Static Preload
A simple method to calculate the static preload in a lumped body multi Degrees of Freedom system
tl,dr:
Final FormulaGiven 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:
$$ V(x) = V_k(x) + V_g(x) $$\(V\) 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
$$ \frac{\partial V}{\partial x}\rvert_{x_0}=0 $$ $$ \frac{\partial V_k}{\partial x}\rvert_{x_0}+\frac{\partial V_g}{\partial x}\rvert_{x_0}=0 $$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 \( \Delta l\)
$$ V_k(x) = \sum_{i=1}^{s} \frac{1}{2} k_i \Delta l_i^2(x) $$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:
$$ \Delta l(x) \approx \Delta l(0) + \frac{\partial \Delta l }{\partial x}\rvert_{x=0} x +\frac{1}{2}\frac{\partial^2 \Delta l }{\partial x^2}\rvert_{x=0} x^2 =\Delta l_0 + \lambda_k(0)x + \frac{1}{2}\nu_k(0)x^2 $$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.
$$ \Delta l(x)\rvert_{x=0} \approx \Delta l_0 + \lambda_k(0)x = \Delta l_0 + \lambda_{k0} x $$When putting it back into the sum we get:
$$ V_k(x) \rvert_{x=0} \approx \sum_{i=1}^{s} \frac{1}{2} k_i (\Delta l_{i0} + \lambda_{ik0} x )^2 = \sum_{i=1}^{s} \frac{1}{2} k_i (\Delta l_{i0}^2 + \lambda_{ik0}^2 x^2 + 2 \Delta l_{i0} \lambda_{ik0} x ) $$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:
$$ \frac {\partial V_k }{\partial x}\rvert_{x=0}= \sum_{i=1}^s k_i \Delta l_{i0} \lambda_{ik0} $$We can rewrite this in vector form as:
$$ [k_{ph}] = diag(k_1; k_2; ... ; k_s) $$ $$ \vec {\Delta l_0 } = [ \Delta l_{10}; \Delta l_{20}; ... ; \Delta l_{s0}] $$ $$ \vec \lambda_{k} = [\lambda_{1k};\lambda_{2k};...;\lambda_{sk}] $$ $$ \frac{\partial V_g(x)}{\partial x}\rvert_{x=0} = \vec \lambda_{k}^T [k_{ph}] \vec {\Delta l_0 } $$Gravitational energy 1-dof
$$ V_g(x) = g \sum_{j=1}^n m_j h_j(x) $$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:
$$ h(x)\rvert_{x=0} \approx h(x)\rvert_{0}+\frac{\partial h}{\partial x}\rvert_{0}x+\frac{1}{2}\frac{\partial^2 h}{\partial x^2}\rvert_{0} x^2 $$Let’s focus on the lower order terms and we get:
$$ h(x)\rvert_{x=0} \approx h_{0}+\lambda_gx$$In this context \(\lambda_g\) is the linearized displacement of the height at equilibrium.
$$ V_g(x) = g \sum_{j=1}^n m_j (h_{j0}+\lambda_{jg}x) $$Deriving it we get
$$ \frac{\partial V_g(x)}{\partial x}\rvert_{x=0}=g \sum_{j=1}^n m_j \lambda_{jg} $$Let’s put everything in vector form
$$ \vec {m_{ph}} = \begin{bmatrix} m_1 \\ m_2 \\ \vdots \\ m_n \end{bmatrix} $$ $$ \vec \lambda_{g} = \begin{bmatrix} \lambda_{1g} \\ \lambda_{2g}\\ \vdots \\ \lambda_{ng} \end{bmatrix} $$ $$ \frac{\partial V_g(x)}{\partial x}\rvert_{x=0} = g \vec \lambda_{g}^T \vec {m_{ph}} $$Final expression
$$ \frac{\partial V_k}{\partial x}\rvert_{x_0}+\frac{\partial V_g}{\partial x}\rvert_{x_0}=0 $$ $$ \vec \lambda_{k}^T [k_{ph}] \vec {\Delta l_0 } + g \vec \lambda_{g}^T \vec {m_{ph}} = 0 $$To find \( \vec {\Delta l_0 } \) we just rearrange it as:
$$ \vec {\Delta l_0 } = -g(\vec \lambda_{k}^T [k_{ph}])^{-1} \vec \lambda_{g}^T \vec {m_{ph}} = [A]^{-1}\vec b $$N-DOF case
We follow the same procedure but now we have additional dimensions for our \(\vec x\):
$$ \vec x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_N \end{bmatrix} $$So we need to evaluate the gradient of our potential energy funcion for the equilibrium:
$$ \frac{\partial V(\vec x)}{\partial \vec x}\rvert_{\vec x= \vec 0} = \begin{bmatrix} \frac{\partial V(\vec x)}{\partial x_1} \\ \frac{\partial V(\vec x)}{\partial x_2} \\ \vdots \\ \frac{\partial V(\vec x)}{\partial x_N} \end{bmatrix}\rvert_{\vec x= \vec 0} =\vec 0$$ $$ \nabla V(\vec x)\rvert_{\vec x= \vec 0} = \nabla V_k(\vec x)\rvert_{\vec x= \vec 0} + \nabla V_g(\vec x)\rvert_{\vec x= \vec 0} = \vec 0$$Elastic energy
$$ V_k (\vec x)=\frac{1}{2}\sum_{i=1}^{s}k_i\Delta l_i^2(\vec x) = \frac{1}{2} \vec{\Delta l_0}^T (\vec x) k_{ph} \vec{\Delta l_0(\vec x)} $$with
$$ k_{ph}=diag(k1,k2,...,ks)$$ $$ \vec{\Delta l_0}(\vec x) = \begin{bmatrix} \Delta l_{10}(\vec x) \\ \Delta l_{20}(\vec x) \\ \vdots \\ \Delta l_{N0}(\vec x) \end{bmatrix} $$As usual we expand our \(\Delta l(\vec x)\) near the equilibrium:
$$ \vec{\Delta l(\vec x)} \rvert_{\vec x=\vec 0} \approx \vec {\Delta l(\vec 0)} + \frac{\partial \Delta l }{\partial \vec x}\rvert_{\vec x= \vec 0} \vec x +\frac{1}{2}\frac{\partial^2 \Delta l }{\partial x^2}\rvert_{\vec x= \vec 0} \vec x^2 = \vec{\Delta l_0} + [\lambda_k(0)]\vec x + \frac{1}{2}\vec x^T[\nu_k(0)]\vec x $$and we simplify it as:
$$ \vec{\Delta l(\vec x)}\rvert_{\vec x=\vec 0} \approx \vec{\Delta l_0} + [\lambda_k(0)] \vec x = \vec{\Delta l_0} + [\lambda_{k0}] \vec x $$Plugging it into the initial equation we get:
$$ V_k (\vec x) \rvert_{\vec x=\vec 0} \approx \frac{1}{2} (\vec {\Delta l_{0}} + [\lambda_{k0}] \vec x)^T [k_{ph}] (\vec {\Delta l_{0}} + [\lambda_{k0}] \vec x) = $$ $$ = \frac{1}{2}( \vec x^T[\lambda_{k0}]^T [k_{ph}] [\lambda_{k0}] \vec x + \vec {\Delta l_{0}}^T[k_{ph}] \vec {\Delta l_{0}} + \vec {\Delta l_{0}}^T[k_{ph}] [\lambda_{k0}] \vec x + \vec x^T[\lambda_{k0}]^T [k_{ph}] \vec {\Delta l_{0}} ) $$Like for our 1-DOF case the only terms that survive the gradient are the mixed ones:
$$ \nabla V_k\rvert_{\vec x= \vec 0}= \frac{\partial V_k}{\partial \vec x} (\vec x) \rvert_{\vec x=\vec 0} \approx \frac{1}{2} (\vec {\Delta l_{0}}^T[k_{ph}] [\lambda_{k0}] + [\lambda_{k0}]^T [k_{ph}] \vec {\Delta l_{0}} ) = [\lambda_{k0}]^T [k_{ph}] \vec {\Delta l_{0}} $$N.b.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:
$$ V_g(\vec x) = g \vec h(\vec x)^T \vec {m_{ph}} $$with
$$ \vec m_{ph}= \begin{bmatrix} m_1 \\ m_2 \\ \vdots \\ m_n \end{bmatrix} $$ $$ \vec h(\vec x) = \begin{bmatrix} h_1(\vec x) \\ h_2(\vec x) \\ \vdots \\ h_n(\vec x) \end{bmatrix} $$Expanding the height we get:
$$ \vec h(\vec x)\rvert_{\vec x=\vec 0} \approx \vec h(\vec x)\rvert_{\vec 0}+\frac{\partial \vec h}{\partial \vec x}\rvert_{\vec 0}\vec x+\frac{1}{2}\frac{\partial^2 h}{\partial \vec x \partial \vec x}\rvert_{\vec 0} \vec x^2 = \vec h_0 + [\lambda_k(0)]\vec x+ \frac{1}{2}\vec x^T[\nu_k(0)]\vec x$$Picking just the linear and terms we get:
$$ \vec h(\vec x)\rvert_{\vec x=\vec 0} \approx \vec h_{0}+[\lambda_g]\vec x$$Transposing and putting it back in the original energy expression we get:
$$ V_g(\vec x)\rvert_{\vec 0} = g ( \vec h_0^T + \vec x^T [\lambda_g]^T)\vec {m_{ph}} $$Deriving it we get
$$ \nabla V_g\rvert_{\vec x= \vec 0} = \frac{\partial V_g(\vec x)}{\partial \vec x}\rvert_{x=0}=g [\lambda_g]^T \vec{m_{ph}} $$