- Email: [email protected]
Solutions to exercises
Solutions to exercises
Chapter I Ex. 1.1
Initial yield stress (r
(a)
= Py 2.89 kN A--'o = 12.5 mm ~ = 231.2 MPa
Al Strain at initial yield ey = lo
Elastic modulus E =
(b)
0.0563 50
(aDo 231.2MPa = 205.33 GPa ey = 0.001 12
R-value
R
11.41 In i2.5 = 1.88 12.5 x 50 In 11.41 x 57.5
113 In----wo
In w~
wl Ex. 1.2 Instant area: A =
P Stress:
tr =
--
Aolo l
Pl =
A
Ao lo
At 4% elongation, PA IA O'A ~---
Ao lo
1.59 kN 52 = ...... x -- = ll8.11MPa 14 50
eA = In (1~0) = 0 . 0 3 9 2
176
0.00112
At 8% elongation, PB IB 0rB ~--.
Ao lo
~--.
1.66 kN 54 --- = 128.06 MPa 14 50
en = In (/~0) = 0 . 0 7 6 9
Therefore, In (~A) - In (~B) n~--=
......
In (eA) - In (6B) K =
O'A
=0.12
= 174.21 MPa
Ex. 1.3
O" =
K6nk m
6=In
kl =
~2 =
(l) ~
=In
0.5 x 10 -3 60 50 x 10 -3 60
cq = 459 MPa,
(55)
=0.0953
/55 x 10 -3 = 0.1515 x 10 -3 s -1
/55 • 10 -3 = 15.15 x 10 -3 S-1 o2 = 480 MPa
Aor = O'2 -- O'l = 21 MPa
A P = A o A o l l , = 0.27kN Ex. 1.4 By plotting the data in the engineering stress-engineering strain, true stress and true strain curves, the following properties are obtained. Initial yield stress: Ultimate tensile strength: True strain at maximum load: Total elongation: Strength coefficient, K: Strain hardening index, n:
156 MPa 294 MPa 0.24 45.4% 530 MPa 0.24
Solutions to exercises
177
0"0 I 11JI, v
cj'OI
8u!u~od lmal,q ~a~ttgfo so!u.rlO~lq 8L[ mm 8s
=
8"0 x 8 x 8 = ~ x 17"6 x g ' 9 = ~ 1 7 6 1 7 6=
's
~ml tuoJj
:ssou'4o!tO leU!d I ' ~ "x~[
Je),deqo 0"1.I
S't-
0"~-
I
cj'~-
I
0"s
I
%/~.0 I , /
~/IF
q,S
5WF~/
OL'~
-
(U!eJ|$)6OI SA($$eJ),s)6o"1 u!eJi$ eruj. 0s
OtfO
OEO
00"0
Ot'O
I
I
I
,,,
f
O OS OOt 00~
|
o~
ooe oss oo~ o~ u!eJ~,s-sseJls em.l. u!mls "6u3 frO
IfO
s
I
~'0 I
0
t'O
I
I
l O~ ~
i ooL O~t .~
.
oo~ r 0 ~
~
00s 0~;s
u!mls-sseJls 5u.ueeu!Su3
The principal strains"
El
=
In
(9.4
8mm
] = 0.161
e2 =
(6.5 ram') = In \ 8 m m ] -0.208
6'3 =
- - ( e l + 6'2) =
0.047
Strain ratio: r /5 =
-0.208
-ez
=
=
-1.29
0.161
Stress ratio: a =
2~+1 2+~
= -2.225
Membrane stresses: effective strain, g
=
1 + fl + fl2)e I = 0.218
effective stress, 5 = 600 (0.008 + ~-)0.22 = 432.57 M P a or I =
~ / l _ oa~ + a 2
-- 151.3 M P a
o2 = aOl = - 3 3 6 . 6 M P a trl + t r 2 + t r 3
Hydrostatic stress trh =
3
= --61.7 M P a
Deviatoric stresses: ?
tr I = trl =
o2
=
tr3
-
-
Oh =
151.3
-- (--61.7)
=
213
MPa
trh = --274.9 M P a Oh =
61.7
MPa.
Check for flow rule: el 0.161 " 7 = 2 1 3 = 0.000 756 o" 1 82
-0.208 -274.9
= 0.000 756
f3 0.047 -- =------ = 0.00076 cr~ 61.7 Solutions to exercises
179
Plastic work of deformation, tO
f0.218
=
vol
600 ( 0 . 0 0 8 + g)0.22 dg = 600 (0.2261.22 -- O. 0081.22)
1.22
,10 = 78.77 x 106 Jim 3
vol = 51.2 m m 3 = 51.2 x 10 -9 m 3 w = 4.03 J
Ex. 2.2 Diameter of Mohr circle (trl - 0 3 ) Yielding with a von Mises criterion, (0" l -- 0"2) 2 + (0"2 -- 0-3) 2 + (0"3 -- (71) 2
2
=trf
Replacing r
+a3 2
we get that 2 -
=
~/3
Ex. 2.3 Effective stress, i
"
-- 0"2) 2 + (0"2 -- 0"3) 2 + (0"3 -- O"1)2
2
=trf
Therefore, for o'2=0"3=0 m
(7 = ( 7 1 = O f
From flow rule, El
/~2
E3
crI
02
%
t
2 3
'
,
o'1
a2
3
Therefore, 2
180
Mechanics of Sheet Metal Forming
t or 3 =
r -----
3
Effective strain, ,
,
,
Ex. 2 . 4
(a)
Stress and strain ratios al = 400 MPa, 9a =
I
and
~r2 = 200 MPa fl=O
Therefore, 62=0 63 -----"-61
(b)
=~
61 63
Under fluid pressure, - 2 5 0 MPa. 61
- - = - 1 will not change. 63
This can be proven from the flow rule, trl = 4 0 0 - 250 = 150 MPa, tr3 = 0 -
tr2 = 2 0 0 - 250 = - 5 0 MPa
250 = - 2 5 0 MPa
Mean stress and deviatoric stresses: 150- 50O'h =
3
250
= - 5 0 MPa,
o~ = 150 - ( - 5 0 ) = 200 MPa, o~ = - 2 5 0 - ( - 5 0 ) = - 2 0 0 MPa 61
63
therefore, El E3
This is because yielding is independent of the average stress on an element. Therefore, the strains are also independent. Ex. 2.5
Major strain, 61 ~--"ln(12/lO) = Minor strain, e2 = iS.el =
0.182 0.182
0.182 -0.182
Solutions to exercises
181
Thickness strain, 83 = - ( 1 +/~)81 Effective strain, ~ = ~/(4/3)(1 + fl + fl2)81 Effective stress, 5" = 850e-~ Strain ratio, a = (2~ + 1)/(2 + fl) Major stress, ai = "#lJ1 - a + a 2 Minor stress, tYlO~ Thickness, t = t exp e3
-0.365 0.365 723 1 723 723 0.833
,,
0 0.211 662 MPa -1 382 MPa - 3 8 2 Mpa 1.2mm
Ex. 2.5
Strain/Stress state
al .
.
Plane stress
.
.
a2 .
Plane stress
Of
Of
,/g
,,5
2 ~af
plane train Plane stress
81
E2
83
,,
----
pure shear
0"3
.
af
0
81
--81
1 --~of 43
0
el
0
-'81
Of
0
el
81
-2e~
biaxial tension
Chapter 3 Ex. 3.1
(c) Principal strains el = l n ( 6 ".l.m . . .m. .) 5.0ram E3 = -
e2 = In (45"8mm) - 0 " 0 4 1 . 0=m m
= 0.199,
(el "4-82)=--0.158 82
/3 = - - = -0.21 =~
t~=
81
2~+1
2+#
= 0.324 (ratio of stress)
Effective strain ~" = ~/~ (1 + ~1 + , 2 ) e , = 0 . 2 1 Tensions K.~ TI = trlt = ~/1 "t~ +~t ~ t ~ 600 x 0.210.22 = ~/1 _. ().324 + 0.32"42
+
t3) l)
x 0.8 x exp(-(1 - 0 . 2 1 ) x 0.199) = 329.3(kN/m)
T2 = c~Tl = 0.324 x 248.6 = 108.2(kN/m) 182 Mechanics of Sheet Metal Forming
Ex. 3.2 (a)
Constant thickness
.,,
~1 -- o'2 Tmax --"
~ O'1
2
Therefore O"
gm~
(b) Uniaxial tension
~" =
1
+ oe2) Crl =
Crl - - o'2 Tmax ~'~
2
Or1
Crl ~'~ - -
2
Therefore a .
=2
Vmax
(c) Plane strain fl=O
1
oe=~
=~
Of -F Of2) O'1 = tYl - - 0 " 3
al
2
2
~m~
T
tyl
Therefore, t7 .....
~
~r~
~'m~ Ex. 3.3
(a) 0.015, 0.015
Location Strains: Strain ratio ~: Stress ratio ~ =
I 2 f, l + l
2+/3
9
(c)
(b) 0.050, 0.000 0
0.150, -0.100 -0.667
I/2
-0.251
Solutions to exercises
183
Thickness strain - ( 1 + fl)el" Thickness: t = to exp(e3): 300 Stress o"I ~/i "- O~ "+" t~ 2" Tension:
-0.030 0.485
-0.050 0.476
-0.050 0.476
300
346
262
(MPa)
146
165
125
(kN/m)
Ex. 3.4
WD
Vol
=
fo
"rid'g=
U l+n
fo
K (eo + g)d~" K
(E 0 + ~--)n+l [g
l+n
6 0 0 x 106
{(6o + ~,+l _ ~g+l}
{0.041"22 -- 0.011.22} ---~7.90 x
1.22
106(j/m3)
Work done in unit mass:
WD
7.9 x 10 6
kg
7850
- 1.01 x 1030)
Temperature increase:
ATCe = J J .... =
AT=
Ce
1.01 x 103
2.2~
0.454 x 10 3
Chapter 4 Ex. 4.1
rl =
2K to ~/~
2)~ ~el
( )n 2_2_e
~v/~l
exp(-el)
T~ exp(-el) = 2Kto
x 340 x 103 2 x 700 x 106 x 0.8 x 10 -3
By trial and error, el
0.08 0.547
0.07 0.536
(mm)
0.06 0.523
Therefore, et = 0 . 0 6 184 Mechanics of Sheet Metal Forming
= 0.5258
Ex. 4.2 Maximum wall tension is when el = n. TI max - -
2Kt~ n V ~~/3) exp(-n) --~n
Tension at centre and at binder: T~,o =
Tl,max
Ti,B =
=
71"
exp(~ -~)
= 1 266 x V~
Tl,max 71"
exp (0.15 ~- )
0.2
exp(-0.2) =
~
et
exp(-el)
1 (el) ~ e x p ( - e l ) = 1.266 (0"2)02 exp(-0.2) = 0.469 el "
0.02 0.448
0.03 0.481
0.04 0.505
0.025 0.466
0.026 0.469
el = 0.026 Tlo=T,n=2x600xO.8xlO3(__~__ , , ......... ~/~ 267 B=2x0115=890kN,
0.026
)0.2
exp(-0.026) = 267(kN)
2B=1780kN
Ex. 4.3 Side-wall tension: 1.266T1,o = 338(kN). Punch force: 676(kN) Blank-holder force 1780 = -- 2.6 Punch force 676
Ex. 4.4 Angle of wrap from O to A is 0 where: 600-10 sin 0 = 2000 - 10' 0 = 17.25 ~ = 0.301 (rad) Tension at mid-point, Tl,o = 2 • 400 • 0.8 • 103 (__. ~ ~c~ .0.025
)0.17
exp(-0.025) = 197(kN/m)
Tension at A, T I , A - - T I , O x e 0"1•
--
197 x 1.03 = 203(kN/m) Solutions to exercises
185
At B, angle of wrap is 60 degrees (1.047 rad), TI,B -" Tl,o x e 0"1x1"047 = 197 x 1.11 = 219(kN/m)
Chapter 5 Ex. 5.1 For the tensile strip, the load is Lo
P = triAl = K(eo + e)nAO-~l = K(eo + e)ntowoexp(-e) 10P
Ktowo Oe
= 0
[(eo+e) ~exp(-e)]
Oe = n(eo + e) n-1 e x p ( - e ) - (eo + e) n e x p ( - e ) = 0
n --- E 0 - F E
E=n--EO
Ex. 5.2 For the tensile strip,the load is n Lo P = ~IAl ----K61Ao-~l = K~Ttowoexp(-el)
For the narrow section, at m a x i m u m load e = n, and Pmax = Ktowon n cxp(-n l)
= 750 x I06 x 1.2 x 12.4 x I0-6(0.22) -~
exp(-0.22) = 6.42(kN)
For the wider section, the strain in A is e IA. P = Ke~At0w0 exp(-eiA) at m a x i m u m load,
P = Ktowoe~A e x p ( - e l A ) = 750 x 106 x 1.2 x 12.5 x 10-6(elA) -'22 exp(--elA) = 6.42(kN) Therefore, f ( e l A ) = e_0.221Aexp(--elA) = ,-z-70.22 12.4 022 " exp(--0.22) = 0.571 IL.D
186
Mechanics of Sheet Metal Forming
By iteration, elA
f(flA)
0.2 0.18 0.16 0.17
0.5746 0.5728 0.5694 0.571
Therefore, eta = 0.17, and the 20 mm length becomes: l = 20exp(elA) = 23.7(mm) Ex.
5.3
From Equation 5 13, dAo = - 0 . 2 = - 0 . 0 2 (n-
e u ) ~-'
~
9
Ao
'_ndAo
=
Ao
10
~/0.02n
(n - 6u) 2 = 0.02n n 2 -- 2 n e u + e u2 = O . 0 2 n
n 2 _ (2eu +O.02)n + e u2 =
0
(2eu + 0.02) 4- ~/i2eu + 0.02) 2 - 4 e u2
2 (26u + 0.02) 4- ~/0.08tu + 0 . ~ 2 0.3 0.25 0.2-
0.15 0.1
--
0.05 ,=
0
I
I
I
I
0.05
0.1
0.15
0.2
0.25
Ex. $.4
As the load transmitted is equal in each section, P = o'aAa -
o'bAb
Ab
Solutions
to exercises
187
But, cr = 9m
B~ m
Ab B ~ n
Bea = A--aa
~a = (A~bab)'/m
eb, Oreb:(A-~b) '/m ~a
In terms of elongation, the total elongation is
8L = 81a+ 81b = l (ela + 8L = l (kla
v=
8t
8L,
where
elb)8t
+ Elb) = l (Ela + ~lb) = / k l a
( 1 + (A~bb)l/m)
Therefore, 11 Ela --" / [ 1 + (Aa/Ab)l/m] ' ~lb = /[1 +
(Ab/Aa) l/m]
Ex. 5.5
For both elements, the material has zero plastic strain so that the effective stress is = 600(0.004) 0.2 = 198.9MPa crlA = 198.9 MPa,
trlB =
O'lA
fo
=
198.9
0.995
= 199.9 MPa
Strain ratio is a, then from Equation 2.18(b), ~=
(~/'1-a
+ a2) or,
o"
~/1 - a
(7
+or 2
.'. ~/1 - ~ + r o2B = r
=O'IB --"
f0
= fo = 0.995,
a = 0.99
= 0.99 x 199.9 = 197.9(MPa)
Chapter 6 Ex. 6.1
(a)
Determine the limiting elastic curvature
where
I= 188
wt 3 12
Mechanics of Sheet Metal Forming
At limit of elastic bending,
O"l =
O'f
Limiting elastic curvature:
Me
I
S
tot E
250 X 106 X 50 X 10 -3 X (2 x 10-3) 2
y
6
6 8.33
= E/=
200 x 109 • =~
50
= 1.25 m-
10 -3
X
= 8.33(Nm)
X
(2
x
10-3) 3
i2
R =800mm
(b) Construct a moment curvature diagram Elastic moment: M e = E l ( l ) wt 2
Plastic moment: Mp = S--~- = 12.5 Nm Elastic plastic moment, >
but before the moment reaches fully plastic moment Mp. From Figure 6.11, the material is in plastic deformation above Ye, and elastic below Ye. Strain
s =
--
E
=
=~
Ye
Ye =
s
E(1/p)
Equilibrium equation in bending, t12 M = 2 ~wy dy = 2w E .1o
y~
g y2 dy + S
p \ 3 ]o + S
ydy
= w--~--(3- m 2) Ye
1
(c)
1
1
where - = - - . - - (see Equation 6.23) p Pe m 9 Using the approximate relation (,) S A ~ =-3---Et p = 400mm
=~
---
Et
=
(1)
1.88
m-
= 2.5m-' l
Solutions to exercises
189
Final curvature 1 - = 2 . 5 - 1.88 = 0.625 m -l P p = 1.6m 9 Using the diagram in (b) trf Ye = = 0.5 X 1 0 - 3
E(1/p)
/ Ep, (y3) - (y2) T O +trf
M=2w A
(1)
'1
= 11.46 Nm
yo
AM=I.72m-
= E1
l
p = 1.28 m I..... I Ex. 6.2
Initial radius of curvature and wrap: po=~
l
1 and 0 o = I n = 2 z r 27r Po
Springback due to unloading: 1
1 = 00 = E't p~ /90 The gap between the ends of the strip after unloading is given by gap = poA0 =
~l ( - ~ 73S ~po0o )
=
2zrl ( 3 S )
312S 2rr t E'
gap= Ex. 6.3
E' =
E 200 = = 219.8 GPa 1 - 112 1 -- 0.3 2
2 tr ~ ~
2 ( 2 ~0.22 = ~ 600 ~ e / = 715e~
190
Mechanics of Sheet Metal Forming
MPa
~ - " 2zr
0.2(2 x 1 0 - 3 ) 2.22 IPI
--'~
= 3.94 x 10 -8
..........
21"22 • 2 . 2 2
from Equation 6.27
From Equation 6.26 M=
InK'
3.94 x 10 -s x 715 x 106
=
0.080.22
= 49.1Nm
For elastic unloading
A
= E'I
-49.1
12
219.8 x 109
0.2(2 x 1 0 - 3 ) 3
= 1.675 m -1
Final curvature is 1.675 = 10.8 m -1
0.08
and the final radius of curvature is
1
..... = 92 mm 10.8
Ex. 6.4
Springback in an elastic perfectly plastic material:
1
E' t
P0 Springback between aluminum and steel can be compared as follows:
A (llP)Ifllpo)I~ '
B
SAI
S^~IE'A~ ~
A (I/p) / (I/~) Isteel
#
~---
Ssteel / Este~i
Esteel X
Ssteel
EAI
Since the material is work-hardening, an approximation can be made by assuming the average stress to be S = K e n at the respective bend ratio. Therefore, B~--"
A (llP)l(ll~)IAl A (1/p)/(1/p0)ISter
190 x ---- = 0.98 (e)-0.06 530e 0"26 75 205e 0.2
"~--
For p / t = 10, the strain is e = y / p = t / 2 p = 0.05 B = 1.173
For p / t = 5, the strain is e = y / p = t / 2 p = 0.1 B = 1.125
Solutions to exercises 191
Alternative solutions The springback for a work-hardening material can be derived as 6
t
n
Therefore, the springback ratio between aluminum and steel is 6
n
2+ B
t
IAI
_..
6 (2.~.'n) (~-~) (~) nlStee'
:o~
2.26
--
,9o ( , ~o~-o~o
2.2
2 + nsteel 2 + n AI
x
KAI Ksteel
x
-0.06 1.007 ( ~ )
For p / t = 10, B = 1.007 (0.05)-o'o6= 1.205 For p / t = 5, B = 1.007 (0.1) -o.o6 = 1.156
Chapter 7 Ex. 7.1 In the hole expansion process the equilibrium equations are
ri) r
m
2T Then for T~ = --~ at r = r0:
Ex. 7.2 For shells,
dTo dr
T o - T~ =0 r m
For nosing, To = - T . Therefore,
dTo
T + T~
dr
r
dTo
dr
T4~+ T
r
=0
192 Mechanics of Sheet Metal Forming
Esteel EA!
(+)
0.2-0.26
Integrating the above equation yields the following: In (T~ + T-')
io~. =
r ]rir
-In
In(r, +r--) - 1 . ~
- -1.
(r)
~
r
Ex. 7.3
In flaring the equilibrium equations, given the boundary conditions T~ = 0 at ro, are
T,
=
- T i n (-~)
To = "T+ T,/,= "T(I - In(r--~)) Therefore, the limit of T~ is given by - T at =-T
T, = -'In (~)
=~
In ( ;ro )
ri
=1=~
( ;ro )
= e
Then r can move between ri and the maximum ro = r~e.
! r'
(a)
From Equation 8.11
hf=~-
-1
(b) Max punch force ,f
Fd = 2rrrito ltrf In I
(r0) ri §
= 37.5 mm
oxp( )
0.1 x 30000 } (0.1 x n') Fd = 2n" x 2 5 x 1.2 {3501n ( 5 ~ ) + . . . . . . . . exp = 57(kN) rr x 50x 1.2 2 Ex. 8.2
The maximum punch load will be lower for n = 0.2 than for n = 0.1. The curve also moves to the fight as shown below. Solutions to exercises
193
n=0.1 n=0.2
I
~aX
iI
\
I
I I I 1
~h
,
h~
Ex. 8.3 For a small section of the sheet, assume the sheet slides down. Tool pressure per unit width: p=2Tsin
(do) -~-
T+dT+gP=
=Td0 T
dT = - t t TdO dT T
In
= - I t dO
(T2)~ =
- "2" ~'Tt
T2 = Tl exp ( --2
T2- T, = T, [exp (--~-/~)- 1]
T+dT
194 Mechanics of Sheet Metal Forming
~)
Ex. 8.4
(a) Volume of metal,
rr ~to = rrdftfhf + rr ~to do = 120mm Total drawn ratio =
120mm 62mm
Limiting drawn ratio
= 1.94
(LDR) LDR < exp (1/) LDR < exp (0.6) = 1.82
Therefore, the can cannot be drawn in one operation. (b)
If the draw ratios are equal for the two drawing operations,
do di
di df
di = ~/i20 x 62 = 86 mm
Height from first draw:
ri h i - - "~
-\ri/
- 1
/
=20ram
(c) Punch force, neglecting blank holding force and die friction: 2n'rito
Fd= -------af l n ( ~ ) = (d)
21.6kN
Cup height at the finish of second draw: hf==
rf 2
ro --
-1
=42.6mm
(e) The full height is obtained by ironing the middle section.
(0 Saving from reduced thickness: rrd~ rrl202 {0.41 - 0.33} = 905 mm 3 Volume of saving per blank = - - ~ {t1983 - - t1998} " 4 For 100 billion blanks, volume saved = 905 x 102 m 3 Cost of metal saved = 5 x 2800 x 905 x 102 = $1.25 billion
Solutions to exercises 195
Chapter 9 Ex. 9.1 Using the relation in Section 9.1.3, the effective stress is 6.4 3 5" = {(61/3) 2 + 4} (61/50) 2 = 620 MPa 16 1.2 and the effective strain is, = 2In (61/50) = 0.40
Ex. 9.2 The principal radius of curvature Pl at and near the tangent point is 50 mm. The principal tensions at the tangent point, from Section 9.2.1, are T# = 374 MPa and
To =
aT# = 0.89 x 374 = 333 MPa
As indicated in Section 9.2, in the unsupported region, p = 0, and therefore,
ro Pl
P2
i.e.
374
02 =
333
x 50 = - 5 6 mm
Ex. 9.3 From Equations 9.4 and 9.6, it may be shown that --=1+ t
From Equation 9.1, the effective strain is
= -et
= In (to~t) = In [1 +
(h/a) 2]
and the membrane stress is tr# = tro = ~ = 350e-~
MPa
From Equation 9.5, the bulging pressure is 2tr#t p
~
2 x 350~"~
1.2 x 10 -3
....
P
a 2 [1 +
(h/a) 2]/2h"
[1 +
(h/a) 2]
where a2[1 + (h/a) 2] p= 2h
196
Mechanics of Sheet Metal Forming
MPa
For a = 50mm, and computing in the range, 0 < h < 45, we obtain the characteristic shown below. (The maximum pressure occurs at a membrane strain of approximately 0.2 (~ = 0.4). As indicated in Figure 9.3, for a ductile material this is probably less than the forming limit curve in biaxial tension and it is expected that the sheet would deform beyond this maximum pressure and fail at a higher strain under a falling pressure gradient.)
10
_
O.
t~ ta
6-
4-
Q..
_
I
0
I
10
20 Height, hmm
I
I
30
40
C h a p t e r 10
Ex. 10.1 The stress after bending over the block without tension is, from Equation 6.17, 78 x 109 x 1.85 x 10 -3 Ett o'1=4"~-~ =-I= 120 MPa 2x0.6
On the outer (tension) side of the sheet, the additional stress to cause yielding as in Figure 10.2(d), is 1 8 0 - 120 = 60 MPa. The tension to be applied to add this stress is T = AtYlt = 60 x 106 x 1.85 x 10 -3 = 111 k N / m The fully plastic tension, as in Figure 10.2(e), is
Ty = St = 180 x 106 x 1.85 x 10 -3 = 333 k N / m
Solutions to exercises
197
Ex. 10.2
The limiting elastic radius of curvature is, from Equation 6.19, E't 78 x 109 X 1.85 • 10 -3 = 0.401 m P e = 2S = 2 x 1 8 0 x 106
From Equation 10.2, the moment prior to applying the tension is 78 x 109 (1.85 x 10-3) 3 Mo =
12 x 0.6
= 68 N m / m
From Equation 10.6, when m = 0, the moment is M = Mo/2 = 34 N m / m
From Equation 10.7, the tension for m = 0 is
I ~} { __~
T=Ty
1-4----.~ ~
=333
1- 4x0.6
As the loading is elastic, the change in curvature on unloading is, from Equation 6.30, 34 x 12 A
= E'I
78 • 109 (1.85
x
= -0.826 10-3) 3
The final curvature is 1
0.826 = 0.84 m 1
0.6
and the final radius of curvature is 1/0.84 = 1.19 m
i
.. L~ ~,
L. <,~
L~ ,,~
(a)
(b)
(c)
198 Mechanics of Sheet Metal Forming
'.......'iiii i i i i i i i ....................... l ..................... L~ =~
(d)
L.
<,~
(e)
The stress distribution is shown diagrammatically above. The elastic, plastic stress state under the tension and moment, similar to that in Figure 10.2(d), is shown in (a). Unloading the tension is equivalent to subtracting the uniform stress in (b)to give a distribution in (c) in equilibrium with the moment. Subtracting the elastic bending stress in (d) occurs when the moment is released to give the residual stress in (e).
Ex. 10.3 From Equation 10.12, the slope of the stress-strain curve for steel is 0.2 x 700 x 0.0120.2 0.012
= 4.82 GPa
and for aluminium, is 0.2
x
400
x
0.012 0.2
0.012
= 2.75 GPa
From Equation 10.11, the springback, i.e. the change in curvature on unloading, is, for steel, A
(1) st~l
=
4.821 220 2.5
= -0.0088 m -1
and for aluminium, A
(1) ~.
=
2.751 = -0.0141 m 78 2.5
The final curvature for steel is ( 1 / 2 . 5 ) - 0.0088 = 0.391 giving a radius of curvature of 2.56 m. For aluminium, the final curvature is ( 1 / 2 . 5 ) - 0.0141 = 0.386, giving a radius of curvature of 2.59. Clearly the springback is very small and similar for both metals. The result is independent of thickness and depends on the ratio of the slope of the stress-strain curve to the elastic modulus, which are not very different for these two cases. Ex. 10.4 From Equation 10.20, the tension after the first bend is 0.34 } T = To + AT = 0.6Sto + 4 ~ o (1 + 0.62) = Sto 0.6 + -~--
Solutions to exercises
199
The thickness reduction is, from Equation 10.21,
to
2bto ~
t
o.o 2b
The thickness downstream of the first bend is
The stress in the sheet downstream of the first bend is trl T
S
St
Evaluating in the range given, we obtain the diagrams below. 1 0.80.6u) 0 . 4 0.2-
0
3
I
I
I
I
I
I
4
5
6
7
8
9
10
Bend ratio, b (=r/t )
o's1
.o
'~ 10 Ir C
9c-~- 5
O
3
I 4
i 5
i 6
i 7
Bend ratio, b (=r/t)
200 Mechanics of Sheet Metal Forming
I 8
10
Chapter 11 Ex. 11.1
L
(a) For the frictionless case The current radius of the tube at the comer is r. The tube wall thickness will be uniform. From volume constancy
(
" ) =~Rto " r)+~r
t 2(R
' / / / / / !
re-arranging t =
I
r I _1
4
r
.
.
.
.
For the plane strain condition = ~---~e0=
In
2[
2
cro=-~K e o + ~ l n
and 5" = (4
-
( 4 - zr) R)]n ~r "'
Internal pressure is given by p=
trot
2toK
--
r
+ --~ In ( 4 - ( 4 ~ 7r) R)ln
q/3r [4 _.-i~~,Jr).~] leo
r
(b) For the full sticking case The relation between the comer radius and the comer thickness can be obtained from the incremental relation 7t
7t
7/;~"' \ r +
t dr + -~(r - dr)(t - dt) = ~rt f~(4
l) dr yr
r
fotdt t
t ~ - - t o ( ~ ) (r4 - 1 )
Solutionsto exercises 201
Circumferential stress 2 [ e o + ~2 ( _7r )4 _I 1n O'l'-~K
(R)] ~
The internal pressure is given by p
~
tTlt
r
,,
2t0 r ( 4-1) [
VC~r (-~)
K eo+~
2 (4)
~--1
In
(R)] n
Ex. 11.2
The minimum comer radius is limited either by necking or by the maximum fluid pressure available. Since
t --
4
eo = - e t = In to
t
e0 = n = 0.2 for local necking of tube wall, therefore,
r=
17mm
t=3.27,
The internal pressure required to form comer radius r is given in Ex. 11.1. Substituting the material properties and die geometry into the equation, we obtain: r(mm) 38.00 37.00 36.00 35.00 34.00 33.00 32.00 31.00 30.00 29.00
t(mm) 3.45 3.44 3.44 3.43 3.42 3.41 3.40 3.39 3.38 3.37
tro(MPa) 566.85 568.86 570.84 572.78 574.70 576.58 578.44 580.27 582.07 583.84
p(MPa) 51.52 52.96 54.47 56.08 57.77 59.56 61.46 63.48 65.63 67.92
Therefore, the minimum comer radius is 31 mm, and is limited by the maximum pressure available. Ex. 11.3
In order to calculate the initial tube thickness, we use volume constancy.
2(tl +t2) 2
L +
rr
zr
= - Rto
202 Mechanics of Sheet Metal Forming
where L = (R - r)
i,a
/.
t o = -1~ 2[ ( t l +~t 2 ) ( R _ r ) + r t 2 ]
L,
tl
The forces that existed at each end of the wall can be calculated as Fl=trltl=~K
eo+~ln
The final internal pressure at the current forming radius can be found from: F2=pr
thus
p=--
F2 r
p To calculate the average friction coefficient from force balance
tZavpL = F2- Fl /Zav= F2-pL FI
where q = p
Ex. 11.4 (a) For the free ends, there is no stress in the axial direction, tr2 = O. For a thin cylinder, o3 = 0. el
= In ( ~ o )
In(t~)=- 89In(~o) ?o = Effective stress and strains are
"~ = trl andS-el Solutions to exercises
203
The internal pressure is calculated as trlt p= r
p=Kto
ro
In
r ro
To calculate the expanded radius at instability, let dp = 0. This leads to --
r0
--- - - or r 3
=
roe2n/3
(b) For restricted ends, there is no strain in the axial direction, e2 = 0.
El "- In ( F ~ ) E3 --'---El In ( t ~ =) (- rIn~ ) t
ro
to
r
Effective stress and strains are: if" =-.~-.~0'l and ~" = ~3e I
The internal pressure is calculated as trlt p= r
oro[In
p=
K--~-
To calculate the expanded radius at instability, let dp = 0. This leads to
in
204
= ~ orr=r0
en/2
Mechanics of Sheet Metal Forming
Chapter I Ex. 1.1
Initial yield stress (r
(a)
= Py 2.89 kN A--'o = 12.5 mm ~ = 231.2 MPa
Al Strain at initial yield ey = lo
Elastic modulus E =
(b)
0.0563 50
(aDo 231.2MPa = 205.33 GPa ey = 0.001 12
R-value
R
11.41 In i2.5 = 1.88 12.5 x 50 In 11.41 x 57.5
113 In----wo
In w~
wl Ex. 1.2 Instant area: A =
P Stress:
tr =
--
Aolo l
Pl =
A
Ao lo
At 4% elongation, PA IA O'A ~---
Ao lo
1.59 kN 52 = ...... x -- = ll8.11MPa 14 50
eA = In (1~0) = 0 . 0 3 9 2
176
0.00112
At 8% elongation, PB IB 0rB ~--.
Ao lo
~--.
1.66 kN 54 --- = 128.06 MPa 14 50
en = In (/~0) = 0 . 0 7 6 9
Therefore, In (~A) - In (~B) n~--=
......
In (eA) - In (6B) K =
O'A
=0.12
= 174.21 MPa
Ex. 1.3
O" =
K6nk m
6=In
kl =
~2 =
(l) ~
=In
0.5 x 10 -3 60 50 x 10 -3 60
cq = 459 MPa,
(55)
=0.0953
/55 x 10 -3 = 0.1515 x 10 -3 s -1
/55 • 10 -3 = 15.15 x 10 -3 S-1 o2 = 480 MPa
Aor = O'2 -- O'l = 21 MPa
A P = A o A o l l , = 0.27kN Ex. 1.4 By plotting the data in the engineering stress-engineering strain, true stress and true strain curves, the following properties are obtained. Initial yield stress: Ultimate tensile strength: True strain at maximum load: Total elongation: Strength coefficient, K: Strain hardening index, n:
156 MPa 294 MPa 0.24 45.4% 530 MPa 0.24
Solutions to exercises
177
0"0 I 11JI, v
cj'OI
8u!u~od lmal,q ~a~ttgfo so!u.rlO~lq 8L[ mm 8s
=
8"0 x 8 x 8 = ~ x 17"6 x g ' 9 = ~ 1 7 6 1 7 6=
's
~ml tuoJj
:ssou'4o!tO leU!d I ' ~ "x~[
Je),deqo 0"1.I
S't-
0"~-
I
cj'~-
I
0"s
I
%/~.0 I , /
~/IF
q,S
5WF~/
OL'~
-
(U!eJ|$)6OI SA($$eJ),s)6o"1 u!eJi$ eruj. 0s
OtfO
OEO
00"0
Ot'O
I
I
I
,,,
f
O OS OOt 00~
|
o~
ooe oss oo~ o~ u!eJ~,s-sseJls em.l. u!mls "6u3 frO
IfO
s
I
~'0 I
0
t'O
I
I
l O~ ~
i ooL O~t .~
.
oo~ r 0 ~
~
00s 0~;s
u!mls-sseJls 5u.ueeu!Su3
The principal strains"
El
=
In
(9.4
8mm
] = 0.161
e2 =
(6.5 ram') = In \ 8 m m ] -0.208
6'3 =
- - ( e l + 6'2) =
0.047
Strain ratio: r /5 =
-0.208
-ez
=
=
-1.29
0.161
Stress ratio: a =
2~+1 2+~
= -2.225
Membrane stresses: effective strain, g
=
1 + fl + fl2)e I = 0.218
effective stress, 5 = 600 (0.008 + ~-)0.22 = 432.57 M P a or I =
~ / l _ oa~ + a 2
-- 151.3 M P a
o2 = aOl = - 3 3 6 . 6 M P a trl + t r 2 + t r 3
Hydrostatic stress trh =
3
= --61.7 M P a
Deviatoric stresses: ?
tr I = trl =
o2
=
tr3
-
-
Oh =
151.3
-- (--61.7)
=
213
MPa
trh = --274.9 M P a Oh =
61.7
MPa.
Check for flow rule: el 0.161 " 7 = 2 1 3 = 0.000 756 o" 1 82
-0.208 -274.9
= 0.000 756
f3 0.047 -- =------ = 0.00076 cr~ 61.7 Solutions to exercises
179
Plastic work of deformation, tO
f0.218
=
vol
600 ( 0 . 0 0 8 + g)0.22 dg = 600 (0.2261.22 -- O. 0081.22)
1.22
,10 = 78.77 x 106 Jim 3
vol = 51.2 m m 3 = 51.2 x 10 -9 m 3 w = 4.03 J
Ex. 2.2 Diameter of Mohr circle (trl - 0 3 ) Yielding with a von Mises criterion, (0" l -- 0"2) 2 + (0"2 -- 0-3) 2 + (0"3 -- (71) 2
2
=trf
Replacing r
+a3 2
we get that 2 -
=
~/3
Ex. 2.3 Effective stress, i
"
-- 0"2) 2 + (0"2 -- 0"3) 2 + (0"3 -- O"1)2
2
=trf
Therefore, for o'2=0"3=0 m
(7 = ( 7 1 = O f
From flow rule, El
/~2
E3
crI
02
%
t
2 3
'
,
o'1
a2
3
Therefore, 2
180
Mechanics of Sheet Metal Forming
t or 3 =
r -----
3
Effective strain, ,
,
,
Ex. 2 . 4
(a)
Stress and strain ratios al = 400 MPa, 9a =
I
and
~r2 = 200 MPa fl=O
Therefore, 62=0 63 -----"-61
(b)
=~
61 63
Under fluid pressure, - 2 5 0 MPa. 61
- - = - 1 will not change. 63
This can be proven from the flow rule, trl = 4 0 0 - 250 = 150 MPa, tr3 = 0 -
tr2 = 2 0 0 - 250 = - 5 0 MPa
250 = - 2 5 0 MPa
Mean stress and deviatoric stresses: 150- 50O'h =
3
250
= - 5 0 MPa,
o~ = 150 - ( - 5 0 ) = 200 MPa, o~ = - 2 5 0 - ( - 5 0 ) = - 2 0 0 MPa 61
63
therefore, El E3
This is because yielding is independent of the average stress on an element. Therefore, the strains are also independent. Ex. 2.5
Major strain, 61 ~--"ln(12/lO) = Minor strain, e2 = iS.el =
0.182 0.182
0.182 -0.182
Solutions to exercises
181
Thickness strain, 83 = - ( 1 +/~)81 Effective strain, ~ = ~/(4/3)(1 + fl + fl2)81 Effective stress, 5" = 850e-~ Strain ratio, a = (2~ + 1)/(2 + fl) Major stress, ai = "#lJ1 - a + a 2 Minor stress, tYlO~ Thickness, t = t exp e3
-0.365 0.365 723 1 723 723 0.833
,,
0 0.211 662 MPa -1 382 MPa - 3 8 2 Mpa 1.2mm
Ex. 2.5
Strain/Stress state
al .
.
Plane stress
.
.
a2 .
Plane stress
Of
Of
,/g
,,5
2 ~af
plane train Plane stress
81
E2
83
,,
----
pure shear
0"3
.
af
0
81
--81
1 --~of 43
0
el
0
-'81
Of
0
el
81
-2e~
biaxial tension
Chapter 3 Ex. 3.1
(c) Principal strains el = l n ( 6 ".l.m . . .m. .) 5.0ram E3 = -
e2 = In (45"8mm) - 0 " 0 4 1 . 0=m m
= 0.199,
(el "4-82)=--0.158 82
/3 = - - = -0.21 =~
t~=
81
2~+1
2+#
= 0.324 (ratio of stress)
Effective strain ~" = ~/~ (1 + ~1 + , 2 ) e , = 0 . 2 1 Tensions K.~ TI = trlt = ~/1 "t~ +~t ~ t ~ 600 x 0.210.22 = ~/1 _. ().324 + 0.32"42
+
t3) l)
x 0.8 x exp(-(1 - 0 . 2 1 ) x 0.199) = 329.3(kN/m)
T2 = c~Tl = 0.324 x 248.6 = 108.2(kN/m) 182 Mechanics of Sheet Metal Forming
Ex. 3.2 (a)
Constant thickness
.,,
~1 -- o'2 Tmax --"
~ O'1
2
Therefore O"
gm~
(b) Uniaxial tension
~" =
1
+ oe2) Crl =
Crl - - o'2 Tmax ~'~
2
Or1
Crl ~'~ - -
2
Therefore a .
=2
Vmax
(c) Plane strain fl=O
1
oe=~
=~
Of -F Of2) O'1 = tYl - - 0 " 3
al
2
2
~m~
T
tyl
Therefore, t7 .....
~
~r~
~'m~ Ex. 3.3
(a) 0.015, 0.015
Location Strains: Strain ratio ~: Stress ratio ~ =
I 2 f, l + l
2+/3
9
(c)
(b) 0.050, 0.000 0
0.150, -0.100 -0.667
I/2
-0.251
Solutions to exercises
183
Thickness strain - ( 1 + fl)el" Thickness: t = to exp(e3): 300 Stress o"I ~/i "- O~ "+" t~ 2" Tension:
-0.030 0.485
-0.050 0.476
-0.050 0.476
300
346
262
(MPa)
146
165
125
(kN/m)
Ex. 3.4
WD
Vol
=
fo
"rid'g=
U l+n
fo
K (eo + g)d~" K
(E 0 + ~--)n+l [g
l+n
6 0 0 x 106
{(6o + ~,+l _ ~g+l}
{0.041"22 -- 0.011.22} ---~7.90 x
1.22
106(j/m3)
Work done in unit mass:
WD
7.9 x 10 6
kg
7850
- 1.01 x 1030)
Temperature increase:
ATCe = J J .... =
AT=
Ce
1.01 x 103
2.2~
0.454 x 10 3
Chapter 4 Ex. 4.1
rl =
2K to ~/~
2)~ ~el
( )n 2_2_e
~v/~l
exp(-el)
T~ exp(-el) = 2Kto
x 340 x 103 2 x 700 x 106 x 0.8 x 10 -3
By trial and error, el
0.08 0.547
0.07 0.536
(mm)
0.06 0.523
Therefore, et = 0 . 0 6 184 Mechanics of Sheet Metal Forming
= 0.5258
Ex. 4.2 Maximum wall tension is when el = n. TI max - -
2Kt~ n V ~~/3) exp(-n) --~n
Tension at centre and at binder: T~,o =
Tl,max
Ti,B =
=
71"
exp(~ -~)
= 1 266 x V~
Tl,max 71"
exp (0.15 ~- )
0.2
exp(-0.2) =
~
et
exp(-el)
1 (el) ~ e x p ( - e l ) = 1.266 (0"2)02 exp(-0.2) = 0.469 el "
0.02 0.448
0.03 0.481
0.04 0.505
0.025 0.466
0.026 0.469
el = 0.026 Tlo=T,n=2x600xO.8xlO3(__~__ , , ......... ~/~ 267 B=2x0115=890kN,
0.026
)0.2
exp(-0.026) = 267(kN)
2B=1780kN
Ex. 4.3 Side-wall tension: 1.266T1,o = 338(kN). Punch force: 676(kN) Blank-holder force 1780 = -- 2.6 Punch force 676
Ex. 4.4 Angle of wrap from O to A is 0 where: 600-10 sin 0 = 2000 - 10' 0 = 17.25 ~ = 0.301 (rad) Tension at mid-point, Tl,o = 2 • 400 • 0.8 • 103 (__. ~ ~c~ .0.025
)0.17
exp(-0.025) = 197(kN/m)
Tension at A, T I , A - - T I , O x e 0"1•
--
197 x 1.03 = 203(kN/m) Solutions to exercises
185
At B, angle of wrap is 60 degrees (1.047 rad), TI,B -" Tl,o x e 0"1x1"047 = 197 x 1.11 = 219(kN/m)
Chapter 5 Ex. 5.1 For the tensile strip, the load is Lo
P = triAl = K(eo + e)nAO-~l = K(eo + e)ntowoexp(-e) 10P
Ktowo Oe
= 0
[(eo+e) ~exp(-e)]
Oe = n(eo + e) n-1 e x p ( - e ) - (eo + e) n e x p ( - e ) = 0
n --- E 0 - F E
E=n--EO
Ex. 5.2 For the tensile strip,the load is n Lo P = ~IAl ----K61Ao-~l = K~Ttowoexp(-el)
For the narrow section, at m a x i m u m load e = n, and Pmax = Ktowon n cxp(-n l)
= 750 x I06 x 1.2 x 12.4 x I0-6(0.22) -~
exp(-0.22) = 6.42(kN)
For the wider section, the strain in A is e IA. P = Ke~At0w0 exp(-eiA) at m a x i m u m load,
P = Ktowoe~A e x p ( - e l A ) = 750 x 106 x 1.2 x 12.5 x 10-6(elA) -'22 exp(--elA) = 6.42(kN) Therefore, f ( e l A ) = e_0.221Aexp(--elA) = ,-z-70.22 12.4 022 " exp(--0.22) = 0.571 IL.D
186
Mechanics of Sheet Metal Forming
By iteration, elA
f(flA)
0.2 0.18 0.16 0.17
0.5746 0.5728 0.5694 0.571
Therefore, eta = 0.17, and the 20 mm length becomes: l = 20exp(elA) = 23.7(mm) Ex.
5.3
From Equation 5 13, dAo = - 0 . 2 = - 0 . 0 2 (n-
e u ) ~-'
~
9
Ao
'_ndAo
=
Ao
10
~/0.02n
(n - 6u) 2 = 0.02n n 2 -- 2 n e u + e u2 = O . 0 2 n
n 2 _ (2eu +O.02)n + e u2 =
0
(2eu + 0.02) 4- ~/i2eu + 0.02) 2 - 4 e u2
2 (26u + 0.02) 4- ~/0.08tu + 0 . ~ 2 0.3 0.25 0.2-
0.15 0.1
--
0.05 ,=
0
I
I
I
I
0.05
0.1
0.15
0.2
0.25
Ex. $.4
As the load transmitted is equal in each section, P = o'aAa -
o'bAb
Ab
Solutions
to exercises
187
But, cr = 9m
B~ m
Ab B ~ n
Bea = A--aa
~a = (A~bab)'/m
eb, Oreb:(A-~b) '/m ~a
In terms of elongation, the total elongation is
8L = 81a+ 81b = l (ela + 8L = l (kla
v=
8t
8L,
where
elb)8t
+ Elb) = l (Ela + ~lb) = / k l a
( 1 + (A~bb)l/m)
Therefore, 11 Ela --" / [ 1 + (Aa/Ab)l/m] ' ~lb = /[1 +
(Ab/Aa) l/m]
Ex. 5.5
For both elements, the material has zero plastic strain so that the effective stress is = 600(0.004) 0.2 = 198.9MPa crlA = 198.9 MPa,
trlB =
O'lA
fo
=
198.9
0.995
= 199.9 MPa
Strain ratio is a, then from Equation 2.18(b), ~=
(~/'1-a
+ a2) or,
o"
~/1 - a
(7
+or 2
.'. ~/1 - ~ + r o2B = r
=O'IB --"
f0
= fo = 0.995,
a = 0.99
= 0.99 x 199.9 = 197.9(MPa)
Chapter 6 Ex. 6.1
(a)
Determine the limiting elastic curvature
where
I= 188
wt 3 12
Mechanics of Sheet Metal Forming
At limit of elastic bending,
O"l =
O'f
Limiting elastic curvature:
Me
I
S
tot E
250 X 106 X 50 X 10 -3 X (2 x 10-3) 2
y
6
6 8.33
= E/=
200 x 109 • =~
50
= 1.25 m-
10 -3
X
= 8.33(Nm)
X
(2
x
10-3) 3
i2
R =800mm
(b) Construct a moment curvature diagram Elastic moment: M e = E l ( l ) wt 2
Plastic moment: Mp = S--~- = 12.5 Nm Elastic plastic moment, >
but before the moment reaches fully plastic moment Mp. From Figure 6.11, the material is in plastic deformation above Ye, and elastic below Ye. Strain
s =
--
E
=
=~
Ye
Ye =
s
E(1/p)
Equilibrium equation in bending, t12 M = 2 ~wy dy = 2w E .1o
y~
g y2 dy + S
p \ 3 ]o + S
ydy
= w--~--(3- m 2) Ye
1
(c)
1
1
where - = - - . - - (see Equation 6.23) p Pe m 9 Using the approximate relation (,) S A ~ =-3---Et p = 400mm
=~
---
Et
=
(1)
1.88
m-
= 2.5m-' l
Solutions to exercises
189
Final curvature 1 - = 2 . 5 - 1.88 = 0.625 m -l P p = 1.6m 9 Using the diagram in (b) trf Ye = = 0.5 X 1 0 - 3
E(1/p)
/ Ep, (y3) - (y2) T O +trf
M=2w A
(1)
'1
= 11.46 Nm
yo
AM=I.72m-
= E1
l
p = 1.28 m I..... I Ex. 6.2
Initial radius of curvature and wrap: po=~
l
1 and 0 o = I n = 2 z r 27r Po
Springback due to unloading: 1
1 = 00 = E't p~ /90 The gap between the ends of the strip after unloading is given by gap = poA0 =
~l ( - ~ 73S ~po0o )
=
2zrl ( 3 S )
312S 2rr t E'
gap= Ex. 6.3
E' =
E 200 = = 219.8 GPa 1 - 112 1 -- 0.3 2
2 tr ~ ~
2 ( 2 ~0.22 = ~ 600 ~ e / = 715e~
190
Mechanics of Sheet Metal Forming
MPa
~ - " 2zr
0.2(2 x 1 0 - 3 ) 2.22 IPI
--'~
= 3.94 x 10 -8
..........
21"22 • 2 . 2 2
from Equation 6.27
From Equation 6.26 M=
InK'
3.94 x 10 -s x 715 x 106
=
0.080.22
= 49.1Nm
For elastic unloading
A
= E'I
-49.1
12
219.8 x 109
0.2(2 x 1 0 - 3 ) 3
= 1.675 m -1
Final curvature is 1.675 = 10.8 m -1
0.08
and the final radius of curvature is
1
..... = 92 mm 10.8
Ex. 6.4
Springback in an elastic perfectly plastic material:
1
E' t
P0 Springback between aluminum and steel can be compared as follows:
A (llP)Ifllpo)I~ '
B
SAI
S^~IE'A~ ~
A (I/p) / (I/~) Isteel
#
~---
Ssteel / Este~i
Esteel X
Ssteel
EAI
Since the material is work-hardening, an approximation can be made by assuming the average stress to be S = K e n at the respective bend ratio. Therefore, B~--"
A (llP)l(ll~)IAl A (1/p)/(1/p0)ISter
190 x ---- = 0.98 (e)-0.06 530e 0"26 75 205e 0.2
"~--
For p / t = 10, the strain is e = y / p = t / 2 p = 0.05 B = 1.173
For p / t = 5, the strain is e = y / p = t / 2 p = 0.1 B = 1.125
Solutions to exercises 191
Alternative solutions The springback for a work-hardening material can be derived as 6
t
n
Therefore, the springback ratio between aluminum and steel is 6
n
2+ B
t
IAI
_..
6 (2.~.'n) (~-~) (~) nlStee'
:o~
2.26
--
,9o ( , ~o~-o~o
2.2
2 + nsteel 2 + n AI
x
KAI Ksteel
x
-0.06 1.007 ( ~ )
For p / t = 10, B = 1.007 (0.05)-o'o6= 1.205 For p / t = 5, B = 1.007 (0.1) -o.o6 = 1.156
Chapter 7 Ex. 7.1 In the hole expansion process the equilibrium equations are
ri) r
m
2T Then for T~ = --~ at r = r0:
Ex. 7.2 For shells,
dTo dr
T o - T~ =0 r m
For nosing, To = - T . Therefore,
dTo
T + T~
dr
r
dTo
dr
T4~+ T
r
=0
192 Mechanics of Sheet Metal Forming
Esteel EA!
(+)
0.2-0.26
Integrating the above equation yields the following: In (T~ + T-')
io~. =
r ]rir
-In
In(r, +r--) - 1 . ~
- -1.
(r)
~
r
Ex. 7.3
In flaring the equilibrium equations, given the boundary conditions T~ = 0 at ro, are
T,
=
- T i n (-~)
To = "T+ T,/,= "T(I - In(r--~)) Therefore, the limit of T~ is given by - T at =-T
T, = -'In (~)
=~
In ( ;ro )
ri
=1=~
( ;ro )
= e
Then r can move between ri and the maximum ro = r~e.
! r'
(a)
From Equation 8.11
hf=~-
-1
(b) Max punch force ,f
Fd = 2rrrito ltrf In I
(r0) ri §
= 37.5 mm
oxp( )
0.1 x 30000 } (0.1 x n') Fd = 2n" x 2 5 x 1.2 {3501n ( 5 ~ ) + . . . . . . . . exp = 57(kN) rr x 50x 1.2 2 Ex. 8.2
The maximum punch load will be lower for n = 0.2 than for n = 0.1. The curve also moves to the fight as shown below. Solutions to exercises
193
n=0.1 n=0.2
I
~aX
iI
\
I
I I I 1
~h
,
h~
Ex. 8.3 For a small section of the sheet, assume the sheet slides down. Tool pressure per unit width: p=2Tsin
(do) -~-
T+dT+gP=
=Td0 T
dT = - t t TdO dT T
In
= - I t dO
(T2)~ =
- "2" ~'Tt
T2 = Tl exp ( --2
T2- T, = T, [exp (--~-/~)- 1]
T+dT
194 Mechanics of Sheet Metal Forming
~)
Ex. 8.4
(a) Volume of metal,
rr ~to = rrdftfhf + rr ~to do = 120mm Total drawn ratio =
120mm 62mm
Limiting drawn ratio
= 1.94
(LDR) LDR < exp (1/) LDR < exp (0.6) = 1.82
Therefore, the can cannot be drawn in one operation. (b)
If the draw ratios are equal for the two drawing operations,
do di
di df
di = ~/i20 x 62 = 86 mm
Height from first draw:
ri h i - - "~
-\ri/
- 1
/
=20ram
(c) Punch force, neglecting blank holding force and die friction: 2n'rito
Fd= -------af l n ( ~ ) = (d)
21.6kN
Cup height at the finish of second draw: hf==
rf 2
ro --
-1
=42.6mm
(e) The full height is obtained by ironing the middle section.
(0 Saving from reduced thickness: rrd~ rrl202 {0.41 - 0.33} = 905 mm 3 Volume of saving per blank = - - ~ {t1983 - - t1998} " 4 For 100 billion blanks, volume saved = 905 x 102 m 3 Cost of metal saved = 5 x 2800 x 905 x 102 = $1.25 billion
Solutions to exercises 195
Chapter 9 Ex. 9.1 Using the relation in Section 9.1.3, the effective stress is 6.4 3 5" = {(61/3) 2 + 4} (61/50) 2 = 620 MPa 16 1.2 and the effective strain is, = 2In (61/50) = 0.40
Ex. 9.2 The principal radius of curvature Pl at and near the tangent point is 50 mm. The principal tensions at the tangent point, from Section 9.2.1, are T# = 374 MPa and
To =
aT# = 0.89 x 374 = 333 MPa
As indicated in Section 9.2, in the unsupported region, p = 0, and therefore,
ro Pl
P2
i.e.
374
02 =
333
x 50 = - 5 6 mm
Ex. 9.3 From Equations 9.4 and 9.6, it may be shown that --=1+ t
From Equation 9.1, the effective strain is
= -et
= In (to~t) = In [1 +
(h/a) 2]
and the membrane stress is tr# = tro = ~ = 350e-~
MPa
From Equation 9.5, the bulging pressure is 2tr#t p
~
2 x 350~"~
1.2 x 10 -3
....
P
a 2 [1 +
(h/a) 2]/2h"
[1 +
(h/a) 2]
where a2[1 + (h/a) 2] p= 2h
196
Mechanics of Sheet Metal Forming
MPa
For a = 50mm, and computing in the range, 0 < h < 45, we obtain the characteristic shown below. (The maximum pressure occurs at a membrane strain of approximately 0.2 (~ = 0.4). As indicated in Figure 9.3, for a ductile material this is probably less than the forming limit curve in biaxial tension and it is expected that the sheet would deform beyond this maximum pressure and fail at a higher strain under a falling pressure gradient.)
10
_
O.
t~ ta
6-
4-
Q..
_
I
0
I
10
20 Height, hmm
I
I
30
40
C h a p t e r 10
Ex. 10.1 The stress after bending over the block without tension is, from Equation 6.17, 78 x 109 x 1.85 x 10 -3 Ett o'1=4"~-~ =-I= 120 MPa 2x0.6
On the outer (tension) side of the sheet, the additional stress to cause yielding as in Figure 10.2(d), is 1 8 0 - 120 = 60 MPa. The tension to be applied to add this stress is T = AtYlt = 60 x 106 x 1.85 x 10 -3 = 111 k N / m The fully plastic tension, as in Figure 10.2(e), is
Ty = St = 180 x 106 x 1.85 x 10 -3 = 333 k N / m
Solutions to exercises
197
Ex. 10.2
The limiting elastic radius of curvature is, from Equation 6.19, E't 78 x 109 X 1.85 • 10 -3 = 0.401 m P e = 2S = 2 x 1 8 0 x 106
From Equation 10.2, the moment prior to applying the tension is 78 x 109 (1.85 x 10-3) 3 Mo =
12 x 0.6
= 68 N m / m
From Equation 10.6, when m = 0, the moment is M = Mo/2 = 34 N m / m
From Equation 10.7, the tension for m = 0 is
I ~} { __~
T=Ty
1-4----.~ ~
=333
1- 4x0.6
As the loading is elastic, the change in curvature on unloading is, from Equation 6.30, 34 x 12 A
= E'I
78 • 109 (1.85
x
= -0.826 10-3) 3
The final curvature is 1
0.826 = 0.84 m 1
0.6
and the final radius of curvature is 1/0.84 = 1.19 m
i
.. L~ ~,
L. <,~
L~ ,,~
(a)
(b)
(c)
198 Mechanics of Sheet Metal Forming
'.......'iiii i i i i i i i ....................... l ..................... L~ =~
(d)
L.
<,~
(e)
The stress distribution is shown diagrammatically above. The elastic, plastic stress state under the tension and moment, similar to that in Figure 10.2(d), is shown in (a). Unloading the tension is equivalent to subtracting the uniform stress in (b)to give a distribution in (c) in equilibrium with the moment. Subtracting the elastic bending stress in (d) occurs when the moment is released to give the residual stress in (e).
Ex. 10.3 From Equation 10.12, the slope of the stress-strain curve for steel is 0.2 x 700 x 0.0120.2 0.012
= 4.82 GPa
and for aluminium, is 0.2
x
400
x
0.012 0.2
0.012
= 2.75 GPa
From Equation 10.11, the springback, i.e. the change in curvature on unloading, is, for steel, A
(1) st~l
=
4.821 220 2.5
= -0.0088 m -1
and for aluminium, A
(1) ~.
=
2.751 = -0.0141 m 78 2.5
The final curvature for steel is ( 1 / 2 . 5 ) - 0.0088 = 0.391 giving a radius of curvature of 2.56 m. For aluminium, the final curvature is ( 1 / 2 . 5 ) - 0.0141 = 0.386, giving a radius of curvature of 2.59. Clearly the springback is very small and similar for both metals. The result is independent of thickness and depends on the ratio of the slope of the stress-strain curve to the elastic modulus, which are not very different for these two cases. Ex. 10.4 From Equation 10.20, the tension after the first bend is 0.34 } T = To + AT = 0.6Sto + 4 ~ o (1 + 0.62) = Sto 0.6 + -~--
Solutions to exercises
199
The thickness reduction is, from Equation 10.21,
to
2bto ~
t
o.o 2b
The thickness downstream of the first bend is
The stress in the sheet downstream of the first bend is trl T
S
St
Evaluating in the range given, we obtain the diagrams below. 1 0.80.6u) 0 . 4 0.2-
0
3
I
I
I
I
I
I
4
5
6
7
8
9
10
Bend ratio, b (=r/t )
o's1
.o
'~ 10 Ir C
9c-~- 5
O
3
I 4
i 5
i 6
i 7
Bend ratio, b (=r/t)
200 Mechanics of Sheet Metal Forming
I 8
10
Chapter 11 Ex. 11.1
L
(a) For the frictionless case The current radius of the tube at the comer is r. The tube wall thickness will be uniform. From volume constancy
(
" ) =~Rto " r)+~r
t 2(R
' / / / / / !
re-arranging t =
I
r I _1
4
r
.
.
.
.
For the plane strain condition = ~---~e0=
In
2[
2
cro=-~K e o + ~ l n
and 5" = (4
-
( 4 - zr) R)]n ~r "'
Internal pressure is given by p=
trot
2toK
--
r
+ --~ In ( 4 - ( 4 ~ 7r) R)ln
q/3r [4 _.-i~~,Jr).~] leo
r
(b) For the full sticking case The relation between the comer radius and the comer thickness can be obtained from the incremental relation 7t
7t
7/;~"' \ r +
t dr + -~(r - dr)(t - dt) = ~rt f~(4
l) dr yr
r
fotdt t
t ~ - - t o ( ~ ) (r4 - 1 )
Solutionsto exercises 201
Circumferential stress 2 [ e o + ~2 ( _7r )4 _I 1n O'l'-~K
(R)] ~
The internal pressure is given by p
~
tTlt
r
,,
2t0 r ( 4-1) [
VC~r (-~)
K eo+~
2 (4)
~--1
In
(R)] n
Ex. 11.2
The minimum comer radius is limited either by necking or by the maximum fluid pressure available. Since
t --
4
eo = - e t = In to
t
e0 = n = 0.2 for local necking of tube wall, therefore,
r=
17mm
t=3.27,
The internal pressure required to form comer radius r is given in Ex. 11.1. Substituting the material properties and die geometry into the equation, we obtain: r(mm) 38.00 37.00 36.00 35.00 34.00 33.00 32.00 31.00 30.00 29.00
t(mm) 3.45 3.44 3.44 3.43 3.42 3.41 3.40 3.39 3.38 3.37
tro(MPa) 566.85 568.86 570.84 572.78 574.70 576.58 578.44 580.27 582.07 583.84
p(MPa) 51.52 52.96 54.47 56.08 57.77 59.56 61.46 63.48 65.63 67.92
Therefore, the minimum comer radius is 31 mm, and is limited by the maximum pressure available. Ex. 11.3
In order to calculate the initial tube thickness, we use volume constancy.
2(tl +t2) 2
L +
rr
zr
= - Rto
202 Mechanics of Sheet Metal Forming
where L = (R - r)
i,a
/.
t o = -1~ 2[ ( t l +~t 2 ) ( R _ r ) + r t 2 ]
L,
tl
The forces that existed at each end of the wall can be calculated as Fl=trltl=~K
eo+~ln
The final internal pressure at the current forming radius can be found from: F2=pr
thus
p=--
F2 r
p To calculate the average friction coefficient from force balance
tZavpL = F2- Fl /Zav= F2-pL FI
where q = p
Ex. 11.4 (a) For the free ends, there is no stress in the axial direction, tr2 = O. For a thin cylinder, o3 = 0. el
= In ( ~ o )
In(t~)=- 89In(~o) ?o = Effective stress and strains are
"~ = trl andS-el Solutions to exercises
203
The internal pressure is calculated as trlt p= r
p=Kto
ro
In
r ro
To calculate the expanded radius at instability, let dp = 0. This leads to --
r0
--- - - or r 3
=
roe2n/3
(b) For restricted ends, there is no strain in the axial direction, e2 = 0.
El "- In ( F ~ ) E3 --'---El In ( t ~ =) (- rIn~ ) t
ro
to
r
Effective stress and strains are: if" =-.~-.~0'l and ~" = ~3e I
The internal pressure is calculated as trlt p= r
oro[In
p=
K--~-
To calculate the expanded radius at instability, let dp = 0. This leads to
in
204
= ~ orr=r0
en/2
Mechanics of Sheet Metal Forming