Problem
The surface x3 = 0 of a large, flat body made of a material having elastic Young’s modulus
E, Poisson ratio ν, and coefficient of thermal expansion α is subjected to a rapid thermal shock. 1 The results in either case can be summed up by the following observations:
• In a thin layer on and immediately below the surface x3 = 0:
– the temperature rapidly changes by an amount “ΔT”, where ΔT ≡ Tafter −Tbefore ;
– the surface x3 = 0 remains tractionfree.
• In the thick substrate beneath the surface layer: .
– the temperature changes negligibly: ΔT = 0; .
– the total strain remains very small: ij = 0ij . For
Assuming that the response of the material during the shock is isotropic linear thermal/elastic: 1. Argue why the inplane strain components 11, 22, and 12 in the thin surface layer should match the corresponding values in the substrate; namely, 11 = 22 = 12 = 0. (Hint: what would happen to the displacement field components ui if any of these strain components were discontinuous across the interface separating surface layer from substrate?)
2. Evaluate all components of the stress tensor σij in the substrate.
3. Evaluate all components of the stress tensor σij in the thin surface layer. Give “physical” reasons to justify the sign of any nonzero stress components, relative to the sign of ΔT.
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4. Evaluate all components of the strain tensor ij in the surface layer. Give “physical” reasons to justify the sign of any nonzero strain components, relative to the sign of ΔT.
5. Suppose that yielding of the material begins whenever the Mises stress measure σ reaches the value σy, which is the tensile yield strength of the material. Calculate the maximum value of Δ| | T which can be applied to the surface layer without having the Mises stress exceed the value “σy”.
Recall that in an isotropic linear thermal/elastic material, the relation between temperature change, ΔT, stress components, σij , and strain components, ij , is
You may consult your [correct] answer to Problem 2, above, in order to find an appropriate “inverted” form giving the stress components σij in terms of the strain components ij and the temperature change, ΔT
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Note: In this and in many other problems, you have information about some (but not all!) of the stress components, and about some (but not all!) of the strain components. In this case, you must engage in “handtohand combat” with the set of linear equations comprising the stress/strain/temp erature relations, obtaining new expressions for those components that you do not know, a priori, in terms of those components that you do know .....
Solution
The procedure is similar to what was discussed in class for elastic constitutive relations without thermal effects. The idea is to eliminate 3 from the given equation and k=1 σkk solve for σij . To do this, take the sum of both sides of the given equation as follows
1. The strain components are determined by the derivatives of displacement components. Since the displacement should be continuous in the interface separating surface layer from substrate, the in-plane strain components in the thin surface layer should correspond to those in the substrate, too. Therefore, 11 = 22 = 12 = 0.
2. The problem stipulates that the thick substrate undergoes negligible total strain and . . temperature change. It means ΔT = 0 and ij(substrate) = 0ij . The constitutive equation (3) gives the relation between stress components and strain components. In the equation, σij(substrate) can be calculated by substituting ij(substrate) and ΔT(which are all zeros).
3. We know that the surface layer has a boundary condition at surface x3 = 0 which remains traction-free (tn = 0).
The in-plane components 11, 22, and 12 are zero. By the equation (3),
In order to justify the sign of the non-zero stress components, σ11 and σ22, we can imagine that, when the temperature increases, the surface layers tends to extend while the substrate tends to prevent it from deforming by shrinking it. Therefore the non-zero stress components would be compression stresses. And similarly, when the temperature decreases, the two in-plane components are tensile stresses. The non-zero components σ11 and σ22 always have opposite sign from that of the change of temperature(ΔT).
4. By the equation (2) and the in-plane strain continuity( 11 = 22 = 12 = 0 ),
Since all the components of the stress tensor in equation (8) are known, the strain tensor can be calculated by the equation (2).
