## Deformable Body Mechanics

 $\sigma$ $\mathrm{N/m^2}$ tension $\varepsilon$ elongation $\varepsilon_q$ lateral strain $E$ $\mathrm{N/m^2}$ modulus of elasticity (material value) $F_{ZUG}$ $\mathrm{N}$ traction force, normal force $F_T$ $\mathrm{N}$ shearing force (tangential) $A$ $\mathrm{m^2}$ area $V$ $\mathrm{m^3}$ volume $l$ $\mathrm{m}$ initial length, length of a body $l’$ $\mathrm{m}$ final length $d$ $\mathrm{m}$ initial thickness $d’$ $\mathrm{m}$ final thickness $\mu$ Poisson’s ratio $\nu$ inverse Poisson’s ratio, transverse direction $K$ $\mathrm{N/m^2}$ compressive modulus $\kappa$ $\mathrm{Pa^{-1}}$ compressibility $\Delta p$ $\mathrm{N/m^2}$ change of pressure $G$ $\mathrm{N/m^2}$ shear modulus, modulus of torsion $\tau$ $\mathrm{N/m^2}$ shear stress $\gamma$ $\mathrm{rad}$ shift, shear, shear angle $\varphi$ $\mathrm{rad}$ torsion angle $r$ $\mathrm{m}$ radius $M$ $\mathrm{Nm}$ torque $M_T$ $\mathrm{Nm}$ torque in torsion $D$ $\mathrm{Nm/rad}$ torsional coefficient, spring constant $I_{POL}$ $\mathrm{rad/m^4}$ polar moment of inertia

### Elongation

$$\Delta l = \left| l – l’ \right| \qquad \varepsilon = \frac{\Delta l}{l}$$

$$\sigma = \frac{F_{ZUG}}{A} \qquad \sigma = \frac{\mathrm{d} F_{ZUG}}{\mathrm{d} A}$$

Hooke’s law, for the elastic range only:

$$\sigma = E \, \epsilon$$

$$E(\sigma) = \frac{\mathrm{d}\sigma}{\mathrm{d}\varepsilon} \qquad E = \sigma \, \frac{l}{\Delta l} \qquad E = \frac{F_{ZUG}}{A} \, \frac{l}{\Delta l}$$

### Lateral Strain

Change of length and thickness.

$$\Delta d = d – d’ \qquad \mu = – \frac{ \displaystyle\frac{\Delta d}{d} }{ \displaystyle\frac{\Delta l}{l} } \qquad \mu = \frac{1}{\nu}$$

$$\varepsilon_q = – \frac{1}{\mu} \, \varepsilon \qquad \varepsilon_q = – \nu \, \varepsilon \qquad \varepsilon_q = \frac{\Delta d}{d}$$

Relative change of volume:

$$\frac{\Delta V}{V} = \varepsilon \, \left( 1 – 2 \, \nu \right) \qquad \frac{\Delta V}{V} = \frac{\Delta l}{l} + \frac{2 \, \Delta d}{d}$$

### All sides Compression

$$\frac{\Delta V}{V} = \frac{\sigma}{K} \qquad \frac{\Delta V}{V} = \frac{\Delta p}{K}$$

$$\mathrm{d}V = – \frac{1}{K} \, V \;\mathrm{d}p$$

$$K = \frac{1}{\kappa} \qquad K = \frac{E}{3 \,( 1 – 2 \, \nu)} \qquad \frac{\Delta V}{V} = 3 \, \varepsilon \, (1 – 2 \,\nu)$$

$$\nu = \frac{1}{\mu} \qquad \varepsilon = \frac{\sigma}{E} \qquad \sigma = – \Delta p$$

### Shear

$$\tau = G \, \gamma \qquad \tau = \frac{F_T}{A} \qquad \tau = \frac{G \, a}{d}$$

$$G(\tau) = \frac{\mathrm{d}\tau}{\mathrm{d}\gamma} \qquad G = \frac{E}{2 \, (1 + \nu)}$$

### Torsion

$$\varphi = \frac{2 \, l \, M}{\pi \, G \, r^4 } \qquad D = \frac{\pi \, G \, r^4}{2 \, l}$$

$$M = D \, \varphi \qquad \vec{M} = D \, \varphi \, \vec{e}_\omega$$

$$\ddot{\varphi} + \frac{D}{J} \, \varphi = 0 \qquad T = 2 \, \pi \, \sqrt{\frac{D}{J}} \qquad \varphi = \frac{M_T \, l}{G \, I_{POL}}$$

Cylinder:

$$\tau = \frac{M_T}{0.196 \, d^3} \qquad \varphi = \frac{M_T \, l}{0.089 \, G \, d^4}$$