Calculation of upsetting force

A cylindrical specimen made of annealed 1112 steel has a diameter of 200 mm and is 125 mm high. It is upset, at room temperature, by open-die forging with flat dies to a height of 50 mm. Assuming that the coefficient of friction is 0.2, calculate the upsetting force required at the end of the stroke. Use the average-pressure formula.

Question

Calculation of upsetting force

A cylindrical specimen made of annealed 1112 steel has a diameter of 200 mm and is 125 mm high. It is upset, at room temperature, by open-die forging with flat dies to a height of 50 mm. Assuming that the coefficient of friction is 0.2, calculate the upsetting force required at the end of the stroke. Use the average-pressure formula.

Accepted Answer

The average-pressure formula, from Eq. (6.18), is given by

[latex]p_{av}\simeq Y_{f}\left\lgroup1+\frac{2\mu r}{3h} \right\rgroup .[/latex]

where Y has been replaced by [latex]Y_{f}[/latex] because the workpiece material is strain hardening. From Table 2.3,

Typical Values for K and n in Eq. (2.11)([latex]\sigma =K\epsilon ^n[/latex] )at Room Temperature
Material	K (MPa)	n
Aluminum, 1100-O	180	0.20
2024-T4	690	0.16
5052-O	210	0.13
6061-O	205	0.20
6061-T6	410	0.05
7075-O	400	0.17
Brass, 7030, annealed	895	0.49
85–15, cold rolled	580	0.34
Bronze (phosphor), annealed	720	0.46
Cobalt-base alloy, heat treated	2070	0.50
Copper, annealed	315	0.54
Molybdenum, annealed	725	0.13
Steel, low carbon, annealed	530	0.26
1045 hot rolled	965	0.14
1112 annealed	760	0.19
1112 cold rolled	760	0.08
4135 annealed	1015	0.17
4135 cold rolled	1100	0.14
4340 annealed	640	0.15
17–4 P-H, annealed	1200	0.05
52100, annealed	1450	0.07
304 stainless, annealed	1275	0.45
410 stainless, annealed	960	0.10

we find that K = 760 MPa and n = 0.19. The absolute value of the true strain is

[latex]\epsilon _{1}=ln\left\lgroup\frac{125}{50} \right\rgroup =0 .9162,[/latex]

and therefore

[latex]Y_{f}=K \epsilon^{n}_{1} = (760)(0.9162)^{0.19} = 747 [/latex]MPa

The final height of the specimen, [latex]h_{1}[/latex], is 50 mm. The radius [latex]r [/latex] at the end of the stroke is found from volume constancy:

[latex]\frac{\pi (200)^2}{4}\cdot 125[/latex]= [latex]πr^{2}_{1} [/latex] . 50 and [latex]r_{1}[/latex]= 158 mm

Thus

[latex]p_{av}\simeq 747\left[1+\frac{(2)(0.2)(158)}{(3)(50)} \right] [/latex]

[latex]p_{av} = 1062[/latex] MPa

The upsetting force is

[latex]F = (1062 × 10^{6})π . (0.158)^2[/latex]

[latex]F = 8.32 × 10^7 [/latex]N

Note that Fig. 6.9b can also be used to solve this problem.

Question 6.1: Calculation of upsetting force A cylindrical specimen made o...

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