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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=7Xpu8RVV51k7/24/2019 Statics and Strength of Materials Intro Beam Analysis
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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$ternal and7eaction orces on simpl" supported beamstructures!
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%eam theor"
7eaction calculations s"mmetrical loading
5EE0
1EE0 1EE0
1m 1m
If a beam is loaded s"mmetricall" we can simpl"divide the total load b" the number of supports!
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%eam theor"
-ow to /alculate reaction forces 71 ( 75
3,E0 5EE0 1EE0
5m1m
1m1m71 75
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or the purpose of calculation change one of thesupports to a point of pivot!
If this were a point of pivot each force would be
creating a rotational force about this point of pivot! +rotational force as we know from earlier is a moment!
Chat we see here is the sum of the 3 anti6clockwisemoments are e*ualised b" the one anti6clockwise# thusin static e*uilibrium!
3,E0 5EE0 1EE0
5m1m
1m1m
71
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+t this point due to the beam being in e*uilibriumwe can appl" our Nrules of equilibrium.
The sum of all horiBontal forces ; E
The sum of all vertical forces ; E
The sum of all moments ; E
+s a moment is de.ned as force x distance then b"anal"sing our beam)6
71$, ; 1EE$5O5EE$3O3,E$4
,71 ; 5EEO@EEO14EE
,71 ;55EE
71 ;55EE8,;44E0
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+t this point we can update our diagram!
0ow we appl" our rules of e*uilibrium again!The sum of all horiBontal forces ; E
The sum of all vertical forces ; E
The sum of all moments ; E
This time we would isolate the vertical forces!
3,E0 5EE0 1EE0
5m1m
1m1m
44E0 75
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+s we are onl" dealing with forces acting verticall"and the beam is in e*uilibrium# it then stands that)6
71O75; 3,EO 5EEO 1EE
+s we have Kust found 71 ; 44E0 Then we know)6
44E0 O75 ; 3,EO 5EEO1EE
This then lets us .nd 75 b")6
75; 3,EO5EEO1EE P 44E 75; @,EP 44E
75; 51E0
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ree bod" diagram Ce can now take this information and construct a
free bod" diagram)6
There are no linear measurements of length in this!%!D! as it is drawn to scale e!g! 5Emm)1m
If "ou wish to double check "our answers "ou could
appl" vector anal"sis to the forces and it shouldtotal E0
3,E0 5EE0 1EE0
44E0 51E0
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Shear force diagram
+n" force acting perpendicular to the neutral a$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
$ample!
+s there are no horiBontal loads Hramust be E0
To .nd a! asthe beam being in e*uilibrium we can appl" ourNrules of equilibrium.
The sum of all horiBontal forces ; E
The sum of all vertical forces ; E
The sum of all moments ; E
7a
Ma
-7a
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+s the beam is onl" supported at one end we can sa" that
7a ; ,E O
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ree bod" and Shear force diagram
,E0
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Tr" the beam tutorial*uestions!
reaction calculations
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Uniforml" distributed oads U!D!!Ls Uniforml" distributed loads are a load that is spread
acrossa region of the beam!
%est e$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$ample
+s before we use rules of e*uilibrium!
The sum of all moments ; E
7a$, ; 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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Section Selection 6 $ample
(. )etermine maximum bending moment rom %M diagram# ma$! 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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Section Selection 6 $ample
3cm67.9508.51.486
yIZ ===