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What work has to be performed to make a hoop out of a steel band of length l=2.0m, width h=6.0cm, and thickness delta=2.0mm? The process is assumed to proceed within the elasticity range of the material. |
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Answer» <P> Solution :The work done to make a loop out of a steel band appears as the elastic energy of the loop and may be calculated from the same.If the length of the band is `l`, the radius of the loop `R=(l)/(2pi)`. Now consider an element ABCD of the loop. The elastic energy of this element can be calculated by the same sort of arguments as used to derive the formula for internal bending moment. Consider a fibre at a distance z from the neutral surface PQ. This fibre experiences a force p and undergoes an extension `ds ` where `ds=Zdvarphi`, while `PQ=s=Rdvarphi`. Thus strain `(ds)/(s)=Z/R`. If `alpha` is the cross sectional area of the fibre, the elastic energy associated with it is `1/2E(Z/R)^2Rdvarphialpha` SUMMING over all the fibres we get `(EIvarphi)/(2R)sumalphaZ^2=(Eldvarphi)/(2R)` For the whole loop this gives, USING `INT dvarphi=2pi`, `(Elpi)/(R)=(2EIpi^2)/(l)` Now `I=underset(-delts//2)overset(delta//2)intZ^2hdZ=(hdelta^3)/(12)` So the energy is `1/6(pi^2Ehdelta^3)/(l)=0*08kJ` 
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