Work done in stretch forming

A 400 mm-long sheet with a cross sectional area of $300 \ mm^2$ (see Fig. 7.31) is stretched with a force, F, until $α = 20°$ . The material has a true-stress–true-strain curve $σ = 700 × 10^6 \epsilon^{0.3}$ . (a) Find the total work done, ignoring end effects and bending. (b) What is $α_ {max}$ before necking begins?

Question

Work done in stretch forming

A 400 mm-long sheet with a cross sectional area of [latex]300 mm^2[/latex] (see Fig. 7.31) is stretched with a force, F, until [latex] α = 20°[/latex] . The material has a true-stress–true-strain curve [latex] σ = 700 × 10^6 \epsilon^{0.3}[/latex]. (a) Find the total work done, ignoring end effects and bending. (b) What is [latex] α_ {max}[/latex]before necking begins?

Accepted Answer

a. Note that because the cross section is very small compared to the length of the part, this operation is equivalent to stretching a piece of metal from 400 mm to a length of [latex]a +b[/latex]. For [latex]α= 20°[/latex], the final length is calculated to be [latex] L_{f}= 441.4[/latex] mm and the true strain is

[latex] \epsilon =\ln \left\lgroup\frac{L_{f}}{L_{\omicron }} ight group=\ln \left\lgroup\frac{441.4}{400} ight group=0.0975 [/latex]

The work done per unit volume (see Section 2.12) is

[latex]u= \int_{0}^{0.0975}{\sigma } d\epsilon =700 imes 10^{6} \int_{0}^{0.0975}{\epsilon ^{0.3}}d\epsilon [/latex]

[latex]u=700 imes 10^{6}\left[\frac{\epsilon ^{1.3}}{1.3} ight] ^{0.0975}_{0} [/latex] [latex]u=26 .1 imes 10^{6}[/latex]mN/[latex]m^{3}[/latex]

Since the volume of the workpiece is

V[latex]=\frac{400}{1000} imes \frac{300}{10^{6}} [/latex]

V=[latex]0.00012 m^{3} [/latex]

the work done is

Work =[latex] (u) (V) = 3133 [/latex] Nm

b. The necking limit for uniaxial tension is given by Eq. (7.2). Thus,

[latex]\epsilon =n=0.3 [/latex]

[latex]\ln \left\lgroup\frac{L_{f}}{L_{\omicron }} ight group=0.3;\frac{L_{f}}{L_{\omicron }} =1.349 [/latex]
[latex]L_{max}=400 imes 1.349 =539[/latex] mm

Therefore, [latex]a+ b = 539[/latex] mm, and from similar triangles, we obtain

[latex] a^{2} – 250^{2}=b^{2} – 150^{2}[/latex]

or

[latex]a^{2}-b^{2}=200^{2}; \ a-b=\frac{200^{2}}{539} =74.21 [/latex]

Hence, [latex]a = 306.6[/latex] mm and [latex]b = 232.4[/latex] mm Thus,

[latex]\cos \alpha \frac{250}{306.6} =0.815 [/latex] or [latex] \alpha _{max}=35.4^{\omicron }[/latex]

Question 7.4: Work done in stretch forming A 400 mm-long sheet with a cros...

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