Solution manual 10 12

Chapter 10
10-1 The XYZ Special Alloy Fabrication Company has an order to roll a 4x4x15 in.
billet of a nickel-base super alloy to ½ in. thickness. They tried rolling it with a 5%
reduction per pass, but it split on the fourth pass. At a meeting the process engineer
suggested using a lower reduction per pass, the consultant suggested applying forward
and backward tension and the shop foreman was in favor of heavier reductions per pass.
With whom, if any do you agree. Defend your position.
Solution: ∆ ≈ h/√(Rrho) = 4/√[(6)(0.5)(4)] = 3.6 which is much larger than 1. To avoid centerline
cracking, D should be lowered. Therefore increase the reduction per pass as suggested by the
foreman. (Front and back tension are impractical on a 15 inch long billet in a 12 in. diameter mill
and if possible it woud only raise the level of hydrostatic tension at the centerline.) Note using Eq
(10-5) gives ∆ = 3.56.
10-2 The Mannesmann process for making tubes from cylindrical billets is illustrated
in Figure 10.23. It involves passing the billet between two non-parallel rolls adjusted for
a very small reduction onto a mandrel positioned in the middle. Explain why the axial
force on the mandrel is low and why the mandrel, which is long and elastically flexible,
follows the center of the billet.
Solution: With a very small potential reduction, the contact length between the rolls and
bar is very small, so ∆ is large. Therefore, there is hydrostatic tension at the center,
making the required piercing force quite small. The mandrel tends to stay at the center
where the hydrostatic tension is greatest, and therefore the required piercing force the
least.
10-3 (a) Show that as r and α approach zero equations 10.11 and 10.12 reduce to
φ = 1 + ∆/4 for plane strain and φ = 1+ ∆/6 for axisymmetry.
(b) What percent error does this simplification introduce at r =
0.5 and α = 30°?
Solution: For plane strain, Eq. (10-11) gives f = 1 + (1/2)tanα/εh, and Eq. (10-3) gives
∆ = (2/r -1)sinα. Substituting r = 1 - exp(-εh) - 1 -[1 - εh + εh
2/2 - εh
3/3! - ....],
r ≈ εh for small εh and sinα ≈ tanα for small α.∆ ≈ (2/εh -1)tanα, or 1/εh = 1/2 + ∆/(2tanα).
Substituting into Eq. (10-11),
Chapt 10 - Deformation-zone Geometry 163
Figure 10.23 The
Mannesman process for
making tubes.

φ ≈ 1 + (1/4)(tanα + ∆) but tanα << ∆ for small α, so φ ≈ 1 + ∆/4
For axisymmetry, Eq.(10-12) gives φ = 1 + (2/3)tanα/εh and Eq. (10-4) gives ∆ = (sinα/r)(1 +
√(1-r)2 ≈ 4 sinα/r ≈ 4sina/eh for small r. Substituting 1/εh ≈ ∆/(4sinα) into Eq. (10-11b) and
letting sinα ≈ tanα, φ = 1 + ∆/6
b) For plane strain: φ = 1 + (1/2)tan30/ln[1/(1-0.5)]= 1.416
φ = 1 + (1/4)(2/.5 -1)sin30 = 1.375. the error is (1.416- 1.375)/1.416 = 2.9%
For axisymmetry: φ = 1 + (2/3)tan30/ln[1/(1-0.5)]= 1.556
φ = 1 + (1/6)(sin30/0.5)(1 - √(1-0.5)2 = 1.486 the error is ( 1.556 - 1.486)/ 1.556 = 4.5%
At lower reductions the error is much less
10-4 Consider wire drawing with a reduction of r = 0.25 and α = 6°.
(a) Calculate the ratio of average cross-sectional area to the contact area between
wire and die.
(b) What is the value of ∆?
(c) For strip drawing with α = 6°, what reduction would give the same value of ∆?
(d) For the reduction in (c), calculate the ratio of of average cross-sectional area to
the contact area and compare with your answer to (a).
(e) Explain why for the same ∆, the frictional drag is greater for axisymmetric
drawing.
Solution: a) The mean cross-sectional area is Am = (Ao + Af)/2 and
r = (Ao - Af)/Ao so Af = Ao(1-r).
The contact area is Ac = π(Ro + Rf)[(Ro + Rf)2/tan2α + (Ro - Rf)2]1/2
Ac = π(Ro + Rf)(Ro - Rf){1/tan2α + 1]1/2 =
Ac = π(Ro
2 -Rf
2)[cos2α/sin2α +1]1/2= (Ao-Af)[(cos2α+sin2α)/sin2α]1/2Ac = (Ao-
Af)/sinα =Aor/sinα.
The ratio Ac/Am = [(Ao-Af)/sinα].2/[Ao(2-r)] = 2r/[(2-r)sinα = 2(0.25)/[(2-0.25)sin 6°] = 2.733
b) ∆ = (sin6°/0.25)(1+√(1-.25)2 = 1.46
c) Eq. (10-3) gives ∆ = (2-r)sinα/r = 1.46 so 1.46r = (2-r)(.1045);r = 0.134
d) Using Eq (10-6), Ac/Am = 2r/[(2-r)sinα] = 2(.134)/[(1.866)(.1045)] = 1.374
e) For the same D and a, there is much greater contact area with axisymmetric drawing
and since the stress acting normal to the work-die interface is about the same, the normal force
and thus the frictional force is much greater.
10.5 Some authorities define ∆ as the ratio of the length of an arc through the
middle of the deformation zone and centered at the apex of the cone or wedge formed by
extrapolating the die to the contact length.
(a) Show that for wire drawing with this definition ∆=α(1+1−r)2
/r.
(b) Calculate the ratio of ∆ from this definition to the value of ∆ from Eq. 10-4 for
(i) α = 10°, r = 0.25, (ii) α = 10°, r = 0.50, (iii) α = 45°, r = 0.50,
(iv) α = 90°, r = 0.50.

Solution: a) Let L be the distance from the apex to the arc as shown.
sinα = (Ro + Rf)/(2L) or L = (Ro + Rf)/(2sinα)
The arc length = 2Lα = α(Ro + Rf)/(2sinα)
The contact length is (Ro + Rf)/sinα so
∆ = arclength/contact length = α(Ro + Rf)sinα/[sinα(Ro - Rf)] =
∆ = α(Ro + Rf)/(Ro - Rf)
r = (πRo
2 - πRf
2)/(πRo
2) = 1 - (Rf/Ro)2; Rf = Ro√(1-r)
∆ = α(1 + √1-r)/(1 - √(1-r ) = (α/r)(1 + √(1-r)2.
b) ∆= α(1+√(1-r)2 ∆ = sinα(1+√(1-r)2 % difference
α=10°, r=0.25 2.43 2.419 0.5
α=10°, r=0.50 1.017 1.012 0.5
α=45°, r=0.50 4.577 4.121 10.
α=90°, r=0.25 9.155 5.828 36.3
10.6 Backofen developed the following equation for the mechanical efficiency, η,
during wire drawing: η=[1+C(∆−1)+µtanα]−1
for ∆ > 1, where C is an empirical
constant equal to about 0.12 and µ is the friction coefficient. Note that the term C(∆ -1)
represents wr/wi and the term µ/tanα represents wf/wi so η = (1 + wr/wi +wf/wi)-1
=
(wa/wi)-1
.
(a) Evaluate this expression for µ = 0.05 and α = 0.3 for 2·≤ α ≤ 20°. Note
that for small angles ∆ < 1 so the redundant strain term, C(∆ -1) is zero.
(b) Plot η vs. α and find the optimum die angle, α*.
(c) Derive an expression for α* as a function of µ and r. Does α* increase
or decrease with µ? with r?

Solution: Substituting µ = 0.05, C = 0.12 and r = 0.3 and ∆ = (1 + √(1-r)2sinα/r
=(1 + √(1-.3)2sinα/.3 = 11.244sinα into η = [1 + C(∆-1) + µ/tanα]-1;
η = [1 + 0.12(11.244sinα -1) + 0.05/tanα]-1 = [0.88 +1.349sinα + 0.05/tanα]-1.
Now substituting α = 2,4,6,8,10,12,14 and 16°
For α = 2°, η = 0.424; for α = 4°, η = 0.592; for α = 6°, η = 0.668; for α = 8°, η = 0.7024;
for α = 10°, η = 0.7153; for α = 12°, η = 0.7164; for α = 14°, η = 0.7107; for α = 16°, η =
0.7011. Trial and error gives α* = 11.2°, η* = 0.717. Plotting:
b) d(1/η)/dα = 0.12[(1+√(1-r 2cosα/r] - µcsc2α = 0.
cosαsin2α = µr/[0.12 (1+√(1-r )2]. Substituting r = 0.3, α = 11.2 as found in part a.
As µ increases, α* increases. As r increases, α* increases.
10-7 A slab of annealed copper with dimensions 2.5 cm x 1.5 cm x 8 cm is to be
compressed between rough platens until the 2.5 cm is reduced to 2.2 cm. Describe, with a
sketch, as fully as possible the resulting inhomogeneity of hardness.
Solution:

To the first approximation, the deformation will be in plane strain. The slip-line field has dead metal
caps along the platens and wedges of undeforming material at the ends. These should be regions of
low hardness. The high hardness region is expected in the active regions of the slip-line field
10-8 A high-strength steel bar must be cold reduced from a diameter of 2 cm to 1.2 cm
by drawing. A number of schedules have been proposed. Which schedule would you
choose to avoid drawing failure and minimize the likelihood of centerline bursts? Assume
η = 0.5 and give your reasoning.
(a) a single pass through a die of semi-angle 8°.
(b) two passes (2.0 cm to 1.6 cm and 1.6 to 1.2 cm) through a die of semi-angle 8°.
(c) Three passes (2.0 cm to 1.72 cm, 1.72 cm to 1.44 cm and 1.44 cm to 1.2 cm.) through
a die of semi-angle 8°.
(d) Four passes (2.0 cm to 1.8 cm, 1.8 cm to 1.6 cm, 1.6 cm to 1.4 cm and 1.4 cm to 1,.2
cm) through a die of semi-angle 8°.
(e, f, g, h) Same reductions as in schedules a, b , c and d except through a die of semi-
angle 16°.
Solution: Use as low a ∆ as possible to avoid center-line bursts, i.e. a low die angle (8°) and as high
a reduction as possible without drawing failure. Neglecting work hardening, the highest strain
(reduction) per pass is εmax = η = 0.5
The total strain needed is 2ln(1/.65) = 0.86 so two passes are required. Therefore choose schedule B.
10.9 Determine ∆ from the data of Hundy and Singer (Figure 10.6) for reductions of
2.3%, 6.5%, 13.9% and 26%. Make a plot of I.F. versus ∆.
Solution: Taking ∆ = [(2-r)/2)√[ho/(rR)]
For r ∆ IF = Hs/Hc
0.80 0.212 1
0.60 0.286 1
0.26 0.540 1
0.139 0.789 1
0.097 0.966 80/72 = 1.111
0.065 1.20 79/68=1.162
0.046 1.44 70/58=1.441
0.023 2.06 1
plotting

0 1 2
∆
IF
1.3
1.2
1.1
1.0
Chapter 11
11.1. (a) Explain why inclusion-shape control is of much greater importance in high-
strength steels than in low-carbon steels.
(b) Explain why inclusion-shape control improves the transverse and though-thickness
properties, but has little effect on the longitudinal ones.
Solution:a) The higher stresses necessary for forming are more likely to cause fracture.
The toughness generally decreases as the stress level rises.
b) Without shape control, the inclusions are elongated in the rolling direction. In a
transverse direction test they form a much larger area perpendicular to the tensile axis and
therefore an easy fracture path.With shape control they tend to be spherical so the area
perpendicular
11-2 Figure 11.1 shows that for a given sheet material, greater reductions are possible
before edge cracking if square edges are maintained than if the edges are round. Explain
why.
Solution: For square edges, the stress state is nearly plane-strain compression. The
tensile force in the rolling direction required to cause the necessary elongation is about
half of the compressive stress.
For rounded edges, the middle of the edge sees very little compression so its elongation
(which must be the same as the center) requires almost pure tension in the rolling
direction.
11.3 (a) Wrought iron has a high toughness when stressed parallel to the prior working
direction, but a very low toughness when stressed perpendicular to the prior working
direction. Explain why.
(b) When wrought iron was a commercial product, producers claimed its corrosion
resistance was superior to steel. What is the basis of this claim?
Solution: a) The highly elongated silicates inclusions formed easy fracture paths for
cracks parallel to the rolling direction.
b) The insoluble silicate inclusions tended to block the corrosion, requiring corrosion
to follow a more tortuous path.
11.4 (a) With the same material, die angle and reduction central bursts may occur in
drawing but not in extrusion. Explain why this may be so.

(b) Why may a given material be rolled to strains much higher than the fracture strain in
a tension test?
Solution: There is a higher hydrostatic tension in drawing.
b) The stress state in rolling is compressive, so porosity doesn't grow as in tension.
11.5 Figure 11.18 is a Kuhn-type forming limit diagram for upsetting a certain grade of
steel. The line gives the combination of strains that cause cracking.
(a) Superimpose on this diagram the strain path that leads to failure at point P.
(b) What differences in test variables would lead to cracking at point S?
Figure 11.18 Figure for Problem 11.5
Solution:a)
P
S
ε2
ε1
0.3
0.2
0.1
0.20.4
b) Failure may occur at a lower compressive strain because of more friction or because
he height-diameter ratio is lower.
Chapter 12
12.1 An old shop-hand has developed a simple method of estimating the yield strength
of steel. He carefully bends the strip with his hands and to a given radius, releases it and
dotes whether it has taken a permanent set. He repeats the process until it does take a
permanent set. A strip 0.25in thick, 1 in wide and 10 in long first takes a permanent set at
a radius of 10 in Estimate the yield strength.

Solution: ε = t/(2ρ) = σ/E; σ = Et/(2ρ) = 30x106
(0.25/(2x10) = 350,000 psi
12-2 A coiler is being designed for a cold-rolling line of a steel mill. The coil diameter
should be large enough so that coiling involves only elastic deformation. The sheet to be
coiled is 1 mm thick, 2 m wide and has a yield strength of 275 MPa.
a) What is the minimum diameter of the coiler?
b) For this diameter find the horsepower consumed by coiling at 30 m/s.
Solution:
a) Assuming the Mises criterion, σx = 1.15Y (plane strain), but
σx = Eex/(1-υ2
) and ex = t/(2R) = t/D
1.15Y = (Et/D)/(1-υ2
), D = 207x109
x0.001)/(1.15x275x106
x0.91) = 72 cm.
b) Again, σx = 1.15Y and σx = Eex/(1-υ2
). The stored elastic energy per volume,
w, is w = (1/2)σxex = E'ex
2
where E' = E/(1-υ2
).
ex varies with position, ex = z/R = 2z/D, where z is the distance from the neutral plane.
Taking L as the length coiled, b as the width and t the thickness, the total elastic energy,
W, is
W = 2Lb∫wdz = 2(1/2)E'exLb∫(2z/D)2
dz where the integration is between z =0 (mid
plane) and z = t/2. (The factor 2 is to account for the material between z = -t/2 and z = 0).
Integrating, W = 4LbE'(t/2)3
/(3D2
) = LbE't3
/(6D2
)
The work rate is Wv where v is the velocity = L/t', where t' is the time on roll a length L.
Substituting, W =vLbE't3
/(6D2
) = 30x2x207x109
(0.0001)3
/([0.91x0.0722
) = 2.6Mw
12-3 In many applications the minimum thickness of a sheet is determined by its
stiffness in bending. Then aluminum is substituted for steel to save weight, its thickness
must be greater.
a) By what factor must the thickness increased?
b) What weight saving would be achieved?
c) Would the weight saving be greater, less or unchanged if both sheets were corrugated
instead of being flat?
Hint: For elastic bending, the deflection, δ , is given by δ=A
FL3
′Ewt
, where F is the Force
L is the span length, ′E is the plane-strain modulus, w the sheet width and t the sheet
thickness. For steel ′E = 220 GPA and ρ = 7.9 Mg/m3
. For aluminum ′E = 73 GPA and
ρ = 2.7 Mg/m3
.δ
Solution: a)A constant stiffness depends on the end conditions, load distribution, etc.
However for constant F, d, b, and L, Et3
= constant, so EAltAl
3
= Esttst
3
, tAl/tst = (Es/EA)1/3
= 31/3
= 1.44;
b) wtAl/wtst = (ρAl/ρst )(tAl/tst) = (ρAl/ρst)(Est/EAll)1/3
= (2.7/7.9)(30x106/10x106) = 0.493
wt saving = (1-0.493) = 50.7%
c) If the panel were curved, the potential weight saving would be much less than in
For the flat sheet, the additional material is at the extreme position relative to the neutral
plane, where it is most effective in stiffening the panel. With a curved panel, however,
much of the additional material would be nearer the neutral plane, where it is much less

effective. Therefore the % increase to achieve the same thickness would be much larger
and the weight saving less.
12.4 For some designs, the minimum sheet thickness is controlled by the ability to
absorb energy elastically in bending without any plastic deformation. In this case, what
weight saving can be achieved by substituting aluminum (Y = 25ksi) for steel (Y =
35ksi) . Use the data in Problem 12-3.
Solution: The energy absorbed is U = ∫Fdδ but dδ = [AL3
/(E'wt3
)]dF, where δ is the
deflection, L is the span, t is the thickness, w is the width, E' = E/(1-υ2
) is the plane-strain
modulus, F is the force, and A is a constant that depends on the load distribution and the
support. Integrating, U = ∫[AL3
/(E'wt3
)]FdF = ALAlF2
/(2E'wt3
)
For constant U, A, L, and w, F2/(E't3
) must also be constant. Therefore, assuming that u
is the same for steel and aluminum, FAl
2
/(EAltAl
3
) = Fst2
/(Esttst
3
).
(a) But the force is limited by yielding. Up to ant at yielding, the stress at the surface is
given by σ = Mc/I where the bending moment, c = t/2 and I = wtAl/12 so σ = 6M/(wt2)
or M = swt2/6. Realizing that at yielding σ = Y, and F is proportional to M, F is
proportional to Yt2. Substituting in a,
(YAl
2
tAl4)/(EAltAl
3
) = (Yst
2
tst4)/(Esttst
3
);
(tAl/tst) = (EAl/Est)(Yst/YAl)2
The weight, W, is proportional to the thickness, t, times the density, r, so
(WAl/Wst) = (ρAl/ρst)(tAl/tst) = (ρAl/ρsst)(EAl/Est)(Yst/YAl) =
(2.7/7.9)(10/30)(35/25) = 0.223
12-5 An aluminum sheet, 1 mm thick is to be bent to a final radius of curvature of 75
mm. The plastic portion of the stress strain curve is approximated by 175 + 175ε MPa.
Accounting for springback, what radius of curvature must be designed into the tools if the
loading is:
a) pure bending?
b) tensile enough so that the mid-plane is stretched 2% in tension?
Solution: a) ex = z/R, σx = C1 + C2ex (assuming Tresca). Assuming w = 1,
M = 2_t/2σxzdz = 2_t/2(C1 + C2z/R )zdz
M = 2[(C1/2)(t/2)2
+ (C2/3)(1/R)(t/2)3
]
On unbending, ∆σx = E'∆ex, and ∆ex = z∆(1/R)
∆M = 2_t/2∆σxzdz = 2_t/2E'∆(1/R)z2
dz = (2E'/3)'∆(1/R)(t/2)3
M + ∆M =0, so 2[(C1/2)(t/2)2
+ (C2/3)(1/R)(t/2)3
] +
(2E'/3)'∆(1/R)(t/2)3
= 0
∆(1/R) = -[3C1 + C2(1/R)t]E't.
Since ex + ∆εx = e'x, (1/R) + ∆(1/R) = (1/R') = 1/3; ∆(1/R) = 1/3 - (1/R)
1/3 - (1/R) = -[3C1 + C2(1/R)t]E't; (1/R)(1-C2/E') = 1/3 + 3C1/(E't)
1/R = [1/3 + 3C1/(E't)]/(1-C2/E') =
= (1/3 +3x25x103/(11x106
x 0.04)]/[1 -25x103/11x106) = 0.505, R =1.98in.
b) ex = z/R + 0.02, σx = C1 + C2ex. Now σx varies

continuously across the section so, integrating from z = -t/2 to + t/2,
M = _[C1 + C2(z/R + 0.02]zdz = _[C1 + 0.02C2)zdz + _C2(1/R)z2
dz =
(1/2)(C1 + 0.02C2)[(t/2)2
-(-t/2)2
] + (1/3)C2(1/R)[(t/2)3
-(-t/2)3
] = (2/3)C2(1/R)(t/3)3
As in a) above, ∆M = 2E'∆(1/R)(t/2)/3
and M + ∆M = 0, so -C2(1/R) = E'∆(1/R).
Again '∆(1/R) = 1/3 - 1/R or -C2(1/R) = E'(1/3 - 1/R)
-(C2 + E')(1/R) = E'/3; 1/R = (E'/3)/E' - C2 = (11x106/3)/(11x106
- 25x103
)
1/R = 1.002/3 = 2.993 in.
[Note that tension has almost eliminated the springback]
12.6 What fraction of the cross section remains elastic in Prob. 12-5a?
σx = E'ex and σx = C1 + C2exso
ex = C1/(E' - C2) = 25x103
/(11x106 - 25x103
) = 2.278x10-3
Solution: At yielding, ex = z/R, so at the elastic-plastic interface, z = Rex =
1.98x2.278x10-3
= 4.51x10-3
in. Fraction elastic = 4.51x10-3/0.02 = 22.6%
12.7 It has been suggested that the residual hoop stress in a tube can be found by
slitting a short length of tube longitudinally and measuring the diameter, d, after slitting
and comparing this with the original diameter, d0. A stress distribution must be
assumed. Two simple stress distributions are suggested by Figure 12.11. For a copper
tube, d0. = 25 mm and d = 25.12 mm, t = 0.5 mm. Assume E = 110 GPA and υ = 0.30.
Find the residual stress at the surface using both assumptions about the stress
distribution.
residualstress
positioninwall
outside inside
residualstress
positioninwall
outside inside
a. b.
-
Figure 12-11 Assumed stress distributions for Problem 12-7.
Solution: 7 M = -∆M = -2_t/2∆sxzdz, (assuming w = 1) and ∆σx = E'∆ex
= E'z∆(1/R) so M = 2t/2E'z∆(1/R)zdz = -2E'∆(1/R)(t/2)3
/3
but also M = 2t/2σxzdz
Case A: σx = s's (a constant where σs refers to the surface). Then
M = 2σ's(t/2)2/2 = σ's(t/2)2
Equating,

σ's(t/2)2
2πτr2
dr=
0
R
∫ 2πτR3
/3= -2E'∆(1/R)(t/2)3
/3; σ's= -(2/3)E'∆(1/R)(t/2)
With do = 1, d = 1.005, Ro = 1/2 and R = 1.005/2
D(1/R) = 1/R - 1/Ro = -9.95x10-3, E' = 16x106/(1-0.32); t/2 = 0.1 so
σ's = -(2/3)(16x106/0.91)(-9.95x10-3)(0.01) = 1166 psi
Case B: σx = [z/(t/2)] s's so M = 2_t/2[s's/(t/2)] z2
dz =
(2/3)[σ's/(t/2)](t/2)3
= (2/3)σ's(t/2)2
Equating, (2/3)σ's(t/2)2
= -(2/3)E'∆(1/R)(t/2)
σ's= - E'∆(1/R)(t/2) = (-16x106/0.91)(-9.95x10-3)(0.01) = 1750 psi
12.8 A round bar (radius R and length L) was plastically deformed in torsion until it
the entire cross section has yielded. Assume an ideally plastic material with a shear
strength, k, and a shear modulus, G. When unloaded it untwisted by an amount ∆θ
(radians)
(a) Derive an expression for the level of residual stress, ′τ , as a function of the radial
position, r, and R, G and k.
(b) Find the relative springback, ∆θ/L in terms of G, k and R.
Solution: For equilibrium, T + ∆T = 0 so (2/3)„kR3
= -(„/2)G∆θR4
/L
a) The residual stress, τ' = τ + ∆τ= k + ∆τ but
∆t = G∆θ = Gr∆θ/L and ∆θ/L = -(4/3)k/(GR) so τ' = k[1- (4/3)(r/R)]
(At the surface where r = R, τ's = -k/3)
b) T = 2πτr2
dr=
0
R
∫ 2πτR3
/3,
∆T = 2π(r/R)∆γsGr2
dr
0
R
∫ =2π(R3
/3)∆γsG 2π(r/R)∆γsGr2
dr
0
R
∫ =2π(R3
/3)∆γsG
Equating, ∆θ/L =∆γs = (4/3)τ/G
0.1 m, E[‘
12-9 A plate was bent to a radius of curvature, R. After springback the radius was ′R .
Later the plate was etched, removing the outer surface. How would you expect the last
radius ′′R to compare with R and ′R ?
Solution: The plate will continue to unbend so R" > R' > R. After unloading, the plate
undergoes some springback (R' > R) and the outside surface is under compression and the
inside tension. Removing these layers will cause an unbalance of the bending moment
that can be corrected only if the strip further unbends (R" > R') causing the outside
compression in the new outside surface and tension in the new inside surface.
12-10 Consider bending of a strip, 80 mm wide and 1.0 mm thick. The stress-strain
relation in the elastic region is σ = 210ε GPa and in the plastic region it is σ = 250 MPa.
(a) What is the limiting curvature to which the strip can be bent without yielding.
(b) If the strip is bent to a radius of curvature of 500 mm, what is the radius when it is
released. Assume bending by a pure bending moment.
Solution: a) ε = t/2ρ; ρ = t/2ε; Substituting εmax = (250/210)x10-3
= 0.00119 and
t = 0.001; ; ρ = 0.42 m
b) Using equation 12-10, 1/r’ = 1/r -3σ0/(tE’) = 1/0.5 – 3x250x106
√(4/3)/
[0.001(210x109
/0.91) = 1.75, r’ = 0.57

12-11 A steel sheet, 1.5 m wide, and 1.0 mm thick is bent to a radius of 100 mm.
Assume no strain hardening, no friction and no tension. The effective stress-strain
relation is σ=650(0.015+ε)0.20
. E = 210 GPa and υ = 0.29. Find the radius of curvature
after unloading.
Solution: From eq. 12-10 1/r’ = 1/r – [6/(2+n)](K’/E’)(t/2r)n
(1/t). Substituting r = 0.1 m,
E’ =(210x109
/0.91) , K’ = 650x106
Pa, t = 0.001 m, n = 0.2,

Solution manual 10 12

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (10)

Similar to Solution manual 10 12

Similar to Solution manual 10 12 (20)

Recently uploaded

Recently uploaded (20)

Solution manual 10 12