Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question

2 Answers

Zero strain implies zero internal stress but you can still have external stress or volume forces applied on it. The equation of motion of continuum mechanics are: $$\rho\partial_t^2u_i = C_{ijkl}\nabla_j\nabla_ku_{l} + f_i$$ with $u(\vec{x},t)$ the displacement to the equilibrium position, $C_{ijkl}$ the stiffness matrix, $\rho$ the mass density, $f_i = \nabla_j\tau_{ij}$ the volume forces derived from an applied external (as in not a consequence of Hooke's law and the natural elasticity of your solid) stress $\tau_{ij}$.

share|improve this 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.

share|improve this answer
I see... but in the linear elasticity case, zero strain does produce zero stress, right? – Paul Feb 19 at 21:58
@Paul You are right. However, the OP is asking whether he "can always assume ...", which is not true for all metterials such as those that exhibit elastic hysteresis loop. – JKL Feb 19 at 22:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.