What is Equivalent Stiffness of a Beam?

[boeing_737]

Hi all,

I am trying to understand the concept of equivalent stiffness of a beam. As I see it, the equivalent stiffness is the stiffness of a linear spring that would deflect the same amount under the same load. For a cantilever beam with a load P and a deflection \delta at the free end, if we just look at the deflection of the free end, and somehow lump the stiffness and elastic properties of the beam into the term k_{eq}, it's motion will be the same as a linear spring with stiffness k_{eq} when the same load P is applied to the spring.

My vibrations textbook mentions the k_{eq} for a cantilever with a moment applied to the free end as \frac{EI}{L}. Assuming that the spring being considered is a linear torsional spring, how do we interpret this? I am thinking it goes something like - 'The equivalent stiffness of a cantilever beam with a moment at the free end is the stiffness of a linear torsional spring that would coil by an angle say \theta when the same moment is applied to it.' Now, is \theta the same as the tip deflection \delta or is it the local slope at the free end ie \theta \approx \tan(\theta) = \frac{dy}{dx}

Thanks a lot for the help!

PS - Any suggestions for books that explain the equivalent stiffness concept well?

 

[AlephZero]

\theta is the local slope at the free end.

Of course there is also a displacement at the free end, which you can find from the fact that the curvature of the beam is constant. The geometrical relationship between the slope and the displacement at the free end will depend on the length of the beam, but not on E or I.

 

[boeing_737]

\theta is the local slope at the free end.

Of course there is also a displacement at the free end, which you can find from the fact that the curvature of the beam is constant. The geometrical relationship between the slope and the displacement at the free end will depend on the length of the beam, but not on E or I.

Sorry for the late response.

Thanks for clarifying that. Also, does that mean that if we know the moment acting at a location x, we can calculate the slope of the curved beam at that point due to the moment?

 

[nvn]

boeing_737: Regarding post 3, yes, it does. Regarding post 1, you have a cantilever having an applied tip moment, M. As AlephZero mentioned, the cantilever tip rotational stiffness is equivalent to a linear torsional spring at the cantilever tip having spring constant k_theta = E*I/L. The cantilever tip slope is theta = M/k_theta.

Also, the cantilever tip transverse deflection stiffness is equivalent to a linear translational spring at the cantilever tip having spring constant k = 2*E*I/L^2. The cantilever tip transverse deflection is delta = M/k.