# Finding increased length after deformation

1. Nov 29, 2009

### Dell

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

=$$\int$$K*sin2(x)dx from 0->2pi

=K/2*$$\int$$1-cos(2x)dx

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

=K*pi

where does the R come from? i realise that the change in length should be dependant on the radius, but mathematically how do i come to that?

2. Nov 29, 2009

### Mapes

You're working with $\epsilon_\theta$, an angular strain. When you integrate it, you get an increase in angle, not length. Know what I mean?

3. Nov 29, 2009

### Dell

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??

4. Nov 29, 2009

### Mapes

Right. The equation doesn't know that the material is connected in a hoop. To fix this, use the differential element for segment length, $R\,d\theta$.

5. Nov 29, 2009