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The time derivative of engineering shear strain. A Newtonian simple-shear stress is $\eta \dot{\gamma }$. This scalar is twice the off-diagonal symmetric rate-of-strain-tensor component.

Strain measures local changes of length and angle relative to a chosen reference configuration. For a small displacement field $\mathbf{u}(\mathbf{X})$, the infinitesimal strain tensor isIts diagonal entries are fractional length changes along the coordinate axes. Off-diagonal entries describe changes of angle, with engineering shear strain equal to twice the corresponding tensor component.

A uniform translation has zero strain because its displacement gradient vanishes. An infinitesimal rigid rotation also has zero strain because its displacement gradient is antisymmetric. Finite rotations require finite strain theory, since the linear formula is then insufficient. In tensile testing, the simple ratio of extension to original gauge length is engineering strain; it coincides with the small axial component in the appropriate limit but is not a complete three-dimensional description.

A strain energy density measures stored elastic energy per unit volume. The volume convention must be specified. For a hyperelastic material, write $W(\mathbf{F})$ per unit volume of the reference configuration, where $\mathbf{F}$ is the deformation gradient. The total stored elastic energy iswith $\mathbf{P}$ the first Piola-Kirchhoff stress tensor. If $J=\det \mathbf{F}>0$, the density per current volume is $w=W/J$. At controlled temperature, the corresponding isothermal description uses a suitable free energy.

In uniform small shear strain $\gamma$, a linear elastic material with shear modulus $G$ stores $W=G{\gamma }^2/2$. Differentiating with respect to $\gamma$ gives the shear stress $G\gamma$. Integrating over the specimen gives the work recoverable on elastic unloading. The same construction turns a local material law into a problem in the calculus of variations for a bent or stretched body, subject to its boundary conditions.

A three-point flexural test loads a specimen through one central contact while two separated supports carry the reactions. In a small-deflection simply supported beam model, an incremental central force and the resulting midspan displacement determine the midspan compliance of a simply supported beam. With a known second moment of area, that response yields a flexural modulus.^[ref]

The apparatus must distinguish specimen deflection from machine travel. The manufacturer's fixture description uses adjustable lower anvils and a central loading member; a direct midspan plunger can transmit deflection to an extensometer. A load cell supplies the force record.^[ref] The crosshead displacement can also contain deformation of the frame and contacts.

For an original readout example, stipulate span $L=0.120\mathrm{m}$, width $b=12\mathrm{mm}$ and thickness $h=4\mathrm{mm}$, so $I=6.4\times 10^{-11}{\mathrm{m}}^4$. Suppose a linear loading interval gives direct midspan compliance $C_s=0.020mm/N$, while machine travel gives $C_m=0.025mm/N$. Applying the same beam relation to the two readouts givesThe apparent modulus is twenty percent lower. The numbers are chosen to isolate the displacement-channel effect; they do not characterize a particular fixture. Inferring a material Young's modulus further assumes the beam's linear, homogeneous bending model is adequate. Contact indentation, support motion and transverse shear strain can otherwise alter the measured compliance. The parent simply supported beam supplies the deflection derivation and support-comparison figure.

A circular elastic layer of thickness $h\ll a$ bonded between rigid faces develops shear strain $r\varphi /h$ under a relative twist $\varphi$. Integrating the shear stress times its moment arm gives $K_V=G{J}_p/h$. A small relative tilt produces longitudinal strain $y\theta /h$; when lateral deformation is constrained, the corresponding normal traction gives $K_H=MI/h$. The polar second moment of area is $J_p=\pi a^4/2$ and the diametric second moment of area is $I=\pi a^4/4$. Edge relaxation limits this thin-layer approximation, especially in a nearly incompressible material.

A circular shaft twisted uniformly through small end angle $\varphi$ over length $L$ has local engineering shear strain $r\varphi /L$ at radius $r$. Its shear stress is $Gr\varphi /L$, so the integrated torque is $C={\int }_{A}r(Gr\varphi /L)dA=G{J}_p\varphi /L$, using the polar second moment of area. The restoring torque has the opposite sign to the imposed twist. The stored elastic energy is $C\varphi /2=G{J}_p{\varphi }^2/(2L)$. The circular cross-section is important: general noncircular sections can warp and have a different torsion constant.

Viscoelasticity combines recoverable elastic response with dissipative and time-dependent response. Two specimens may react similarly to a rapid pull but behave differently when the same load is maintained. In a material creep test, constant stress produces evolving strain; in a stress relaxation test, fixed strain produces evolving stress.

The Maxwell model makes the distinction explicit by placing an elastic spring and viscous element in series. In simple shear strain it obeysAfter a small step to fixed ${\gamma }_0$, its shear stress decays as $\tau (t)=G{\gamma }_0{e}^{-t/\lambda }$ with viscoelastic relaxation time $\lambda =\eta /G$. Under constant shear stress ${\tau }_0$, it instead continues to deform, with $\gamma (t)={\tau }_0/G+{\tau }_{0}t/\eta$. These are predictions of this particular model: a Kelvin-Voigt model has bounded creep and different short-time behavior. Small-amplitude oscillatory shear measures the same competition through storage modulus and loss modulus, rather than assuming that one constant dynamic viscosity characterizes the material.