This document contains solutions to problems from Chapter 16 related to sheet metal forming. Key points: 1) The values of L/L0 were calculated for brass and aluminum based on given strain equations and input values. 2) The contour of ε = 0.04 was plotted on a forming limit diagram for values of ε1 vs ε2 using von Mises and Tresca criteria. 3) It was explained that the dashed line in a figure showing the wrinkling limit would change for a material with an R-value of 2 compared to the original material.

1. Chapter 16
16-1 Referring to Figure 16.5, determine the values of L/Lo for brass and aluminum.
Solution: L = (1+er)dx.
0
L0
∫ Substituting dx = (1+e)dr0, (1+er)2
dro=Σ(1+er)2
∆ro.
0
L0
∫
Taking ∆ro = 0.25 in, For aluminum, L = (0.25)[(1.165)2
+(1.18)2
+(1.22)2
+(1.165)2
+
(1.31)2
+ (1.36)2
+1.40)2
+(1.31)2
+(1.13)2
] = 3.19 in.
For brass, L = (0.25)[(1.23)2
+(1.26)2
+(1.295)2
+(1.31)2
+(1.36)2
+ (1.40)2
+(1.49)2
+(1.43)2
]
= 3.64 in.
16-2 Plot the contour of ε = 0.04 on a plot of ε1 vs. ε2 (i.e. on a forming limit diagram.)
Solution: Taking ε=ε1[(4/3)(1+ρ+ρ2
)]−1/2
so ε1=ε[(4/3)(1+ρ+ρ2
)]1/2
and ε2=ρε1
for von Mises and ε1−ε2=εfor Tresca,,ε1/ε and ε2/ε can be calculated for various
values of ρ.ε2/ε
ε1/ε
ε2/ε
Tresca
von Mises
10
1
-1
16.3 Show how the dashed line in Figure 16-6 would change for a material having an
R-value of 2.
Solution:
ε1
w rinkling
p o s sib le
unia x ia l
te ns io n
ε2
w rinkling lim it fo r R = 2

2. 16.4 A sheet an HSLA steel having a tensile strength of 450 MPa and a yield strength
of 350 MPa is to be drawn over a 90° bend. What is the maximum permissible coefficient
of friction?
Solution: Su/Y>exp(µθ); µ < ln(Su/Y)/θ. ln(Su/Y)/θ =ln(450/350)/(π/2) = 0.16. µ < 0.16
16.5 Consider the deep drawing of a flat bottom cylindrical cup. Sketch the strain path
for an element initially on the flange halfway between the perifery and die lip.
Solution:
ε1
ε1
0
16-6 Calculate the total drag force per length of a draw bead attributable to plastic
bending if the bend angle entering and leaving the draw bead is 45° and the bend angle in
the middle of the draw bead is 90°. Assume the bend radii are 10 mm and the sheet
thickness is 1 mm. Let the yield strength be Y and neglect strain hardening.
Solution: There are 3 bends and 3 unbends of 45°. The total work for each bend is
W = FL = 2(wL) σεdz
0
t/2
∫ where w is the width of the sheet and L is the distance.
Substituting ε = z/ρ, FL =2Y(wL/ρ) zdz
0
t/2
∫ = (2Yw/.01)(.005)2
/2. The drag force per
width, F = 0.005Y. For six bend-unbends, the total drag force is .03Y.
16.7 Calculate the drag force attributable to friction in Problem 16-6, assuming a
friction coefficient of 0.10.
Solution: Thru the 1st
bend. ∆F = F2 – F1 =F1(1 + exp(µθ). Taking F1 =0.005Y, and
θ = 6π/4, ∆F =0.013Y
16.8 Carefully sketch the strain signature for the conical cup drawing in Figure 16.20.
Figure 16.20 Drawing of a conical cup

3. Solution:
D
C
B
A
ε1
ε2
.
16-9 Why isn’t the bending included in the drag in regions A-B and B-C in Example
16-1?
Solution: The strains are so small that the work in bending is negligible compared to the
frictional work.
16.9 It has been noted that when a cup is drawn with a hemispherical punch,
greater depths can be achieved before failure when the punch spins. Explain why this
might be so.
Solution: Friction doesn’t act in a direction toward the pole