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:

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