Discussion
Which of the following statement(s) is/are true about the state of a body in plane strain condition?
P: All the points in the body undergo displacements in one plane only, for example the x-y plane, leading to \(\varepsilon_{zz} = \gamma_{xz} = \gamma_{yz} = 0\).
Q: All the components of stress perpendicular to the plane of deformation, for example the x-y plane, of the body are equal to zero, i.e. \(\sigma_{zz} = \tau_{xz} = \tau_{yz} = 0\).
R: Except the normal component, all the other components of stress perpendicular to the plane of deformation of the body, for example the x-y plane, are equal to zero, i.e. \(\sigma_{zz} \ne 0\), \(\tau_{xz} = \tau_{yz} = 0\).
Answer: D
The true statements are P and R.
Plane strain means the body deforms only in one plane, say x-y: the out-of-plane strains are zero, \(\varepsilon_{zz} = \gamma_{xz} = \gamma_{yz} = 0\). That is statement P.
Zero shear strains \(\gamma_{xz}\) and \(\gamma_{yz}\) give zero shear stresses \(\tau_{xz}\) and \(\tau_{yz}\). But stopping the normal strain \(\varepsilon_{zz}\) requires a normal stress: from Hooke's law, \(\varepsilon_{zz} = 0\) gives \(\sigma_{zz} = \nu(\sigma_{xx} + \sigma_{yy})\), which is generally not zero. That is statement R.
Statement Q, with \(\sigma_{zz}\) also zero, describes plane stress, the state of a thin plate, not plane strain. So 'Q only' and 'P and Q' are wrong, and 'P only' leaves out R, which is also true.
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