ch10
DESCRIPTION
structures notesTRANSCRIPT
similarly transform strain
Hooe!s law
• Discuss and use theories to predict the failure of a
material
)ransformation
• As e0plained in Chapter "#"6 general state of strain
in a /ody is represented /y a com/ination of *
components of normal strain 7ε x , ε y , ε z 86 and *
components of shear strain 7γ xy, γ xz , γ yz 8#
• %train components at a pt determined /y using
strain gauges6 which is measured in specified
directions#
components of normal strain 7ε x , ε y8 and one
component of shear strain6 γ xy#
.
• :ote that the normal strains are produced /y
changes in length of the element in the x and y
directions6 while shear strain is produced /y the
relative rotation of two adacent sides of the
element#
strain#
prevent the simultaneous occurrence of plane strain
and plane stress#
ratio6 condition of τ xz < τ yz < =
re(uires γ xz < γ yz < =#
%ign Convention
defined in Chapter "#"#
cause elongation along the x
and y a0es
• %hear strain γ xy is positive if the interior angle A?@
/ecomes smaller than 9=°#
%ign Convention
• %imilar to plane stress6 when measuring the normal and shear strains relative to the x’ and y’ a0es6 the angle θ will /e positive provided it follows the curling of the right-hand fingers6 counterclocwise#
:ormal and shear strains
:ormal and shear strains
we add all elongations together#
• 5rom '(n "#"6 the normal strain along the line dx’ is
ε x’ <δ x’ /dx’. Bsing '(n 1=-16
θ γ θ ε θ ε δ cossincos' dydydx x xy y x ++=
( )210cossinsincos 22
:ormal and shear strains
• )o get the transformation e(uation for γ x’y’6 consider
amount of rotation of each of the line segments dx’
and dy’ when su/ected to strain components#
:ormal and shear strains
• Bsing '(n 1=-1 with α < δ y’/ δ x’,
• As shown6 dy’ rotates /y an amount β .
:ormal and shear strains
cos 7θ 9=°8 < − sin θ 6
• )hus we get
θ γ θ θ ε ε β
2
2
cossincos
90sin90cos90sin
β α γ
:ormal and shear strains
• Bsing trigonometric identities sin "θ = " sinθ cosθ ,
cos2 θ = (1 + cos2 θ 8" and sin"θ cos"θ < 16 we
rewrite '(ns 1=-" and 1=-. as
( )5102sin 2
2cos 22
' -θ γ
ε xy y x y x
x + −
−
−=
:ormal and shear strains
• f normal strain in the y direction is re(uired6 it can
/e o/tained from '(n 1=- /y su/stituting 7θ 9=°8 for θ . )he result is
( )6102sin 2
2cos 22
' -θ γ
ε xy y x y x
y − −
$rincipal strains
• >e can orientate an element at a pt such that the
element!s deformation is only represented /y
normal strains6 with no shear strains#
• )he material must /e isotropic6 and the a0es along
which the strains occur must coincide with the a0es
that define the principal a0es#
• )hus from '(ns 9-. and 9-6
( )8102tan - y x
$rincipal strains
( )910 222
+
−
± +
= xy y x y x γ ε ε ε ε ε
( )1110 222
22 plane-in
( )10102tan -
,a0imum in-plane shear strain
( )1210 2
ε +
,$?2)A:)
• Due to $oisson effect6 the state of plane strain is not a state of plane stress6 and vice versa#
• A pt on a /ody is su/ected to plane stress when the surface of the /ody is stress-free#
• $lane strain analysis may /e used within the plane of the stresses to analyze the results from the gauges# 2emem/er though6 there is normal strain that is perpendicular to the gauges#
,$?2)A:)
• )he state of strain at the pt can also /e represented
in terms of the ma0imum in-plane shear strain# n
this case6 an average normal strain will also act on
the element#
shear strain and its associated average normal
strains is .° from the element representing the
principal strains#
10. Strain Transformation
E$A#PLE 10.1
A differential element of material at a pt is su/ected to
a state of plane strain ε x < ==71=-386 ε y < − *==71=-386
which tends to distort the element as shown#
Determine the e(uivalent strains acting on an element
oriented at the pt6 clocwise *=° from the original
position#
• %ince θ is counterclocwise6 then θ < E*=°6 use
strain-transformation '(ns 1=- and 1=-36
( ) ( )
( ) ( ) ( )( )
( ) ( )( )
ε
ε
• %ince θ is counterclocwise6 then θ < E*=°6 use
strain-transformation '(ns 1=- and 1=-36
( ) ( )( )
( ) ( )( )
γ
• %train in the y’ direction can /e o/tained from '(n
1=-4 with θ < E*=°# However6 we can also o/tain ε y’
( ) ( )
( ) ( ) ( )( )
( ) ( )( )
shown /elow#
10. Strain Transformation
E$A#PLE 10.2
A differential element of material at a pt is su/ected to
a state of plane strain defined /y ε x < E*=71=-386
ε y < "==71=-386 γ xy < =71=-386 which tends to distort the
element as shown# Determine the principal strains at
the pt and associated orientation of the element#
"
positive counterclocwise6 from the
x a0is to the outward normals on
each face of the element#
( )
°°−=
°=°+°−°−=
−− =
− = −
−
xy y x y x
the element in the x’ direction /y applying '(n 1=-
with θ < E.#1.°# )hus
( ) ( ) ( )
( ) ( )
ε
ε
$rincipal strains
Hence ε x’ < ε "# >hen su/ected to the principal strains6
the element is distorted as shown#
10. Strain Transformation
E$A#PLE 10.'
A differential element of material at a pt is su/ected to
a state of plane strain defined /y ε x < E*=71=-386
ε y < "==71=-386 γ xy < =71=-386 which tends to distort the
element as shown# Determine the ma0imum in-plane
shear strain at the pt and associated orientation of the
element#
:ote that this orientation is .° from that shown in
'0ample 1=#" as e0pected#
E$A#PLE 10.' %SOLN&
,a0imum in-plane shear strain
Applying '(n 1=-116
( )
E$A#PLE 10.' %SOLN&
,a0imum in-plane shear strain
)hus tends to distort the element so that the right angle
/etween dx’ and dy’ is decreased 7positive sign
convention8#
γ
on the element determined from '(n 1=-1"
)hese strains tend to
( ) ( )66 107510 2
*.
• Advantage of using ,ohr!s circle for plane strain
transformation is we get to see graphically how the
normal and shear strain components at a pt vary
from one orientation of the element to the ne0t#
• 'liminate parameter θ in '(ns 1=- and 1=-3 and
rewrite as ( ) ( )
xy x
γ ε ε
$rocedure for Analysis
a/scissa represents the normal strain ε 6 with
positive to the right6 and the ordinate represents half
the value of the shear strain6 γ "6 with positive
downward#
• Bsing positive sign convention for ε x6 ε y6 and γ xy6
determine the center of the circle C 6 which is located
on the ε a0is at a distance ε avg < 7ε x ε v8" from the
origin#
$rocedure for Analysis
Construction of the circle
• $lot the reference pt A having coordinates 7ε x, γ xy/2.
)his pt represents the case for which the x’ a0is
coincides with the x a0is# Hence θ < =°# • Connect pt A with center C
of the circle and from the
shaded triangle determine
• %etch the circle#
$rocedure for Analysis
$rincipal strains
• $rincipal strains ε 1 and ε " are determined from the circle as the coordinates of pts B and D 7γ < =8#
*
$rocedure for Analysis
must /e in this same direction6
from the element!s reference a0is x to the x’ a0is#
• >hen ε 1 and ε " are indicated as /eing positive as
shown earlier6 the element shown here will elongate
in the x’ and y’ directions as shown /y the dashed
outline#
$rocedure for Analysis
,a0imum in-plane shear strain
• Average normal strain and half the ma0imum in-plane shear strain are determined from the circle as the coordinates of pts E and F #
• ?rientation of the plane on which and ε avg act can /e
determined from the circle /y calculating "θ s1 using trigonometry# )his angle is measured clocwise from the radial reference lines CA to CF #
planein
ma0
$rocedure for Analysis
θ ps16 must /e in this same
direction6 from the element!s
%trains on ar/itrary plane
• :ormal and shear strain components ε x’ and γ x’y’ for
a plane specified at an angle θ 6 can /e o/tained
from the circle using trigonometry to determine the
coordinates of pt P #
$rocedure for Analysis
%trains on ar/itrary plane
• )o locate P 6 the nown angle θ of the x’ a0is is measured on the circle as "θ # )his measurement is made from the radial reference line CA to the radial reference line CA to CP # 2emem/er that measurements for "θ on the circle must /e in the same direction as for the x’ a0is#
."
%tate of plane strain at a pt represented /y the
components ε x < "=71=-386 ε y < E1=71=-386 and
γ xy < 1"=71=-38# Determine the principal strains
and the orientation of the element#
.*
esta/lished as shown# :ote
/e directed downward so that
counterclocwise rotations of
the element correspond to
around the circle6 and vice
versa# Center of the circle is located on the a0is at
( ) ( ) ( )66 105010
reference pt A 7θ < =°8 has
coordinates F"=71=-386 3=71=-38G#
5rom shaded triangle6 radius
of circle is CA
$rincipal strains
)he ε coordinates of pts B and D represent the principal strains# )hey are
( )
element is oriented
counterclocwise #*°# )his also defines the direction of ε 1#
)he deformation of the element is also shown#
( )
°= −
=
%tate of plane strain at a pt represented /y the
components ε x < "=71=-386 ε y < E1=71=-386 and
γ xy < 1"=71=-38# Determine the ma0imum in-plane
shear strains and orientation of the element#
.
Half the ma0imum in-plane shear strain and average
normal strain are represented /y the coordinates of
( )
( ) ( )
E$A#PLE 10., %SOLN&
,a0imum in-plane shear strain
)o orientate the element6 determine the clocwise angle "θ s1 from the circle6
( )
°= °−°=
%tate of plane strain at a pt represented /y an
element having the components ε x < E*==71=-386
ε y < E1==71=-386 and γ xy < 1==71=-38# Determine the
state of strain on an element oriented "=° clocwise
from this reported position#
esta/lished as shown#
the ε a0is at
Coordinates of reference pt A is F*==71=-386 =71=-38G#
2adius CA determined from shaded triangle6
( ) ( ) ( ) ( )6622 108.1111050200300 −−
radial line CP 6 "7"=°8 < .=° clocwise6 measured from
CA 7θ < =°8# Coordinates of pt P 7ε x!6 γ x’y’ "8 are
o/tained from the geometry of the circle#
)hus ( )
:ormal strain ε y’ can /e determined from the ε
coordinate of pt Q on the circle# >hy
As a result of these strains6 the
element deforms relative to the
x’ 6 y’ a0es as shown#
( ) 66 ' 103.911043.13cos8.111200
along the x’, y’ and z’ a0es as shown#
• Bse ,ohr!s circle to determine
ma0imum in-plane shear strain for
the x’-y’ 6 x’-z’ and y’-z’ planes#
(10.+ ABSOLUTE #A$I#U# SHEAR STRAIN
ma0imum shear strain is
determined from the circle
having the larges radius#
( )
same sign6 the largest circle has
a radius of R < 7γ x’z’ 8ma0"#
• )his value represents the
larger than the ma0imum
max'' maxmax
principal in-plane strains of
minmax
'' max
,$?2)A:)
• &eneral *-D state of strain at a pt can /e represented /y an element oriented so that only three principal strains act on it#
• 5rom this orientation6 the orientation of the element representing the a/solute ma0imum shear strain can /e o/tained /y rotating the element .° a/out the a0is defining the direction of int#
• )he a/solute ma0imum shear strain will /e larger that the ma0imum in-plane shear strain whenever the in-plane principal strains have the same sign6 the a/solute ma0imum shear strain will act out of the plane#
(10.+ ABSOLUTE #A$I#U# SHEAR STRAIN
10. Strain Transformation
E$A#PLE 10.
$lane of strain at a pt is represented /y the strain
components ε x < .==71=-386 ε y < "==71=-386
γ xy < 1=71=-38# Determine the ma0imum in-plane
shear strain and the a/solute ma0imum shear strain#
Bsing ,ohr!s circle method6 center of circle is on the
ε -a0is at
%ince γ xy" < 471=-386 reference pt has coordinates A
FE.==71=-386 471=-38G# 2adius of circle is
( ) ( )66 1010010
( )
element orientations a/out each of the x’ , y’ and z’
a0es are shown# >e see that the principal in-plane
strains have opposite signs6 and ma0imum in-plane
shear strain is also the a/solute
ma0imum shear strain
specimen using an electrical-resistance strain
gauge#
• 5or general loading on a /ody6 the normal strains at
a pt are measured using a cluster of * electrical-
resistance strain gauges#
are called strain rosettes#
• :ote that only the strains in the plane of the gauges
are measured /y the strain rosette# )hat is 6the
normal strain on the surface is not measured#
10., STRAIN ROSETTES
'(n 1=-" to each gauge
• >e determine the values of ε x6 ε y γ xy /y solving the
three e(uations simultaneously#
++=
++=
++=
• 5or rosettes arranged in the .° pattern6 '(n 1=-13 /ecomes
• 5or rosettes arranged in the 3=° pattern6
'(n 1=-13 /ecomes
10. Strain Transformation
E$A#PLE 10./
%tate of strain at pt A on /racet is measured using
the strain rosette shown# Due to the loadings6 the
readings from the gauges give ε a < 3=71=-386
ε b < 1*71=-386 and ε c < "3.71=-38# Determine the
in-plane principal strains at the pt and the directions
in which they act#
'sta/lish x a0is as shown6 measure the
angles counterclocwise from the x a0is
to center-lines of each gauge6 we have
( )
%olving '(ns 7186 7"8 and 7*8 simultaneously6 we get
)he in-plane principal strains can also /e o/tained
directly from '(n 1=-14# 2eference pt on ,ohr!s circle
is A F3=71=-386 E4.#71=-38G and center of circle6 C is on
the ε a0is at ε a"# < 1*71=-38#
5rom shaded triangle6 radius is
E$A#PLE 10./ %SOLN&
( ) ( ) ( ) ( )6
622
102.119
105.7460153
)he in-plane principal strains are thus
Deformed element is shown dashed# Due to $oisson effect6 element also su/ected to an out-of-plane strain6 in the z direction6 although this value does not influence the calculated results#
E$A#PLE 10./ %SOLN&
&eneralized Hooe!s law
• ,aterial at a pt su/ected to a state of tria0ial
stress6 with associated strains#
• >e use principle of superposition6 $oisson!s ratio
7ε lat = υε lon#86 and Hooe!s law 7ε = σ E 8 to relate
stresses to strains6 in the unia0ial direction#
• >ith σ x applied6 element elongates in the x
direction and strain is this direction is
E x
x σ
&eneralized Hooe!s law
• >ith σ y applied6 element contracts with a strain ε I! x
in the x direction6
• Jiewise6 >ith σ z applied6 a contraction is caused
in the z direction6
( )[ ]
( )[ ] ( )
+−=
+−=
+−=
&eneralized Hooe!s law
• f we apply a shear stress τ xy to the element6
e0perimental o/servations show that it will deform
only due to shear strain γ xy# %imilarly for τ xz an( γ xy,
τ yz and γ yz # )hus6 Hooe!s law for shear stress and
shear strain is written as
( )1910 111
• >e stated in chapter *#4
• 2elate principal strain to shear stress6
• :ote that since σ x = σ y = σ z < =6 then from '(n
1=-16 ε x = ε y < =# %u/stitute into transformation
'(n 1=-196
2elationship involving E, υ 6 and G
• @y Hooe!s law6 γ xy < τ xy/G# %o ε max = τ xy / "G#
• %u/stitute into '(n 1=-"1 and rearrange to o/tain
'(n 1=-"=#
stresses σ x, σ y, σ z .
• %ides of element are (x, (y and dz 6 and after stress
application6 they /ecome 71 ε x8dx6 71 ε ydy6
71 ε z8dz 6 respectively#
10.- #ATERIAL"PROPERT RELATIONSHIPS
• Change in volume per unit volume is the
Kvolumetric strainL or dilatation e#
• Bsing generalized Hooe!s law6 we write the
dilatation in terms of applied stress#
( ) ( ) dz dydxdz dydxV z y x −+++= ε ε ε δ 111
( )2210 - z y x dV
V e ε ε ε
δ ++==
uniform pressure p of a li(uid6 pressure is the same
in all directions#
• As shear resistance of a li(uid is zero6 we can
ignore shear stresses#
• )hus6 an element of the /ody is su/ected to
principal stresses σ x = σ y = σ z = –p# %u/stituting into
'(n 1=-"* and rearranging6
Dilatation and @ul ,odulus
• )his ratio 7 pe8 is similar to the ratio of linear-elastic
stress to strain6 thus terms on the 2H% are called
the volume modulus of elasticity or the /ul
modulus# Having same units as stress with
sym/ol 6
• 5rom '(n 1=-"6 theoretical ma0imum value of
$oisson!s ratio is therefore υ < =##
• >hen plastic yielding occurs6 υ < =# is used#
( ) ( )2510
213 -
,$?2)A:)
• >hen homogeneous and isotropic material is su/ected to a state of tria0ial stress6 the strain in one of the stress directions is influence /y the strains produced /y all stresses# )his is the result of the $oisson effect6 and results in the form of a generalized Hooe!s law#
• A shear stress applied to homogenous and isotropic material will only produce shear strain in the same plane#
• ,aterial constants6 E, G and υ are related mathematically#
=
normal strain6 not shear strain#
• )he /ul modulus is a measure of the stiffness of a
volume of material# )his material property provides
an upper limit to $oisson!s ratio of υ < =#6 which
remains at this value while plastic yielding occurs#
10. Strain Transformation
E$A#PLE 10.10
Copper /ar is su/ected to a uniform loading along its
edges as shown# f it has a length a < *== mm6 width
b < = mm6 and thicness ! < "= mm /efore the load
is applied6 determine its new length6 width6 and
thicness after application of the load#
)ae E c+ < 1"= &$a6 υ c+ < =#*.#
"
E$A#PLE 10.10 %SOLN&
@y inspection6 /ar is su/ected to a state of plane
stress# 5rom loading6 we have
Associated strains are determined from generalized
Hooe!s law6 '(n 1=-1;
( )
( )
( ) ( ) ( )
( )
( )
( ) ( )
f rectangular /loc shown is su/ected to a uniform
pressure of p < "= $a6 determine the dilatation and
change in length of each side#
)ae E < 3== $a6 υ < =#.#
)he dilatation can /e determined using '(n 1=-"*
( )
( ) ( )[ ]
Hooe!s law6 '(n 1=-1;
( )[ ]
( ) ( )[ ]
)he negative signs indicate that each dimension is
decreased#
• >hen engineers design for a material6 there is a
need to set an upper limit on the state of stress that
defines the material!s failure#
• However6 criteria for the a/ove failure modes is not
easy to define under a /ia0ial or tria0ial stress#
• )hus6 four theories are introduced to o/tain the
principal stresses at critical states of stress#
(10. THEORIES O! !AILURE
ductile material 7e#g#6 steel8 is slipping#
• %lipping occurs along the contact
planes of randomly-ordered crystals
• 'dges of planes of slipping as they appear on the
surface of the strip are referred to as Jder!s lines#
• )he lines indicate the slip planes in the strip6 which
occur at appro0imately .° with the a0is of the strip#
(10. THEORIES O! !AILURE
the strip6 which occur at appro0imately
.° with the a0is of the strip#
• Consider an element6 determine ma0imum shear
stress from ,ohr!s circle6
proposed the ma0imum-shear-stress
( )2610 2
failure will occur out of the plane
• f in-plane principal stresses are of opposite signs6
failure occurs in the plane
2 maxσ
τ = ma0
for plane stress for any two in-plane principal
( )
σ σ σ σ
σ σ σ σ
strain-energy density#
strain-energy density is written as
( )2810 2
1 -σε =#
332211 2
a/ove e(n
yielding of a ductile material occurs when the
distortion energy per unit volume of the material
e(uals or e0ceeds the distortion energy per unit
volume of the same material when su/ected to
yielding in a simple tension test#
( ) ( )2910
22
1
233121
σ σ σ
• n the case of plane stress6
• 5or unia0ial tension test6 σ 1 < σ M6 σ " < σ * < =
( ) ( ) ( )[ ]2 13
2 32
2 21
υ −+−+−+= E
υ +−+= E
• %ince ma0imum-distortion energy theory re(uires
#d < 7#d 8M6 then for the case of plane or /ia0ial
stress6 we have
graph#
fail#
when the ma0imum principal stress
σ 1 in the material reaches a limiting value that is
e(ual to the ultimate normal stress the material can
sustain when su/ected to simple tension#
• 5or the material su/ected to plane stress
( )31102
1
-ult
ult
have stress-strain diagrams similar in /oth tension
and compression#
• Bse for /rittle materials where the tension and
compression properties are different#
determine the criterion#
• Carry out a unia0ial tensile test to determine the
ultimate tensile stress 7σ ult8t
• Carry out a unia0ial compressive test to determine
the ultimate compressive stress 7σ ult8c
• Carry out a torsion test to determine the ultimate
shear stress τ ult#
• Circle A represents the stress condition σ 1 < σ " < =6
σ * < E7σult8c
σ " < σ * < =
• )he Criterion can also /e represented on a graph
of principal stresses σ 1 and σ " 7σ * < =8#
initiation of yielding6 whereas if it is /rittle6 it is
specified /y fracture#
/etween the crystals that compose the material#
• )his slipping is due to shear stress and the
ma0imum-shear-stress theory is /ased on this
idea#
su/ected to normal stress#
the strain energy that distorts the material6 and not
the part that increases its volume#
• )he fracture of a /rittle material is caused /y the
ma0imum tensile stress in the material6 and not the
compressive stress#
theory6 and it is applica/le if the stress-strain
diagram is similar in tension and compression#
is different in tension and compression6 then
,ohr!s failure criterion may /e used to predict
failure#
/rittle material is difficult to predict6 and so theories
of failure for /rittle materials should /e used with
caution#
%teel pipe has inner diameter of 3= mm and outer
diameter of = mm# f it is su/ected to a torsional
moment of :m and a /ending moment of
*# :m6 determine if these loadings cause failure as
defined /y the ma0imum-distortion-energy theory#
Mield stress for the steel found from a tension test is
σ M < "= ,$a#
(E$A#PLE 10.12 %SOLN&
nvestigate a pt on pipe that is su/ected to a state of
ma0imum critical stress#
throughout the pipe!s length#
distri/utions shown#
(E$A#PLE 10.12 %SOLN&
@y inspection6 pts A and @ su/ected to same state of
critical stress# %tress at A6
( ) ( )
( ) ( ) ( )[ ] ( ) ( )
,ohr!s circle for this state of stress has center located
at
shaded triangle to /e R < 1"4#1
and the in-plane principal
%ince criterion is met6 material within the pipe will not
yield 7KfailL8 according to the ma0imum-distortion-
energy theory#
10. Strain Transformation
(E$A#PLE 10.1+
%olid shaft has a radius of =# cm and made of steel
having yield stress of σ M < *3= ,$a# Determine if the
loadings cause the shaft to fail according to the
ma0imum-shear-stress theory and the ma0imum-
distortion-energy theory#
%tate of stress in shaft caused /y a0ial force and
tor(ue# %ince ma0imum shear stress caused /y
tor(ue occurs in material at outer surface6 we have
( )
( )
( )
at pt A# 2ather than use ,ohr!s circle6 principal
stresses are o/tained using stress-transformation
e(ns 9-
a/solute ma0imum shear stress occur in the plane6
apply '(n 1=-"46
theory#
However6 using the ma0imum-distortion-energy
( ) ( ) ( ) ( ) ( )[ ] ( )
deformations that only occur in a single plane6 then
it undergoes plain strain#
• f the strain components ε x6 ε y6 and γ xy are nown for
a specified orientation of the element6 then the
strains acting for some other orientation of the
element can /e determined using the plane-strain
transformation e(uations#
transformation e(uations#
in a semi-graphical manner using ,ohr!s circle#
• 'sta/lish the ε and γ " a0es6 then compute
center of circle F7ε x + ε y8"6 =G and controlling pt
Fε , γ "G6 /efore plotting the circle#
• 2adius of circle e0tends /etween these two pts
and is determined from trigonometry#
• A/solute ma0imum shear strain e(uals the
ma0imum in-plane shear strain provided the
in-plane principal strains are of opposite signs#
then a/solute ma0imum shear strain will occur out
of plane and is determined from γ ma0 < ε ma0"#
• Hooe!s law can /e e0pressed in * dimensions6
where each strain is related to the * normal stress
components using the material properties E 6 and υ 6 as seen in '(ns 1=-1#
• f E and υ are nown6 then G can /e determined
using G : E F"71 υ G#
• Dilatation is a measure of volumetric strain6 and the
/ul modulus is used to measure the stiffness of a
volume of material#
are nown6 then a theory of failure can /e used
as a /asis for design#
• Ductile materials fail in shear6 and here the
ma0imum-shear-stress theory or the ma0imum-
distortion-energy theory can /e used to predict
failure#
stress#
ma0imum-normal-stress theory or ,ohr!s
• Comparisons are made with the ultimate tensile
stress developed in a specimen#
Hooe!s law
• Discuss and use theories to predict the failure of a
material
)ransformation
• As e0plained in Chapter "#"6 general state of strain
in a /ody is represented /y a com/ination of *
components of normal strain 7ε x , ε y , ε z 86 and *
components of shear strain 7γ xy, γ xz , γ yz 8#
• %train components at a pt determined /y using
strain gauges6 which is measured in specified
directions#
components of normal strain 7ε x , ε y8 and one
component of shear strain6 γ xy#
.
• :ote that the normal strains are produced /y
changes in length of the element in the x and y
directions6 while shear strain is produced /y the
relative rotation of two adacent sides of the
element#
strain#
prevent the simultaneous occurrence of plane strain
and plane stress#
ratio6 condition of τ xz < τ yz < =
re(uires γ xz < γ yz < =#
%ign Convention
defined in Chapter "#"#
cause elongation along the x
and y a0es
• %hear strain γ xy is positive if the interior angle A?@
/ecomes smaller than 9=°#
%ign Convention
• %imilar to plane stress6 when measuring the normal and shear strains relative to the x’ and y’ a0es6 the angle θ will /e positive provided it follows the curling of the right-hand fingers6 counterclocwise#
:ormal and shear strains
:ormal and shear strains
we add all elongations together#
• 5rom '(n "#"6 the normal strain along the line dx’ is
ε x’ <δ x’ /dx’. Bsing '(n 1=-16
θ γ θ ε θ ε δ cossincos' dydydx x xy y x ++=
( )210cossinsincos 22
:ormal and shear strains
• )o get the transformation e(uation for γ x’y’6 consider
amount of rotation of each of the line segments dx’
and dy’ when su/ected to strain components#
:ormal and shear strains
• Bsing '(n 1=-1 with α < δ y’/ δ x’,
• As shown6 dy’ rotates /y an amount β .
:ormal and shear strains
cos 7θ 9=°8 < − sin θ 6
• )hus we get
θ γ θ θ ε ε β
2
2
cossincos
90sin90cos90sin
β α γ
:ormal and shear strains
• Bsing trigonometric identities sin "θ = " sinθ cosθ ,
cos2 θ = (1 + cos2 θ 8" and sin"θ cos"θ < 16 we
rewrite '(ns 1=-" and 1=-. as
( )5102sin 2
2cos 22
' -θ γ
ε xy y x y x
x + −
−
−=
:ormal and shear strains
• f normal strain in the y direction is re(uired6 it can
/e o/tained from '(n 1=- /y su/stituting 7θ 9=°8 for θ . )he result is
( )6102sin 2
2cos 22
' -θ γ
ε xy y x y x
y − −
$rincipal strains
• >e can orientate an element at a pt such that the
element!s deformation is only represented /y
normal strains6 with no shear strains#
• )he material must /e isotropic6 and the a0es along
which the strains occur must coincide with the a0es
that define the principal a0es#
• )hus from '(ns 9-. and 9-6
( )8102tan - y x
$rincipal strains
( )910 222
+
−
± +
= xy y x y x γ ε ε ε ε ε
( )1110 222
22 plane-in
( )10102tan -
,a0imum in-plane shear strain
( )1210 2
ε +
,$?2)A:)
• Due to $oisson effect6 the state of plane strain is not a state of plane stress6 and vice versa#
• A pt on a /ody is su/ected to plane stress when the surface of the /ody is stress-free#
• $lane strain analysis may /e used within the plane of the stresses to analyze the results from the gauges# 2emem/er though6 there is normal strain that is perpendicular to the gauges#
,$?2)A:)
• )he state of strain at the pt can also /e represented
in terms of the ma0imum in-plane shear strain# n
this case6 an average normal strain will also act on
the element#
shear strain and its associated average normal
strains is .° from the element representing the
principal strains#
10. Strain Transformation
E$A#PLE 10.1
A differential element of material at a pt is su/ected to
a state of plane strain ε x < ==71=-386 ε y < − *==71=-386
which tends to distort the element as shown#
Determine the e(uivalent strains acting on an element
oriented at the pt6 clocwise *=° from the original
position#
• %ince θ is counterclocwise6 then θ < E*=°6 use
strain-transformation '(ns 1=- and 1=-36
( ) ( )
( ) ( ) ( )( )
( ) ( )( )
ε
ε
• %ince θ is counterclocwise6 then θ < E*=°6 use
strain-transformation '(ns 1=- and 1=-36
( ) ( )( )
( ) ( )( )
γ
• %train in the y’ direction can /e o/tained from '(n
1=-4 with θ < E*=°# However6 we can also o/tain ε y’
( ) ( )
( ) ( ) ( )( )
( ) ( )( )
shown /elow#
10. Strain Transformation
E$A#PLE 10.2
A differential element of material at a pt is su/ected to
a state of plane strain defined /y ε x < E*=71=-386
ε y < "==71=-386 γ xy < =71=-386 which tends to distort the
element as shown# Determine the principal strains at
the pt and associated orientation of the element#
"
positive counterclocwise6 from the
x a0is to the outward normals on
each face of the element#
( )
°°−=
°=°+°−°−=
−− =
− = −
−
xy y x y x
the element in the x’ direction /y applying '(n 1=-
with θ < E.#1.°# )hus
( ) ( ) ( )
( ) ( )
ε
ε
$rincipal strains
Hence ε x’ < ε "# >hen su/ected to the principal strains6
the element is distorted as shown#
10. Strain Transformation
E$A#PLE 10.'
A differential element of material at a pt is su/ected to
a state of plane strain defined /y ε x < E*=71=-386
ε y < "==71=-386 γ xy < =71=-386 which tends to distort the
element as shown# Determine the ma0imum in-plane
shear strain at the pt and associated orientation of the
element#
:ote that this orientation is .° from that shown in
'0ample 1=#" as e0pected#
E$A#PLE 10.' %SOLN&
,a0imum in-plane shear strain
Applying '(n 1=-116
( )
E$A#PLE 10.' %SOLN&
,a0imum in-plane shear strain
)hus tends to distort the element so that the right angle
/etween dx’ and dy’ is decreased 7positive sign
convention8#
γ
on the element determined from '(n 1=-1"
)hese strains tend to
( ) ( )66 107510 2
*.
• Advantage of using ,ohr!s circle for plane strain
transformation is we get to see graphically how the
normal and shear strain components at a pt vary
from one orientation of the element to the ne0t#
• 'liminate parameter θ in '(ns 1=- and 1=-3 and
rewrite as ( ) ( )
xy x
γ ε ε
$rocedure for Analysis
a/scissa represents the normal strain ε 6 with
positive to the right6 and the ordinate represents half
the value of the shear strain6 γ "6 with positive
downward#
• Bsing positive sign convention for ε x6 ε y6 and γ xy6
determine the center of the circle C 6 which is located
on the ε a0is at a distance ε avg < 7ε x ε v8" from the
origin#
$rocedure for Analysis
Construction of the circle
• $lot the reference pt A having coordinates 7ε x, γ xy/2.
)his pt represents the case for which the x’ a0is
coincides with the x a0is# Hence θ < =°# • Connect pt A with center C
of the circle and from the
shaded triangle determine
• %etch the circle#
$rocedure for Analysis
$rincipal strains
• $rincipal strains ε 1 and ε " are determined from the circle as the coordinates of pts B and D 7γ < =8#
*
$rocedure for Analysis
must /e in this same direction6
from the element!s reference a0is x to the x’ a0is#
• >hen ε 1 and ε " are indicated as /eing positive as
shown earlier6 the element shown here will elongate
in the x’ and y’ directions as shown /y the dashed
outline#
$rocedure for Analysis
,a0imum in-plane shear strain
• Average normal strain and half the ma0imum in-plane shear strain are determined from the circle as the coordinates of pts E and F #
• ?rientation of the plane on which and ε avg act can /e
determined from the circle /y calculating "θ s1 using trigonometry# )his angle is measured clocwise from the radial reference lines CA to CF #
planein
ma0
$rocedure for Analysis
θ ps16 must /e in this same
direction6 from the element!s
%trains on ar/itrary plane
• :ormal and shear strain components ε x’ and γ x’y’ for
a plane specified at an angle θ 6 can /e o/tained
from the circle using trigonometry to determine the
coordinates of pt P #
$rocedure for Analysis
%trains on ar/itrary plane
• )o locate P 6 the nown angle θ of the x’ a0is is measured on the circle as "θ # )his measurement is made from the radial reference line CA to the radial reference line CA to CP # 2emem/er that measurements for "θ on the circle must /e in the same direction as for the x’ a0is#
."
%tate of plane strain at a pt represented /y the
components ε x < "=71=-386 ε y < E1=71=-386 and
γ xy < 1"=71=-38# Determine the principal strains
and the orientation of the element#
.*
esta/lished as shown# :ote
/e directed downward so that
counterclocwise rotations of
the element correspond to
around the circle6 and vice
versa# Center of the circle is located on the a0is at
( ) ( ) ( )66 105010
reference pt A 7θ < =°8 has
coordinates F"=71=-386 3=71=-38G#
5rom shaded triangle6 radius
of circle is CA
$rincipal strains
)he ε coordinates of pts B and D represent the principal strains# )hey are
( )
element is oriented
counterclocwise #*°# )his also defines the direction of ε 1#
)he deformation of the element is also shown#
( )
°= −
=
%tate of plane strain at a pt represented /y the
components ε x < "=71=-386 ε y < E1=71=-386 and
γ xy < 1"=71=-38# Determine the ma0imum in-plane
shear strains and orientation of the element#
.
Half the ma0imum in-plane shear strain and average
normal strain are represented /y the coordinates of
( )
( ) ( )
E$A#PLE 10., %SOLN&
,a0imum in-plane shear strain
)o orientate the element6 determine the clocwise angle "θ s1 from the circle6
( )
°= °−°=
%tate of plane strain at a pt represented /y an
element having the components ε x < E*==71=-386
ε y < E1==71=-386 and γ xy < 1==71=-38# Determine the
state of strain on an element oriented "=° clocwise
from this reported position#
esta/lished as shown#
the ε a0is at
Coordinates of reference pt A is F*==71=-386 =71=-38G#
2adius CA determined from shaded triangle6
( ) ( ) ( ) ( )6622 108.1111050200300 −−
radial line CP 6 "7"=°8 < .=° clocwise6 measured from
CA 7θ < =°8# Coordinates of pt P 7ε x!6 γ x’y’ "8 are
o/tained from the geometry of the circle#
)hus ( )
:ormal strain ε y’ can /e determined from the ε
coordinate of pt Q on the circle# >hy
As a result of these strains6 the
element deforms relative to the
x’ 6 y’ a0es as shown#
( ) 66 ' 103.911043.13cos8.111200
along the x’, y’ and z’ a0es as shown#
• Bse ,ohr!s circle to determine
ma0imum in-plane shear strain for
the x’-y’ 6 x’-z’ and y’-z’ planes#
(10.+ ABSOLUTE #A$I#U# SHEAR STRAIN
ma0imum shear strain is
determined from the circle
having the larges radius#
( )
same sign6 the largest circle has
a radius of R < 7γ x’z’ 8ma0"#
• )his value represents the
larger than the ma0imum
max'' maxmax
principal in-plane strains of
minmax
'' max
,$?2)A:)
• &eneral *-D state of strain at a pt can /e represented /y an element oriented so that only three principal strains act on it#
• 5rom this orientation6 the orientation of the element representing the a/solute ma0imum shear strain can /e o/tained /y rotating the element .° a/out the a0is defining the direction of int#
• )he a/solute ma0imum shear strain will /e larger that the ma0imum in-plane shear strain whenever the in-plane principal strains have the same sign6 the a/solute ma0imum shear strain will act out of the plane#
(10.+ ABSOLUTE #A$I#U# SHEAR STRAIN
10. Strain Transformation
E$A#PLE 10.
$lane of strain at a pt is represented /y the strain
components ε x < .==71=-386 ε y < "==71=-386
γ xy < 1=71=-38# Determine the ma0imum in-plane
shear strain and the a/solute ma0imum shear strain#
Bsing ,ohr!s circle method6 center of circle is on the
ε -a0is at
%ince γ xy" < 471=-386 reference pt has coordinates A
FE.==71=-386 471=-38G# 2adius of circle is
( ) ( )66 1010010
( )
element orientations a/out each of the x’ , y’ and z’
a0es are shown# >e see that the principal in-plane
strains have opposite signs6 and ma0imum in-plane
shear strain is also the a/solute
ma0imum shear strain
specimen using an electrical-resistance strain
gauge#
• 5or general loading on a /ody6 the normal strains at
a pt are measured using a cluster of * electrical-
resistance strain gauges#
are called strain rosettes#
• :ote that only the strains in the plane of the gauges
are measured /y the strain rosette# )hat is 6the
normal strain on the surface is not measured#
10., STRAIN ROSETTES
'(n 1=-" to each gauge
• >e determine the values of ε x6 ε y γ xy /y solving the
three e(uations simultaneously#
++=
++=
++=
• 5or rosettes arranged in the .° pattern6 '(n 1=-13 /ecomes
• 5or rosettes arranged in the 3=° pattern6
'(n 1=-13 /ecomes
10. Strain Transformation
E$A#PLE 10./
%tate of strain at pt A on /racet is measured using
the strain rosette shown# Due to the loadings6 the
readings from the gauges give ε a < 3=71=-386
ε b < 1*71=-386 and ε c < "3.71=-38# Determine the
in-plane principal strains at the pt and the directions
in which they act#
'sta/lish x a0is as shown6 measure the
angles counterclocwise from the x a0is
to center-lines of each gauge6 we have
( )
%olving '(ns 7186 7"8 and 7*8 simultaneously6 we get
)he in-plane principal strains can also /e o/tained
directly from '(n 1=-14# 2eference pt on ,ohr!s circle
is A F3=71=-386 E4.#71=-38G and center of circle6 C is on
the ε a0is at ε a"# < 1*71=-38#
5rom shaded triangle6 radius is
E$A#PLE 10./ %SOLN&
( ) ( ) ( ) ( )6
622
102.119
105.7460153
)he in-plane principal strains are thus
Deformed element is shown dashed# Due to $oisson effect6 element also su/ected to an out-of-plane strain6 in the z direction6 although this value does not influence the calculated results#
E$A#PLE 10./ %SOLN&
&eneralized Hooe!s law
• ,aterial at a pt su/ected to a state of tria0ial
stress6 with associated strains#
• >e use principle of superposition6 $oisson!s ratio
7ε lat = υε lon#86 and Hooe!s law 7ε = σ E 8 to relate
stresses to strains6 in the unia0ial direction#
• >ith σ x applied6 element elongates in the x
direction and strain is this direction is
E x
x σ
&eneralized Hooe!s law
• >ith σ y applied6 element contracts with a strain ε I! x
in the x direction6
• Jiewise6 >ith σ z applied6 a contraction is caused
in the z direction6
( )[ ]
( )[ ] ( )
+−=
+−=
+−=
&eneralized Hooe!s law
• f we apply a shear stress τ xy to the element6
e0perimental o/servations show that it will deform
only due to shear strain γ xy# %imilarly for τ xz an( γ xy,
τ yz and γ yz # )hus6 Hooe!s law for shear stress and
shear strain is written as
( )1910 111
• >e stated in chapter *#4
• 2elate principal strain to shear stress6
• :ote that since σ x = σ y = σ z < =6 then from '(n
1=-16 ε x = ε y < =# %u/stitute into transformation
'(n 1=-196
2elationship involving E, υ 6 and G
• @y Hooe!s law6 γ xy < τ xy/G# %o ε max = τ xy / "G#
• %u/stitute into '(n 1=-"1 and rearrange to o/tain
'(n 1=-"=#
stresses σ x, σ y, σ z .
• %ides of element are (x, (y and dz 6 and after stress
application6 they /ecome 71 ε x8dx6 71 ε ydy6
71 ε z8dz 6 respectively#
10.- #ATERIAL"PROPERT RELATIONSHIPS
• Change in volume per unit volume is the
Kvolumetric strainL or dilatation e#
• Bsing generalized Hooe!s law6 we write the
dilatation in terms of applied stress#
( ) ( ) dz dydxdz dydxV z y x −+++= ε ε ε δ 111
( )2210 - z y x dV
V e ε ε ε
δ ++==
uniform pressure p of a li(uid6 pressure is the same
in all directions#
• As shear resistance of a li(uid is zero6 we can
ignore shear stresses#
• )hus6 an element of the /ody is su/ected to
principal stresses σ x = σ y = σ z = –p# %u/stituting into
'(n 1=-"* and rearranging6
Dilatation and @ul ,odulus
• )his ratio 7 pe8 is similar to the ratio of linear-elastic
stress to strain6 thus terms on the 2H% are called
the volume modulus of elasticity or the /ul
modulus# Having same units as stress with
sym/ol 6
• 5rom '(n 1=-"6 theoretical ma0imum value of
$oisson!s ratio is therefore υ < =##
• >hen plastic yielding occurs6 υ < =# is used#
( ) ( )2510
213 -
,$?2)A:)
• >hen homogeneous and isotropic material is su/ected to a state of tria0ial stress6 the strain in one of the stress directions is influence /y the strains produced /y all stresses# )his is the result of the $oisson effect6 and results in the form of a generalized Hooe!s law#
• A shear stress applied to homogenous and isotropic material will only produce shear strain in the same plane#
• ,aterial constants6 E, G and υ are related mathematically#
=
normal strain6 not shear strain#
• )he /ul modulus is a measure of the stiffness of a
volume of material# )his material property provides
an upper limit to $oisson!s ratio of υ < =#6 which
remains at this value while plastic yielding occurs#
10. Strain Transformation
E$A#PLE 10.10
Copper /ar is su/ected to a uniform loading along its
edges as shown# f it has a length a < *== mm6 width
b < = mm6 and thicness ! < "= mm /efore the load
is applied6 determine its new length6 width6 and
thicness after application of the load#
)ae E c+ < 1"= &$a6 υ c+ < =#*.#
"
E$A#PLE 10.10 %SOLN&
@y inspection6 /ar is su/ected to a state of plane
stress# 5rom loading6 we have
Associated strains are determined from generalized
Hooe!s law6 '(n 1=-1;
( )
( )
( ) ( ) ( )
( )
( )
( ) ( )
f rectangular /loc shown is su/ected to a uniform
pressure of p < "= $a6 determine the dilatation and
change in length of each side#
)ae E < 3== $a6 υ < =#.#
)he dilatation can /e determined using '(n 1=-"*
( )
( ) ( )[ ]
Hooe!s law6 '(n 1=-1;
( )[ ]
( ) ( )[ ]
)he negative signs indicate that each dimension is
decreased#
• >hen engineers design for a material6 there is a
need to set an upper limit on the state of stress that
defines the material!s failure#
• However6 criteria for the a/ove failure modes is not
easy to define under a /ia0ial or tria0ial stress#
• )hus6 four theories are introduced to o/tain the
principal stresses at critical states of stress#
(10. THEORIES O! !AILURE
ductile material 7e#g#6 steel8 is slipping#
• %lipping occurs along the contact
planes of randomly-ordered crystals
• 'dges of planes of slipping as they appear on the
surface of the strip are referred to as Jder!s lines#
• )he lines indicate the slip planes in the strip6 which
occur at appro0imately .° with the a0is of the strip#
(10. THEORIES O! !AILURE
the strip6 which occur at appro0imately
.° with the a0is of the strip#
• Consider an element6 determine ma0imum shear
stress from ,ohr!s circle6
proposed the ma0imum-shear-stress
( )2610 2
failure will occur out of the plane
• f in-plane principal stresses are of opposite signs6
failure occurs in the plane
2 maxσ
τ = ma0
for plane stress for any two in-plane principal
( )
σ σ σ σ
σ σ σ σ
strain-energy density#
strain-energy density is written as
( )2810 2
1 -σε =#
332211 2
a/ove e(n
yielding of a ductile material occurs when the
distortion energy per unit volume of the material
e(uals or e0ceeds the distortion energy per unit
volume of the same material when su/ected to
yielding in a simple tension test#
( ) ( )2910
22
1
233121
σ σ σ
• n the case of plane stress6
• 5or unia0ial tension test6 σ 1 < σ M6 σ " < σ * < =
( ) ( ) ( )[ ]2 13
2 32
2 21
υ −+−+−+= E
υ +−+= E
• %ince ma0imum-distortion energy theory re(uires
#d < 7#d 8M6 then for the case of plane or /ia0ial
stress6 we have
graph#
fail#
when the ma0imum principal stress
σ 1 in the material reaches a limiting value that is
e(ual to the ultimate normal stress the material can
sustain when su/ected to simple tension#
• 5or the material su/ected to plane stress
( )31102
1
-ult
ult
have stress-strain diagrams similar in /oth tension
and compression#
• Bse for /rittle materials where the tension and
compression properties are different#
determine the criterion#
• Carry out a unia0ial tensile test to determine the
ultimate tensile stress 7σ ult8t
• Carry out a unia0ial compressive test to determine
the ultimate compressive stress 7σ ult8c
• Carry out a torsion test to determine the ultimate
shear stress τ ult#
• Circle A represents the stress condition σ 1 < σ " < =6
σ * < E7σult8c
σ " < σ * < =
• )he Criterion can also /e represented on a graph
of principal stresses σ 1 and σ " 7σ * < =8#
initiation of yielding6 whereas if it is /rittle6 it is
specified /y fracture#
/etween the crystals that compose the material#
• )his slipping is due to shear stress and the
ma0imum-shear-stress theory is /ased on this
idea#
su/ected to normal stress#
the strain energy that distorts the material6 and not
the part that increases its volume#
• )he fracture of a /rittle material is caused /y the
ma0imum tensile stress in the material6 and not the
compressive stress#
theory6 and it is applica/le if the stress-strain
diagram is similar in tension and compression#
is different in tension and compression6 then
,ohr!s failure criterion may /e used to predict
failure#
/rittle material is difficult to predict6 and so theories
of failure for /rittle materials should /e used with
caution#
%teel pipe has inner diameter of 3= mm and outer
diameter of = mm# f it is su/ected to a torsional
moment of :m and a /ending moment of
*# :m6 determine if these loadings cause failure as
defined /y the ma0imum-distortion-energy theory#
Mield stress for the steel found from a tension test is
σ M < "= ,$a#
(E$A#PLE 10.12 %SOLN&
nvestigate a pt on pipe that is su/ected to a state of
ma0imum critical stress#
throughout the pipe!s length#
distri/utions shown#
(E$A#PLE 10.12 %SOLN&
@y inspection6 pts A and @ su/ected to same state of
critical stress# %tress at A6
( ) ( )
( ) ( ) ( )[ ] ( ) ( )
,ohr!s circle for this state of stress has center located
at
shaded triangle to /e R < 1"4#1
and the in-plane principal
%ince criterion is met6 material within the pipe will not
yield 7KfailL8 according to the ma0imum-distortion-
energy theory#
10. Strain Transformation
(E$A#PLE 10.1+
%olid shaft has a radius of =# cm and made of steel
having yield stress of σ M < *3= ,$a# Determine if the
loadings cause the shaft to fail according to the
ma0imum-shear-stress theory and the ma0imum-
distortion-energy theory#
%tate of stress in shaft caused /y a0ial force and
tor(ue# %ince ma0imum shear stress caused /y
tor(ue occurs in material at outer surface6 we have
( )
( )
( )
at pt A# 2ather than use ,ohr!s circle6 principal
stresses are o/tained using stress-transformation
e(ns 9-
a/solute ma0imum shear stress occur in the plane6
apply '(n 1=-"46
theory#
However6 using the ma0imum-distortion-energy
( ) ( ) ( ) ( ) ( )[ ] ( )
deformations that only occur in a single plane6 then
it undergoes plain strain#
• f the strain components ε x6 ε y6 and γ xy are nown for
a specified orientation of the element6 then the
strains acting for some other orientation of the
element can /e determined using the plane-strain
transformation e(uations#
transformation e(uations#
in a semi-graphical manner using ,ohr!s circle#
• 'sta/lish the ε and γ " a0es6 then compute
center of circle F7ε x + ε y8"6 =G and controlling pt
Fε , γ "G6 /efore plotting the circle#
• 2adius of circle e0tends /etween these two pts
and is determined from trigonometry#
• A/solute ma0imum shear strain e(uals the
ma0imum in-plane shear strain provided the
in-plane principal strains are of opposite signs#
then a/solute ma0imum shear strain will occur out
of plane and is determined from γ ma0 < ε ma0"#
• Hooe!s law can /e e0pressed in * dimensions6
where each strain is related to the * normal stress
components using the material properties E 6 and υ 6 as seen in '(ns 1=-1#
• f E and υ are nown6 then G can /e determined
using G : E F"71 υ G#
• Dilatation is a measure of volumetric strain6 and the
/ul modulus is used to measure the stiffness of a
volume of material#
are nown6 then a theory of failure can /e used
as a /asis for design#
• Ductile materials fail in shear6 and here the
ma0imum-shear-stress theory or the ma0imum-
distortion-energy theory can /e used to predict
failure#
stress#
ma0imum-normal-stress theory or ,ohr!s
• Comparisons are made with the ultimate tensile
stress developed in a specimen#