl# the 3 e$uilibrium

e$uations to determine the un2nownsE.ample

F-% o segment % can be used to determine

the internal loadings at provided the reactions o

the pins are 2nown

l


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Members=rocedure or 0nal#sisSupport &eactions

-eore the member is cut or sectioned(determine the members support reactions E$uilibrium e$uations are used to solve or

internal loadings during sectioning o the

members I the member is part o a rame or

machine( the reactions at its connectionsare determined b# the methods used in 6!6

) * I l F


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Members=rocedure or 0nal#sisFree8-od# %iagrams

>eep all distributed loadings( couplemoments and orces acting on themember in their e.act locations( then passan imaginar# section through the member(perpendicular to its a.is at the point theinternal loading is to be determined

0ter the session is made( draw the F-% othe segment having the least loads

) * I t l F


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Members=rocedure or 0nal#sisFree8-od# %iagrams

Indicate the "( #( " components o the orceand couple moments and the resultantcouple moments on the F-%

I the member is subjected to a coplanar

s#stem o orces( onl# N( V and M act at thesection

%etermine the sense b# inspection? i not(assume the sense o the un2nown loadings

) * I t l F


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Members=rocedure or 0nal#sisE$uations o E$uilibriumMoments should be summed at the section

about the a.es passing through the centroid orgeometric center o the members cross8sectional area in order to eliminate the un2nownnormal and shear orces and thereb#( obtaindirect solutions or the moment components

I the solution #ields a negative result( the senseis opposite that assume o the un2nownloadings

) * I t l F


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Members The lin2 on the bac2hoe

is a two orce member

It is subjected to bothbending and a.ial loadat its center

-# ma2ing the memberstraight( onl# an a.ialorce acts within themember

) * I t l F


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MembersE.ample )!*

The bar is ;.ed at its end and is

loaded! %etermine the internalnormal

orce at points - and C!

) * I t l F


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MembersSolutionSupport &eactions

F-% o the entire bar -# inspection( onl# normal orceA# acts at the ;.ed support

A. @ A and A" @ A

BD F# @ A? 24 4- @ A

4- @ 24

) * I t l F


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MembersSolutionF-% o the sectioned bar

4o shear or moment acton the sections sincethe# are not re$uired ore$uilibrium

Choose segment 0- and%C since the# contain theleast number o orces

) * I t l F


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MembersSolutionSegment 0-

BD F# @ A? 24 4- @ A4- @ 24

Segment %CBD F# @ A? 4C G24@ A

4C @ G24

) * I t l F


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MembersE.ample )!HThe circular shat is subjected to three

concentrated tor$ues! %etermine theinternal

tor$ues at points - and C!

) * I t l F


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MembersSolutionSupport &eactions Shat subjected to onl# collinear tor$ues

D M. @ A?

8*A4!m B *4!m B HA4!m T% @ A

T% @ H4!m

) * Internal Forces


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MembersSolutionF-% o shat segments 0- and C%

) * Internal Forces


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MembersSolutionSegment 0-

D M. @ A? 8*A4!m B *4!m T- @ A

T- @ 4!m

Segment C%

D M. @ A? TC H4!m@ A

TC @ H4!m

) H Shear and


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)!H Shear andMoment E$uations

and %iagrams-eams structural members designed tosupport loadings perpendicular to their a.es

-eams straight long bars with constant

cross8sectional areas0 simpl# supported beam is pinned at one end

and roller supported at

the other0 cantilevered beam is

;.ed at one end and ree

at the other

) H Shear and


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and %iagrams For actual design o a beam( appl#8 Internal shear orce 5 and the bending

moment M anal#sis

8 Theor# o mechanics o materials8 0ppropriate engineering code to determine

beams re$uired cross8sectional area

5ariations o 5 and M obtained b# the methodo sections

raphical variations o 5 and M are termed asshear diagram and bending moment diagram

) H Shear and


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and %iagrams Internal shear and bending momentunctions generall# discontinuous( or theirslopes will be discontinuous at points

where a distributed load changes or whereconcentrated orces or couple momentsare applied

Functions must be applied or each

segment o the beam located betweenan# two discontinuities o loadings Internal normal orce will not be

considered

) H Shear and


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and %iagrams'oad applied to a beam actperpendicular to the beams a.isand hence produce onl# an internal

shear orce and bending momentFor design purpose( the beams

resistance to shear( and particularl#

to bending( is more important thanits abilit# to resist a normal orce

) H Shear and


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and %iagramsSign Convention To de;ne a positive and negative shear

orce and bending moment acting on the

beam=ositive directions are denoted b# an

internal shear orce that causes

cloc2wise rotation o the member onwhich it acts and b# an internal momentthat causes compression or pushing onthe upper part o the member

) H Shear and


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and %iagramsSign Convention0 positive moment

would tend to bendthe member i it wereelastic( concave

upwards'oadings opposite to

the above areconsidered negative

) H Shear and


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and %iagrams=rocedure or 0nal#sisSupport &eactions

%etermine all the reactive orces andcouple moments acting on the beam&esolve them into components acting

perpendicular or parallel to the

beams a.is

) H Shear and


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and %iagrams=rocedure or 0nal#sisShear and Moment &eactions

Speci# separate coordinates . having an

origin at the beams let end and e.tending toregions o the beams between concentratedorce andJor couple moments or where thereis no continuit# o distributed loadings

Section the beam perpendicular to its a.is ateach distance . and draw the F-% o one othe segments

) H Shear and


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and %iagrams=rocedure or 0nal#sisShear and Moment &eactions

5 and M are shown acting in their positivesense

The shear 5 is obtained b# summing theorces perpendicular to the beams a.is

The moment M is obtained b# summingmoments about the sectioned end o thesegment

) H Shear and


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and %iagrams=rocedure or 0nal#sisShear and Moment %iagrams

=lot the shear diagram /5 versus .1 and the

moment diagram /M versus .1 I computed values o the unctions describing

5 and M are positive( the values are plottedabove the . a.is( whereas negative values

are plotted below the . a.is

Convenient to plot the shear and the bendingmoment diagrams below the F-% o the beam

) H Shear and


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and %iagramsE.ample )!)%raw the shear and bending moments

diagrams or the shat! The support at 0 is athrust bearing and the support at C is a

journal bearing!

) H Shear and


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and %iagramsSolutionSupport &eactions

F-% o the shat

) H Shear and


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and %iagramsSolution

m xkN M M

kN V F y

.5.2;0

5.2;0

==∑

==∑↑+


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) H Shear and


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and %iagramsSolutionShear diagram internal shear orce is alwa#s

positive within the shat 0- Kust to the right o -( the

shear orce changes sign andremains at constant valueor segment -C

Moment diagram Starts at "ero( increases

linearl# to - and thereoredecreases to "ero

) H Shear and


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and %iagramsSolutionraph o shear and

moment diagrams is

discontinuous at points oconcentrated orce ie( 0( -(C

0ll loading discontinuous

are mathematical( arisingrom the ideali"ation o aconcentrated orce andcouple moment

) 3 &elations between


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)!3 &elations between%istributed 'oad( Shear and

Moment%istributed 'oadConsider beam 0% subjected to an

arbitrar# load w @ w/.1 and a series oconcentrated orces and moments

%istributed load assumed positive whenloading acts downwards



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Moment%istributed 'oad 0 F-% diagram or a small

segment o the beam havinga length L. is chosen at point. along the beam which is notsubjected to a concentrated

orce or couple moment 0n# results obtained will not

appl# at points oconcentrated loadings



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Moment%istributed 'oad The internal shear orce and

bending moments shown on theF-% are assumed to act in thepositive sense

-oth the shear orce and

moment acting on the right8hand ace must be increased b#a small( ;nite amount in order to2eep the segment in e$uilibrium



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Moment%istributed 'oad The distributed loading has been replaced

b# a resultant orce LF @ w/.1 L. that actsat a ractional distance 2 /L.1 rom theright end( where A 2 *

( )[ ]2

)()(

0)()(;0

)(

0)()(;0

xk xw xV M

M M xk x xw M xV M

x xwV

V V x xwV F y

∆−∆=∆

=∆++∆∆+−∆−=∑

∆−=∆

=∆+−∆−=∑↑+



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Moment

%istributed 'oad

Slope o the @ 4egative o

shear diagram distributed loadintensit#

Slope o @ Shear moment diagram

V dx

dM

xwdx

dV

=

−= )(



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Moment%istributed 'oad 0t a speci;ed point in a beam( the slope o the

shear diagram is e$ual to the intensit# o the

distributed load Slope o the moment diagram @ shear

I the shear is e$ual to "ero( dMJd. @ A( a pointo "ero shear corresponds to a point o

ma.imum /or possibl# minimum1 momentw /.1 d. and 5 d. represent di7erential area

under the distributed loading and sheardiagrams



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Moment

%istributed 'oad

Change in @ 0rea under

shear shear diagram

Change in @ 0rea under

moment shear diagram

∫

∫

=∆

−=∆

Vdx M

dx xwV

BC

BC )(



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Moment

%istributed 'oad Change in shear between points - and C is

e$ual to the negative o the area under the

distributed8loading curve between thesepoints

Change in moment between - and C ise$ual to the area under the shear diagram

within region -C The e$uations so not appl# at points where

concentrated orce or couple moment acts



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Moment

Force F-% o a small segment

o the beam

Change in shear isnegative thus the shear

will jump downwardswhen F acts downwardson the beam

F V F y −=∆=∑↑+ ;0



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Moment

Force F-% o a small segment o

the beam located at thecouple moment

Change in moment is

positive or the momentdiagram will jump upwardsMO is cloc2wise

O M M M =∆=∑ ;0



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Moment

E.ample )!<

%raw the shear and moment diagrams or

thebeam!



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Moment

Solution

Support &eactions

F-% o the beam



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MomentSolutionShear %iagram5 @ B*AAA at . @ A

5 @ A at . @ HSince d5Jd. @ 8w @ 8AA( a straight negative

slopingline connects the end points

)!3 &elations between


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Moment

Solution

Moment %iagram

M @ 8*AAA at . @ A

M @ A at . @ H

dMJd. @ 5( positive #et linearl# decreasing rom

dMJd. @ *AAA at . @ A to dMJd. @ A at . @ H

internal forces.ppt

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