Compatibility and displacements of a strain field: 22-Mec-A7 Advanced Strength of Materials, May 2016, Q6

Question 6: Compatibility and displacements of a strain field (equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. ε_x = c(−4.5x²+10.5y²), ε_y = c(1.5x²−7.5y²), γ_xy = 1.5bxy; displacements vanish at the origin; c = 2.5.

Find. (a) the compatibility relation b(c); (b) u and v at (3, 7).

Approach. Enforce the 2-D strain compatibility equation to relate b and c, then integrate the strain–displacement relations, using the origin condition (and vanishing rigid-body rotation) to fix the integration functions.

Compatibility — part (a). The single 2-D compatibility condition is $$\frac{\partial^2\varepsilon_x}{\partial y^2}+\frac{\partial^2\varepsilon_y}{\partial x^2}=\frac{\partial^2\gamma_{xy}}{\partial x\,\partial y}$$ Substituting the derivatives (21c + 3c on the left, 1.5b on the right): $$21c+3c=1.5b \;\Rightarrow\; \boxed{b=16c}$$
Integrate ε_x for u. $$u=\int\varepsilon_x\,dx=c\!\left(-1.5x^3+10.5xy^2\right)+f(y)$$
Integrate ε_y for v. $$v=\int\varepsilon_y\,dy=c\!\left(1.5x^2y-2.5y^3\right)+g(x)$$
Use the shear strain to fix f and g. With b = 16c so that 1.5b = 24c, $$\gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=21cxy+3cxy+f'(y)+g'(x)=24cxy$$ so f′(y) + g′(x) = 0; both are constants (rigid-body terms). With u = v = 0 and no rotation at the origin, f = g = 0.
Evaluate at (3, 7) with c = 2.5. $$u=2.5(-1.5\cdot27+10.5\cdot3\cdot49)=2.5(1503)=\boxed{3757.5}$$ $$v=2.5(1.5\cdot9\cdot7-2.5\cdot343)=2.5(-763)=\boxed{-1907.5}$$ (values in the length units implied by the constant c and the coordinates).

Question 6 — results
Quantity	Value
Compatibility relation	b = 16c
u(3, 7)	3757.5
v(3, 7)	−1907.5

Topic. Strain compatibility and recovery of the displacement field from strains.

Key relations. Compatibility ∂²ε_x/∂y² + ∂²ε_y/∂x² = ∂²γ_xy/∂x∂y; strain–displacement ε_x = ∂u/∂x, ε_y = ∂v/∂y, γ_xy = ∂u/∂y + ∂v/∂x.

Why this works. A strain field cannot be arbitrary: three strains derive from only two displacement functions, so they must satisfy one compatibility equation, which here fixes b = 16c. Once compatibility holds, integrating the two normal-strain relations recovers u and v to within functions of the other variable; the shear-strain relation then constrains those functions to a rigid-body motion, removed entirely by the pinned, non-rotating origin. The result is a single-valued, continuous displacement field.

Common pitfall. Solving part (b) with an incompatible field (b ≠ 16c) — the shear-strain equation then cannot be satisfied and no continuous displacement exists. Also, forgetting that the integration “constants” are functions of the other coordinate.

Source. Ugural & Fenster Ch. 2 (strain, compatibility); Boresi & Schmidt Ch. 2.

Question 6 of 8: Compatibility and displacements of a strain field

Question 6: Compatibility and displacements of a strain field (equal value)