Find. (a) the friction-hill pressure profile at the end of the stroke; (b) the minimum press power (kW); (c) the effect of doubling ram speed; (d) the overall process efficiency and ways to improve it.
Approach. Get the constant ram speed from the geometry and stroke time, find the instantaneous radius/height/strain rate at the END of the stroke (where force is largest and governs press sizing), compute the hot-working flow stress there, apply the sticking-friction average-pressure formula to get force and power, then repeat with doubled speed and compare.
Figure — friction-hill pressure profile across the diameter at the end of the stroke ($h=h_f=0.10$ m, $r=0.141$ m). Sticking friction gives a LINEAR "tent": $P=Y(1+(r-|x|)/h)$, peaking at the centre ($P_{max}\approx270$ MPa) and falling to the uniaxial flow stress $Y\approx112$ MPa at the free edges. The shaded rectangle (blue, height $Y$) is the work to deform the material homogeneously; the triangular excess above it (orange) is the extra work consumed overcoming interfacial friction.
Constant ram speed and end-of-stroke geometry.
$$v=\frac{h_0-h_f}{\Delta t}=\frac{0.20-0.10}{2}=0.05\ \text{m/s (constant)}$$
By volume constancy $r^2h=r_0^2h_0$, the radius at the end of the stroke ($h=h_f$) is
$$r_f=r_0\sqrt{\frac{h_0}{h_f}}=0.10\sqrt{\frac{0.20}{0.10}}=0.10\sqrt{2}\approx0.1414\ \text{m}$$
Both the contact radius AND the pressure ratio grow as the cylinder squashes down, so force (and hence required press capacity) is largest at the END of the stroke — that instant governs the minimum press size.
(b) Flow stress at the end of the stroke. Strain rate at constant ram speed is $\dot\varepsilon=v/h$, largest when $h$ is smallest:
$$\dot\varepsilon_f=\frac{v}{h_f}=\frac{0.05}{0.10}=0.5\ \text{s}^{-1}\qquad Y=C\dot\varepsilon_f^{\,m}=120(0.5)^{0.10}\ \text{MPa}$$
$$\boxed{Y\approx112.0\ \text{MPa}}$$
(b) Average die pressure, sticking friction (cylindrical), and force. From the formula sheet, $P_{avg}\cong Y\!\left(1+\dfrac{r}{3h}\right)$:
$$P_{avg}=112.0\left(1+\frac{0.1414}{3(0.10)}\right)=112.0(1.4714)\approx164.7\ \text{MPa}$$
$$F=P_{avg}\times\pi r_f^2=164.7\ \text{MPa}\times\pi(0.1414\ \text{m})^2$$
$$\boxed{F\approx10.35\ \text{MN}}$$
(b) Minimum press power. Power is the force-velocity product at the governing (end-of-stroke) instant:
$$P_{power}=F\times v=10.35\times10^6\ \text{N}\times0.05\ \text{m/s}$$
$$\boxed{P_{power}\approx518\ \text{kW}}$$
The press must be rated at least this large to complete the forging without stalling near the end of the stroke.
(c) Effect of doubling ram speed. Doubling $v$ to 0.10 m/s doubles the end-of-stroke strain rate to $1.0\ \text{s}^{-1}$, which raises the flow stress by the strain-rate-sensitivity factor $2^{m}=2^{0.10}\approx1.072$ (from $Y\propto\dot\varepsilon^m$), and hence raises force by the same 7.2%. Power is force times velocity, so it scales by $2\times1.072\approx2.14$:
$$Y'=120(1.0)^{0.10}=120\ \text{MPa}\qquad F'\approx11.10\ \text{MN}\qquad P_{power}'\approx1110\ \text{kW}$$
$$\boxed{\text{Power more than doubles (to} \approx1110\ \text{kW, a factor of} \approx2.14\text{)}\text{, and the required press capacity must increase accordingly}}$$
(The stroke time also halves, to 1 s, since the same displacement is covered in half the time.)
(d) Overall efficiency. Define efficiency as the ratio of the ideal (frictionless, homogeneous) flow stress to the actual average die pressure at the governing instant:
$$\eta=\frac{Y}{P_{avg}}=\frac{112.0}{164.7}$$
$$\boxed{\eta\approx0.68\ (68\%)}$$
32% of the work goes to overcoming sticking friction rather than useful deformation. To improve efficiency: lubricate the dies to break sticking friction and move toward the much lower sliding-friction pressure multiplier; use a lower aspect ratio (larger $h$ relative to $r$, e.g. multi-step/incremental forging instead of one large reduction) to keep $r/h$ small throughout the stroke; polish/coat dies to reduce the friction coefficient; and preheat the dies closer to the workpiece temperature to reduce chilling-driven flow-stress increases at the die interface.
Part
Result
(a) Friction-hill profile
Linear tent, $P_{max}\approx270$ MPa (centre) to $Y\approx112$ MPa (edges)
(b) Minimum press power
≈518 kW
(c) Doubled ram speed
Power → ≈1110 kW (×2.14); stroke time halves
(d) Overall efficiency
≈68%
Check: press sizing is evaluated at the END of the stroke ($h=h_f$), where both the contact radius and the $r/h$ friction ratio are largest, so force (and therefore the instantaneous power demand at constant ram speed) is maximum there — this is the standard governing condition used to size a forging press for a single-blow, constant-velocity stroke. The efficiency in (d) is likewise evaluated at that same instant rather than as a stroke-averaged integral.
