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CHAPTER 1STRESS AND STRAIN
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1. Todays Objectives:
Students will be able to:
Explain some of the important principles of statics.
se the principles to determine internal resultant loadin!s in abody.
Explain the concepts of normal" shear" bearin! and thermal
stress.
Topics:
Introduction Main Principles of Statics
Stress
Normal Stress Shear Stress Bearing Stress Thermal Stre
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4
Overview of #echanics
1.1 Introduction
Mechanics : The study of how bodies react to forces acting on them
$%&%' (O'%ES
(Things that do not change shape
Statics : The study of bodies
in an e!uilibrium
'E)O$#*(+E (O'%ES
(Things that do change shape)+%'S
Mechanics of Materials :
The study of the relationships
between the externalloads
applied to a deformable body and
the intensity of internal forces
acting within the body"
Incompressible #ompressible
$ynamics :
%" ,inematics& concerned
with the geometric aspects
of the motion
'" ,inetics& concerned
with the forces causing the
motion"
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5
External +oads
1.2 Main Principles of Statics
External +oads
Surface )orces- caused by direct contact of one body with
the surface of another"
(ody )orce- deeloped when one body e)erts a force on
another body without direct physical contactbetween the bodies"
* e"g earth+s graitation (wei!ht
concentrated force
linear distributed load" w-s
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Axial Load Normal Stress Shear Stress
Bearing Stress Allowable Stress Deformation of Structural under Axial Load Statically indeterminate roblem
Thermal Stress
STRESS AND STRAIN
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!echanics of material is a study of the
relationshi between the external loads aliedto a deformable body and the intensity ofinternal forces acting within the body"
Stress # the intensity of the internal force on aseci$c lane %area& assing through a oint"
Strain # describe the deformation by changes in
length of line segments and the changes in theangles between them
Stress And Strain
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Type of Stress
1.1 Introduction
Normal Stress : stress which acts perpendicular, or normal to, the
(- cross section of the load*carrying member"
: can be either compressie or tensile" Shear Stress : stress which acts tangent to the cross section of
(. the load*carrying member" : refers to a cutting*li/e action"
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Normal Stress' the intensity of force' or force er unit area' acting
normal to A
A ositi(e sign will be used to indicate a tensile stress%member in tension&
A negati(e sign will be used to indicate a comressi(e stress%member in comression&
Normal Stress and Normal Strain
= P / A
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(a)
(b)
Stress ( ) = Force (P) Cross Section (A)
Unit: Nm - N0mm'or MPa
N0m' or Pa
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Examples of *xially +oaded (ar
1.4 Axial Loading Normal Stress
*ssumptions :
%" 1niform deformation: Bar
remains straight before and
after load is applied, and
cross section remains flat orplane during deformation
'" In order for uniform
deformation, force /be
applied along centroidal a)isof cross section #
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AP
AP
AFFFA
!
=
=
==+
dd2
12
*vera!e 0ormal Stress 'istribution
"3 aerage normal stress at any point
on cross sectional area
P3 internal resultant normal force
A3 cross*sectional area of the bar
1.4 Axial Loading Normal Stress
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/rocedure of *nalysis
1se e!uation of "3P0Afor cross*sectional area of a member when
section sub4ected to internal resultant force /
Internal Loading
Section memberperpendicularto its longitudinal a)is at ptwhere normal stress is to be determined
$raw free*body diagram 1se e!uation of force e!uilibrium to obtain internal a)ial
force /at the section
$etermine member+s )*sectional area at the section #ompute aerage normal stress "3P0A
Average Normal Stress
1.4 Axial Loading Normal Stress
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Examle )")*
Two solid cylindrical rods AB and B+ arewelded together at B and loaded as shown"
,nowing that d)#-.mm and d/#/.mm' $nda(erage normal stress at the midsection of %a&rod AB' %b& rod B+"
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Examle )"/
Two solid cylindrical roads AB and B+ are
welded together at B and loaded as shown",nowing that d) # -. mm and d/ # 0. mm'$nd the a(erage normal stress in the midsection of %a& rod AB' %b& rod B+"
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Normal strain' is the elongation orcontraction of a line segment er unitof length
L # elongation
Lo #length
= L / Lo
strainnormal5=
=
* L=
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Examle )"-*Determine the corresonding strain for a bar of
length L#."1..m and uniform cross sectionwhich undergoes a deformation #)0.).21m"
6
6
6
%78 %8 m '78 %8 m m5 8 688 m
'78 %8 '78
/.
@
= = =
=
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1.4A cable and strut assembly AB+ suorts a (ertical load3#)/4N" The cable has an e5ecti(e cross sectional area of)1.mm6' and the strut has an area of -7.mm6"
%a& +alculate the normal stresses in the cable and strut"
%b& If the cable elongates )")mm' what is the strain8
%c& If the strut shortens ."-9mm' what is the strain8
Stress and Strain Example
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1.5The bar shown has a s:uare cross section%/.mm x 7.mm& and length' L#/";m" If anaxial force of 9.4N is alied along the
centroidal axis of the bar cross sectional area'determine the stress and strain if the bar endu with 7m length"
70kN 70kN
2.8m
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Tensile test is an exeriment to determinethe load2deformation beha(ior of thematerial"
Data from tensile test can be lot into stressand strain diagram"
Examle of test secimen
2 note the dog2bone geometry
The Stress-StrainDiaram
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Stress2Strain Diagrams
A number of imortant mechanical
roerties of materials that can be deduced
from the stress2strain diagram are illustratedin $gure abo(e"
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3oint >2A # linear relationshi between stress
and strain
3oint A# roortional limit %3L&The ratio of stress to strain in this linear region
of stress2strain diagram is called?oung !odulus or the !odulus of Elasticity gi(en
At oint A2B'secimen begins yielding"
3oint B# yield oint 3oint B2+# secimen continues to elongate without any increase in stress" Its
refer as erfectly lastic @one 3oint + # stress begins to increase 3oint +2D# refer as the @one of strain hardening 3oint D# ultimate stressstrength secimen
begins to nec42down 3oint E# fracture stress
= < PL
Unit: MPa
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3oint > to A
3oint + to D
3oint D to E
At oint E
Normal or engineeringstresscan be determined
by di(iding the alied load by the secimen
original cross sectional area"True stress is calculated using the actual cross
sectional area at the instant the load is
measured"
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Some of the materials li4e aluminum %ductile&'does not ha(e clear yield oint li4esstructural steel" Therefore' stress (aluecalled the o5set yield stress' ?Lis usedin line of a yield oint stress"
As illustrated' the o5set yield stress isdetermine by Drawing a straight line that best $ts the data in initial %linear&
ortion of the stress2strain diagram Second line is then drawn arallel to the original line but o5set by
seci$ed amount of strain The intersection of this second line with the stress2strain cur(e determine the o5set yield stress" +ommonly used o5set (alue is ."../."/C
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Brittle material such as ceramic and glass
ha(e low tensile stress (alue but high in
comressi(e stress" Stress2strain diagram for
brittle material"
33
El i i d Pl i i
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Elasticityrefers to the roerty of a material suchthat it returns to its original dimensions after
unloading " Any material which deforms when sub=ected to load
and returns to its original dimensions when unloaded
is said to be elastic"
If the stress is roortional to the strain' the materialis said to be linear elastic' otherwise it is non-linear
elastic" Beyond the elastic limit' some residual strainorpermanent strainswill remain in the material uon
unloading " The residual elongation corresonding to the
ermanent strain is called thepermanent set"
Elasti!it" and Plasti!it"
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!" amo#nt o$ %t&ain '!i! i% &"o"&" #+on #n,oain- i%a,," t!" ",a%ti( &"(o)"&.
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Poisson#s Ratio$ hen an elastic' homogenous and isotroic material
is sub=ected to uniform tension' it stretches axially
but contracts laterally along its entire length" Similarly' if the material is sub=ected to axial
comression' it shortens axially but bulges outlaterally %sideways&"
The ratio of lateral strain to axial strain is a constant
4nown as the 3oissons ratio'
where the strains are caused by uniaxial stress only 2(e sign is used since longitudinal elongation
%ositi(e strain& causes lateral contraction %negati(estrain& and (ice (ersa"
axial
lateral
#
=
3
&aia,
,on-it#ina,
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Examle )"1
A ). cm diameter steel rod is loaded with ;1/ 4N bytensile forces" ,nowing that the E#/.9 F3a and #."/G' determine the deformation of rod diameter
after being loaded"
Solution
in rod' #
Lateral strain'
38
MPa
m
Nx
A
p9"%8:
%"8(;
%
%8
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Exercises ))" A steel ie of length L#)"/ m' outside diameter d/#)0.mm and
inside diameter d)#)).mm is comressed by an axial force 3#
1/.4N"The material has modulus of elasticity E# /..F3a and3oissonHs Ratio v # ."-."Determine *
a& the shortening' % ans *2."700 mm&
b& the lateral strain'J lateral%ans* ))-"Gx).21&
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/" A hollow circular ost AB+ as shown in Kigure / suorts a load3)#9"0 4N acting at the to" A second load 3/ is uniformly
distributed around the ca late at B" The diameters andthic4nesses of the uer and lower arts of the ost are dAB#-/
mm' tAB# )/mm' dB+09 mm and tB+#Gmm' resecti(ely"
a& +alculate the normal stress' ABin the uer art
of the ost" %ans* G"G0 !3a&
b& If it is desired that the lower art of the ost
ha(e the same comressi(e stress as the uer
art' what should be the magnitude of the load 3/8
%ans * 3/#14N&
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-" A standard tension test is used to determine theroerties of an exerimental lastic" The testsecimen is a )0 mm diameter rod and it issub=ected to a -"0 4N tensile force" ,nowing that an
elongation of )) mm and a decrease in diameter of."1/ mm are obser(ed in a )/. mm gage length"Determine the modulus of elasticy' the modulus ofrigidity' and 3oissonHs ratio of the material"
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Shear Stress
A force actingparallel or tangentialto a section ta4enthrough a material %i"e" in the planeof the material& is called
a shear forceThe shear force intensity' i"e" shear force di(ided by thearea o(er which it acts' is called the average shear stress,
# shear stress M # shear force
A # cross2sectional area
Shear stress arises as a result of the direct action of forcestrying to cut through a material' it is 4nown as direct shear
forceShear stresses can also arise indirectly as a result of tension'
torsion or bending of a member"
A
&=
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Deending on the tye of connection' a connectingelement %bolt' ri(et' in& may be sub=ected tosingle shearor double shearas shown"
Ri(et in Single Shear
;
'dP
A&
==
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Ri(et in Double Shear
Examle )"G
Kor the )/ mm diameter bolt shown in the bolted =ointbelow' determine the a(erage shearing stress in the bolt"
''
'
;
(' d
P
d
P
A
&
===
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A
F
A
P==ae
Single Shear
A
F
A
P
'a(e ==
$ouble Shear
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Shear Strain
The e5ect of shear stress is to distortthe shae ofa body by inducing shear strains
The shear strain'is a measure of the angulardistortion of the body"
%units* degrees' radians&
L
L
x=
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%earin Stress
Bearing stress is also 4nown as a contact stress
Bearing stress in shaft 4ey
Bearing stress in ri(et and lat
r'L
M
L'
rM
A
P
(
(
'
'(===
td
P(=
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Examle /".
A unch for ma4ing holes in steel lates is shownin the $gure" Assume that a unch ha(ing
diameter d#/. mm is used to unch a hole in an ;mm lates' what is the a(erage shear stress in thelate and the a(erage comressi(e stress in theunch if the re:uired force to create the hole is 3# )).4N"
"P
20 mm
8 mm
4
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Shear &od'l's
It also 4nown as Shear !odulus of Elasticity or the!odulus of Rigidity"
Malue of shear modulus can be obtained from thelinear region of shear stress2strain diagram"
The modulus young %E&' oissonHs ratio%& and themodulus of rigidity %F& can be related as
)= Unit : Pa
%(' +=
$)
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Because of the change in the dimensions of a bodyas a result of tension or comression' the (olume ofthe body also changes within the elastic limit"
+onsider a rectangular arallel ied ha(ing sidesa' b and c in the x' y and @ directions' resecti(ely"
(ol'me Chane
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The tensile force 3 causes an axial elongation of aand lateral contractions of b and c in the x' y'and @ directions resecti(ely" ence'
Initial (olume of body' Mo# abc
Kinal (olume' Mf # %a P a&%b 2 b&%c 2 c abc%) P &%) 2&/
nitia,bo
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Exanding and neglecting higher orders of %since is (erysmall&'
Kinal (olume' Mf# abc%) P 2 /&
+hange in (olume'
M # Kinal Molume 2 Initial Molume# abc%) P 2 /& 2 abc# abc%) P
2 /
2 )&
# abc%2 / Mo%) 2 / &
ence'
0
'%(
'%(
=
=
$
&&
o
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Isotroic material is sub=ected to general triaxialstress x' yand @"
Since all strain satisfy QQ )' so (# xP yP @
x#
y#
@ #
[ ](% *x$
+
[ ](% x*$
+
[ ](% *x$
+
('%
*x#$
++
=
1
, 2 1
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am+," 2.1
titani#m a,,o ba& !a% t!" $o,,o'in- o&i-ina, im"n%ion%: =10m = 4m an 6 = 2m. !" ba& i% %#b"t" to %t&"%%"% = 14 N an
= N a% iniat" in $i-#&" b",o'. !"&"mainin- %t&"%%"% (6 6an 6) a&" a,, 6"&o. L"t = 1 kNan= 0.33 $o& t!" titani#m a,,o.(a)"t"&min" t!" !an-"% in t!" ,"n-t! $o&
an 6.
(b) "t"&min" t!" i,atation .
6
14 N14 N
N
N
2
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Allo)a*le Stress
Alied load that is less than the load the member can fully suort"%maximum load&
>ne method of secifying the allowable load for the design or analysis of amember is use a number called the Kactor of Safety %KS&"
Allowable2Stress Design
allo+
fail
F
FFS=
; 1
FS
or
FS
*ield
allo+
*ield
allo+
==
3
Stati!all" Indeterminate Axiall"
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Stati!all" Indeterminate Axiall"+oaded &em*er If a bar is $xed at both ends, as shown in
$g" %a&' two un4nown axial reactionsoccurs' and the force e:uilibriume:uation becomes
8P>>
28>
?B
y
=+
=+
8B0? =
n t!i% a%" t!" ba& i% a,," %tatia,,in"t"&minat" %in" t!" ">#i,ib&i#m">#ation a&" not %#$$ii"nt to "t"&min"t!" &"ation%.
t!" &",ation%!i+ b"t'""n t!" $o&"% atin- ont!" ba& an it% !an-"% in ,"n-t! a&" kno'n a%$o&"i%+,a"m"nt &",ation%
t!" &",ati" i%+,a"m"nt o$ on" "n o$ t!" ba&'it! &"%+"t to t!" ot!"& "n i% ">#a, to 6"&o
%in" t!" "n% %#++o&t% a&" $i". ?"n"
St ti ll I d t i t A i ll + d d
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Stati!all" Indeterminate Axiall" +oaded&em*er ,!ont
?@
P5,8B0? == 8B? =+
?#
#BB?
?#
#BB?
#BB?#?
#BB?#?
5
5>>
5
?@
?@
5>>
?@
5>
?@
5>
8
?@
5>
?@
5>
=
=
=
=
+=
+=
=
%5
5>P
>5
5>P
5
5>>P
?#
#BB
B?#
#BB
?#
#BB
B
@"a,i6in- t!at t!" int"&na, $o&" in %"-m"nt A i% B; an in %"-m"nt AC
t!" int"&na, $o&" i% D;C. !"&"$o&" t!" ">#ation an b" '&itt"n a%
=
=
+=
+=
5
5P>
5
5>P
5
55>P
5
5
5
5>P
?#B
?#B
?#
?##BB
?#
?#
?#
#B
B
B??B >P>,8P>> ==+
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Example 2.2:
S l ti
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=A ? B
=B ?
> 8 > > '8 %8 N 8 %
> '8 %8 >
, ( ) ................( )
( )
+ = + =
=
( ) ( )
B ?
? B
? ?# B #B
?
' ' ' '
? B
8 88%m
8 88%m
> 5 > 5 8 88%m?@ ?@
> 8 ;m >B 8 8 ;m > 8 8
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Example 2.:
S l i
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Sol!tion: =y ? # @
#
=? @
> 8 > > > %7 %8 N 8 %
##B M 8
> 8 ; %7 %8 8 ' > 8 ; 8 '
, ( ) ................( )
( . ) ( )( . ) ( . ) ...........( )
+ = + + =
+ =
+ + =
!" a++,i" ,oa 'i,, a#%" t!" !o&i6onta, ,in"A mo" to in,in" ,in" EAEE
# @? @
# @ ? @
? @# @
? @# @
# ? @
# #$ ? ?B @ @>7 7 7
st st st
# ?7 7
st st
8 < 8 ;
8 ; 8 58 7 8 7
% 7 %8 @ ' 7 %8 @ ' 7 %8 @
> 8 7 > 8 78 7
% 7 %8 @ ' 7 %8 @
. .
. .
..
. .
.
. .
. .. . .
( . ) ( . ).
. .
=
=
=
= +
= +
= +
=
@7
st
= = =# ? @
= =? @
# =
# ? @
> 8 78 7
' 7 %8 @
== == %8 > %8 %8 > %8 %8 >
%8 %8 > %8 %8 >>
== == %8
> 8 => 8 => e! =
( . ).
.
.
.
. . ................. ( )
+
= +
+ =
= +
=? # @> 8 > > > %7 %8 N 8 %( ) ( )+ = + + =
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y ? # @
#
=? @
# ? @
=? # @
?
> 8 > > > %7 %8 N 8 %
## M 8
> 8 ; %7 %8 8 ' > 8 ; 8 '
> 8 => 8 => e! =
Substitutee! = oe! %
> > > %7 %8 N 8 %
>
, ( ) ................( )
( . ) ( )( . ) ( . ) ...........( )
. . ................. ( )
( ) int ( )
( ) ................( )
+ + +
+ =
+ + =
= +
+ + ==
? @ @
=? @
=?
@
=@ ?
8 => 8 => > %7 %8 8
% => % => %7 %8
%7 %8 % =>>
% =
> %% 7=< %8 > e! ;
( . . ) ( )
. . ( )
( ) .
.
. ( ) ....................... ( )
+ + =
+ =
=
=
=? @
= =? ?
= =? ?
=
?
=
Substitutee! ; oe! '
> 8 ; %7 %8 8 ' > 8 ; 8
> 8 ; = %8 8 ; %% 7=< %8 > 8
> 8 ; = %8 ; 6%7 %8 8 ;> 8
9 6%7%8>
8 ' 8'/N o e! =
> 8 => 8 =>
8 = 7% %8 8 = ' 8' %8
= ;6' /N
Re . int ( )
.
. ( )
. ( ) . ( )
.
Re . int ( )
. .
. ( . ( ) . ( . )
.
=
=
=
= =
=
= +
= +
=
Thermal Stress
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A change in temerature can cause material to change itsdimensions"
If the temerature increases' generally a material exands'whereas if the temerature decreases' the material willcontract"
If this is the case' and the material is homogenous andisotroic' it has been found from exeriment that thedeformation of a member ha(ing a length L can be calculatedusing the formula
T#TLhere #linear coecient of thermal exansion %unit*
)+& T#change in temerature L#original length of the member
T#change in length of the member
Thermal Stress
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Fi"n: =1210/A
Example 2.":
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>>>
8>
B?
C
===+
Sol!tion:
?B 8= !" !an-" in ,"n-t! o$ t!" ba&i% 6"&o (b"a#%" t!" %#++o&t% onot mo")
?B T >( )+ =
o "t"&min" t!" !an-" in,"n-t! &"mo" t!" #++"& %#++o&to$ t!" ba& an obtain a ba& i%$i" at t!" ba%" an $&"" toi%+,a" at t!" #++"& "n.o t!" ba& 'i,, ",on-at" b an
amo#nt G '!"n on,t"m+"&at#&" !an-" i% atin-n t!" ba& %!o&t"n% b anamo#nt G;'!"n on, t!" &"ationi% atin-
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?B T >
T >
6
'
;
'
; '
8
>5T5 8?@
> %%' %8 68 =8 % 8
8 8% '88 %8
> %= 6 %8
8 8% '88 %8
> = 6 %8 8 8% '88 %8
9 '/N
( )
( )( )( )
. ( )
( ).
. ( )
. . ( )
.
+ =
=
=
=
=
= =
'
> 9 '/N9'MPa
? 8 8%
.;
. = =
"&a-" no&ma, t!"&ma, %t&"%%:
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Example 2.#
Fi"n:
6st
6al
st
al
%' %8 #
'= %8 #
@ '88 %8 Pa
@ 9= % %8 Pa
/
/
.
=
=
=
=
Sol'tion
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Sol'tion=
y st al> 8 '> > 8 %8 N 8 e! %, ( ) ......... ( )+ = + =
st al
st st T st >
al al T al >
st T st > al T al >
st al
st al
6 st' :
6
e! '
> 5 > 5T5 T5
? @ ? @
> 8 '7%' %8 8 '7% < %8 = ;7 %8
'7% ='9 %8 '86 6 = ;7 %8 % '% %8 >
: :;9 %8 > = ;7 %8 % '% %8 > % < %8
% 67>
( . ))( . )
( . ) ( . )
( . ) ( . ). .
. ) ( . ). . . .
. . . .
.
=
=
=
=
; :al
%8
=al
%8 % '% %8 >
: :;9 %8
%67
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=st al
= =
al al= =
al al
=al
al
al
=st
Substitute e! = o e! %
'> > 8 %8 N 8
' %67 ;'% 96 %8
> %'' %'' %67
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TRIAL )
Determine the reactions at A and B for the steelbar and loading shown' assuming a close $t at both
suorts before the loads are alied"
n%'"& @= 323 kN @b= 577kN
"t"&min" t!" &"ation% at 4an C$o&
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"t"&m n" t!" &"at on% at 4 an C $o&t!" %t"", ba& an ,oain- %!o'n a%%#min-a ,o%" $it at bot! %#++o&t% b"$o&" t!",oa% a&" a++,i".
o," $o& t!" &"ation at 4#" to
a++,i" ,oa% an t!" &"ation $o#n at C.
@">#i&" t!at t!" i%+,a"m"nt% #" to
t!" ,oa% an #" to t!" &"#nant&"ation b" om+atib," i.". &">#i&" t!att!"i& %#m b" 6"&o.
o," $o& t!" i%+,a"m"nt at C#" tot!" &"#nant &"ation at C.
HLUHN:
Aon%i"& t!" &"ation at Ca% &"#nant&","a%" t!" ba& $&om t!at %#++o&t an%o," $o& t!" i%+,a"m"nt at C#" to
t!" a++,i" ,oa%.
HLUHN: o," $o& t!" i%+,a"m"nt at C #" to t!" a++,i"
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o," $o& t!" i%+,a"m"nt at C#" to t!" a++,i",oa% 'it! t!" &"#nant on%t&aint &","a%"
$$ALP
LLLL
AAAA
PPPP
i ii
ii
:
5
;='%
'6;=
'6'%
=
;
=
='%
%8%'7"%
m%78"8
m%8'78m%8;88
N%8:88N%86888
==
====
====
====
o," $o& t!" i%+,a"m"nt at C#" to t!"&"#nant on%t&aint
( )
==
==
==
==
i
,
ii
ii!
,
$
!
$A
LP-
LL
AA
!PP
=
'%
'6'
'6%
'%
%8:7"%
m=88"8
m%8'78m%8;88
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@">#i&" t!at t!" i%+,a"m"nt% #" to t!" ,oa% an #" tot!" &"#nant &"ation b" om+atib,"
( )
/N799N%8799
8%8:7"%%8%'7"%
8
=
=:
==
=
=
=+=
,
,
!L
!
$
!
$
;in t!" &"ation at 4#" to t!" ,oa% an t!" &"ation atC
/N='=
/N799/N688/N=888
=
+==
A
A*
!
!F
/N799
/N='=
=
=
,
A
!
!
TRIAL /
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TRIAL /
Two cylindrical rods' +D made of steel %E#/.. F3a& and
A+ made of aluminum %E#9/ F3a&' are =oined at + and
restrained by rigid suorts at A and D" Determine
%a& the reactions at A and D %RA#0/"G4N' RD# ;9") 4N&
%b& The deection of oint + %.".;1 mm&
7
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TRIAL -
At room temerature %/)o+& a ."0 mm ga existsbetween the ends of the rods shown" At a later time
when the temerature has reached )1..+' determine
%a&The normal stress in the aluminum rod %a#2)0."1
!3a&
%b&The change in length of the aluminum rod %a# ."-1G
mm&
9
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