Finding increased length after deformation

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Dell
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Diagram of a circular hoop centred at the origin of x and y axes, with radius R marked and angle theta measured from the positive x-axis.


the strain, as a function of the angle is K*sin2(x)

now i know that the change in length is the integral of the strain

=[tex]\int[/tex]K*sin2(x)dx from 0->2pi

=K/2*[tex]\int[/tex]1-cos(2x)dx

=K/2*(2pi - 0.5*sin(4pi) )

=K*pibut the answer says K*pi*R

where does the R come from? i realize that the change in length should be dependent on the radius, but mathematically how do i come to that?
 
You're working with [itex]\epsilon_\theta[/itex], an angular strain. When you integrate it, you get an increase in angle, not length. Know what I mean?
 
thats what i thought, but what does that mean, are there not still 360 degrees?? does it mean that each radian is now longer than 1/2pi of the circumference of the original circle, sort of like the length of an arc??
 
Right. The equation doesn't know that the material is connected in a hoop. To fix this, use the differential element for segment length, [itex]R\,d\theta[/itex].