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Engineering
General Engineering
Radial contraction of a stretched hyperelastic cylinder
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[QUOTE="SerArthurRamShackle, post: 5447601, member: 589252"] I can write, for the moment that: 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πr'[SUP]2[/SUP]h'W 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πf(x)λxW/(λ - 1) Essentially I have replaced an explicit functional dependence of r on x with an implicit one and have continued as normal so that I may perform the relevant differentiation when I do know it and can then plug it in. but λ = (h+ x)/h so: 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πf(x)((h+ x)/h)xW/((h+ x)/h) - 1) 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πf(x)(h+ x)xW/(h + x - h) 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πf(x)(h+ x)xW/x 2L = mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP] - 2πf(x)(h+ x)W hence d/dx (2L) = d/dx(- 2πf(x)(h+ x)W) and so we just use product rule: d/dx (2L) = d/dx(-2πf(x)hW - 2πf(x)xW) d/dx (2L) = -2πf'(x)hW - 2πf'(x)xW - 2πf(x)W d/dv (2L) = d/dv(mv[SUP]2[/SUP] + (πr[SUP]2[/SUP]hρ)v[SUP]2[/SUP]) d/dv (2L) = 2mv + 2πr[SUP]2[/SUP]hρv d/dt d/dv (2L) = d/dt(2mv + 2πr[SUP]2[/SUP]hρv) = 2ma + 2πr[SUP]2[/SUP]hρa finally:d/dx (L) - d/dt d/dv (L) = 0 -πf'(x)hW - πf'(x)xW - πf(x)W - ma - πr[SUP]2[/SUP]hρa = 0 ma + πr[SUP]2[/SUP]hρa = -πf'(x)hW - πf'(x)xW - πf(x)W a(m + πr[SUP]2[/SUP]hρ) = -πf'(x)hW - πf'(x)xW - πf(x)W a = (-πf'(x)hW - πf'(x)xW - πf(x)W)/(m + πr[SUP]2[/SUP]hρ) This also worries me because I'm not sure it should be negative. [/QUOTE]
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Engineering
General Engineering
Radial contraction of a stretched hyperelastic cylinder
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