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Compression and stretch viewed on rotated faces. To isolate shear, take equal small principal strains ${ϵ}_{xx}=-e$, ${ϵ}_{yy}=+e$. A square with sides at $45^{\circ }$ to these directions becomes a rhombus: adjacent side lengths are unchanged to first order, but their right angle changes by $2e$. In axes rotated through $45^{\circ }$, the same strain tensor has zero diagonal entries and off-diagonal entries of magnitude $e$.

The pure shear from opposite principal strains therefore has engineering shear strain $\gamma =2e$, not $e$. Equal opposite principal stresses $(-S,+S)$ give purely tangential traction of magnitude $S$ on those rotated faces. HenceIf the compression and stretch are not equal and opposite, an additional dilatational strain remains; their deviatoric part still supplies shear.

The shaft's torsional stiffness. At radius $r$, an end twist $\varphi$ over length $L$ gives tangential displacement $r\varphi$ and shear angle $r\varphi /L$. The local shear stress is $Gr\varphi /L$. An annular area $dA=2\pi rdr$ contributes torque $r\tau dA$, soThus torsion of a solid circular shaft givesHere $J_p$ is the polar second moment of area, in ${\mathrm{m}}^4$, not a mass moment of inertia. This $C$ is the transmitted torque magnitude; the shaft's restoring couple opposes the twist. The formula assumes small elastic shear and a circular uniform section.

Take downward deflection $y(x)$ positive, with the support level $y=0$. The rectangular section second moment of area about the horizontal neutral axis isFor small slopes the curvature relation from part (b), with the stated sign convention, isIntegrating twice givesThe support boundary conditions $y(0)=y(L)=0$ give $C_2=0$ and $C_1=q{L}^3/24$. HenceThis is a downward displacement; its upward component is $-y(x)$. As checks, reflection symmetry gives $y^{\prime }(L/2)=0$, and the maximum downward deflection isThe units are length: $\rho g/(E{b}^2)$ has units ${\mathrm{m}}^{-3}$. The width $a$ cancels because both weight and bending resistance scale with it. The small-slope approximation requires $\rho g{L}^3/(E{b}^2)\ll 1$; shear strain is also neglected, as appropriate for a slender elastic beam.

The infinitesimal strain tensor $\mathrm{diag}(-e,e)$ compresses one principal direction and stretches the perpendicular one by the same amount. Rotating the coordinate axes through $45^{\circ }$ turns it into a tensor with zero normal strains and off-diagonal entries of magnitude $e$. The engineering shear strain is twice the tensor shear component, so its shear angle has magnitude $\gamma =2e$. Likewise principal stresses $(-S,S)$ give zero normal stress and shear stress of magnitude $S$ on those rotated faces. The shear modulus law $S=G\gamma$ therefore gives $S=2Ge$.

Engineering shear strain is the first-order change of the right angle between material lines initially directed along two perpendicular axes. For a displacement $\mathbf{u}$, ${\gamma }_{xy}={\partial }_y{u}_x+{\partial }_x{u}_y=2{\epsilon }_{xy}$, where ${\epsilon }_{xy}$ is the tensor strain component. Slender Euler-Bernoulli beam equation models neglect the transverse shear strain contribution to deflection.

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.