statics and strength of materials intro beam analysis

7/24/2019 Statics and Strength of Materials Intro Beam Analysis

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Statics and Strength of

materialsDr1T 34


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Introduction

This Unit is designed to enable candidates to develop knowledgeand understanding of the principles and laws that relate to Staticsand Material Strength that underpin so much of more advancedstudies in Mechanical ngineering!

The Unit will also provide "ou with an opportunit" to stud" thewa"s in which a variet" of methods such as nodal anal"sis# vectoranal"sis and method of section can lead to the same results andthat material properties are e$tremel" important within design!

%" the end of the Unit "ou will be e$pected to solve static andstrength of material problems using the concepts and theorems"ou have learned! &ou will also be e$pected to sketch vector#shear force and bending moment diagrams!

If "ou have studied Statics earlier in "our education# the earl"parts of this Unit will provide "ou with an opportunit" to revise theStatics concepts and theorems "ou have learnt in previouscourses!


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Introduction 'nowledge# Understanding ( Skills

This development of the knowledge# understanding and skillsduring the deliver" of the unit will take place in the in the formof the following two main areas)

Solve problems relating to static e*uilibrium!

Solve problems relating to compressive# tensile and shearloading on materials!

+ssessment

The formal assessment for this Unit will consist of a singleassessment paper lasting no more than 1!, hours! Theassessment will be conducted under closed book conditions inwhich "ou will not be allowed to take notes# te$tbooks etc! into

the assessment! -owever# "ou will be allowed to use a scienti.ccalculator! &ou will sit this assessment paper at the end of theUnit!

/andidates stud"ing towards an -0/ in Mechanical ngineeringwill also have to answer *uestions on Statics as part of the .nalgraded unit e$amination!


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earning 2utcomes

2utcome 1) Solve problems relating to static e*uilibrium

Knowledge and/or skills to be developed

e$ternal reactions

compressive and tensile forces in plane frames

instabilit" and redundanc"compressive and tensile forces in practical engineeringapplications

simpl" supported beams

cantilever beams

shear force diagrams

bending moment diagrams magnitude and position of ma$imumbending stress compressive and tensile


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earning 2utcomes

2utcome 5) Solve problems relating to compressive# tensile

and shear loading on materials!Knowledge and/or skills to be developed Stress Strain Stress6Strain 7elationship

Modulus of lasticit" Direct Shear StressTorsional Shear Stress Single8Multiple Shear 9lane /onditions 0eutral +$is

Second Moment of +rea 9arallel +$is Theorem 9olar Moments of +rea %ending *uationTor*ue *uation %eam Selection


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Materials re*uired

9lease bring the following materials to all remainingclasses)

/alculator Set s*uares# ruler and protractor

9encil


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Importance to industr"

The principles of force anal"sis taught in this unit areapplicable to man" industries such as)6

/ivil ngineering

Structural engineering

+rchitecture

Surve"ing

Medical

Dental

+reo

0autical


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$ample of real engineering

http)88www!"outube!com8watch:v;7??,1k
http://www.youtube.com/watch?v=7Xpu8RVV51khttp://www.youtube.com/watch?v=7Xpu8RVV51khttp://www.youtube.com/watch?v=7Xpu8RVV51khttp://www.youtube.com/watch?v=7Xpu8RVV51k


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$amples of engineering failure

http)88www!"outube!com8watch:v;b-4v@?nA?p>

http)88www!"outube!com8watch:v;$B


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Ch" should we consider forceswhen designing a product:

To see if a structure is going to fail i!e! break or deform

To select components

To .nd weaknesses


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Metric PrefxesMetric prefxesare prett" eas" to understand and ver" hand" formetric conversions! &ou dont have to know the nature of a unit toconvert# for e$ample# from kilo-unitto mega-unit! +ll metric pre.$es

are powers of 1E! The most commonl" used pre.$es are highlighted inthe table!Most people even in the countries where metric s"stem is used onl"know the most important metric pre.$es like kilo and milli! The" arever" hand" for understanding metric conversions!

Prexis Symbol Factor

tera T 1E15 ; 1#EEE#EEE#EEE#EEE

giga F 1EG ; 1#EEE#EEE#EEE

mega M 1E@ ; 1#EEE#EEE

kilo k 1E3 ; 1#EEE

hecto h 1E5 ; 1EE

deka da 1E1 ; 1E

deci d 1E61 ; E!1

centi c 1E65 ; E!E1

milli m 1E63 ; E!EE1

micro H 1E6@ ; E!EEE#EE1

nano n 1E6G ; E!EEE#EEE#EE1


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orce anal"sisTo represent forces vectors we use arrows)

The length represents the magnitudeThe angle represents the direction

+dding vectors cannot be done using the siBe onl"!

50 N30

+ 20 N45

=

48 N

54


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orce anal"sis

The sum of two vectors can be obtained using thegraphicalparallelogram rule:

9lace the ends of two vectors together

/omplete the parallelogram

Diagonal represents the sum of vectors

7emember to use a suitable scale for the vectors

a

b

a+b


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orce anal"sisor two forces to be in e*uilibrium the" have to)

%e e*ual in siBe

-ave lines of action that pass through the same point beconcurrent

+ct in e$actl" opposite directions

In equilibrium Not in equilibrium


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orce anal"sisor three forces to be in e*uilibrium the" have to)

+ll be in the same plane coplanar

-ave lines of action that pass through the same point beconcurrent

Five no resultant force in an" direction


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orce anal"sis*uilibrium for three forces can be obtainedusing the graphical triangle rule:

Draw an arrow to represent one of the forces

Take the forces in the se*uence the" occur e!g!

clockwise! Draw the ne$t arrow from the head of theprevious!

/omplete the triangle!

F2

F1

Resultant


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orce anal"sis*uilibrium for more than three vectors can beobtained using the graphicalpolygon rule:

Draw an arrow to represent one of the forcesTake the forces in the se*uence the" occur e!g!

anticlockwise! Draw the ne$t arrow from the head of the previous! /omplete the pol"gon!

F1 F2

F3

F4

Resultant

F1

F2F3

F4


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7esolving forces

It is *uite often useful to know what the horiBontal andvertical components of a force are)

F

Horizontal om!onent

"ertialom!onent

If we know and J)

= sinFFv

= cosFFh2

h

2

v

2 FFF +=


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Tr" vector tutorial*uestions


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7ules of *uilibrium

Chen anal"sing static problems we must .rst

assume static e*uilibrium!

This allows us to appl" the rules that

The sum of all horiBontal forces ; EThe sum of all vertical forces ; E

The sum of all moments ; E

These are known as The rules of equilibrium


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7eactionsChat happens when "ou place "our hand against the walland push:

Chen e$ternal forces are applied to an obKect from theoutside# internal forces are induced in the obKect tomaintain e*uilibrium!

Internal #ores

$%ternal

#ore$%ternal

#ore

$%ternal

#ore

Internal #ores

$%ternal

#ore


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7eactionsInternal forces are produced in reaction to the applicationof e$ternal forces and are called reactions!

Chen an obKect is at rest on the ground# the weight of theobKect must be counter6balanced b" an opposing force togive e*uilibrium!

&ei'(t

Reation

Chat would happen if

the reaction wasnLt there:

http)88www!"outube!com8watch:v;$B


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Moments

or e*uilibrium a structure not onl" needs a resultant toprevent motion but also to prevent rotation!

Moments of forces cause rotation about an a$is)

F

)%is at

ri'(t an'les

r

dFMoment =


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Moments

The moment of a force is de.ned as being the product ofthe force and its perpendicular distance r from the a$isto the line of action of the force!

The SI unit for the moment is the 0ewton metre 0m!

)%is at

ri'(t an'les

F

r

dFMoment =


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Moments

Moments have a direction) /lockwise positive

+nticlockwise negative

It doesnLt matter if "ou regard clockwise as positive or

negative as long as "ou stick with "our convention!

F

*

lo,-ise

F

*

)ntilo,-ise


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free bod" diagram

The previous diagram is known as a free bod" diagram!Imagine an element of the s"stem to be isolated from the

rest of the s"stem! Draw a diagram of this element and show all of the forces

acting on it!

Draw all of the e$ternal loads as well as the reactive forces!

Use a pol"gon of forces to determine the e*uilibrium force!

7epeat for all of the elements in the s"stem!


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Determinate beamsIn Statics we look at how weight or moreaccuratel" force eects dierent structures of

dierent materials!

or this section we shall look at e$ternal and7eaction orces on simpl" supported beamstructures!


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%eam theor"

7eaction calculations s"mmetrical loading

5EE0

1EE0 1EE0

1m 1m

If a beam is loaded s"mmetricall" we can simpl"divide the total load b" the number of supports!


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%eam theor"

-ow to /alculate reaction forces 71 ( 75

3,E0 5EE0 1EE0

5m1m

1m1m71 75


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or the purpose of calculation change one of thesupports to a point of pivot!

If this were a point of pivot each force would be

creating a rotational force about this point of pivot! +rotational force as we know from earlier is a moment!

Chat we see here is the sum of the 3 anti6clockwisemoments are e*ualised b" the one anti6clockwise# thusin static e*uilibrium!

3,E0 5EE0 1EE0

5m1m

1m1m

71


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+t this point due to the beam being in e*uilibriumwe can appl" our Nrules of equilibrium.

The sum of all horiBontal forces ; E

The sum of all vertical forces ; E


+s a moment is de.ned as force x distance then b"anal"sing our beam)6

71$, ; 1EE$5O5EE$3O3,E$4

,71 ; 5EEO@EEO14EE

,71 ;55EE

71 ;55EE8,;44E0


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+t this point we can update our diagram!

0ow we appl" our rules of e*uilibrium again!The sum of all horiBontal forces ; E



This time we would isolate the vertical forces!

3,E0 5EE0 1EE0

5m1m

1m1m

44E0 75


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+s we are onl" dealing with forces acting verticall"and the beam is in e*uilibrium# it then stands that)6

71O75; 3,EO 5EEO 1EE

+s we have Kust found 71 ; 44E0 Then we know)6

44E0 O75 ; 3,EO 5EEO1EE

This then lets us .nd 75 b")6

75; 3,EO5EEO1EE P 44E 75; @,EP 44E

75; 51E0


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ree bod" diagram Ce can now take this information and construct a

free bod" diagram)6

There are no linear measurements of length in this!%!D! as it is drawn to scale e!g! 5Emm)1m

If "ou wish to double check "our answers "ou could

appl" vector anal"sis to the forces and it shouldtotal E0

3,E0 5EE0 1EE0

44E0 51E0


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Shear force diagram

+n" force acting perpendicular to the neutral a$iswill induce a shearing tendenc"! +s engineers wehave to be able to *uantif" this tendenc" asshearing forces! These forces can be calculated atan" point among the beam however it is conventionfor these forces to be displa"ed graphicall" on ashear force diagram!

Chen plotted as past of a beam anal"sis the sfdeasil" shows areas of concern with regard tostructural integrit"!

Chen we plot a S! !D! it is done so in directrelationship with the free bod" diagram# i!e! it is

plotted directl" below appl"ing the same scale forbeam length!


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Shear force diagram contLd

7emember tode.ne basic signconvention)

E E

S

F

F

(+ve)

F

F

(-ve)

3,E0 5EE0 1EE0

44E0 51E0

44E0

GE0

611E0

6

51E0


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/antilever beams

Cith cantilever beams there is onl" one support

Ce can anal"se the force arrangement at the .$ed

end!


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/antilever beams

$ample!

+s there are no horiBontal loads Hramust be E0

To .nd a! asthe beam being in e*uilibrium we can appl" ourNrules of equilibrium.

The sum of all horiBontal forces ; E



7a

Ma

-7a


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+s the beam is onl" supported at one end we can sa" that

7a ; ,E O


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ree bod" and Shear force diagram

,E0


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Tr" the beam tutorial*uestions!

reaction calculations


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Uniforml" distributed oads U!D!!Ls Uniforml" distributed loads are a load that is spread

acrossa region of the beam!

%est e$ample of this is the weight o the beam itsel It is measured in N/mand is represented b" man" arrows

or a bmpyline!

Chen calculating moments with the UD# treat it as apoint loadin the middleof the region with the sm o

the entire load! e!g! or a 1Em long beam with a UD of ,k08m# it can

be counted as a ,Ek0 point load ,m from the ends!

The shear force and bending moment diagrams# it lookslike the following)

Shear forceDiagram %ending MomentDiagram


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U!D! e$ample

+s before we use rules of e*uilibrium!


7a$, ; 1EE$5O5EE$3O3,E$4O5,$,$5!,

,7a ; 5EEO@EEO14EEO315!,

,7a ;5,15!,

7a ;5,15!,8,;,E5!,0

3,E0 5EE0 1EE0

5m1m

1m1m

7a 7e

5,08m


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+s we are onl" dealing with forces acting verticall"and the beam is in e*uilibrium# it then stands that)6

7aO7e; 3,EO 5EEO 1EE O5,$,

+s we have Kust found 7a ; ,E5!,0 Then we know)6

,E5!,0 O7e ; 3,EO 5EEO1EE O5,$,

This then lets us .nd 7e b")6 7e; 3,EO5EEO1EEO15, P ,E5!,

7e;


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%ending moment calculations

+t this point we need to anal"se the rotational forces at allpoints of interest along the beam length!

Ce do this b" working from one end of the beam to the other#moving our datum point as we progress! i!e! anal"se forceswith respect to point we anal"sing!

In this process we must appl" sign convention to our forces!

In the case of working from the 7ight hand side to the left#forces to the right of the point of anal"sis acting up arepositive and down are negative!

This is best illustrated b" a diagram!


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%ending moment calculations

6

O

d

d

o Chen working from 7!-!S! we ignore an" forces to the leftof our point of interest!


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Corked $ample

%eam anal"sis

rom before we know that 7a;,E5!,0 ( 7e;5


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%ending moment calculations working from the right hand side

%mQ; 5


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%eam Selection due to beamstresses!

The further slides should be used after the completionof

Stress and strain

/ross sectional anal"sis


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T"pes of Stress

There are T-7 basic t"pes of stress)

Direct stress

Shear stress

%ending stress

Most components are subKected to one or a combinationof these stresses!


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Direct Unia$ial Stress

Ma" be tensile orcompressive!

ine of action of theforce is normal to the

cross section beingconsidered!

Stress is distributeduniforml" over thecross section!

F

F/rossSectional +rea#

"

A

F=


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Shear stress

F

F

/rossSectional +rea#"

ine of action of theload is parallel to thecross section beingconsidered!

Ma$ Shear orce S

comes from shear forcediagram!

A

Fs=


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%ending stress

M M

Neutral Axis of

Section

%ending stressis the result ofbendingmomentswithin thecomponent!

%ending occursaround theneutral a$is ofa componentLscross section!


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%ending stress

M M%ending stressvaries across across section)

bothcompressive andtensile on an"cross section#

ma$imum atpoints furthestfrom the neutrala$is#

Bero at the

.o!

sur#ae in

tension

/ero stress

on neutral

!lane

ottom sur#ae

in om!ression


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ngineerLs theor" of bending

or a beam that has been bent into an arc the followingrelationship e$ists!

7

"

%ending moment

%ending stressModulus of elasticit"

7adius of neutral a$is

Distance to neutral a$is

Second moment of area

R

E

yI

M=

=


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ngineerLs theor" of bending

#b$ %ending Stress

%; %ending Moment

y; Distance from 0eutral+$is to outer surface

&; Second Moment of

+rea of the cross section

"

I

Myb =


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Feometrical properties of sections

X X

d

X X

d

b

The most commonformulae for &:

The 5ndMomentof +rea is thedistribution ofarea about theneutral a$is# =6=!

/an be derivedthrough standard

formulae# fromtables# or from a/+D s"stem!

4

64dI xx

=

12

3bdI xx =


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Section modulus

Since Iand yareconstantsdependent on thesection geometr"#it is oftenconvenient toe$press them in

terms of thesection modulus Zi!e)

The bending equationcan be re'ritten

therefore as:

y

IZ=

Z

Mb =


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+llowable stress

To design for strengthwe need an allowablestress based on thematerial selection

and a factor ofsafet"!

a= "llowable Stress

y=#ield Stress o chosen

material

Chen selecting a section siBe# the designershould aim for)

SOF

y

a..

=

ab


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actor of Safet" !o!S!

It is not normal or safe to design a component with acalculated stress level e*ual to the "ield or ultimate stressvalue for the material!

actors of Safet" are used to allow a margin of safet"!

These factors are used to account for wear and tear in use#assumptions in calculations etc!


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actor of Safet" !o!S!

In calculating factors of safet"

or static loads and ductilematerials use "ield stress or

where appropriate E!5R proofstress!

or static loads and brittle

materialsthat have no wellde.ned "ield point use theultimate stress!


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2ccasionall"# allowable stresses are pre6calculated with no need to use a !2!S!

+llowable stress


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Standard Sections

Man" materials come in a range of siBes of standard crosssection

These standard sections should be used whenever possibleas the" tend to be)

/heaper

%e more readil" available Data such as allowable stress# weight# &values are readil"

available from catalogues and data books


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/ommon standard sections

S*uare hollow

7ectangular hollow

7ound tube

I8Universal beam# Koist +ngle

/hannel


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Sample data table


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Section Selection 6 $ample

$etermine the si%e o &'( beam or theapplication shown below!

10kN

1000mm


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Section Selection 6 $ample$etermine vale o

maximm bendingmoment

)!$raw ree bodydiagram

*!$raw shear orcediagram

+!$raw bendingmoment diagram

F

= 10 ,Nm

10kN

R=10,N

1m


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(. )etermine maximum bending moment rom %M diagram# ma$! bending moment ;1E '0m

*. +hoose component material and grade Structural steel Frade 43

,. )etermine maximum allo'able bending stress

1,E M08m5rom data table

. +alculate required alue of section modulus / Use the bending e*uation b; M8 remember b; a

; M8b ; 1E1E381,E1E@; @!@@1E6, m3; @@!@< cm3

Therefore re*uired section modulus ; ,,!,- cm+

0. 1elect si2e of section Select an NI section with a value of @@!@< cm3or above

from the standard section tables!


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3cm67.9508.51.486

yIZ ===

statics and strength of materials intro beam analysis

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