Finding increased length after deformation

[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

=[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 dependant on the radius, but mathematically how do i come to that?

 

[Mapes]

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?

 

[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??

 

[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, [itex]R\,d\theta[/itex].

 

[Dell]

thanks,, could you please take a look at this thread too

https://www.physicsforums.com/showthread.php?t=358770