REINFORCED STRUT

Construction Design & Examples

Reinforced Strut

INTRODUCTION

Struts are often used for connecting two bodies in 1 DOF (x), while keeping the other DOFs free. In many cases the function of a stiff element with two hinges is desired. This is achieved by reinforcing the midsection of the strut.

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The poles of the hinges are located at the center of the thin sections, so to keep the same motion profile the strut length increases slightly. The buckling force is increased significantly, while the remaining stiffness in the motion directions is kept relatively unchanged.

In theory the midsection has 4 unconstrained DOFs, which leads to internal resonances. This can affect the controlability of the mechanism in high frequency applications.

Design parameters

\lambda =\frac l{L_P}                  \gamma =\frac d D

L_P=L+l

L_0=L+2l

Deformation characteristics

u_z=\frac{F_z}{C_z}

u_x{\approx}\frac{u_z^2}{2L_P}

Stiffness

C_x=\frac 1{2\gamma ^2\lambda -\gamma ^2-2\lambda }{\cdot}\frac{\mathit{E\pi }d^2}{4L_P}

C_y=C_z=\frac 1{\left(8\lambda ^3-12\lambda ^2+6\lambda \right)\left(1-\gamma ^4\right)+\gamma ^4}{\cdot}\frac{3\mathit{E\pi }d^4}{16L_P^3}

K_x=\frac 1{-2\gamma ^4\lambda +2\lambda +\gamma ^4}{\cdot}\frac{\mathit{G\pi }d^4}{32L_P}

K_y=K_z=\frac 1{-2\gamma ^4\lambda +2\lambda +\gamma ^4}{\cdot}\frac{\mathit{E\pi }d^4}{64L_P}

Stress

\sigma _{\mathit{max}}=\left(1+\lambda \right){\cdot}\frac{16L_PF_z}{\pi d^3}

Force limits

F_{x,\mathit{buckle}}=\frac 1{4\lambda ^2}{\cdot}\frac{E\pi ^3d^4}{64L_P^2}  or  F_{x,\mathit{buckle}}=\frac 1{\gamma ^4\left(1-2\lambda \right)^2}{\cdot}\frac{E\pi ^3d^4}{64L_P^2}

Whichever buckles first: First eq. if:  4\lambda ^2>\gamma ^4\left(1-2\lambda \right)^2

F_{z,\mathit{max}}=\frac{\sigma _y\pi d^3}{16L_P(1+\lambda )}

Maximum deflection

u_{z,\mathit{max}}=\frac{F_{z,\mathit{max}}}{C_z}=\frac{\left(8\lambda ^3-12\lambda ^2+6\lambda \right)\left(1-\gamma ^4\right)+\gamma ^4}{\left(1+\lambda \right)}{\cdot}\frac{\sigma _yL_P^2}{3\mathit{Ed}}

Figure S-shape deformation

Reinforced Strut

Design guidelines

Keep:   \frac 1{10}<\lambda <\frac 1 4   and   \frac 1{10}<\gamma <\frac 1 2

Typical:   \lambda =\frac 1 6   and   \gamma =\frac 1 3

Then:

C_x=2.5{\cdot}\frac{\mathit{E\pi }d^2}{4L_P}          K_x=3.0{\cdot}\frac{G\mathit{\pi d}^4}{32L_P}

C_y=1.4{\cdot}\frac{3\mathit{E\pi }d^4}{16L_P^3}          K_y=3.0{\cdot}\frac{\mathit{E\pi }d^4}{64L_P}

C_z=1.4{\cdot}\frac{3\mathit{E\pi }d^4}{16L_P^3}          K_z=3.0{\cdot}\frac{\mathit{E\pi }d^4}{64L_P}

F_{x,\mathit{buckle}}=9.0\ast\frac{\pi ^3Ed^4}{64L_P^2}

Pros & Cons

High buckling resistance

Predictable stiffness

Internal resonances

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