# Solid

Corresponding Wikipedia article: Solid

## Continuum

The solid is a state of matter and a special type of continuum.

General continuum Solid
Transfer of matter can exist Transfer of matter does not exist
Isotropic density Anisotropic deformation
Isotropic pressure Anisotropic stress
Barotropic fluid Elasticity

The stress is a volumetric energy density of the directed interaction of particles in contrast to the chaotic thermal interaction. In reference to the faces of elementary cubic volume, the stress can be:

• Normal (perpendicular to a face like a usual pressure).
• Tangential (parallel to a face). It's caused by the anisotropy of normal stress.

The material properties comprise the stress limits:

• Elastic limit is a condition of plasticity.
• Proportionality of deformation and stress.
• Yield strength is a point where the particles energy is sufficient to liquefy the body at low temperature. This effect is typical for the metals.
• Fatigue and ultimate strength is a point where the particles energy is sufficient to break them apart at low temperature. Evaporation does not occur, because the stress drops rapidly after the fracture.

The volumetric isotropic stress $$P$$ is related to the relative volumetric deformation $$\frac{\Delta V}{V}$$ of a body with density $$\rho$$ through the bulk modulus $$K$$: $P=K\frac{\Delta \rho}{\rho}=K\frac{\Delta V}{V}\tag{1}$

The volumetric deformation is related to the one-dimensional isotropic linear deformation $$\Delta L$$: $\frac{\Delta V}{V}=\left(1+\frac{\Delta L}{L}\right)^3-1\approx 3\frac{\Delta L}{L}\tag{2}$

The linear (normal) elastic deformation is determined by Hooke's law: $E\frac{\Delta L}{L}=P_\parallel-2\mu P_\perp\tag{3}$ $$E$$ is Young's modulus;
$$\mu$$ is Poisson's ratio of longitudinal compression to transversal stretching;
$$P_\parallel$$ is a longitudinal normal stress;
$$P_\perp$$ is a transverse normal stress along one of two axes.

The bulk modulus at $$P=P_\parallel=P_\perp$$: $K=\frac{E}{3(1-2\mu)}\tag{4}$

The shear (tangential) deformation is represented by deformation of a square section within a linearly deformable object. $tan\;\alpha=\frac{1-\mu\frac{\Delta L}{L}}{1+\frac{\Delta L}{L}}\approx\left(1-\mu\frac{\Delta L}{L}\right)\left(1-\frac{\Delta L}{L}\right)\approx 1-\frac{\Delta L}{L}(1+\mu)\tag{5}$

The shear angle $$\theta$$ is related to angle $$\alpha$$ so: $\theta=\frac{\pi}{2}-2\alpha\tag{6}$ $tan\;\alpha=tan\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\approx 1-\theta\tag{7}$

The consequence from (5) and (7) is: $\theta=\frac{\Delta L}{L}(1+\mu)\tag{8}$

The shear deformation at small angles is: $\theta=\frac{\Delta L_G}{L_G}\tag{9}$

The bulk shear and linear deformations are related to each other through the stress equation: $E\frac{\Delta L}{L}=\frac{E}{1+\mu}\frac{\Delta L_G}{L_G}\tag{10}$

The shear modulus $$G$$ is defined for one-dimensional deformation under shear stress, which is half of the bulk shear stress: $G=\frac{E}{2(1+\mu)}\tag{11}$

## Waves

The propagation speed of longitudinal waves (“Waves”, 5) within solids follows from (1): $c_s=\sqrt{\frac{K_W}{\rho}}\tag{12}$ $$K_W$$ is a wave modulus, which is roughly estimated as the bulk modulus $$K$$.

The modulus of one-dimensional waves (within a thin rod) is Young's modulus $$E$$.

The energy of volumetric wave is the sum of volumetric energy of normal and tangential (shear) stresses along two of three axes (along the wave front). So its modulus is: $K+\frac{2E}{3(1+\mu)}=K+\frac{4G}{3}\tag{13}$

## Rotation

Each particle of a rotating solid has the same angular velocity $$\omega$$. So the moment of inertia $$J$$ can be used to simplify the calculations. The rotational energy without relativistic corrections (see "Mass and inertia") is: $A=\int{\frac{\rho v^2}{2}\mathrm{d}V}=\frac{\omega^2}{2}\int{r^2\rho\mathrm{d}V}=\frac{J\omega^2}{2}\tag{14}$

The equation of power and angular momentum $$L$$ is: $\frac{\mathrm{d}A}{\mathrm{d}t}=J\omega\frac{\mathrm{d}\omega}{\mathrm{d}t}=L\frac{\mathrm{d}\omega}{\mathrm{d}t}\tag{15}$

The equation of torque is: $\mathbf{\overrightarrow{M}}=\frac{\mathrm{d}\mathbf{\overrightarrow{L}}}{\mathrm{d}t}=J\frac{\mathrm{d}\mathbf{\overrightarrow{\omega}}}{\mathrm{d}t}=J\frac{\mathrm{d}\omega}{\mathrm{d}t}\frac{\mathbf{\overrightarrow{L}}}{|\mathbf{\overrightarrow{L}}|}+\mathbf{\overrightarrow{\omega}}_p\times\mathbf{\overrightarrow{L}}\tag{16}$ $$\mathbf{\overrightarrow{\omega}}_p$$ is an angular velocity of precession.

The center of mass is a point, around which an object with mass $$m$$ can rotate without any external torque (moment of force). The moments of inertia forces are balanced at this point: $\mathbf{\overrightarrow{r}}_0=\frac{1}{m}\int{\mathbf{\overrightarrow{r}}\rho\mathrm{d}V}\tag{17}$

The bending of beam and the torsion of shaft are similar to rotation in many ways. As the linear displacement is directly proportional to the distance from center of rotation, so the relative deformation is directly proportional to the distance from axis of bending or torsion. The stress $$P$$ of elastic deformation at distance of $$r$$ from axis can be expressed as: $P=\frac{P_{MAX}}{R}r$ $$P_{MAX}$$ is the maximum stress at the cross-section edge of beam or shaft;
$$R$$ is a distance from the axis to that edge.

The equation of torque integrated over the cross-section $$S$$ is: $M=\int{rP\mathrm{d}S}=\frac{P_{MAX}}{R}\int{r^2\mathrm{d}S}=P_{MAX}\frac{I}{R}=P_{MAX}W$ $$I$$ is a polar moment of inertia or a second moment of area;
$$W$$ is a polar or section modulus of torsion or bending, respectively.