K Matrix - Axial element w/ Spring 1

[jwells65]

How can I formulate an element stiffness matrix for an axial force member (2 d.o.f. only) that has continuous in-line springs attached along its length? (These are not perpendicular to the beam. They are parallel).

I'm having trouble writing a displacement function that describes the displacements along the length of the element that takes into account the spring.

 

[vonlueke]

Let's see. Wouldn't that be as follows?

k_basic = member basic coordinate system (CS) stiffness matrix.
k = truss (rod) member global stiffness matrix.
K = structure global stiffness matrix.
D = structure nodal global displacements.
R = structure nodal global applied external loads.

k_rod = EA/L.
k_spr = ko.
  
k_basic = k_rod + k_spr = (EA/L + ko).
  
r = kd

--Truss-member global stiffness relation. (1)

[r1]            [ c^2  cs  -c^2 -cs ][d1]
[r2] = (k_basic)[ cs   s^2 -cs  -s^2][d2]

(2)

[r3]            [-c^2 -cs   c^2  cs ][d3]
[r4]            [-cs  -s^2  cs   s^2][d4]

where

ko = spring constant of parallel spring.
r = member end-forces array, in global CS.
d = member end displacements array, in global CS.
c = cos(alpha).
s = sin(alpha).
alpha = member end 1 angle from global x axis to member axis.

Note that the truss member basic stiffness matrix, k_basic, is 1 x 1. If there's only one member in your structure, solve the equation P = (k_basic)(delta) for delta and you're done. I.e., delta = P/(k_basic), where P = member nodal basic applied external (axial tensile) loads and delta = relative end displacements in basic (axial) CS.

If, however, there's more than one member in your structure, assemble (add) partitions of each Eq. 2 member global stiffness matrix k into proper node locations of structure global stiffness matrix K, then solve R = KD for D. I.e., D = (K^-1)R. Then backsubstitute D solution partitions back into Eq. 2 member global stiffness equations to solve for r. Good luck.

 

[jwells65]

Thank you for responding.

What has me confused is that the axial force is not constant throughout the member. The springs pull load away from the member along its length based on the displacement of each point.

The external axial force will not be the same at both ends of the member.

So the strain should not be constant, and Delta=Px/AE does not hold. Or am I making it way too complicated?

By the way, this is the model of skin friction along an element.

 

Is your continuous, parallel spring attached to each of the two rod end supports, or is the spring truncated just before reaching each of the rod end supports?

 

[jwells65]

The springs do not connect to the ends. Think of it as an infinite number of springs spaced close together--kind of like a beam on elastic foundation. Except in this case the springs are parallel to its length.

 

[vonlueke]

According to your problem description, you have a bonded lap joint, or to be more precise in this case, we might call it a bonded "cap" joint. I assume your continuous springs are a continuous layer of material bonded to the bar element. The solution herein assumes the "spring" layer is not attached at its ends to the two end supports but, instead, is truncated at the ends before reaching the end supports.

You are correct in saying the continuous springs pull load away from the member. This phenomenon is governed by a second-order, boundary-value, ordinary differential equation that tracks the migration of axial force across the bonded interface (via shear stress on the bond). It's far too lengthy to explain the derivation, but the answer is as follows. Let your bar be made of material 1. Let your truncated "spring" layer (cap) be made of material 2. Material 2 is assumed to be the same width as material 1. The solution herein treats the member as if it were a simply-supported, one-dimensional, axial member that has no transverse displacement. Bending, if any, and buckling are neglected. Let's define the following input parameters. CTE means coefficient of thermal expansion.

L = material 2 length.
w = bar width.
t1 = material 1 thickness.
t2 = material 2 thickness.
E1 = material 1 normal elasticity modulus.
E2 = material 2 normal elasticity modulus.
Ge = adhesive shear elasticity modulus (perhaps 670 MPa if epoxy).
te = adhesive layer thickness.
alpha1 = material 1 CTE.
alpha2 = material 2 CTE.
T = ambient temperature (of both materials) relative to temperature during assembly; i.e., T = T2 - T1, where T2 occurs after T1.
F = axial load applied at each end of material 1 only; tension is positive.

Let
a = E2*t2/(E1*t1 + E2*t2).
b = {(Ge/te)[1/(E1*t1) + 1/(E2*t2)]}^0.5.

Axial force in material 2 as a function of x is as follows, with origin of x axis at left-hand end of material 2, and positive x axis directed toward right-hand end of material 2. Domain of x is 0 to L.

F2(x) = a[w*E1*t1(alpha1-alpha2)T + F]{1 - [sinh(b(L-x))+sinh(bx)]/sinh(bL)}.

Axial force in material 1 is F1(x) = F - F2(x). Shear stress in adhesive is as follows.

tau(x) = ab[E1*t1(alpha1-alpha2)T + F/w][cosh(b(L-x))-cosh(bx)]/sinh(bL).

The material 1 strain at any x would be eps1(x) = F1(x)/(E1*w*t1), and I believe the axial displacement of either material at any x relative to x=0 would be delta(x) = integral[eps1(x)dx], integrated from x = 0 to x. Good luck.

 

[jwells65]

Excellent! Thanks.