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Derive an expression for the elastic energy stored per unit volume of a wire.

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :When a body is stretched, work is done against the <a href="https://interviewquestions.tuteehub.com/tag/restoring-1187260" style="font-weight:bold;" target="_blank" title="Click to know more about RESTORING">RESTORING</a> force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension l and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy.Then , the work done by the force F is equal to the energy gained by the wire. <br/> The work done in stretching the wire by `dl, dW = F dl` <br/> The total work done in stretching the wire from 0 to l is <br/> `W = int_(0)^(l) F dl "" ....(1)` <br/> From Young.s modulus of elasticity, <br/> `Y = F/A xx L/l implies F = (YAl)/L "" ....(2)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> equation (2) in equation (1), we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> <br/> `W = int_(0)^(l) (YA l)/(L) dl` <br/> Since, l is the dummy variable in the <a href="https://interviewquestions.tuteehub.com/tag/intergration-2743881" style="font-weight:bold;" target="_blank" title="Click to know more about INTERGRATION">INTERGRATION</a>, we can change l to l. (not in limits), therefore <br/> `W = int_(0)^(l) (YAl^.)/(L) dl. = (YA)/L ((l.^2)/(2))_0^l = (YA)/L (l^2)/2 = 1/2 ((YAl)/(L)) l=1/2 Fl` <br/> `W = 1/2 Fl` = Elastic potential energy <br/> Energy per unit volume is called energy density <br/> `u = ("Elastic potential energy")/("Volume") = (1/2 Fl)/(AL)` <br/> `1/2 F/A l/L = 1/2 ("Stress" xx "Strain")`.</body></html>


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