Solution manual 13 15

18
Chapter 13 13.1 Show that the 1948 Hill criterion and flow rules predict an angular variation of R as: R θ = H + (2 N - F - G - 4 H )sin 2 θ cos 2 θ F sin 2 θ + G cos 2 θ . Solution : Let the tensile axis be x'. Then σ x = σ x' cos 2 θ, σ y = σ x' sin 2 q and τ xy = σ x' cosθsinθ. Using the flow rules, Eq.(13-7), ε x = λ[H(σ x -σ y ) + G(σ x -σ z )] = λ[H(cos 2 θ -sin 2 θ) +Gcos 2 θ]σ x' ε y = λ[F(σ y -σ z ) + H(σ y -σ x )] = λ[Fsin 2 θ +H(cos 2 θ -sin 2 θ)]σ x' γ xy = 2ε xy = 2λNτ xy = 2λNcosθsinθσ x' ε z = λ[F(σ z -σ y ) + G(σ z -σ x )] = -λ(Fsin 2 θ + Gcos 2 θ)σ x' but ε y' = ε x sin 2 θ + ε y cos 2 θ - γ xy cosθsinθ so R θ = ε y' /ε z = (ε x sin 2 θ + ε y cos 2 θ - γ xy cosθsinθ)/ε z = R θ = {[H(cos 2 θ -sin 2 θ) +Gcos 2 θ]sin 2 θ + [Fsin 2 θ +H(cos 2 θ -sin 2 θ)]cos 2 θ - 2Ncos 2 θsin 2 θ}/[-(Fsin 2 θ + Gcos 2 θ)] = R θ = [H(cos 4 θ+sin 4 θ -2cos 2 θsin 2 θ) +(2N-F-G)cos 2 θsin 2 θ]/(Fsin 2 θ + Gcos 2 θ) Since 1 = (cos 2 θ + sin 2 θ) 2 = cos 4 θ + 2cos 2 θsin 2 θ + sin 4 θ, we can substitute cos 4 θ + sin 4 θ = 1 - 2cos 2 θsin 2 θ, Therefore, R θ = [H + (2N - F -G -4H)cos 2 θsin 2 θ]/(Fsin 2 θ + Gcos 2 θ) 13.2 Using the results of Problem 13-1, derive an expression for N/G in terms of R, P (= R 90 ) and R 45 . Solution : a) for θ = 45 ° , cos 2 θ = sin 2 θ = 1/2, so from Prob 13-1, R 45 = [H +(2N-F-G-4H)/4]/[(F+G)/2] = (2N-F-G)/[2(F+G)] R 45 = (2N/G -F/G - 1)/[2(F/G+1)]; 2N/G = 2(F/G+1)R 45 + F/G +1. Now substitute F/G = R/P; 2N/G = 2(R/P+1) R 45 + R/P +1 or N/G =[2(R+P)R 45 +R + P]/(2P) = (2R 45 +1)(R+P)/(2P) b) Substituting R = 4, P = 2 and R 45 = 2.5, N/G =[6x6]/(2x2) = 9 G + H = 1/X 2 = 1/(71.5 ksi) 2 = 1.96x10 -4 , H/G = R = 4 5G = 1.96x10 -4 , G = 0.392x10 -4 and H = 4G = 1.568x10 -4 ; F + H = 1/Y 2 = 1/(65.3x10 3 ) 2 = 2.35x10 -10 , H/F = P = 2, H = 2F 3F = 2.35x10 -4 , F = 0.783 and H = 2F=1.567x10 -4 N = 9G = 3.521x10 4 . In a tension test along x', (with x' oriented at q from x), σ x = σ x' cos 2 θ, σ y = σ x' sin 2 θ, τ xy = σ x' cosθsinθ, σ z = 0. The yield criterion (eq13-3) reduces to Fsin 4 θ Gcos 4 θ H(cos 2 θ - sin 2 θ) 2 2Ncos 2 θsin 2 θ =1/σ x' 2 ,

Transcript of Solution manual 13 15

Page 1: Solution manual 13 15

Chapter 13

13.1 Show that the 1948 Hill criterion and flow rules predict an angular variation of R

as:

Rθ=H+(2N−F−G−4H)sin2θcos2θ

Fsin2θ+Gcos2θ.

Solution: Let the tensile axis be x'. Then σx = σx'cos2θ, σy = σx'sin2q and

τxy = σx'cosθsinθ. Using the flow rules, Eq.(13-7),

εx = λ[H(σx -σy) + G(σx -σz)] = λ[H(cos2θ -sin2θ) +Gcos2θ]σx'

εy = λ[F(σy -σz) + H(σy -σx)] = λ[Fsin2θ +H(cos2θ -sin2θ)]σx'γxy = 2εxy = 2λNτxy = 2λNcosθsinθσx'

εz = λ[F(σz -σy) + G(σz -σx)] = -λ(Fsin2θ + Gcos2θ)σx' but

εy' = εxsin2θ + εycos2θ - γxycosθsinθ so

Rθ = εy'/εz = (εxsin2θ + εycos2θ - γxycosθsinθ)/εz =

Rθ = {[H(cos2θ -sin2θ) +Gcos2θ]sin2θ + [Fsin2θ +H(cos2θ -sin2θ)]cos2θ -

2Ncos2θsin2θ}/[-(Fsin2θ + Gcos2θ)] =

Rθ = [H(cos4θ+sin4θ -2cos2θsin2θ) +(2N-F-G)cos2θsin2θ]/(Fsin2θ + Gcos2θ) Since 1 =

(cos2θ+ sin2θ)2 = cos4θ+ 2cos2θsin2θ + sin4θ, we can substitute cos4θ+ sin4θ = 1 -

2cos2θsin2θ,

Therefore, Rθ = [H + (2N - F -G -4H)cos2θsin2θ]/(Fsin2θ + Gcos2θ)

13.2 Using the results of Problem 13-1, derive an expression for N/G in terms of R, P (= R90) and R45.

Solution: a) for θ = 45°, cos2θ = sin2θ = 1/2, so from Prob 13-1,R45 = [H +(2N-F-G-4H)/4]/[(F+G)/2] = (2N-F-G)/[2(F+G)]

R45 = (2N/G -F/G - 1)/[2(F/G+1)]; 2N/G = 2(F/G+1)R45 + F/G +1.

Now substitute F/G = R/P; 2N/G = 2(R/P+1) R45 + R/P +1 or

N/G =[2(R+P)R45 +R + P]/(2P) = (2R45+1)(R+P)/(2P)

b) Substituting R = 4, P = 2 and R45 = 2.5, N/G =[6x6]/(2x2) = 9

G + H = 1/X2 = 1/(71.5 ksi)2 = 1.96x10-4, H/G = R = 4

5G = 1.96x10-4, G = 0.392x10-4 and H = 4G = 1.568x10-4;

F + H = 1/Y2 = 1/(65.3x103)2 = 2.35x10-10, H/F = P = 2, H = 2F

3F = 2.35x10-4, F = 0.783 and H = 2F=1.567x10-4

N = 9G = 3.521x104.In a tension test along x', (with x' oriented at q from x),

σx = σx'cos2θ, σy = σx'sin2θ, τxy = σx'cosθsinθ, σz = 0. The yield criterion (eq13-3) reduces to

Fsin4θ Gcos4θ H(cos2θ - sin2θ)2 2Ncos2θsin2θ =1/σx'2,

Page 2: Solution manual 13 15

(F + H)sin4θ + (G + H)cos4θ + 2(N-H)cos2θsin2θ =1/σx'2, Substituting N,F,G & H, σx'

=[2.35sin4θ 1.96cos4θ 3.91cos2θsin2θ]-1/2 x104

13-3 In strip tension tests, the strain ratios, R0 = 2.0, R90 = 2.0, R45 =2.5 and yield strengths , Y0 = 45, Y90= 45, were measured . Using the Hill ’48 criterion calculate Y22.5,

Y45,and Y67.5, and plot Y as a function of θ.Solution: N = (2R45 + 1)(R + P)/[2(R+1)PX2] = 24/[(12)(45)2] = 0.9877x10-3.F = (1/2)[1/Y2 + 1/Z2 -1/X2], substituting Z2 = P(R+1)X2/(R+P) = 6(49)2/4= 3038F = (1/2)[1/452 + 1/3038 -1/452] = 0.1519x10-3

G = (1/2)[1/Z2 + 1/X2 -1/Y2] = (1/2)[1/3038 + 1/452 -1/452]=0.1519x10-3

H= (1/2)[1/X2 + 1/Y2 -1/Z2] = (1/2)[1/452 + 1/452 -1/4000] = 0.4938x10-3

Substituting these value into

Yθ = [H + Fsin2θ + Gcos2θ +(2N-F-G-4h)cos2θsin2θ]-1/2

For θ = 45, cos2θ = sin2θ = 0.5, cos2θsin2θ = 0.25

Y45 = [0.4938x10-3 + 0.5(0.1519x10-3 +0.1519x10-3) + (2x0.9877x10-3 – 2x0.1519x10-3-

4x0.4938x10-3)(0.25)]-1/2= 41.9

For θ = 22.5, cos2θ = 0.8536, sin2θ = 0.1464, cos2θsin2θ = 0.125

Y22.5 = [0.4938x10-3 + (0.1464 +0.8536) (0.1519x10-3) + (2x0.9877x10-3 – 2x0.1519x10-

3- 4x0.4938x10-3)(0.125)]-1/2= 42.4From symmetry of the equations Y67.5 =Y22.5

Plotting:

13.4 A thin wall tube was made from sheet metal by bending the sheet into a cylinder andwelding. The prior rolling direction is the axial direction of the tube. The tube has a diameter of 5.0 in and a wall thickness of 0.025 in. The strain ratios are: R0 = .5, R90 = 0.8, R45 = 1.8 and the yield strength in the rolling direction is 350 MPa. Neglect elastic effects. a) If the tube is capped and loaded under internal pressure, at what pressure will it yield?b) Will the length increase, decrease or remain constant?

Page 3: Solution manual 13 15

c) If the tube is extended in tension, will the volume inside the tube increase, decrease or remain constant?

d) Find the stress in the walls if the tube is capped and filled with water, and pulled in tension to yielding. Solution: a) The stress state in a thin-wall capped tube is σx = σy/2, σz = 0, where σx, σy and σz are the axial, hoop and radial directions respectively.Substituting in the Hill yield criterion,

Rσy2 + P(σy/2)2 + RP(σy/2)2 = P(R+1)X2,

σy2= X2P(R+1)/(R + P/4 +RP/4), Substituting R = 2.5, P = 0.8 and X = 105,

σy2= 0.875X2, σy

= .935x105 psi at yielding.

The pressure P = σx(t/r) = .935x105(0.025/2.5) = 935 psi

b) From the flow rules, (eq. 13-12)εx/εy = [R(σx- σy) + (σx- σz)]/[(R/P)(σy- σz) +R(σy- σx)]

Substituting σx = σy/2, σz = 0,

εx/εy = [R/2 +1]/[(R/P) +R(σy- σx)] = (1-R)/(2R/P +R). Since R > 1, the axial strain is

negative (the tube shortens). εx/εy = (1-2.5)/(2x2.5/.8 +2.5) = -0.171

c) Taking the volume inside the tube as v = πr2L, dv = π(r2dL+ 2Lrdr)dv/v = dL/L + 2dr/r = dεx + 2dεy.

Substituting dεy = 1/(-.171)dεx = -5.8dεx (from part b),

dv/v = -4.8dεx . In tension dεx is positive so dv/v is negative,and the volume decreases.

d) For the volume to remain constant (dv/v = 0), dεy = -(1/2)dεx. But according to the flow rules Eq.

(13-12) with σz = 0

dεy/dεx = [(R/P)σy +R(σy- σx)]/[R(σx- σy) + σx]. Substituting α = σy/σx and dεy = -(1/2)dεx;

-(1/2)[R(1- α) +1] = [(R/P)α +R(α -1)]α(R/P +R -R/2) = R -R/2 -1/2; α = (R-1)/(2R/P+R) = P(R-1)/[R(2+P)]α = σy/sσ = 0.8(1.5)/[2.5(2.8)] = 0.171

Substituting into the yield criterion, Eq. (13-11)

[2.5(0.171)2 +0.8 +2.5x0.8(1-.171)2]σx2 = 0.8X2,

σx = 1.116X = 111,600 psi

13.5 Consider a sheet with planar isotropy (equal properties in all directions in the sheet) loaded under plane stress, (σz = 0).a) Express the ratio, ρ = εy/εx, as a function of the stress ratio, α = σy/σx. b) Write an expression for

σ in terms of α, R and σx. c) Write an expression for

dε in terms of α, R and dεx. Remember that

σ

dε = σxdεx + σydεy + σzdεz .

Solution: 5 a) From the flow rules, Eq. 13-17) with σz = 0,

Page 4: Solution manual 13 15

ρ = εy/εx = [(R + 1)σy -Rσx]/[(R + 1)σx - Rσy] = [(R + 1)α -R]/[(R + 1) - Ra]

b) Using Eq.(13-16) to define effective stress,

σ 2 = σx2[1 -2αR/(R + 1) + α2] or

σ = σx[1 -2αR/(R + 1) + α2]1/2 (Note that for uniaxial tension,

α = 0 so

σ reduces to σx)

c)

σ d

ε = σxdεx + σydεy + σzdεz. Substituting σz = 0, dεy = ρdεx, and

σy = ασx,

σ d

ε = σxdεx(1 + ρα), or d

ε = dεx(σx/»σ)(1 + ρα)

From part a), ρα + 1 = {α[(R + 1)α -R]/[(R + 1) - Rα]} + 1 = ρα + 1 =

[(R + 1) - 2Rα + (R + 1)a2]/[(R + 1) - Rα]

From part b),

σ /σx = [1 -2αR/(R + 1) + α2]1/2,d

ε = dεx(σx/

σ )(1 + ρα)

= dεx [(R + 1) - 2Rα + (R + 1)α2]/[(R + 1) - Ra]/ [1 -2αR/(R + 1) + α2]1/2

(Note that for uniaxial tension (α = 0), this reduces to d

ε = dεx)

13-6 Using the 1948 Hill criterion for a sheet with planar isotropy,a) Derive an expression for α = σy/σx for plane strain (εy = 0) and plane stress, (σz = 0). b) Find the stress for yielding under this form of loading in terms of X, R and P.c) For a material loaded such that dεz = 0 and σz = 0, calculate σx and σy at yielding. d) For a material loaded such that dεx = dεy and σz = 0, calculate σx and σy at yielding.Solution: a) Using the flow rules, Eq. (13-12) with σz = 0 and dεy = 0,

dεy = 0 = (R/P)σy + R(σy - σx); (R + R/P)σy = Rσx,

α = σy/σx =R/(R + R/P) = P/(P + 1)

b) Substituting σz = 0 and σy = σxP/(P + 1) into the yield criterion, Eq. (13-11),

σx2{RP2/(P + 1)2 + P + RP[1-P/(P + 1)]2} = P(R + 1)X2

σx2[RP2 + P(P + 1)2 + P + RP]/(P + 1)2 = P(R + 1)X2

σx2[P(R + P + 1)]/(P + 1) = P(R + 1)X2

σx = X[(R + 1)(P + 1)/ (R + P + 1)]1/2

c) σx = X[(3)(2.5)/4.5]1/2 = 1.291X = 38,700 psi

α = 1.5/2.5 = 0.6

d) From eq. 13-10 Y/X = {[P(R+1)(P+R)]/[R(P+1)(P+R)]}1/2 = {[P(R+1)/[R(P+1)}1/2 =

{[1.5(2+1)]/[2(1.5+1)]}1/2 = 0.948, Y = 9.48x 30,000psi = 28,460 psi

13.7 For a sheet with X = 350 MPa, R = 1.6 and P = 2.0, calculate:a) The yield strength in the y-direction.b) The value of σx at yielding with σz = 0 and εy= 0. c) The values of σx and σy at yielding with σz = 0 and εz= 0.d) The values of σx and σy at yielding with σz = 0 and εx= εy.Solution: a) Consider a uniaxial tension test in the y-direction. σx = σz = 0, and σy = Y at

yielding. Substituting into the yield criterion, Eq.(13-11),

Page 5: Solution manual 13 15

R(σy -σz)2 + P(σz -σsx)2 + RP(σx -σy)2 = P(R + 1)X2,

RY2 + P(0)2 + RPY2 = P(R + 1)X2,

(Y/X)2 = [P(R+1)]/[R(P+1)] or Y = X [P(R+1)]/[R(P+1)]1/2. Substituting

X = 50 ksi, R= 1.6 and P = 2.0, Y = 50[(2x2.6)/(1.6x3)]1/2 = 52.0 ksi.b) For plane-strain tension, εy = 0, with σz = 0, the flow rules,

Eq. (13-12), give 0 = (R/P)σy + R(σy - σx),

0 = σy(1 + P) - Pσx, σy = [P/(P+1)]σx = (2/3)σx

Now substituting this, R = 1.6, P = 2 and σz = 0 into Eq. (13-12),

[1.6(2/3)2 + 2 + 3.2(1/3)2]σx2 = 2x2.6X2

σx = X[2x2.6/3.06666]1/2 = 65.1 ksi.

c) If εz = 0 and σz = 0, the flow rule,(eq 13-12), gives

-(R/P)σy - σx = 0, or σy = -(P/R)σx = -(2/1.6)σx = -1.25σx.

Substituting this, R = 1.6 and P = 2 into the yield criterion,(eq.13-11),

R(σy -σz)2 + P(σz -σx)2 + RP(σx -σy)2 = P(R + 1)X2

[1.6(1.25)2 + 2 + 3.2(2.25)2](σx/X)2 = 2(2.6), (σx/X)2 = 0.251,

σx = 50x(0.251)1/2 = 25.06 ksi, sy= -1.25sx = -31.3 ksi

d) With εx = εy and σz = 0, the flow rules, Eq (13-12), give

R(σx -σy) + σx = (R/P)σy + R(σy -σx), σy = σx[P(2R+1)]/[R(2P+1)] = σx[(1.052x4.2)/(1.6x5) =

σx. Substituting this, R=1.6 and P = 2 into the yield criterion,

R(σy -σz)2 + P(σz -σx)2 + RP(σx -σy)2 = P(R + 1)X2

{1.6(1.05)2 + 2 + 3.2(.05)2}(σx/X)2 = 5.2

σx/X = [1.379]1/2, σx = 58.7 ksi, σy= 1.05σx = 61.6 ksi

13.8 Consider a material with planar isotropy. Calculate the stress ratio, α = σy/σx, for plane strain, εy = 0 using both the ’48 Hill criterion (equation13.18) and the high exponent criterion (equation13.22) with a = 8 for a) R = 0.5b) R = 2.0.Solution: For the ’48 Hill criterion, the flow rules for planar isotropy with εy = 0 give 0 = σy

+R(σx - σy) ,σy/σx, = R/(1+R) so for a) σy/σx = 1/1.5 = 0.333 and for b) 0.666.

For the high exponent criterion the flow rules for planar isotropy with εy = 0 give 0 = σy

a-1 – R(σx -σy )a-1. With R = 0.5, σy7– 0.5(σx -σy )7 = 0, or α7– 0.5(1 - α )7 = 0

Solving by trial and error, α = σy/σx, = 0.4753With R = 0.5, α7– 2(1 - α )7 = 0, Solving by trial and error, α = σy/σx, = 0.525

Page 6: Solution manual 13 15

13.9Using equations 13.25 and 13.26 for planar isotropy with σz = 0, determine the plane

strain-to-biaxial strength ratio,

χ=σx(σx=σy)(σz=0)

σx(σy=σz=0)

, as a function of R. Plot this ratio as a

function of R from R = 0.5 to R = 2 for both the Hill criterion (a = 2) and for a = 6.Solution: For εy = εx, σy = σx so σx/X = [(R+1)/2]1/a = χ. Plotting:

13.10 Marciniak propose a method of biaxially stretching a sheet by using a cylindricalpunch and a spacer sheet with a circular hole as sketched in Figure 13.6. It is often assumed that the stress state is balance biaxial stress (σx = σy) if there is equal biaxial straining (εx= εy), but this isn’t true unless R0 = R90 = R45. In experiments using this technique, it was found that εy/εx = 1.035 for an aluminum sheet with R0 = 0.55 and R90 =0.89. a) Calculate the stress ratio, σy/σx, that would produce εy/εx = 1.035 using both the Hill criterion (a = 2) and for a = 6.b) Calculate the ratio of strains, εy/εx, that would result from equal biaxial tension, σx

= σy, using both the Hill criterion (a = 2) and for a = 6.

Figure 13.6. Setup for biaxial stretchingSolution: a) Using the flow rules, Eq.(13-12) for 1948 Hill with σz = 0,

εy/εx = 1.035 = [(R/P)σy + R(σy - σx)]/[R(σx - σy)+ σx]

1.035 = [(.55/.89)σy + 0.55(σy - σx)]/[0.55(σx - σy)+ σx]

Page 7: Solution manual 13 15

1.035 = [(.618+ 0.55)σy - 0.55σx)]/[1.55σx - σy]

1.035[1.55σx - σy] = [(.618+ 0.55)σy - 0.55σx)]

(1.035x1.55 + 0.55)σx = (.618+ 0.55 + 1.035)σy

σy/σx = (2.203)/(3.3135) = 1.423

Using the flow rules, Eq.(13-24) for the new criterion with σz = 0 and a =8

εy/εx = 1.035 = (R/P)[α7 -P(1-α)7]1 -R(1-α)7]

1.035 = 0.618[α7 -0.89(1-α)7]/[1 -0.55(1-α)7]

[α7 -0.89(1-α)7]/[1 -0.55(1-α)7] = 1.675

Here α = σy/σx must be found by trial and error. However for a values near 1, (1-α)7 is negligible

compared to α7. Therefore a very good approximation is α7 = 1.675, α = 1.6751/7 = 1.076b) Using the flow rules, Eq.(13-12) for 1948 Hill with σz = 0 and

σy = σx, εy/εx = [(R/P)]/[1] = 0.55/0.89 = 0.618

Using the flow rules, Eq.(13-24) for the new criterion with σz = 0, σy = σx, and a = 8

εy/εx = (R/P)[α7 -P(1-α)7]/[1 -R(1-α)7]

= 0.618[α7 -0.89(1-α)7]/[1 -0.55(1-α)7]= 0.618[1 ]/[1] = 0.618

13-11 W. Lode (Z. Phys. v36, 1926, pp. 913-30) proposed two variable for critically testing isotropic yield criteria and their flow rules:

µ=2(σ2−σ3)(σ1−σ3)

−1 and

υ=2(ε2−ε3)(ε1−ε3)

−1.

Plot µ vs υ for the Tresca, von Mises criteria and for equations 13.23 and 13.24 with R = 1 and a = 6. Compare your plot with Figure 13.7 which is based on experimental data (G. I. Taylor and H. Quinney, Phil. Trans. Royal Soc. Ser. A 203 (1931), pp. 323-62.

Page 8: Solution manual 13 15

Figure 13.7. Plot of Lode’s variables for Problem 13-11

Solution: It is easiest to compare these variables for conditions of σ3 = 0, in which case µ =

2σ2/σ1 - 1 = 2α - 1 (where α = σ2/σ1). Substituting ε3 = - ε2 - ε1, υ can be expressed as υ =

2(2ε2+ e1)/(2ε1+ ε2) -1 = 3ε2/(2ε1+ ε2).

Substituting ρ = ε2/ε1, υ = 3ρ/(2+ ρ). We can assume values of α between 0 and 0.5, find µ =

2α - 1, use the flow rules to find ρ, and then calculate υ = 3ρ/(2+ ρ). For von Mises, ρ = (σ2 -σ1/2)/(σ1 -σ2/2) = (2α-1)/(2-α), so

υ = 3ρ/(2+ ρ) = 3[(2α-1)/(2-α)]/[2+(2α-1)/(2-α)] = 3[(2α-1)]/3 = (2α-1) = µ. (For Mises υ = µ)For Tresca, ε2 = 0 for -1 ≤ α ≤ 0, so ρ = 0 and υ = 0.

For the newer criterion, the flow rules (eq. 26) with R= P = 1, give ρ = dε2/dε1 = [σya-1+ (σy-

σx)a-1] [σxa-1+ (σx- σy)a-1],

ρ = [αa-1-(1- α ) a-1]/[1+ (1- α)a-1].

Page 9: Solution manual 13 15

13-12 An automobile part is to be formed from a low-carbon steel. The steel has planar isotropy with R = 1.8. During tryouts, circles, 5 mm diameter, were marked on the sheet before it was press formed. After forming it was found that a circles in a critical region had become an ellipse with major and minor diameters of 4.87 and 6.11 mm. Assume thatσz = 0. a) Calculate the strains in the critical area.b) Find the ratio, σy/σx, assuming Hill’s ’48 criterion.Find the ratio, σy/σx, assuming the high exponent criterion with a = 6.Solution: a ε1 = ln(6.11/5) = 0.2005, ε2 = ln(4.87/5) = -.0263

b) Using the flow rules, Eq.(13-12) for 1948 Hill with σz = 0 and R = P, εy/εx = ρ = [σ2 +

R(σ2 - σ1)]/[R(σ1 - σ2)+ σ1] = [α + R(α - 1)]/[R(1 - α)+1]

[R(1 - α)+ 1]ρ = [α + R(α - 1)]; α[Rρ +1 + R] = R + ρR; α = R(1 + ρ)/[R(1 + ρ) +1] Substituting ρ = -.0263/0.2005 = -0.1314 and R = 1.8,α = 1.8(1 -0.131)/[1.8(1 -0.131) +1] = 1.564/2.564 = 0.610c) From the flow rules, Eq.(13-31),

ρ = dεx/dεy = [σya-1+R(σy -σx)a-1]/[σx

a-1+R(σx -σy)a-1] =

ρ = [αa-1+R(α - 1)a-1]/[1+ R(1 - α)a-1] =

-0.1314 = [a5+1.8(a -1)5]/[1+ 1.8(1 - a)5]

For small α, ρ = [R(α -1)a-1]/[1- R(α -1)a-1];

R(α -1)a-1 = ρ[1- R (α -1)a-1],

R(α -1)a-1(1+ ρ) = ρ, (α -1)a-1 = ρ/[R (1+ ρ)]

(α -1)5 = -0.1314/[1.8(1 -0.1314)] = -0.084; (α -1) = -0.609; α = 0.391[Note the calculated stress ratio depends heavily on which yield criterion is used.]

Chapter 14

Page 10: Solution manual 13 15

Calculate the height-to-diameter ratio for drawing ratios of 1.8, 2.0, 2.25 and 2.5. Assume a constant thickness.

Solution: With constant thickness the surface areas of the blank and final cup are equal. πr12 +

2πr1h = πro2; 2r1h = ro

2 - r12 ;

h/r1 = [(ro/r1)2-1]/2 = [(do/d1)2-1]/2. h/d1 = (1/2) h/r1 = [(do/d1)2-1]/4.

for do/d1 = 1.8, h/r1 = 0.56; for do/d1 = 2, h/r1 = 0.75;

for do/d1 = 2.25, h/r1 =1.016; for do/d1 = 2.5, h/r1 = 1.312.

Calculate the slope of LDR vs.

R for

R = 1n h = 0.75 according to a) Whiteley’s equation, 14.10b) Whiteley’s analysis (equation 14.7) using equation 14.11 for β.

Solution: a) LDR = exp{η[(

R + 1)/2]1/2}. Differentiating,

d(LDR)/d

R = exp{η[(

R + 1)/2]1/2}(1/2)η[(

R + 1)/2]-1/2(1/2).Simplifying and substituting η = 0.75 and

R = 1,

d(LDR)/d

R = (η/4)exp{η[(

R +1)/2]1/2}/[(

R +1)/2]-1/2 = (.75/4)exp(.75)/1 = 0.397

b) with LDR = exp{η[

R /(

R +1)]0.27}d(LDR)/d

R = exp{η[

R /(

R +1)]0.27η(0.27)[2

R /(

R +1)]1-

0.27[(2/(

R +1) - 2

R /(

R +1)2]Now substituting: d(LDR)/d

R = [exp(0.75)](0.75)(0.27)[1- 1/2)] = 0.214

14-3 Show that the frictional force acting on the die lip is approximately equal to πηFd/2.Solution:

Page 11: Solution manual 13 15

Expanding, exp(-µπ/2) = 1- µπ/2 + (µπ/2)2/2! - (µπ/2)3/3! + ... ≈ 1- µπ/2So Fd - Ff ≈ Fdµπ/2

14-4 A typical aluminum beverage can is 5.25in. high and 2.437 in. in diameter with a wall thickness of 0.0005 in. and a bottom thickness of 0.016 in. The starting material is aluminum alloy 3004-H19 that has already been cold rolled over 80%a) What diameter blank is required?b) Is a redrawing step necessary (assume that a safe drawing ratio is 1.8. If so how many redrawing steps are necessary?c) How many ironing stages are required? (Assume a deformation efficiency of 50%)

Solution: a) The volume of the can is v = πd1tfhf + π(d12/4)to where d1 = (27/16) = 2.4375

in., hf = 5.25, to = 0.016 and tf = 0.005in.

The volume of the blank is v = π(do2/4)to

Equating blank and can volumes, π(do2/4)to = πd1tfhf + π(d1

2/4)to.

do = {[d1tfhf + (d12/4)to](4/to)}1/2 = 4.684.

b) If no redrawing step were used, initial drawing ratio would be 4.684/2.375 = 1.92. Yes, a redraw step is required. One should be sufficient.c) Drawing and redrawing without thickness change would produce a can with walls of 0.016 in instead of 0.005 in. This is a thinning strain of ln(16/5) = 1.16. The heavy cold reduction prior to drawing produces a material at the start of drawing that has very little work hardening left. Assume n = 0. Then from Fig 14-16, the max ironing reduction is r = 0.33, which is equivalent to a maximum strain per ironing pass of ln[1/(1-r) = 0.400. A total ironing strain would then require threeseparate stages.

d) After drawing and redrawing the cup height, h2, would be such that h2 = df[(do/df)2 - 1]/4 =

1.637 in. Let us assume that half of this height is attained in the initial draw so after the initial draw h1 = 0.82 in. Let us also assume that each ironing stage produces the same strain, (1.16/3) = .387.

Then After the first ironing h3 = h2exp(.387) = 1.637(1.472) = 2.409, then h4 = 1.637(2.409) =

3.547 and h5 = hf = 1.637(3.547) = 5.22

Then the total stroke would be h1 + h2 + h3 + h4 + h5 = 10.8 in.

Page 12: Solution manual 13 15

(The answer may differ from this depending on how the diameter reduction was distributed between the cupping and redrawing and how the wall thinning was distributed between the ironing stages.)

14-5 In the analysis of deep drawing the work in bending and unbending at the die lip was neglected. a) Derive an expression for the energy expended in bending and unbending as a function of the strain ratio, R, the yield strength, X, the thickness, t, and the die lip radius.Assume planar isotropy.

b) What fraction of the total drawing load might come from this source for a sheet with R = 1, t = 1.0 mm, r = 0.8 mm, do = 50 mm d1 = Solution: a) For a non-workhardening material, the work/volume, dW/dV = σxεx because σz = 0

and εy = 0. [Note that W = ∫(dW/dV)dV where dV = πd1dzdh.]

In bending εx ≈ z/(r + t/2) where z is the distance from the mid-plane and (r + t/2) is the radius

of curvature at the mid-plane. Therefore the total work in bending per unit punch stroke is (dW/dh) =

2πd1∫t/2σxzdz/(r + t/2) = 2πd1σx(t/2)2/[2(r + t/2)]. Unbending requires the same amount of work so

the total bend-unbend work is dW/dh = 2πd1σx(t/2)2/(r + t/2).

The bending and unbending is in plane strain, (εy = 0), so σx = X(R + 1)/√(2R+1)

dW/dh = 2πd1(t/2)2X(R + 1)/√(2R+1)/(r + t/2).

b) for R = 1, t/2 = 0.016 and r = 0.25,

dW/dh = X(2π)(1)(0.016)2(2)/[(√3)(0.25 + 0.016)] = 0.00698Xwhereas the total drawing force is Fd = (1/η)πd1tσfln(d0/d1)

where from Eq.(14-4), σf = X[(1 + R)/(R + 1/2)]1/2 = 1.155X

Fd = 1.155X(1/η)πd1tln(d0/d1) = 1.155X(1/η)π(2)(0.016)ln(2.2) = 0.0915X/η. Assuming η = 0.75, Fd = 0.122X. Therefore the work in bending and unbending represents 0.00698/0.122 = 5.7%

of the total work. (This fraction of course depends on the radius of the die lip relative to the sheet thickness.)

14-6 Significant increases in the limiting drawing ratio can be achieved using pressurized water on the punch side of the blank as shown in Figure 14.22. Limiting drawing ratios of three have been achieved for material that would with conventional tooling have an LDR of 2.1.a) Explain how the pressurized water can increase the drawability.b) Estimate the level of pressure required to achieve LDR = 3 with η = 0.75 and Y = 250 MPa.

Page 13: Solution manual 13 15

Solution: a) Much of the energy required for the deformation work is supplied by the fluid pressureacting on the edges of the flange, so the tensile stress in the wall is reduced.b) The work done by the fluid pressure per unit punch travel, dh, is (dW/dh)fluid = 2πrt(-

dr/dh)P where r is the current outer radius of the flange. Since 2πr1h + πr2 = πro2, 2r1dh = -2rdr

or -dr/dh = r1/r. Substituting (dW/dh)flange = 2πr1tP.

The total rate of work per punch travel is dW/dh)total = (1/η)2πr1tσfln(r/r1)

The force supplied by the punch, Fd = dW/dh)total -(dW/dh)fluid

Fd = (1/η)2πr1tσfln(r/r1) - 2πr1tP = 2πr1t[(1/h)σfln(r/r1)- P]

At the drawing limit, for a non-work hardening material, r = ro, and Fd = σw2πr1t so

σw = (1/η)σfln(r/r1)- P,

For isotropy according to Mises σw = σf = 1.155Y. Substituting:

P = 1.155Y[1- (1/η)ln(r/r1)]

With Y = 40,000, r/r1 = 3 and η = 0.75, P = 21,500 psi

(This is a very high fluid pressure and would require a vessel with very thick walls.)

14-7 Using the procedure in Section 14-7, estimate

∆h/h=42h45−(h0+h90)2h45+(h0+h90)

for a

drawing ratio of 1.8 for materials having the following strain ratios:a) R0 = 1.8, R90 = 2.0, R45 = 1.4b) R0 = 1.2, R90 = 1.4, R45 = 1.6c) R0 = 0.6, R90 = 0.4, R45 = 0.8d) R0 = 0.8, R90 = 0.8, R45 = 0.5For each material calculate

2∆R/R and ∆h/h with Figure 14.14. Solution: For each material, calculate

2∆R/R . Compare the values of ∆h/h and 2DR/√R for each material with Fig. 14-12.

h45/h0 = [1 + (d1/d0)1/(P+1)]/[1 + (d1/d0)1/(Q+1)]

h90/h0 = [1 + (d1/d0)1/(P+1)]/[1 + (d1/d0)1/(R+1)]

2∆R/R = [2h45-(h0 +h90)]/[2h45+(h0 +h90)]

2∆R/R = [2Q-(R + P))]/ [2Q+(R + P))]a) h45/h0 = 0.954 h90/h0 = 0.992

Figure 14.22 Cup drawing with pressurized water on the blank

Page 14: Solution manual 13 15

∆h/h = -0.0216

2∆R/R = -0.151b) h45/h0 =1.0106 h90/h0 = 0.9875

∆h/h = 0.0084

2∆R/R = +0.103c) h45/h0 = 1.0568 h90/h0 = 1.0318

∆h/h = 0.0197

2∆R/R = +0.231d) h45/h0 = .946 h90/h0 = 1.000

∆h/h = -0.0276

2∆R/R = -0.272

14-8 During the drawing of flat bottom cups, the thickness, tw, at the top of the wall is usually greater than t0 at the bottom of the wall as shown in Figure 14.23. Derive an expression for t/t0 in terms of the blank diameter, do, and the cup diameter, d1. Assume isotropy and neglect the blank holder pressure. Find the ratio of tw/t0 for a drawing ratio of 1.8. (Consider the stress state at the outside of the flange.

Solution: At the outside of the blank, the stress state during drawing is uniaxial compression.The compressive strain is ey = ln(d1/do). Then assuming isotropy, ez = ln(tw/to) = (-1/2)ey =

(-1/2)ln(d1/do); tw/to = (d1/do)-1/2. For (do/d1) = 1.8, tw/to = √1.8 = 1.34

Figure 14.23 Wall thickness variation.

Page 15: Solution manual 13 15

Chapter 15

15.1 Derive an expression for the critical strain, ε1*, to produce a diffuse neck as a function of n and the stress ratio, α. Assume the von Mises criterion and loading under constant α. [Hint: start with equation 15.15.]

Solution: Eq.(15-15) gives ε1* = 2n(1 + ρ + ρ2)/[(1+ ρ)(2ρ2 - ρ +2)]

The flow rules for Mises give ρ = dε2/dε1 = (σ2 -σ1/2)/(σ1 -σ2/2) = (2α-1)/(2-α). Now

substituting for ρ,

ε1* = A/B where A = 2n[1+(2α-1)/(2-α)+(2α-1)2/(2-α)2] and

B = [(2α-1)/(2-α)+1][2(2α-1)2/(2-α)2-(2α-1)/(2-α)+2]Expanding and simplifying:

ε1* = 2n[1-α+α2)/(2-α)]/[4α2 - 7α +4)(1+α)]

As a check, note that for α = 0 (uniaxial tension), for α = 1/2 (plane-strain tension), and for α =1 (balanced biaxial tension), ε1* = n.

15.2 In principle one can determine the R-value by measuring the angle, θ, of the neck in a strip tension test. How accurately would one have to measure θ to distinguish between two materials having R-values of 1.6 and 1.8?

Solution:θ = arctan[(R+1)/2]1/2 Substituting R = 1.6, q = 51.89°; and for R = 1.8, q = 51.28°This is a difference of only 0.61°. One would have to measure to an accuracy of about ±0.3° to be sure they were different. Such accuracy in measuring the angle of a neck isn't possible so this method is not used to find R.

15.3 It has been suggested that failures in sheets occur when the thickness strain reaches a critical value. Superimpose a plot of e1 versus e2 for this criterion on Figure 15.8. Adjust the constant so that the curves coincides for plane strain, e2 = 0. How good isthis criterion?Solution: ε3 = C, and ε3 = -ε1-ε2, so ε1 = -C - ε2.

Converting to engineering strains, ε2 = ln(1+e2) or e2 = exp(ε2) -1 and

e1 = exp(ε1) -1 = exp(-C-ε2) -1 = exp[-C- ln(1+e2)] -1

At e2 = 0, -C = ln(e1+1) = ln(1.43) so e1 = 1.43/(1+e2) -1.

Page 16: Solution manual 13 15

15.4 Repeat Problem 15-3 for a failure criterion that predicts failure when the absolutely largest principal strain, |ei|max reaches a critical value. Solution: The criterion is the same as in 15-3 for e2 > 0 where

|ei|max = -e3 (where e3 is negative) and the criterion is the same as in Prob 15-3. For e2 < 0, e1 = |

ei|max so the criterion is e1 = C. (horizontal straight line.) Matching at e2 = 0, C = 0.43. Neither

regime makes any

15.5 Ironing of a two-piece beverage can thins the walls from 0.015 to 0.005 in. thickness. This strain far exceeds the forming limits. Explain why failure doesn’t occur. Solution: This corresponds to ε1 = ln3 = 1.099 or ε1 = 2.0. Although this reduction is in excess of

those indicated on the forming limit diagram for plane strain (Fig 15-7), it must be realized that theforming limit diagrams are determined under plane-stress conditions, (σz = 0) and are therefore

applicable only under those conditions. For ironing, σz is very compressive so the FLDs are not

applicable. (This is also true for extrusion, rolling, etc.)

15.6 The true minimum of an FLD determined by in-plane stretching (without bending) occurs plane strain, ε2 = 0. However forming limit diagrams are determined by stretching strips over a 4-in. diameter dome. Assume that the forming limit is reached when the center of the strip reaches the limit strain for in-plane stretching. If the limit strain is ε1 = 0.30 for in-plane stretching, what values of ε1 and ε2 would be reported for stretching over a 4-in. diameter dome?Solution: On the outside surface both ε2 and ε1 are larger than art the midplane by the bending

strain, ∆ε = (t/2)/R. Substituting R = 51 mm. for a 102 mm. diameter hemispherical punch and t = 0.8 mm, ∆ε = .004 ε2 and ε1 = 0.304 and ε2 = 0.004. (The change is about 1/2%)

Page 17: Solution manual 13 15

15.7 The forming limit diagram for a certain steel is shown in Figure 15.24. a) Show the loading path of a specimen that fails at point N.b) Plot as accurately as possible the locus of points corresponding to uniaxial tension, σ2 = 0. c) Describe qualitatively how the line in (b) would be drawn differently for R = 2.

Solution:

ε

15.8 Frequently the minimum on experimentally determined forming limit diagrams occurs at a slightly positive value of e2 rather than at e2 = 0. Explain why this might be.Solution: Strains are measured on the outside of the bend. When the middle of the sheet is in plane strain, the outside will be in biaxial stretching. Furthermore any changing strain path that starts initially with a little biaxial stretching will move the minimum total strain to the right of plain-=strain conditions as shown in Figure 15-22

15-9 The following points were taken from the forming limit diagram for aluminum alloy 2008; The strain hardening exponent is 0.265 and K = 538 MPa. Plot the forming limits in stress space (i.e. values of σ1 at failure as a function of σ2). Assume isotropy.

Figure 15.234. Schematic forming limit diagram for Problem 15-7.

Page 18: Solution manual 13 15

ε1 ε2

-0.22-0.10 0.00.225 0.100.23 0.23Solution:For each case we can calculate

ε and then

σ . And then σ1 =

σ /√(1+α+α2) and σ2= ασ1 ε1 ε2

ε

σ ρ α σ1 σ2

-0.22 0.636 477 -0.489 0.0147 480 7-0.10 0.419 427 -0.303 0.232 470 1090.0 0.247 371 0 0.5 428 2140.225 0.10 0.310 365 0.444 0.773 402 3110.23 0.23 0.428 430 1.0 1 430 430

σ 1

0 100 200 300 400 500σ2

0

100

200

300

400

500

MPa

MP

a