# Material utilisation optimisation

## Introduction

A high stiffness and a low mass can be important for a moving frame, but which parameter needs to be optimized? Typically there are multiple outputs to display in an FEM environment such a stress, strain, and displacement. The proper parameter will be derived. A strongly simplified two spring and one lever example serves as a case study for series stiffness with lever in between, and shown in Figure 1. Spring 1 and 2 have stiffness:

$$

c_1=\frac{E_1A_1}{L_1}

$$

$$

c_2=\frac{E_2A_2}{L_2}

$$

In which the Young’s modulus is $E$, the cross sectional area is $A$, and the length is $L$. The displacements of the individual springs $u$ are:

$$

u_1=\frac{F}{c_1}=\frac{FL_1}{E_1A_1}

$$

$$

u_2=\frac{Fi}{c_2}=\frac{FiL_2}{E_2A_2}

$$

In which $i$ is the transmission ratio, and $F$ the external force. The total displacement at the external force node $u$ is

$$

u=u_1+iu_2=F \left(\frac{L_1}{E_1A_1}+i^2 \frac{L_2}{E_2A_2} \right)

$$

The total mass of the springs is

$$

m_{tot}=m_1+m_2=A_1L_1\rho1+A_2L_2\rho_2

$$

$$

A_2=\frac{m_{tot}-A_1L_1\rho1}{L_2\rho_2}

$$

The compliance at the external force node $s$ is

$$

s=\frac{u}{F}= \frac{L_1}{E_1A_1}+i^2 \frac{L_2}{E_2A_2}

$$

$$

s=\frac{L_1}{E_1A_1}+i^2 \frac{L_2^2\rho_2}{E_2\left(m_{tot}-A_1L_1\rho_1\right)}

$$

The partial derivative of the compliance to the area $A_1$ is equal to 0

$$

\frac{\partial s}{\partial A_1}=-\frac{L_1}{E_1A_1^2}+i^2 \frac{\rho_1 \rho_2 L_1 L_2^2}{E_2\left(m_{tot}-A_1L_1\rho_1\right)^2}=0

$$

$$

\frac{L_1}{E_1A_1^2}=i^2 \frac{\rho_1 \rho_2 L_1 L_2^2}{E_2\left(m_{tot}-A_1L_1\rho_1\right)^2}

$$

with $i$

$$

i=\frac{\frac{E_2A_2}{L_2}u_2}{\frac{E_1A_1}{L_1}u_1}

$$

yields

$$

\frac{E_1}{\rho_1L_1^2}u_1^2= \frac{E_2}{\rho_2L_2^2}u_2^2

$$

$$

\frac{E_1}{\rho_1}\varepsilon_1^2= \frac{E_2}{\rho_2}\varepsilon_2^2

$$

$$

\frac{\sigma_1 \varepsilon_1}{\rho_1}= \frac{\sigma_2 \varepsilon_2}{\rho_2}

$$

$$

\left(\frac{\textrm{strain energy}}{\textrm{mass}}\right)_1= \left(\frac{\textrm{strain energy}}{\textrm{mass}}\right)_2

$$

**Match of strain energy per unit mass**

Note that the same steps can be taken to get the maximum stiffness with a certain amount of volume. The result is:

$$

\sigma_1 \varepsilon_1= \sigma_2 \varepsilon_2

$$

**Match of strain energy density**

### Principle

To get the maximum stiffness with a certain amount of mass:

**Match of strain energy per unit mass**

### Example

So, to maximize the series stiffness of the two springs shown in Figure 2

$$

\frac{\sigma_1 \varepsilon_1}{\rho_1}= \frac{F_1\Delta L_1}{A_1L_1\rho_1}=\frac{F_2\Delta L_2}{A_2 L_2\rho_2}=\frac{\sigma_2 \varepsilon_2}{\rho_2}

$$

with

$$

\Delta L_1= \frac{F_1 L_1}{E_1 A_1}

$$

and

$$

\Delta L_2= \frac{F_2 L_2}{E_2 A_2}

$$

$$

\frac{F_1^2}{A_1^2 E_1 \rho_1}=\frac{F_2^2}{A_2^2 E_2 \rho_2}

$$

with $F_1=F_2$

$$

A_1^2 E_1 \rho_1=A_2^2 E_2 \rho_2

$$

with $E_1=E_2$ and $\rho_1=\rho_2$ then $A_1=A_2$.

### Parallel springs

The optimum for a parallel set of springs, Shown in Figure 3, can be derived in a similar manner. However, taking the partial derivative of the stiffness or compliance to the area $A_1$ and equating to zero, leads to the minimum stiffness. The maximum stiffness is found at on of the extreme values for $A_1$ either extreme of $A_1$. If

$$

\frac{E_1}{\rho_1 L_1^2}>\frac{E_2}{\rho_2 L_2^2}

$$

Then use all material to increase $A_1$ as much as possible, otherwise use all material to increase $A_2$ as much as possible.

### Developed by University of Twente

- Dannis Brouwer, Jaap Meijaard