Chapter 1_stress and Strain
Transcript of Chapter 1_stress and Strain
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STRESS ANDSTRAIN
http://www.youtube.com/watch?v=S6gu7bzvMqQDAM 21! " M#$A%&$ '#'#(A)*
http://../VISUAL/Stress%20and%20Strain.flvhttp://../VISUAL/Stress%20and%20Strain.flv
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Topics:
• Introduction• Main Principles of Static's
Stress
• Normal Stress• Shear Stress• Bearing Stress• Thermal Stress
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1.1 Introduction
Mechanics : The study of how bodies react to forces acting on them
RIGID BODIES
Things that do not change shape!
Statics : The study of bodies
in an e"uilibrium
DEFORMABLE BODIES
Things that do change shape!FLUIDS
Mechanics of Materials :The study of the relationships
between the external loads
applied to a deformable body and
the intensity of internal forces
acting within the body#
Incompressible $ompressible
%ynamics :
Kinematics concerned
with the geometric aspects
of the motion
(# Kinetics concerned
with the forces causing the
motion#
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1.2 Main Principles of Statics
+ A ,oa- may be -e0e a the comb0e- eect o ete30a, o3ce
act0g o0 the bo-y.
Point Load:A po0t ,oa- o3 co0ce0t3ate- ,oa- o0e whch co0-e3e- to actat a po0t. &0 actua, p3actce4 the ,oa- ha to be -t3bute- ove3ma,, a3ea4 becaue uch ma,, 50e+e-ge co0tact a3e ge0e3a,,y
0ethe3 pob,e 0o3 -e3ab,e.
Distributed Load:A -t3bute- ,oa- o0e whch -t3bute- o3 p3ea- 0 omema00e3 ove3 the ,e0gth o the beam. & the p3ea- u0o3m ".e.at the u0o3m 3ate4 ay w 5% o3 %/mete3 3u0 * t a- to beu0o3m,y -t3bute- ,oa- a0- abb3evate- a D). & thep3ea- 0ot at u0o3m 3ate4 t a- to be 0o0+u0o3m,y-t3bute- ,oa-. 3a0gu,a3 a0- t3apezo-a,,y -t3bute- ,oa-a,, u0-e3 th catego3y.
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Point Load Uniform Load Non-uniformlyload
Figure
Free bodydiagram
(FBD)
w 5%w 5%/m w 5%/m
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Axial Load Normal Stress Shear Stress
Bearing Stress Alloable Stress De!ormation o! Stru"tural under Axial Load Stati"ally indeterminate #roblem
Thermal Stress
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$e"hani"s o! material is a study o! the
relationshi# beteen the external loads a##liedto a de!ormable body and the intensity o!internal !or"es a"ting ithin the body%
Stress & the intensity o! the internal !or"e on as#e"i'" #lane (area) #assing through a #oint%
Strain & des"ribe the de!ormation by "hangesin length o! line segments and the "hanges inthe angles beteen them
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9/741.1 Introduction
• Normal Stress : stress which acts perpendicular) or normal to) the
*! cross section of the load+carrying member#
: can be either compressi,e 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 σ the intensity o! !or"e or !or"e #er unit area a"tingnormal to ∆A
A positive sign ill be used to indi"ate a tensile stress (member in tension)
A negative sign ill be used to indi"ate acompressive stress (member in compression)
= P / A
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"a*
"b*
Stress ( ) = Force (P)
Cross Section (A)
Unit : Nm-² N/mm( or MPa
N/m( or Pa
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Assumptions :
0niform deformation: Bar
remains straight before and
after load is applied) and
cross section remains flat or plane during deformation
(# In order for uniform
deformation) force be
applied along centroidal a1isof cross section $
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A P
A P
A F F F A
!
=
=
=Σ=↑+ ∫ ∫
σ
σ
σ dd2
" 3 a,erage normal stress at any point
on cross sectional area
P 3 internal resultant normal force
A 3 cross+sectional area of the bar
1.4 Axial Loading – Normal StressDAM 21! " M#$A%&$ '#'#(A)*
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• 0se e"uation of " 3 P / A for cross+sectional area of a member when
section sub4ected to internal resultant force
Internal Loading
• Section member perpendicular to its longitudinal a1is at ptwhere normal stress is to be determined
• %raw free+body diagram• 0se e"uation of force e"uilibrium to obtain internal a1ial
force at the section
• %etermine member5s 1+sectional area at the section• $ompute a,erage normal stress " 3 P / A
Average Normal Stress
1.4 Axial Loading – Normal StressDAM 21! " M#$A%&$ '#'#(A)*
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8ee3e0ce: http://e0.w5pe-a.o3g/w5/De,ta9",ette3*
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Exam#le %*
To solid "ylindri"al rods AB and B+ are elded together at Band loaded as shon% ,noing that d)&-.mm and
d/&/.mm 'nd a0erage normal stress at the midse"tion o!*(a) rod AB1 and
(b) rod B+%
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Exam#le %/
To solid "ylindri"al roads AB and B+ are elded
together at B and loaded as shon% ,noing thatd) & -. mm and d/ & 2. mm 'nd the a0erage
normal stress in the mid se"tion o!*
(a) rod AB1 and
(b) rod B+%
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#amp,e 1.!
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Normal strain ε is the elongation or"ontra"tion o! a line segment #er unito! length
∆L & elongation
Lo & length
ε = ") )o* / )o ∆) / )o
strainnormal
6
=δ
=ε
; ∆)= δ
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Exam#le %3*Determine the "orres#onding strain !or a bar o!
length L&.%4..m and uni!orm "ross se"tionhi"h undergoes a de!ormation δ&2.×.54m%
7
7
7
&89 &9 m(89 &9 m m
6 9 799 m
(89 &9 (89
/.
@
−
−
−
δ ×ε = = = ×
= × µ
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#amp,e 1.< :A tee, 3o- 2 mm -amete3 a0- m,o0g ub>ecte- to a0 aa, pu,, o 5%. &t ete0-e- by 2.
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So,uto0 :Damete3 o tee, 3o-4 - = 2 mm @ .2 m
)e0gth4 , = m'3eu3e4 ' = 5%#te0o04 -, = 2.
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Tensile test is an ex#eriment to determinethe load5de!ormation beha0ior o! thematerial%
Data !rom tensile test "an be #lot into stressand strain diagram%
Exam#le o! test s#e"imen
5 note the dog5bone geometry
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6ni0ersal Testing $a"hine 5 e7ui#mentused to sub8e"t a s#e"imen to tension
"om#ression bending et"% loads andmeasure its res#onse
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Stress5Strain Diagrams
A number o! im#ortant me"hani"al #ro#erties o! materials that "an be dedu"ed !rom the stress5strain
diagram are illustrated in 'gure abo0e%
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Stress5Strain Diagrams
!
'o0t C+A = ,0ea3 3e,ato0hp betwee0 t3e a0- t3a0
'o0t A = p3opo3to0a, ,mt "σ')*he 3ato o t3e to t3a0 0 th ,0ea3 3ego0 o t3e+t3a0 -ag3am ca,,e- ou0g Mo-u,u o3 the Mo-u,u o #,atcty gve0
ε
σ
∆
∆Ε
σ E σ')0t: M'aDAM 21! " M#$A%&$ '#'#(A)*
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Stress5Strain Diagrams
!
'o0t A+F = pecme0 beg0 ye,-0g.
'o0t F = ye,- po0t'o0t F+G = pecme0 co0t0ue to e,o0gate wthout a0y 0c3eae 0 t3e. &t3ee3 a pe3ect,y p,atc zo0e'o0t G = t3e beg0 to 0c3eae'o0t G+D = 3ee3 a the zo0e o t3a0 ha3-e00g'o0t D = u,tmate t3e/t3e0gth pecme0 beg0 to 0ec5+-ow0
'o0t # = 3actu3e t3e DAM 21! " M#$A%&$ '#'#(A)*
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9oint : to A
9oint + to D
9oint D to E
At #oint E
Normal or engineering stress "an be determined by di0iding thea##lied load by the s#e"imen original "ross se"tional area%
True stress is "al"ulated using the a"tual "ross se"tional area atthe
instant the load is measured%
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Some o! the materials li;e aluminum (du"tile) does not ha0e"lear
yield #oint li;es stru"tural steel% There!ore stress 0alue "alled theo
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Brittle material su"h as "erami" and glass ha0e lotensile
stress 0alue but high in "om#ressi0e stress% Stress5straindiagram !or brittle material%
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#amp,e 1.6 :
he mm -amete3 cab,e FG ma-e o a tee, wth #=2H'a.$0ow0g that the mamum t3e 0 the cab,e mut 0ot ecee-1BM'a a0- that the e,o0gato0 o the cab,e mut 0ot ecee-6mm4 0- the mamum ,oa- ' that ca0 be app,e- a how0.
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Elasticity re!ers to the #ro#erty o! a material su"hthat it returns to its original dimensions a!ter
unloading% Any material hi"h de!orms hen sub8e"ted to load
and returns to its original dimensions hen unloaded
is said to be elasti"%
I! the stress is #ro#ortional to the strain the materialis said to be linear elastic otherise it is non-linear
elastic% Beyond the elasti" limit some residual strain or permanent strains ill remain in the material u#on
unloading% The residual elongation "orres#onding to the
#ermanent strain is "alled the permanent set %
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I he amou0t o t3a0 whch 3ecove3e- upo0 u0,oa-0g ca,,e- the e,a1t/c 3ecove3y.
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@hen an elasti" homogenous and isotro#i" materialis sub8e"ted to uni!orm tension it stret"hes axially
but "ontra"ts laterally along its entire length% Similarly i! the material is sub8e"ted to axial
"om#ression it shortens axially but bulges outlaterally (sideays)%
The ratio o! lateral strain to axial strain is a "onstant
;non as the 9oissons ratio
here the strains are "aused by uniaxial stress
only
axial
lateral
$ε
ε −=
!6
pa.si 1
sisi y
6
6
b d
b d
@
@
δε ε =
δ δε ε = − = −
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Exam#le %
A . "m diameter steel rod is loaded ith C4/ ;Nby tensile !or"es% ,noing that the E&/. 9a and
ν& .%/ determine the de!ormation o! rod
diameter a!ter being loaded%Solutionσ in rod σ &
Lateral strain
∴
!
MPa
m
N x
A
p;#&9=
!<
&
&9>7(
((
:
==
π
9998:#9&9(9;
;#&9=
:
===
MPa x
MPa
% a
σ ε
#"$"#$ 9998:9(=9al
ε ν ε
999&8
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Exer"ises % A steel #i#e o! length L&%/ m outside diameter d/&2.mm and
inside diameter d)&.mm is "om#ressed by an axial !or"e 9&
4/.;N%The material has modulus o! elasti"ity E& /..9a and
9oissons Ratio v & .%-.%Determine *
a) the shortening G (ans* 5.%322 mm)
b) the lateral strainH lateral (ans* -%x.54)
") the in"rease d/ in the outer diameter and the in"rease d)
in
the inner diameter
(ans* .%. mm and .%./2mm)
d) the in"rease t in the all thi";ness
(ans* .%..//C mm)
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/% A hollo "ir"ular #ost AB+ as shon in Figure / su##orts a load9)&%2 ;N a"ting at the to#% A se"ond load 9/ is uni!ormly
distributed around the "a# #late at B% The diameters andthi";nesses o! the u##er and loer #arts o! the #ost are dAB&-/
mm tAB& /mm dB+ 2 mm and tB+&mm res#e"ti0ely% a) +al"ulate the normal stress JAB in the u##er #art
o! the #ost% (ans* %2 $9a)
b) I! it is desired that the loer #art o! the #ost
ha0e the same "om#ressi0e stress as the u##er
#art hat should be the magnitude o! the load 9/K
(ans * 9/&4;N)
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- A standard tension test is used to determine the
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-% A standard tension test is used to determine the#ro#erties o! an ex#erimental #lasti"% The tests#e"imen is a 2 mm diameter rod and it issub8e"ted to a -%2 ;N tensile !or"e% ,noing that an
elongation o! mm and a de"rease in diameter o!.%4/ mm are obser0ed in a /. mm gage length%Determine the modulus o! elasti"s the modulus o!rigidity and 9oissons ratio o! the material%
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A !or"e a"ting parallel or tangential to a se"tion ta;enthrough a material (i%e% in the plane o! the material) is
"alled a shear force The shear !or"e intensity i%e% shear !or"e di0ided by thearea o0er hi"h it a"ts is "alled the average shear stress, τ
τ & shear stress
V & shear !or"e
A & "ross5se"tional area
Shear stress arises as a result o! the dire"t a"tion o! !or"estrying to "ut through a material it is ;non as dire"t shear
!or"e
Shear stresses "an also arise indire"tly as a result o!
tension torsion or bending o! a member%
A
' =τ
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De#ending on the ty#e o! "onne"tion a "onne"tingelement (bolt ri0et #in) may be sub8e"ted tosingle shear or double shear as shon%
Ri0et in Single Shear
<
(d
P
A
'
π
τ ==
2
Ri0et in Double Shear
((
(
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A
F
A
P ==a,eτ
Single Shear
A
F
A
P
(
a,e ==τ
%ouble Shear
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The e
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Bearing stress is also ;non as a "onta"t stress
Bearing stress in sha!t ;ey1
Bearing stress in ri0et and #lat1
r(L
M
L(
r M
A
P
)
)
(
!(===σ
td
P
)
=σ
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6
So,uto0 :
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It also ;non as Shear $odulus o! Elasti"ity or the$odulus o! Rigidity%
alue o! shear modulus "an be obtained !rom thelinear region o! shear stress5strain diagram%
The modulus young (E) #oissons ratio( ν) and themodulus o! rigidity () "an be related as
γ τ * = 0t : 'aca, o3 'a o3%/m2
!&( ν += % *
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Be"ause o! the "hange in the dimensions o! a bodyas a result o! tension or "om#ression the 0olume
o! the body also "hanges ithin the elasti" limit% +onsider a re"tangular #arallel #i#ed ha0ing sidesa b and " in the x y and M dire"tions res#e"ti0ely%
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The tensile !or"e 9 "auses an axial elongation o! aε and lateral "ontra"tions o! b νε and " νε in the x y andM dire"tions res#e"ti0ely% en"e
Initial 0olume o! body o & ab"
Final 0olume ! & (a P aε)(b 5 b νε)(" 5 " νε)& ab"( P ε)( 5 νε)/
&0ta,bo-y
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Ex#anding and negle"ting higher orders o! ε (sin"e ε is0ery small)
Final 0olume ! & ab"( P ε 5 / νε)
+hange in 0olume
∆ & Final olume 5 Initial olume& ab"( P ε 5 / νε) 5 ab"& ab"( P ε 5 / νε 5 )& ab"(ε 5 / νε)& o ε( 5 / ν)
en"e
6
!(&
!(&
ν σ
ε
ν ε
−=
−=∆
%
'
'
o
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Isotro#i" material is sub8e"ted to general triaxialstress σx σy and σM%
Sin"e all strain satis!y ε QQ so ε0 & εx P εy P εM
εx &
εy &
εM &
[ ]!& + x %
σ σ ν σ +−
[ ]!& x + %
σ σ ν σ +−
[ ]!& + x %
σ σ ν σ +−
!(&
+ x$ %
σ σ σ ν
ε ++−
=
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#amp,e 2.
A tta0um a,,oy ba3 ha the o,,ow0g o3g0a, -me0o0: =1cm y = cm a0- z = 2cm. he ba3 ub>ecte- to t3ee σ
= 1 % a0- σ y = + 6 %4 a 0-cate- 0 gu3e be,ow. he3ema00g t3ee "σz4 τy4 τz a0- τ yz* a3e a,, ze3o. )et # = 16 5%a0- ν = .!! o3 the tta0um a,,oy.
"a*Dete3m0e the cha0ge 0 the ,e0gth o3
∆4 ∆ y a0- ∆z.
"b* Dete3m0e the -,atato04 εv.
z
y
1 %1 %
6 %
6 %
62-/,atat/o0 = pe0gemba0ga0DAM 21! " M#$A%&$ '#'#(A)*
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A##lied load that is less than the load the member "an !ullysu##ort% (maximum load)
:ne method o! s#e"i!ying the alloable load !or the design oranalysis o! a member is use a number "alled the Fa"tor o! Sa!ety(FS)%
Alloable5Stress Design
allo,
fail
F
F FS =
KS L 1
FS
or
FS
+ield
allo,
+ield
allo,
τ
τ
σ
σ ==
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I! a bar is 'xed at both ends, as shon in
'g% (a) to un;non axial rea"tionso""urs and the !or"e e7uilibriume7uation be"omes1
9P??
29?
@B
y
=−+
=Σ↑+
9B/@ =δ
I &0 th cae4 the ba3 ca,,e- tatca,,y0-ete3m0ate4 0ce the equ,b3umequato0 a3e 0ot uce0t to -ete3m0ethe 3eacto0.
I the 3e,ato0hp betwee0 the o3ce act0g o0the ba3 a0- t cha0ge 0 ,e0gth a3e 50ow0 ao3ce+-p,aceme0t 3e,ato0
I the 3e,atve -p,aceme0t o o0e e0- o the ba3wth 3epect to the othe3 e0- equa, to ze3o
0ce the e0- uppo3t a3e e-. e0ce
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@A
P6)9B/@ =δ=δ 9B@ =δ+δ
@$
$BB@
@$
$BB@
$BB@$@
$BB@$@
6
6??
6
@A
@A
6??
@A
6?
@A
6?
9
@A
6?
@A
6?
=
×=
=
=−
+=
+=
=−
&6
6?P
?6
6?P
6
6??P
@$
$BB
B@$
$BB
@$
$BB
B
I 8ea,z0g that the 0te30a, o3ce 0 egme0t AG NKA4 a0- 0 egme0t GF4
the 0te30a, o3ce KF. he3eo3e4 the equato0 ca0 be w3tte0 a
=
=
+=
+=
66P?
6
6?P
6
66?P
6
6
6
6?P
@$B
@$B
@$
@$$BB
@$
@$
@$
$BB
B@@B ?P?)9P?? −==−+
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#amp,e 2.1:
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#amp,e 2.2:
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#amp,e 2.!:
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@ B
B @
? 9 ? ? (9 &9 N 9 &
? (9 &9 ?
, ( ) ................( )
( )
+ → Σ = − − + =
= −
( ) ( )
B @
@ B
@ @$ B $B
@
( (= ( = (
@ B
9 99&m
9 99&m
? 6 ? 6 9 99&m@A @A
? 9 m9 99&m
9 99(8m (99 &9 Nm 9 99(8m (99 &9 Nm
or
? 9 m =(; 9N (
Substitute e" & o e" (
?
/ .
.
.
( . ) ( . ).
. .
( . ) ( . ) . ................( )
( )int ( )
− −
δ =
δ − δ =
− =
− = π × π ×
− =
@ @
@
9 m =(; 9N
? &7 7.N
?B =.N
( . ) ( , )( . ) .
.
.
− − =
=
=
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A "hange in tem#erature "an "ause material to "hange itsdimensions%
I! the tem#erature in"reases generally a material ex#andshereas i! the tem#erature de"reases the material ill"ontra"t%
I! this is the "ase and the material is homogenous andisotro#i" it has been !ound !rom ex#eriment that thede!ormation o! a member ha0ing a length L "an be"al"ulated using the !ormula1
δ T&α(∆ T)L here
α&linear "oe"ient o! thermal ex#ansion (unit* >+°) ∆ T&"hange in tem#erature L&original length o! the member (m or mm) δ T&"hange in length o! the member
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E " %
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Hve0: α=121+6/G°
Eam!"e #$%:Eam!"e #$%:
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???
9?
B@
C
===Σ↑+
So"ution:So"ution:
@B 9=δ he cha0ge 0 ,e0gth o the ba3 ze3o "becaue the uppo3t -o0ot move*
@B T ?( )+ ↑ δ = δ − δ
o -ete3m0e the cha0ge 0,e0gth4 3emove the uppe3 uppo3to the ba3 a0- obta0 a ba3 e- at the bae a0- 3ee to-p,ace at the uppe3 e0-.So the ba3 w,, e,o0gate by a0
amou0t O whe0 o0,ytempe3atu3e cha0ge act0gA0- the ba3 ho3te0 by a0amou0t OK whe0 o0,y the 3eacto0
act0g
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@B T ?
T ?
7
( =
<
( =
< ( =
9
?6T6 9@A
? &&( &9 79 9 & 9
9 9& (99 &9
? & 7 &9
9 9& (99 &9
? 7 &9 9 9& (99 &9
; (.N
( )
( )( )( )
. ( )
( ).
. ( ). . ( )
.
−
−
−
+ ↑ δ = δ − δ
δ − δ =
α∆ − =
× ° − ° − =×
× =
×= × × ×=
(
? ; (.N;(MPa
@ 9 9&
.;
.σ = =
Ave3age 0o3ma, the3ma, t3e:
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y st al? 9 (? ? =9 &9 N 9 e" &, ( ) ......... ( )+ ↑ Σ = + − =
st al
st st T st ?
al al T al ?
st T st ? al T al ?
st al
st al
7 st( =
7
e" (
? 6 ? 6T6 T6
@ A @ A
? 9 (8&( &9 >9 (9 9 (89 9( (99 &9
(: &9 >9 (9
............................... ( )
( ) ( ) ( )( ) ( )
( ) ( ) ( ) ( )
( . )( )( . )( . ) ( )
(
−
−
δ = δ
+ ↑ δ = δ − δδ = δ − δ
δ − δ = δ − δ
α∆ − =α∆ −
× ° − ° − =π ×
× ° − ° al( =
< &9 : 8 &9& > &9 = =
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st al
al al
al al
al
al
al
st
Substitute e" o e" &
(? ? =9 &9 N 9
( &78 >> &9 & (&7? ? =9 &9 N 9& ;7 &9 (
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66
-1 = 22.6 mm a0- -2 = .2 mm
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6 :
To "ylindri"al rods +D made o! steel (E&/.. 9a) and
A+ made o! aluminum (E&/ 9a) are 8oined at + and
restrained by rigid su##orts at A and D% Determine
(a) the rea"tions at A and D1 and (RA&2/%;N RD& C%
;N)
(b) the dee"tion o! #oint +% (.%.C4 mm)
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At room tem#erature (/o+) a .%2 mm ga# existsbeteen the ends o! the rods shon% At a later timehen the tem#erature has rea"hed 4..+ determine
(a) the normal stress in the aluminum rod1 and
(Ja &52.%4 $9a)
(b) the "hange in length o! the aluminum rod
(Ga& .%-4 mm)