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Consider a solid disc of radius $R$ and height $h$ which we glue onto an infinite plane surface. If we now change the temperatur and the assume that

  1. the (linear) thermal expansion coefficient of the substrate vanishes, $\alpha_{substrate} = 0 \mu m /(K m)$,
  2. the (linear) thermal expansion coefficient of the disc is given by $\alpha>0$, and
  3. that the glue hold the disc and the substrate perfectly in place,

does anybody know a formula to describe the (free) surface of the disc? I'm not looking for finite element simulations, but a descriptive formula, which gives me some intuition about the shape. I expect, that the shape depends on all kind of properties, like e.g. Young's module $E$, and the height $h$. However, this problem seems to be fundamental for all kind of engineering work.

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Throughout the majority of the disc, the strains in the radial and hoop directions will be essentially zero, and the strain in the thickness direction will be uniform. Only at the very edge, within about 2h radially from the edge will this picture be disturbed. Within the broad central region, if you solve Hooke’s law 3D equations for this situation , you will find that the thickness strain is $$\epsilon =\frac{(1+\nu)}{(1-\nu)}\alpha \Delta T$$where $\nu$ is the Poisson ratio of the disc material.

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  • $\begingroup$ Will there be shear strain? $\endgroup$
    – aditya_stack
    Dec 13, 2019 at 13:16
  • $\begingroup$ If, by shear strain, you mean the off-diagonal elements of the strain tensor or, alternately, changes in the angles between material line segments originally lined up along the radial, hoop, and thickness directions, then no. Only in close proximity to the edges of the disc (within about 2h radially) will there be such shear strains. $\endgroup$
    – Chet Miller
    Dec 13, 2019 at 21:18
  • $\begingroup$ I am a high school student and do not know anything about tensors. But by shear strain I meant; as the bottom of the disc is fixed and the top is relatively free to move, would the shape of the disc change to be more frustum-like due to thermal stress? $\endgroup$
    – aditya_stack
    Dec 14, 2019 at 5:48
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    $\begingroup$ @aditya_stack The shear strain you refer to only occurs near the very edge of the disc. This shear strain is accompanied by a shear stress at the table surface near the edge. Since the radial stress in the disc is zero at its very edge, this shear stress allows the radial stress to build up within the disk edge region so that, throughout most of the disc, the radial stress is constant and prevents any disc thermal expansion in the radial direction. So throughout most of the disc, the only displacement is uniform, and in the thickness direction. $\endgroup$
    – Chet Miller
    Dec 14, 2019 at 13:26
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    $\begingroup$ @adita_stack Yes, over most of the disc. $\endgroup$
    – Chet Miller
    Dec 14, 2019 at 14:31

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