Chapter 3 - Design for static stiffness

# 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