Calculation of Stiffness constant

Hi John0D1,
Yeah you are correct, each individual element does have its own stiffness coeeficient when related to others. Normally, for simple sub frames k (individual stiffness) can be related to I(second moment area) divied by L, length of element. For a continous member, I/L is multipied by by 0.5. So the stiffness of any element in subframe is its own stiffness divided by the sum of the stiffnesses of the other parts, at the given intersection.
At the higher end, Stiffness gets a bit more complicated, and is subject to a number of equation(my book reference for this is at work!!) but stiffness if a function of E, I and L. Let me know if you need more, and i'll try and help.
Andy

I'll let the big boy's tell me i'm wrong now!!!

The stiffness is defined as the force required to make an element undergo a 1 unit deformation. With beams, logs, and widgets of varying proportions, the stiffness is likely different for each.

Is this the text book question it appears to be?

If you ignore shear deformations, the stiffness of a cantilevered beam is 3EI/L^3 in bending, or EA/L in axial deformation.

Obviously PFC's and UB's have different second moments of inertia and area, so they will have different stiffnesses.

Another small point to add is that if, for example, you were to assess forces in a steel rope, the "stiffness" ('K') required to work out the deformation ('x') due to axial force ('F') or vice-versa would be:

K = F/x

The 'F' and 'x' values would be obtained from the material's stress-strain mechanical properties at yield. Hope this helps,

-- drej --

Thanks for the replys.
I have new topic of interest: Blast Overpressure Loading on industrial structures.

Can anyone guide me to some useful books/web sites that could provide the basics?

Regards,