Substituting for M
p
from Equation 6.21, this may be written as ˙
W
b
= St
2
v 4ρ
1 −
T T
y 2
10.16 As may be seen from Figure 10.6 and Equation 10.13, the strain at the mid-surface is
ε
a
= e
ρ =
t 2ρ
T T
y
This is the average strain for the element shown in Figure 10.8a and therefore during bending it will extend an amount
δ = ε
a
dl =
t 2ρ
T T
y
dl The tension will therefore do work on the element of magnitude
W
t
= T δ
= t
2ρ T
2
T
y
. dl
As indicated, this is done in a time dlv , therefore the rate of doing tensile work is
˙ W
t
= W
t
dlv =
t.v 2ρ
. T
2
T
y
= St
2
v 2ρ
T T
y 2
10.17 Combining Equations 10.16 and 10.17, the total rate of doing plastic work on the element is
˙ W
pl.
= St
2
v 4ρ
1 +
T T
y 2
10.18 It is appropriate to consider that the work done at the bend at A will be greater than that
dissipated by plastic work alone and to assign an efficiency factor, η 1, to account for friction and redundant work. The work balance may be written as
˙ W
ext.
= ˙
W
pl.
η We assume that the external work done arises from a sudden increase in the tension as
shown in Figure 10.8c. This external work done at this point is ˙
W
ext.
= [T + T
− T ] v
= T v
10.19 and from Equations 10.18, 10.19 and 6.21, we obtain
T =
M
p
ηρ 1
+ T
T
y 2
= St
2
4ηρ 1
+ T
T
y 2
= 1
4η T
y
ρt 1
+ T
T
y 2
10.20 It has already been stated that this analysis is an approximate one that applies only when
the neutral axis lies within the sheet and the tension is less than the yielding tension. Nevertheless, it is a useful relation as it shows that the increase in tension is inversely
proportional to the bend ratio and increases with the back tension.
Combined bending and tension of sheet 147
10.5.2 Thickness change during bending
As indicated above, the average strain in the sheet at the point of bending A is ε
a
. As this is a plane strain process, the thickness strain will be, ε
t
= −ε
a
; for small strains we may write
t t
≈ dt
t ≈ ε
t
= −ε
a
= − t
2ρ T
T
y
= 1
2 ρt T
T
y
10.21 Thus there is a thickness reduction which increases linearly with tension and is greater for
small bend ratios.
10.5.3 Friction between the points A and B
An element sliding on the surface between A and B in Figure 10.7 is shown in Figure 10.9. If the thickness of the sheet is small compared with the radius of the mid-surface, the
contact pressure, following Section 4.2.5, is
p =
T ρ
10.22 The equation of equilibrium in the circumferential direction is
T + dT = T + μp dθ1
i.e. dT
T = μ dθ
10.23 If the tension in the sheet after bending at A is T
A
= T + T
, and before unbending at B is T
B
, then integrating Equation 10.22 gives [ln T ]
T
B
T
A
= [μθ]
π 2
n
p T
T + dT
d q r
m p
Figure 10.9 Element of sheet sliding on the surface between A and B in Figure 10.7.
148 Mechanics of Sheet Metal Forming
or T
B
= T
A
exp μ
π 2
10.24 It is assumed that the tension is less than the yielding tension, so there is no change of
thickness during this part of the process.
10.5.4 Unbending at B
During unbending at B there will be an increase in tension and a decrease in thickness as at A above and these can be determined from relations similar to Equations 10.20 and
10.21 using the tension T
B
instead of T . The final tension in the process in Figure 10.7
is T
f
= T
B
+ T
B
.
10.5.5 Worked example drawing over a radius
Sheet is drawn over a radius under plane strain conditions as shown in Figure 10.7. The initial thickness is 2 mm and the tool radius is 8 mm i.e. ρ
= 9 mm. The material has a constant plane strain flow stress of 300 MPa. The back tension is 250 kNm and the friction
coefficient is 0.08. Assuming an efficiency factor of 0.8, determine by approximate methods the thickness of the sheet and the tension at exit.
Solution.
The yield tension at the first bend is T
y
= 300 × 10
6
× 2 × 10
−3
= 600 kN m From Equation 10.2, the increment in tension at the first bend is
T =
300 × 10
6
2 × 10
−3 2
4 × 0.8 × 9 × 10
−3
1 +
250 600
2
= 49 kN m The change in thickness at the first bend is
t 2
= 2
2 × 9
250 600
= 0.046mm i.e.
t = 0.092mm and t = 1.91mm
The tension after the first bend is 250 + 49 = 299 kNm. The tension after the sheet has
moved over the curved surface, from Equation 10.24, is T
= 299 exp 0.08 × π2 = 339 kN m The yield tension at the unbend is
T
y
= 300 × 10
6
× 1.91 × 10
−3
= 573 kN m Combined bending and tension of sheet 149
