Mechanical and Metal Trades HandbookFull description
Theory of Metal Cutting-Mechanics of Metal CuttingFull description
Solution Manual 3rd Ed. Metal Forming: Mechanics and Metallurgy Chapter 1 Determine the principal stresses for the stress state 10 σ ij = −3
4
−3
4
5
2.
2
7
Solution: I1 = 1!"!#=3$% I$ = &'"!3"!#( !) !* !1+ = &1$+% I3 = 3" &*, &* &, 3 &+3 = 11)- σ $$σ$ &1$+σ &11) = . / trial and error solution gi0es σ &= 13.*. Factoring out 13.*% σ $ &,.)+σ ! ).1+ = . Sol0ing- σ1 = 13.*% σ2 = #.#,"% σ3 = 1.1#". 1&$ / "&cm. "&cm. diamet diameter er solid solid shaft is simul simultane taneousl ouslyy su2ect su2ected ed to to an aial aial load load of of , 45 and a tor6ue of * 5m. a. Determine the principal stresses at the surface assuming elastic eha0ior. . Find the the largest shear shear stress. Solution: a. 7he shear stress% τ% at a radius% r% is τ = τsr89 here τsis the shear stress stress at the $ surface 9 is the radius of the rod. 7he tor6ue% 7% is gi0en y 7 = ;$
?'*.#8$($ ! '1+8$($(@18$ = 1.$)% &.+$$ Ma . the largest largest shear stress stress is '1.$$) ! .+$$(8$ = .)$" Ma Ma / long thin&all tue% capped on oth ends is su2ected to internal pressure. During elastic loading% does the tue length increase% decrease or remain constantA Solution: Bet y = hoop direction% = aial direction% and = radial direction. e = e$ = '18E([σ - υ( σ3 + σ1(@ = '18E(?σ$ - υ'$σ$(@ = 'σ$8E('1&$υ) Since u 18$ for metals% e = e$ is positi0e and the tue lengthens. * / sol solid id $&cm $&cm.. dia diame mete terr rod rod is su2 su2ec ecte tedd to to a tens tensil ilee for force ce of * 45. 45. /n /n ide ident ntic ical al rod is su2ected to a fluid pressure of 3" Ma and then to a tensile force of * 45. hich rod eperiences the largest shear stressA Solution: 7he shear stresses in oth are identical ecause a hydrostatic pressure has no shear component. 1&" Consider Consider a long long thin& thin&all all%% " cm cm in in diame diameter ter tue% tue% ith ith a all all thic4ne thic4ness ss of .$" mm that is capped on oth ends. ends. Find the three principal stresses hen it is loaded under a tensile force of * 5 and an internal pressure of $ 4a. Solution: σ = D8*t ! F8'
1
1&+ 7hree strain gauges are mounted on the surface of a part. auge / is parallel to the &ais and gauge C is parallel to the y&ais. 7he third gage% G% is at 3H to gauge /. hen the part is loaded the gauges read auge / 31&+ auge G 3" 1&+ auge C 1 1&+ a. Find the 0alue of γ xy. . Find the principal strains in the plane of the surface. c. S4etch the Mohrs circle diagram. Solution: Bet the G gauge e on the ais% the / gauge on the &ais and the C gauge on 2 2 e e = + γ l x l xy+ l xx l x % here l x x = cose x = 3 = J38$ and l ′ = cos x the y&ais. e x x x x y y x y + = K. Sustituting the measured strains% 3" = 3'J$83($ 1'18$($ ! γ y'J38$('18$( γ y = '*8J38$(L3"&?3−'1'J38$( $!1'18$($@ = $%3) '1&+( . e1%e$ = 'e !ey(8$> ?'e&ey($ ! γ y$@18$8$ = '3!1(8$ > ?'3&1($ ! $3)$@18$8$ .e1 = 3"3'1&+(% e$ = *#'1&+(% e3 = . c( ′′
′′
′
′
′
′
x y
γ/2 εx
ε2
2θ=60°
ε1
ε
εx’ εy
Find the principal stresses in the part of prolem 1&+ if the elastic modulus of the part is $" a and oissonss ratio is .$). Solution: e3 = = '18E(? & ν 'σ1!σ$(@% σ1 = σ$ e1 = '18E('σ1 & ν σ1(- σ1 = Ee18'1&ν ( = $"1)'3"31&+(8'1&.$)$( = #) Ma Sho that the true strain after elongation may e epressed as ε
ln(
=
1
) here r is the 1 r −
reduction of area. ε
ln(
=
1
). 1 r −
Solution: r = '/o&/1(8/o =1 /18/o = 1 Bo8B1. ε = ln?18'1&r(@ / thin sheet of steel% 1&mm thic4% is ent as descried in Eample 1&11. /ssuming that E = is $" a and ν = .$)% ρ = $. m and that the neutral ais doesnt shift. a. Find the state of stress on most of the outer surface. . Find the state of stress at the edge of the outer surface. $
Solution: a. Sustituting E = $"1)% t = .1% ρ = $. and ν = .$) into σ x and σ y
=
E t ν 2
E t =
−
% σ = "+ Ma% % σy = 1+.$ Ma
2 ρ (1 ν ) −
. 5o σy = % so σ y =
ν Et
= "1 Ma
ρ 2
1&1 For an aluminum sheet% under plane stress loading ε = .3 and ε y = .1. /ssuming that E = is +, a and ν = .3% find ε. 2 Solution: ey = '18E('σy&ν σy(% e = '18E('σ ν Εe y ν σ(. Sol0ing for σ% 2 2 σ = ?E8'1&ν (@ey ! ν ey(. Similarly% σy = ?E8'1&ν (@'ey ! ν e(. Sustituting into 2 2 e = '18E('&ν σy&ν σy( = '&ν 8E('E8'1&ν (?ey ! ν ey! ey ! ν e ( = ?&ν (1+ ν )/ 8'1&ν (@'ey ! ey( = .$)'&1.$)8.)1+('.*( = &.1+3 1&11 / piece of steel is elastically loaded under principal stresses% σ 1 = 3 Ma% σ$ = $" Ma and σ 3 = &$ Ma. /ssuming that E = is $" a and ν = .$) find the stored elastic energy per 0olume. Solution: = '18$('σ1e1 ! σ$e$ ! σ3e3(. Sustituting e1 = '18E(?σ1 & ν (σ$ ! σ3(@% e$ = '18E(?σ$ & ν (σ3 ! σ1(@ and e3 = '18E(?σ3 & ν (σ1 ! σ$(@% = 18'$E(?σ1$ ! σ$$ ! σ3$ & $ν (σ$σ3!σ3σ1!σ1σ$(@ = '18'$$"1)(?3$ !$"$ ! $$ '$.$)('&$$" 3$" ! $"!3(@11$ = *N8m3 1.1$
/ sla of metal is su2ected to plane&strain deformation 'e$=( such that σ 1 = * 4si and σ 3 = . /ssume that the loading is elastic and that E = is $" a and ν = .$) '5ote the mied units.( Find a. the three normal strains. . the strain energy per 0olume. Solution: = '18$('σ1e1 ! σ$e$ ! σ3e3( = '18$('σ1e1 ! ! ( = σ1e18$ σ1 = *4si'+.,)Ma84si( = $#+ Ma = e$ = '18E(?σ$ &ν σ1@% σ$ =ν σ1 = .$)$#+ = , Ma e1 = '18E('σ1 &ν σ$( ='18$"13(?$#+&.$)',(@ = .1$1 = '$#+1+('.1$1(8$ = 1+# 4N8m3 Chapter $ a( If the principal stresses on a material ith a yield stress in shear of $ Ma are σ 2 = 1#" Ma and σ 1 = 3" Ma.% hat is the stress% σ3% at yielding according to the 7resca criterionA ( If the stresses in 'a( ere compressi0e% hat tensile stress σ 3 must e applied to cause yielding according to the 7resca criterionA
3
2
2 ρ (1 ν )
Solution: a( σ1 & σ3 = $4% σ3 = $4 σ1 = * & 3" = " Ma. ( σ3 = $4 σ1 = * '3"( = " Ma Consider a +&cm diameter tue ith 1&mm thic4 all ith closed ends made from a metal ith a tensile yield strength of $" Ma. /fter applying a compressi0e load of $ 5 to the ends. hat internal pressure is re6uired to cause yielding according to a( the 7resca criterion. ( the 0on Mises criterionA Solution: a( 7he ratio of the tue diameter to all thic4ness is 0ery large% so it can e treated as a thin all tue. 7he stress caused y the pressure can e found y & and y& direction force alances. From pressure% σ = d8'$t( = + and σy = d8'*t( = 3. 7he stress caused y the aial load is σy = F8'dt( = &$58?<'.+('.1(@= &1.+ Ma% so the total stress% σy = 3 &1.+ Ma a( σ = + = σma is the largest stress% σy = 3 &1.+ Ma and σz = . 7here are to possiilities hich must e chec4ed. i. If σ σy% σ = σmin% yielding ill occur hen +& = O% or =O8+ =$"8+ = .*1+ Ma ii. If σy σ% σy = σmin% and yielding ill occur hen +&'3&1.+( = O% or 3 = O ! 1.+% = 'O!1.+(83 = 3".+83 = 1.11,# Ma Oielding ill occur hen the smaller of the to 0alues is reached% and therefore the smaller one is appropriate. = .*1" Ma ( Sustituting into e6. $ 'in Ma(% $'$"($ = ?+&'3 &1.+(@$ !?'3 &1.+(&@$ ! ?&+@$ 1$" = "*$ ! $$*% p = .*3+ Ma $&3 Consider a ." m&diameter cylindrical pressure 0essel ith hemispherical ends made from a metal for hich k = " Ma. If no section of the pressure 0essel is to yield under an internal pressure of 3" Ma% hat is the minimum all thic4ness according to a( the 7resca criterionA ( the 0on Mises criterionA Solution: / force alance in the hemispherical ends gi0es σ ' =σy( = D8'*t(. / force alance in the cylindrical section gi0es σ = D8'$t(. σy = D8'*t( so this section has the greatest stress. a. σma & σmin = $4% D8$t = $4% t = D8'*4( = 3"'."(8'*"( = ,.#" mm . 'σ8$ & ($ ! ' & σ($! 'σ &σ8$($ = +4 $% '38$(σ$ = +4 $% σ = $4 = D8'$t(% t = D8'*4( hich is identical to part a. t = ,.#" mm ε =
2
2
ε 2 (ε x + y )/
$&* / thin&all tue is su2ected to comined tensile and torsional loading. Find the relationship eteen the aial stress% σ % the shear stress% τ % and the tensile yield strength% Y % to cause yielding according to a( the 7resca criterion% ( the 0on Mises criterion. 2 2 Solution: a( σ , /2 / 2 ) + If σ σ σ σ /2 /2 ) + P % σmin = % so the σ = ±( −( 1 2 2 2 2 7resca criterion predicts yielding hen σ /2 σ /2 )+ τ /2 /2 )+ σ ±( = . If σ −( 2 2 2 2 % σmin = − (σ /2 ) +τ /2 ) +τ % so the 7resca criterion predicts yielding hen 2 (σ $ 2 2 2 $ 18$ ( L$?σ /2 /2 ) + @ !? 2 (σ /2 ) +τ @ = J$O! σ −( *
Consider a plane&strain compression test ith a compressi0e load% F y% a strip idth% w% an indenter idth% b% and a strip thic4ness% t . Qsing the 0on Mises criterion% find: a( ε as a function of ε y. ( σ as a function of σ y. c( an epression for the or4 per 0olume in terms of ε y and σ y. d( an epression in the form of σ y = f(K,ε y ,n) assuming σ K . = 2 2 ε = 2 (ε Solution: a. If ε = % εy = & e ε x + y )/ = = 1.1"* εy 2 2 2 1 / 2 ) [ (σ / 2 )+ ( / 2 0 )+ ( 0 ) = σy81.1"* =( − − − σ σ . σ = % σ = &'18$(σy- σ y σ y y y c. = ;σydεy n σ ε ε /3 /3 /3 (4 /3 ) )= '*83(n!18$ ey =4 =4 K K d. σ y=4 y
$&+ 7he folloing yield criterion has een proposed: ROielding ill occur hen the sum of the to largest shear stresses reaches a critical 0alue. R Stated mathematically ( σ1 - σ 3 ) + ( σ1 - σ $ ) = C if ( σ1 - σ $ ) > ( σ$ - σ 3 ) or ( σ$ - σ 3 ) + ( σ1 - σ $ ) = C if ( σ1 - σ $ ) ≤ ( σ$ σ 3 ) here σ 1 > σ $ > σ 3 , C = 2Y and Y = tensile yield strength. a( Is this criterion satisfactory for an isotropic solid here O is insensiti0e to pressureA Nustify your anser. ( lot the σ = yield locus. S4etch the 7resca yield locus on the same plot c( here σ = % find the 0alues of σ and σ y for i. plane strain% ε = % ith ε > ii. aisymmetric flo ith ε y = ε = ε 8$ and ε > Solution: a( Yes. The value of the left hand sides are not affected if each principal stress is increased the same amount. b)
First find the constant C. Consider an x-direction tension test. At yielding, σx = σ1 = Y,
7herefore (σ1 - σ2)> (σ2 - σ3) so criterion I applies% and C = (σ1 σ3) + (σ1 - σ2) = 2Y. 7herefore C = 2Y. e can also thin4 aout an &direction compression test. /t yielding% σx = σ3 = -Y, σy = σz = σ2 = σ3 = 0 . 7herefore (σ2 - σ3)>(σ1 - σ2)> so criterion II applies% and C = (σ1 - σ3) + (σ2 σ3( = &'&$O( or again C = $O. 5o consider se0eral loading paths: In region A, σx = σ1, σy = σ2, σz= σ3 = 0 and σx >2σy so (σ1 - σ3) >(σ1 - σ2) 7herefore criterion I, (σx - 0) + (σx - σy) = $O, or σx = Y + σy/2 σy = σz = σ2 = σ3 = 0.
In region B, σx = σ1, σy = σ2, σz= σ3 = 0 ut σx <2σy so (σ1 - σ3)<(σ1 - σ2) 7herefore criterion II, (σx - 0) + (σy - 0) = 2Y, or σx = 2Y - σy In region C, σy = σ1, σx= σ2, σz= σ3 = 0 ut σy <2σx so (σ1 - σ3)<(σ1 - σ2) 7herefore criterion II, (σy - 0) + (σx - 0) = 2Y, or σy = 2Y - σx
"
In region D, σy = σ1, σx = σ2, σz= σ3 = 0 and σy >2σx so (σ1 - σ2) >(σ2 - σ3) 7herefore criterion I, (σy - 0) + (σy - σx) = 2Y, or σy = Y + σx/2 In region E, σx = σ1, σy = σ3, σz= σ2 = 0 and (σ1 - σ2) >(σ2 - σ3) 7herefore criterion I, (σx - 0) + (σx - σy) = 2Y, or σx = Y + σy/2 In region f, σx = σ1, σy = σ3, σz= σ2 = 0 so (σ1 - σ2) >(σ2 - σ3) 7herefore criterion I, (σx - 0) + (σx - σy) = 2Y, or σx = Y + σy/2 lotting these in the appropriate regions% and using symmetry to construct the left hand half
!) i.
For plane strain (εy = 0) and εx > 0, 7he normal to the locus is at the corner eteen A and B regions. Goth σx = Y + σy/2 and σx = 2Y - σy must e satisfied. Sol0ing simultaneously% σx = ("/3)Y ut σy = (2/3)Y ##. Axisymmetric flow with εy = εz = -(1/2)εx ith εx > 0, is satisfied e0eryhere in 9egion I, so σx = Y + σy/2, ith (2/3)Y $ σx $ ("/3) $ Consider the stress states 15
3
0
3
10
0
0
0
5
and
10
3
0
3
5
0.
0
0
0
a( Find σ m for each. ( Find the de0iatoric stress in the normal directions for each c( hat is the sum of the de0iatoric stresses for eachA Solution: a( '1" ! 1 ! "(83 = 1 and '1 ! " ! (83 = " ( 1" 1 = "% 1&1 = " 1 = &" and 1&" = "% "&" = % &" = &" c( 7he sum of the de0iatoric stresses oth = .
+
$&, / thin all tue ith closed ends is made from steel ith a yield strength of $" Ma. 7he tue is $ m. long ith a all thic4ness of $ mm. and a diameter of , cm. In ser0ice it ill eperience an aial load of , 45 and a tor6ue of $.# 5m. hat is the maimum internal pressure it can ithstand ithout yielding according to a( the 7resca criterion% ( the 0on Mises criterionA Solution: D8t = * so this can e regarded as a thin&all tue. For this solution% stresses ill e epressed in 4si. F8/ = $8'
2 Y2 = σ
2 2 2 2 2 2 $ y + σ x + (σ x − σ y) + 6τ xy = 2 [ σ y − σ xσy+ σ x ] + 6τ xy
'38$(σ ! *.$** ! ?σ$8* & *.$**σ ! *).,""@18$ = , ?σ$8* & *.$**σ ! *).,""@18$ = #".#"+ &'38$(σ$ σ$8* & *.$**σ ! *).,""@ = ?#".#"+ &'38$(σ@$ = "#3,.) &$$#.$+σ !$.$"σ$ σ$?.$" & $.$"@ ! ?&*.$** !$$#.$+@s ! *).,"" & "#3,.) = $σ$ &$$3.3s ! "+,) = - σ = L$$3.3 > ?$$3.3$ &*$ "+,)@18$8'$$( = "".," > 1+.*)% σ = #$.3* or 3).3+% 7he smaller 0alue is correct 7hen = '$t8d(σ = 3).3+'$."83( = 1.31$ 4si 5o e must chec4 to see hether σ$ P . Sustituting / = *.$**% τ = $.,$) and σ = 3).3+ into σ$ = '38*(σ ! /8$ & '18$(?σ$8* & /σ ! /$ ! *τy$@18$ σ$ = '38*(3).3+ ! *.$**8$ & '18$(?3).3+$8* & *.$**3).3+ ! *.$**$ ! *$.,$) $@18$ = 31.*,. 7herefore the solution for σ$ P is appropriate. / τ for a( pure shear. ( uniaial tension% and c( plane $&) Calculate the ratio of σ strain tension. /ssume the 0on Mises criterion. Solution: a( σ1 = τ% σ$ = 0% σ3 = &τ% σ = L?τ$ ! '$τ) $ + τ$@8$18$% σ 8τ = J3 ( σ 8τ = $ c( σ1 = τ% σ$ = τ8$% σ3 = % σ = L?'τ8$($ ! τ$ !'τ8$($@8$18$% σ 8τ = J'38$( m a x
/ material yields under a iaial stress state% σ 3 = &'18$(σ 1% σ $ = . a( /ssuming the 0on Mises criterion% find d ε 18 d ε $. hat is the ratio of τ / at yieldingA Solution: dε18dε$ = ?σ1 & 'σ$ ! σ3(8$@8?σ$ & 'σ3 ! σ1(8$@ = ?σ1 & '0&σ18$(8$@8? & '&σ18$ &σ1(8$@ = '"8*(8'38*( = "83 m a x
$&11 / material is su2ected to stresses in the ratio% σ 1 % σ $ = .3σ 1 and σ 3 = &."σ 1. Find the ratio of σ 18Y at yielding using the a( 7resca and ( 0on Mises criteria. Solution: a( For 7resca% σ 1 – '&."σ 1( = O% σ 1 O= $83 ( For 0on Mises% L?'.3!."($ ! '&." 1($ ! '1&.3($@8$18$σ 1 = O% σ 1 O= .## $&1$ / proposed yield criterion is that yielding ill occur hen the diameter of the largest Mohrs circle plus half the diameter of the second largest Mohrs circle reaches a critical 0alue. lot the yield locus in σ 1 0s. σ $ in σ 3 = space. Solution: Di0ide stress space into regions ith different conditions for yielding. 7o e0aluate C% consider an &direction tension test. /t yielding σ = O% σy = % 7he diameters of the to largest Mohrs circle are O. O = O8$ = C. C = 38$O
$.13 Ma4e plot of ε 1 0ersus ε $ for a constant le0el of ε = .1 according to a. 0on Mises. . 7resca. 2 1 21 / = ρ ε [ (4 /3 ) ( 1 ρ ρ ) ] / so ε ε [ (4 /3 ) ( 1 ρ ρ ) ] and ε = + + = + + Solution: 7a4ing ε 2 1 1 1 / can e calculated for 0arious ε / and ε − = for 7resca%% ε for 0on Mises and ε 2 ε 1 2 1 / 0alues of ρ. ε 2 ε −
ε11 / ε ε
10 1
Tresca
von Mises
-10 -1
0
1 10
ε2 / ε
CV/7E9 3 hen a rass tensile specimen% initially ."" in. in diameter% is tested% the maimum load of 1"% ls as recorded at an elongation of *W. hat ould the load e on an identical tensile specimen hen the elongation is $WA Solution: n = εma load = ln'1!ema load( = ln'1.*( = .3+". 3 σma load = sma load '1!ema load( = '1$%(8.$('1.*( = ,*1 . Gut also σma load = X'.3+"(.3+" = .+)3$X. E6uating and sol0ing for X% X = ,*1 38.+)3$ = 1$1%. /t $W elongation% ε = ln'1.$( = .1,$3. σ = 1$1%'.1,$3(.3+" = +"%. s = +"%81.$ = "*%1,. F = "*%'.$( = 1. , ls. )
3&$ During a tension test the tensile strength as found to e 3* Ma. 7his as recorded at an elongation of 3W. Determine n and K if the approimation σ K = applies. Solution: n = εma load = ln'1!ema load( = ln'1.3( = .$+$. σma load = sma load '1!ema load( = 3*'1.3( = **$ Ma. Gut also σma load = X'..$+$(.$+$ = .#*X. X = **$8.#* = +$# Ma. Y + for a metal stretched in tension to / (n ε 3.3 Sho that the plastic or4 per 0olume is σ 1 1 if σ . = ε k + / (n 1 ε Solution: = ;σ1dε1 = ;4 ε1ndε1 = 4 ε1n!18'n!1( = 4 ε1ε1n8'n!1( = σ 1 1 ε 1
3.* For plane&strain compression 'Figure 3.11( a. Epress the incremental or4 per 0olume% d% in terms of σ and d ε and compare it ith d = σ1dε1 ! σ$dε$ ! σ3dε3. n . If σ % epress the compressi0e stress% as a function of σ1% K and n. = ε k Solution: a. ith εy = and σ = % d = σ3dε. σy = σ8$% σ =% $ $ $ 18$ σ = L?'σy & σ( !'σ σ( !'σ σy( @8$ = L?'&σ8$($ !'&σ($ ! '&σ8$($@8$18$ = '38*(σ 2 2 21 2 1 /2 2 / [ ( 2 / 3 ) (d ε ε ε ] = { (2 / 3 ) [ (− ε )+ 0 ε } = '*83(18$dε d e d d d d = + + x y + z) x z] 18$ σ d = '38*(σ'*83( dε = 'σdε 1 /2 1 /2 n 1 /2 /2 n ( 4 / 3 ) σ = ( 4 / 3 ) k ε = ( 4 / 3 ) k ( 4 / 3 ) . σ = '*83('n!1(8$en. z= 3." 7he folloing data ere otained from a tension test: Boad Min. 5ec4 true true corrected dia. radius strain stress true stress σ (MPa) '45( 'mm( 'mm( σ (MPa) ,.+) Z $#. ,.13 Z .133 "$ "$ 3*." #.+$ Z *.+ +.,+ Z 3,.3 "."" 1.3 $).$ 3.,1 1., a. Compute the missing 0alues . lot oth σ and σ 0s. ε on a logarithmic scale and determine K and n. c. Calculate the strain energy per 0olume hen ε = .3". Solution: a( Boad Min. dia. '45( 'mm( ,.+) $#. ,.13 3*." #.+$ *.+ +.,+ 1
5ec4 radius 'mm( Z Z Z Z
true strain
true stress
a89
corrected true stress
σ (MPa)
.133 .$+3 .*#3
"$ #"* 1))
σ
"$ +"* 1))
(MPa)
3,.3 $).$
"."" 3.,1
1.3 1.,
.)#, 1.+"
1#1# $"+1
.$+ 1.+
1+31 $1
3.+ Consider a steel plate ith a yield strength of * 4si% Ooungs modulus of 31+ psi and a oissons ratio of .3 loaded under alanced iaial tension. hat is the 0olume change% ∆[8[% 2ust efore yieldingA Solution: /t yielding σ1 = σ$ = *% psi% σ3 = . e1 = e$ = '18E(?σ1 υσ1@% e3 = '18E(?& $υσ1@- ∆080 = e1 ! e$ ! e3 = 'σ1 8E(?$&* υ@ = .1#1&3. 3.#
7he strain&hardening of a certain alloy is etter approimated y . Determine the true strain at nec4ing in terms of / σ = /?1&&ep'&Gε (@ than y σ = k ε and G. Solution: σ = /?1&&ep'&Gε (@ =dσ8dε = /Gep'&Gε (- / = /'G!1(ep'&Gε (- ε = ln'1!G(8G 3&, Epress the tensile strength% in terms of / and G for the material in rolem 3. Solution: σ ma load = /L1\ep?&G( ln'1!G(8G)] = /?1!'1!G(@ = /'$!G(7ensile strength = σ ma load ep'ε ) =/'$!G(ep?ln'1!G(8G@ = /'$!G('1!G(18G 3&) / metal sheet undergoing plane&strain tensile deformation is loaded to a tensile stress of 3 Ma. hat is the ma2or strain if the effecti0e stress&strain relationship is 0 . σ = ε 6 5 0( 0 . 0 1 5+ ) MaA Solution: .$$ 18.$$ = .1""- ε = J'*83( ε = σ =J'*83( σ =+"'.1"! ε ( - ε =?J'38*('3(8+" &.1"@ .1#)