In solid mechanics, can I always assume that if an object undergoes no strain, then no stress is applied to it? I think it's true only because I can't seem to find a counter-example.
[Physics] Does zero strain always imply zero stress
elasticitysolid mechanicsstress-strain
Related Solutions
1. Yes, the relation $$\mathrm{stress}=d(\mathrm{strain\,energy\,density})/d(\mathrm{strain})$$ holds for all elastic bodies, not just linearly elastic bodies. This equation implies that all differential work goes into elastic strain energy, which holds even for nonlinearly elastic materials (e.g., hyperelastic materials). However, the equation wouldn't apply to plastic deformation, for example, in which substantial amounts of work are converted to heat and expended through the formation of crystal defects.
2. Regarding the intuition behind this equation, we can say that any way to add energy to a system involves two parameters (called thermodynamic conjugate variables): a generalized force and a generalized displacement. The first term is intensive; i.e., if you doubled the system size, then the generalized force would stay the same. The second term is extensive; if you doubled the system size, then this term would also double.
The simplest example of a generalized force and displacement is an actual force $F$ and displacement $x$ and the familiar equations $w=\boldsymbol{F\cdot x}$ and $dw=F\,dx$ for the work $w$. Another example is the pressure $P$ and volume $V$: $dw=-P\,dV$, with the minus sign appearing because pressure is compressive. Note how a gradient in pressure, the intensive variable, drives a shift in volume, the extensive variable. This effect is common for all of these pairs, whose units invariably multiply to give units of energy.
(This framework applies even to heating: the system energy $U$ increases with $T\,dS$, where gradients in temperature $T$ drive shifts in the entropy $S$. Here again, the units multiply to give units of energy.)
Yet another example of a conjugate pair is the stress and strain. Well actually, this isn't entirely true. If you look at the units, you'll see that the product of stress and strain has units of volumetric energy. So we can work with the elastic strain energy density or what you call above the strain energy function $W$, or we can work in terms of energy by multiplying by the volume, as in the fundamental relation for a first-order closed system under a general mechanical load: $dU=T\,dS+\boldsymbol{\bar{\sigma}} V\,d\boldsymbol{\bar{\epsilon}}$, where $\boldsymbol{\bar{\sigma}}$ and $\boldsymbol{\bar{\epsilon}}$ are the stress and strain tensors, respectively. (If the load is pressure, or equitriaxial compressive stress, then we recover the familiar $dU=T\,dS-P\,dV$.)
3. As for deriving your starred equation, I checked Nye's Physical Properties of Crystals and Ugural & Fenster's Advanced Strength and Applied Elasticity, and they proceed as you do: define the increase in strain energy from a uniaxial load applied to a differential element and then build up to the complete 3D case. For an isotropic material (which obeys generalized Hooke's Law), for example, Ugural & Fenster obtain a strain energy density of $$W=\frac{1}{2E}\left(\sigma_{x}^2+\sigma_{y}^2+\sigma_{z}^2\right)-\frac{\nu}{2E}\left(\sigma_{x}\sigma_y+\sigma_{y}\sigma_z+\sigma_{x}\sigma_z\right)+\frac{1}{2G}\left(\tau_{xy}^2+\tau_{yz}^2+\tau_{xz}^2\right).$$
Does the strain induces stress or does the stress induces strain? Or is it both ways?
When we are dealing with mechanical stress and strain, which is what we deal with most frequently, it is normally the stress resulting from loading that induces strain.
On the other hand, if we are dealing with thermal stress, strain can induce stress if the body is constrained from expansion or contraction.
For example, take a metal bar and place the ends between two fixed surfaces. Heat the bar. When the bar attempts to elongate it will be constrained from doing so by the fixed surfaces resulting in compressive stress. Or take a heated bar and attach the ends to fixed surfaces. When the bar cools and attempts to shorten the constraints will prevent it from doing so, resulting in tensile stress.
Hope this helps.
Best Answer
Zero strain does not always imply zero stress and visa versa. There are matterials that display stress-strain, $\sigma-\epsilon,$ hysteresis behaviour. In matterials like this, when you start loading them, they behave normally, i.e increasing the stress increases the strain. However, when you start to unload them (remove the load), instead of the stress becoming zero when the strain becomes zero, the matterial has some residual stress applied to it! Similary, if you repeat the cycle, although the stress becomes zero the strain retains a permanent value, i.e. the matterial remains permanently deformed! These are very interesting elastic properties of such matterials. Stress-Strain hysteresis phenomena are very well known and are discussed extensively in literature.