Metal Forming Mechanics and Metallurgy Hosford Caddell 3rd solution-manual.pdf

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   $$σ$ &1$+σ &11) = . / trial and error solution gi0es σ &= 13.*. Factoring out 13.*% σ $ &,.)+σ ! ).1+ = . Sol0ing- σ1  = 13.*% σ2 = #.#,"% σ3 = 1.1#". 1&$ / "&cm. "&cm. diamet diameter er solid solid shaft is simul simultane taneousl ouslyy su2ect su2ected ed to to an aial aial load load of of , 45 and a tor6ue of * 5m. a. Determine the principal stresses at the surface assuming elastic eha0ior.  . Find the the largest shear shear stress. Solution: a. 7he shear stress% τ% at a radius% r% is τ  = τsr89 here τsis the shear stress stress at the $ surface 9 is the radius of the rod. 7he tor6ue% 7% is gi0en y 7 = ;$ ?'*.#8$($ ! '1+8$($(@18$ = 1.$)% &.+$$ Ma  . the largest largest shear stress stress is '1.$$) ! .+$$(8$ = .)$" Ma Ma / long thin&all tue% capped on oth ends is su2ected to internal pressure. During elastic loading% does the tue length increase% decrease or remain constantA Solution: Bet y = hoop direction%  = aial direction% and  = radial direction.   e = e$ = '18E([σ - υ( σ3 + σ1(@ = '18E(?σ$ - υ'$σ$(@ = 'σ$8E('1&$υ) Since u  18$ for metals% e  = e$ is positi0e and the tue lengthens. * / sol solid id $&cm $&cm.. dia diame mete terr rod rod is su2 su2ec ecte tedd to to a tens tensil ilee for force ce of * 45. 45. /n /n ide ident ntic ical al rod is su2ected to a fluid pressure of 3" Ma and then to a tensile force of * 45. hich rod eperiences the largest shear stressA Solution: 7he shear stresses in oth are identical ecause a hydrostatic pressure has no shear component. 1&" Consider Consider a long long thin& thin&all all%% " cm cm in in diame diameter ter tue% tue% ith ith a all all thic4ne thic4ness ss of .$" 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 $ 4a. Solution: σ = D8*t ! F8'
1

1&+ 7hree strain gauges are mounted on the surface of a part. auge / is parallel to the &ais and gauge C is parallel to the y&ais. 7he third gage% G% is at 3H to gauge /. hen the part is loaded the gauges read auge / 31&+ 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 Mohrs circle diagram. Solution: Bet the G gauge e on the  ais% the / gauge on the &ais 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&ais. e  x   x   x x   y y  x y + = K. Sustituting 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 prolem 1&+ if the elastic modulus of the part is $" a and oissonss ratio is .$). Solution: e3 =  = '18E(? & ν  'σ1!σ$(@% σ1 = σ$ e1 = '18E('σ1 & ν σ1(- σ1 = Ee18'1&ν ( = $"1)'3"31&+(8'1&.$)$( = #) Ma Sho that the true strain after elongation may e epressed 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 descried in Eample 1&11. /ssuming that E  = is $" a and ν  = .$)% ρ = $. m and that the neutral ais doesnt 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. Sustituting E = $"1)% t = .1% ρ = $. and ν  = .$) into σ    x  and σ    y

=

 E t  ν  2

 E t  =

−

% σ = "+ Ma% % σy = 1+.$ Ma

2 ρ (1 ν  ) −

 . 5o σy = % so σ  y =

ν   Et 

 = "1 Ma

 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(. Sustituting 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 Ma% σ$ = $" Ma and σ 3 = &$ Ma. /ssuming that E  = is $" a and ν  = .$) find the stored elastic energy per 0olume. Solution:  = '18$('σ1e1 ! σ$e$ ! σ3e3(. Sustituting 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$" ! $"!3(@11$ = *N8m3 1.1$

/ sla of metal is su2ected 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 mied units.( Find a. the three normal strains.  . the strain energy per 0olume. Solution:  = '18$('σ1e1 ! σ$e$ ! σ3e3( = '18$('σ1e1 !  ! ( = σ1e18$ σ1 = *4si'+.,)Ma84si( = $#+ Ma  = e$ = '18E(?σ$ &ν σ1@% σ$ =ν σ1 = .$)$#+ = , Ma e1 = '18E('σ1 &ν σ$( ='18$"13(?$#+&.$)',(@ = .1$1  = '$#+1+('.1$1(8$ = 1+# 4N8m3 Chapter $ a( If the principal stresses on a material ith a yield stress in shear of $ Ma are σ 2 = 1#" Ma and σ 1 = 3" Ma.% 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" = " Ma.  ( σ3 = $4  σ1 = *  '3"( = " Ma Consider a +&cm diameter tue ith 1&mm thic4 all ith closed ends made from a metal ith a tensile yield strength of $" Ma. /fter applying a compressi0e load of $ 5 to the ends. hat internal pressure is re6uired to cause yielding according to a( the 7resca criterion. ( the 0on Mises criterionA Solution: a( 7he ratio of the tue diameter to all thic4ness is 0ery large% so it can e treated as a thin all tue. 7he stress caused y the pressure can e found y & and y& direction force alances.  From pressure% σ = d8'$t( = + and σy = d8'*t( = 3. 7he stress caused y the aial load is σy = F8'dt( = &$58?<'.+('.1(@= &1.+ Ma% so the total stress% σy = 3 &1.+ Ma a( σ = + = σma is the largest stress% σy = 3 &1.+ Ma and σz = . 7here are to  possiilities hich must e chec4ed. i. If  σ  σy% σ = σmin% yielding ill occur hen +& = O% or =O8+ =$"8+ = .*1+ Ma 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,# Ma Oielding ill occur hen the smaller of the to 0alues is reached% and therefore the smaller one is appropriate.  = .*1" Ma  ( Sustituting into e6. $&# 'in Ma(% $'$"($ = ?+&'3 &1.+(@$ !?'3 &1.+(&@$ ! ?&+@$ 1$" = "*$ ! $$*% p = .*3+ Ma $&3 Consider a ." m&diameter cylindrical pressure 0essel ith hemispherical ends made from a metal for hich k  = " Ma. If no section of the pressure 0essel is to yield under an internal pressure of 3" Ma% hat is the minimum all thic4ness according to a( the 7resca criterionA ( the 0on Mises criterionA Solution: / force alance in the hemispherical ends gi0es σ ' =σy( = D8'*t(. / force alance in the cylindrical section gi0es σ  = D8'$t(. σy = D8'*t( so this section has the greatest stress. a. σma & σmin = $4% D8$t   = $4% t = D8'*4( = 3"'."(8'*"( = ,.#" mm  . 'σ8$ & ($ ! ' & σ($! 'σ &σ8$($ = +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 tue is su2ected to comined tensile and torsional loading. Find the relationship eteen the aial 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 epression for the or4 per 0olume in terms of ε  y and σ  y. d( an epression 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 folloing yield criterion has een proposed: ROielding ill occur hen the sum of the to 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 anser.  ( 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. aisymmetric 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 aout 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 eteen  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 e0eryhere 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 tue ith closed ends is made from steel ith a yield strength of $" Ma. 7he tue is $ m. long ith a all thic4ness of $ mm. and a diameter of , cm. In ser0ice it ill eperience an aial load of , 45 and a tor6ue of $.# 5m. hat is the maimum 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 tue. For this solution% stresses ill e epressed 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

O2 = σ y2  − σ σy + σ x2 + 3τ y2 Sustituting σy = σ8$ ! /%   σ8$ ! /T U$ & σ  8$ ! /T Uσ !σ! 3τy$ & O$ =  σ$'18* &18$ ! 1( ! σ'/ & /( ! '/$ ! 3τy$& O$( =  '38*(σ$ ! '/$ ! 3τy$& O$( =  σ$ ! G =  here G = '*83('/$ ! 3τy$& O$( Substituting G = '*83('*.$**$ ! 3$.,$)T$& $$( = &$.), σ$ = $.),% σ = *".,% σ = d8'$τ(%  = '$τ8d(s = *".,'$."83( = 1."$, 4si b)

For Tresca, we must find the principal stresses.

> '18$(?'σ & σy($ ! *τy$@18$ Sustituting σy = σ8$ !/% σ1%$ = '38*(σ ! /8$ > '18$(?'σ8$ & /($ ! *τy$@18$ = '38*(σ ! /8$ > '18$(?σ$8* & /sσ ! /$ ! *τy$@18$ σ1 = '38*(σ ! /8$ ! '18$(?σ$8* & /σ ! /$ ! *τy$@18$ σ$ = '38*(σ ! /8$ & '18$(?σ$8* & /σ ! /$ ! *τy$@18$ 7here are to possiilities: σ$ P % and σ$  . 1st assume that s$ P . 7hen σ1 &  = O% '38*(σ ! /8$ ! '18$(?σ$8* & /σ ! /$ ! *τy$@18$ = O '38*(σ ! /8$ ! '18$(?σ$8* & /σ ! /$ ! *τy$@18$ = O '38*(σ ! *.$**8$ ! '18$(?σ$8* & *.$**σ ! *.$**$ ! *$.,$)T$@18$ = * σ1%$ = 'σ ! σy(8$

#

'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 . Sustituting / = *.$**% τ = $.,$) and σ = 3).3+ into σ$ = '38*(σ ! /8$ & '18$(?σ$8* & /σ ! /$ ! *τy$@18$ 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. ( uniaial 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 iaial 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 su2ected 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 Mohrs circle plus half the diameter of the second largest Mohrs 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 to largest Mohrs circle are O. O = O8$ = C. C = 38$O

,

σy + (1/2)(σx) = (3/2)Y

σy + (1/2)(σy-σx) = (3/2)Y σy - (1/3)σx =Y

(2/3)σy + (1/3)(σx) = Y

σy σx + (1/2)(σ y) = (3/2)Y (2/3)σx + (1/3)(σy) = Y

σx + (1/2)(σ x-σy) = (3/2)Y σx - (1/3)σy =Y

σx (σx+σy)+(1/2) σx = (3/2)Y

σx + (2/3)σy =Y

(σx+σy)+(1/2)(-σy) = (3/2)Y

(2/3)σx + σy =Y

$.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 maimum load of 1"% ls 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. , ls. )

3&$ During a tension test the tensile strength as found to e 3* Ma. 7his as recorded at an elongation of 3W. Determine n and K  if the approimation σ  K  = applies. Solution: n = εma load = ln'1!ema load( = ln'1.3( = .$+$. σma load = sma load '1!ema load( = 3*'1.3( = **$ Ma. Gut also σma load = X'..$+$(.$+$ = .#*X. X = **$8.#* = +$# Ma. 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. Epress 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  % epress the compressi0e stress% as a function of σ1% K  and n. = ε  k  Solution: a. ith εy =  and σ = % d = σ3dε. σy = σ8$% σ =% $ $ $ 18$ σ = L?'σy & σ(  !'σ  σ(  !'σ  σy( @8$  = L?'&σ8$($ !'&σ($ ! '&σ8$($@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  = 'σdε 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 folloing data ere otained 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% Ooungs modulus of 31+ psi and a oissons ratio of .3 loaded under alanced iaial 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 approimated y . Determine the true strain at nec4ing in terms of / σ  = /?1&&ep'&Gε (@ than y σ  = k  ε  and G. Solution: σ  = /?1&&ep'&Gε (@ =dσ8dε = /Gep'&Gε (- / = /'G!1(ep'&Gε (- ε  = ln'1!G(8G 3&, Epress the tensile strength% in terms of / and G for the material in rolem 3&#. Solution: σ ma load = /L1\ep?&G( ln'1!G(8G)] = /?1!'1!G(@ = /'$!G(7ensile strength = σ ma load ep'ε )  =/'$!G(ep?ln'1!G(8G@ = /'$!G('1!G(18G 3&) / metal sheet undergoing plane&strain tensile deformation is loaded to a tensile stress of 3 Ma. hat is the ma2or strain if the effecti0e stress&strain relationship is 0 . σ  = ε  6 5 0( 0 . 0 1 5+ )  MaA Solution: .$$ 18.$$  = .1""- ε = J'*83( ε = σ =J'*83( σ =+"'.1"! ε ( - ε =?J'38*('3(8+" &.1"@ .1#)

11