# Leaf spring/flexure: Reinforced

## Construction Design & Examples

#### Introduction

Flexures or leaf springs can be used for play and friction free motion. A downside is stiffness and to minimize the needed force, the flexures are made slender and thin. Smart reinforcement of flexures and leaf springs can help to keep the needed motion-force minimal while the flexure or leaf spring is made thicker, which is beneficial for its carrying stiffness and easier to manufacture which will decrease the manufacturing costs.

#### Design parameters

$\lambda=\frac{L_s}{L_0}$

$0<\lambda<\frac{1}{2}$

$L_P=L_{RF}+L_s=\left(1-\lambda\right)L_0$

$\gamma=\frac{t}{T}$

$0<\gamma<1$

#### Deformation characteristics

$u_z=\frac{F}{C_z}$
$u_x=\frac{u_z^2}{2L_0\left(1-\lambda\right)}$

#### Stiffness for s- and c-shape deformation

$C_x=\frac{1}{2\lambda\left(1-\gamma\right)+\gamma}\cdot\frac{Etb}{L_0}$
$C_y=\frac{1}{2\lambda\left(4\lambda^2-6\lambda+3\right)\left(1-\gamma\right)+\gamma}\cdot\frac{Etb^3}{L_0^3}$ (s-shape deformation)
$C_y=\frac{1}{2\lambda\left(4\lambda^2-6\lambda+3\right)\left(1-\gamma\right)+\gamma}\cdot\frac{Etb^3}{4L_0^3}$ (c-shape deformation)
$C_z=\frac{1}{2\lambda\left(4\lambda^2-6\lambda+3\right)\left(1-\gamma^3\right)+\gamma^3}\cdot\frac{Ebt^3}{{L_0}^3}$ (s-shape deformation)
$C_z=\frac{1}{2\lambda\left(4\lambda^2-6\lambda+3\right)\left(1-\gamma^3\right)+\gamma^3}\cdot\frac{Ebt^3}{4{L_0}^3}$ (c-shape deformation)

$K_x=\frac{1}{2\lambda\left(1-\gamma^3\right)+\gamma^3}\cdot\frac{Gbt^3}{3L_0}$
$K_y=\frac{1}{2\lambda\left(1-\gamma^3\right)+\gamma^3}\cdot\frac{Ebt^3}{12L_0}$ (c-shape deformation)
$K_z=\frac{1}{2\lambda\left(1-\gamma\right)+\gamma}\cdot\frac{Etb^3}{12L_0}$ (c-shape deformation)

#### Force limits (buckling)

When a force in x-direction is applied buckling can occur, for the buckling shape depicted above the buckling force is:

$F_{x\ buckle}=\frac{\pi^2EI_s}{4L_sL_0}$

See Area moment of inertia for more information on $I$.

#### Design guidelines

Keep $\frac{1}{10}<\lambda<\frac{1}{3}$ and $\frac{1}{10}<\gamma<\frac{1}{2}$

Typical $\lambda=\frac{1}{6}$ and $\gamma=\frac{1}{5}$

Then:
$C_x=2.1\cdot\frac{Etb}{L_0}$
$C_y=1.3\cdot\frac{Etb^3}{L_0^3}$
$C_Z=1.4\cdot\frac{Ebt^3}{L_0^3}$

$K_x=3.0\cdot\frac{Gbt^3}{3L_0}$
$K_y=3.0\cdot\frac{Ebt^3}{12L_0}$
$K_z=2.1\cdot\frac{Etb^3}{12L_0}$

$F_{x\ buckle}=\frac{1}{3}\frac{\pi^2Ebt^3}{\left(2L_s\right)^2}$

#### Normalized stiffness increase due to reinforcement

The graphs shown below indicate the normalized stiffness increase with respect to the non-reinforced case ($\lambda=0.5$).

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