Static Preload

A simple method to calculate the static preload in a lumped body multi Degrees of Freedom system

tl,dr:

💡 Tip

Final Formula Given a system of $s$ springs and $n$ bodies in $N$ degrees of freedom, we can calculate the static preload of our springs at equilibrium by:

$$\vec {\Delta l_o} = [A]^{-1}[b]$$

with $\vec {\Delta l_o}$ as the vector of our $s$ ordered solution, with

$$[A]=[\lambda_k]^T[k_{ph}]$$

and

$$[b]=-g[\lambda_g]^T\vec m_{ph}$$

$[\lambda_k]$ is our standard lagrangian matrix for the springs, $[\lambda_g]$ instead is built by selecting the components of vertical displacement from $[\lambda_m]$ (lagrangian matrix of the inertial terms) and $\vec m_{ph}$ as the list of masses in column form.

Proof 1-DOF case

Let’s start by defining our problem. $x$ is our free coordinate, so we have:

$V$ is the potential energy of our system and it can be divided into an elastic component $V_k$ and a gravitational component $V_g$ .

The equilibrium is defined as the point $x_0$ so that

To simplify our math let’s suppose that $x_0=0$ (it’s a simple change of coordinates that will help in the N-dof case).

The first step is to derive the components of the potential energy:

Elastic energy 1-dof

$V_k$ describes the energy stored in an ideal linear spring with Hooke’s coefficient $k$ subject to a deformation $\Delta l$

We can ignore the index for now and express $\Delta l (x)$ as $\Delta l(x) = l(x) - l_r$ , with $l(x)$ as the total length of our spring and $l_r$ as the initial length of the spring at resting position, thus giving us the convention of positive sign for tension and negative sign for compression of our spring.

We can expand it as:

We are going to square and derive this expression, so every term of $x$ with power greater or equal than two $\{ x^n : n\geq 2 \}$ is going to show up as a function of $x^{n-1}$ and when evaluated at $x=0$ they simply dissapear.

When putting it back into the sum we get:

Let’s evaluate the derivative in respect to $x$ :

$\Delta l_0^2$ is a constant and gets eliminated by the derivative, the other sqare term becomes $\lambda_{k0}^2 x$ and goes to zero when evaluated at $x=0$ , so only the mixed term $2\Delta l_o \lambda_{k0} x$ survives and the derivative becomes:

We can rewrite this in vector form as:

Gravitational energy 1-dof

This expresses the total gravitational energy in our system supposing $n$ bodies with mass $m_j$ and height $h_j$ measured from an arbitrary system of coordinates.

Applying the same concepts as the elastic term we get:

Let’s focus on the lower order terms and we get:

In this context $\lambda_g$ is the linearized displacement of the height at equilibrium.

Deriving it we get

Let’s put everything in vector form

Final expression

To find $\vec {\Delta l_0 }$ we just rearrange it as:

N-DOF case

We follow the same procedure but now we have additional dimensions for our $\vec x$ :

So we need to evaluate the gradient of our potential energy funcion for the equilibrium:

Elastic energy

with

As usual we expand our $\Delta l(\vec x)$ near the equilibrium:

and we simplify it as:

Plugging it into the initial equation we get:

Like for our 1-DOF case the only terms that survive the gradient are the mixed ones:

ℹ️ Note

For $x,y \in \mathbb{R}\_{n,1}$ and $[S] \in \mathbb{R}_{n,n}$ .

Then $x^T[S]y$ is just a scalar. So:

$$(x^T[S]y)^T=x^T[S]y$$

If $[S]$ is symmetric:

$$x^T[S]y=(x^T[S]y)^T=y^T[S]^Tx=y^T[S]x$$

Gravitational Energy

This must be straightforward now if you understood the past passages but for completeness let’s write them down:

with

Expanding the height we get:

Picking just the linear and terms we get:

Transposing and putting it back in the original energy expression we get:

Deriving it we get

Final formula