Chapter 1 - Stress and Strain

7/26/2019 Chapter 1 - Stress and Strain

1/72

CHAPTER 1STRESS AND STRAIN


2/72

2


3/72

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


4/72

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"


5/72

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


6/72

Axial Load Normal Stress Shear Stress

Bearing Stress Allowable Stress Deformation of Structural under Axial Load Statically indeterminate roblem

Thermal Stress

STRESS AND STRAIN


7/72

!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


8/728

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"


9/72

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


10/72

(a)

(b)

Stress ( ) = Force (P) Cross Section (A)

Unit: Nm - N0mm'or MPa

N0m' or Pa


11/72

11

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 #


12/72

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



13/72

13

/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



14/72

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+"


15/72


16/72

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+"


17/72

Normal strain' is the elongation orcontraction of a line segment er unitof length

L # elongation

Lo #length

= L / Lo

strainnormal5=

=

* L=


18/72

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

/.

@

= = =

=


19/72

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


20/72

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


21/72

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

28


22/72


23/72

Stress2Strain Diagrams

A number of imortant mechanical

roerties of materials that can be deduced

from the stress2strain diagram are illustratedin $gure abo(e"

30


24/72

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

31


25/72

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"

31


26/72

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

32


27/72

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


28/72

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"

34


29/72

!" amo#nt o$ %t&ain '!i! i% &"o"&" #+on #n,oain- i%a,," t!" ",a%ti( &"(o)"&.

35


30/72

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,


31/72

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


32/72

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&

39


33/72

/" 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&

40


34/72

-" 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"

49


35/72

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

&=

41


36/72

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&

==

42


37/72

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

&

===

43


38/72

A

F

A

P==ae

Single Shear

A

F

A

P

'a(e ==

$ouble Shear


39/72

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=

44


40/72

%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(=

45


41/72

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


42/72

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

%(' +=

$)

48


43/72

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

58


44/72

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

59


45/72

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


46/72

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


47/72

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


48/72

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"


49/72

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


50/72

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>> ==+


51/72

Example 2.2:

S l ti


52/72

=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


53/72

Example 2.:

S l i


54/72

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 %( ) ( )+ = + + =


55/72

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


56/72

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


57/72

Fi"n: =1210/A

Example 2.":


58/72

>>>

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-


59/72

?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&"%%:


60/72

Example 2.#

Fi"n:

6st

6al

st

al

%' %8 #

'= %8 #

@ '88 %8 Pa

@ 9= % %8 Pa

/

/

.

=

=

=

=

Sol'tion


61/72

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


62/72

=st al

= =

al al= =

al al

=al

al

al

=st

Substitute e! = o e! %

'> > 8 %8 N 8

' %67 ;'% 96 %8

> %'' %'' %67


63/72

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&


64/72

"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"


65/72

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


66/72

@">#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 /


67/72

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


68/72


69/72


70/72

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


71/72


72/72

Documents

Chapter 1 - Stress and Strain