Strain Compatibility and Displacement Field: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q5

Question 5: Strain Compatibility and Displacement Field (20 marks)

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. $\varepsilon_x=c(-9x^2+21y^2)$, $\varepsilon_y=c(3x^2-15y^2)$, $\gamma_{xy}=3bxy$; displacements vanish at the origin; for part (b), $c=10$.

Find. (a) the constraint linking $b$ and $c$; (b) $u$ and $v$ at $(2,5)$.

Approach. Apply the 2-D Saint-Venant compatibility equation to fix $b$ in terms of $c$. Then integrate the strain–displacement relations, using the origin conditions (and no rigid-body rotation) to eliminate the integration functions.

Compatibility. The 2-D condition is $\dfrac{\partial^2\varepsilon_x}{\partial y^2}+\dfrac{\partial^2\varepsilon_y}{\partial x^2}=\dfrac{\partial^2\gamma_{xy}}{\partial x\,\partial y}$: $$42c+6c=3b\;\Rightarrow\;\boxed{b=16c}.$$
Integrate for $u$. From $\varepsilon_x=\partial u/\partial x$, $$u=\int c(-9x^2+21y^2)\,dx=c\left(-3x^3+21xy^2\right)+f(y).$$
Integrate for $v$. From $\varepsilon_y=\partial v/\partial y$, $$v=\int c(3x^2-15y^2)\,dy=c\left(3x^2y-5y^3\right)+g(x).$$
Fix the integration functions. Requiring $\partial u/\partial y+\partial v/\partial x=\gamma_{xy}=48c\,xy$ gives $f'(y)+g'(x)=0$; with $u(0,0)=v(0,0)=0$ and no rigid-body rotation, $f=g=0$.
Evaluate at $(2,5)$ with $c=10$. $$u=10\big(-3(8)+21(2)(25)\big)=\boxed{10\,260},\qquad v=10\big(3(4)(5)-5(125)\big)=\boxed{-5\,650}$$ (in the same length units as the coordinates).

Quantity	Result
Compatibility constraint	$b=16c$
$u(2,5)$	$10\,260$
$v(2,5)$	$-5\,650$

Topic: Strain compatibility and recovery of the displacement field by integration.

Key relations: Compatibility $\varepsilon_{x,yy}+\varepsilon_{y,xx}=\gamma_{xy,xy}$; strain–displacement $\varepsilon_x=u_{,x}$, $\varepsilon_y=v_{,y}$, $\gamma_{xy}=u_{,y}+v_{,x}$.

Why this works: Three strains derive from only two displacement functions, so they cannot be independent—the compatibility equation is exactly the condition that a single-valued, continuous $(u,v)$ exists. Once $b=16c$ makes the field compatible, integrating the strain definitions reconstructs $u$ and $v$ up to rigid-body terms, which the point conditions at the origin remove.

Common pitfall: Treating the integration “constants” as true constants—they are functions $f(y)$, $g(x)$ that must be pinned by the shear-strain relation and the boundary conditions. Forgetting the cross term in $\gamma_{xy}$ (both $u_{,y}$ and $v_{,x}$) leaves the constraint under-determined.

Source: Ugural & Fenster, Advanced Strength and Applied Elasticity 4th ed., Ch. 2–3 (compatibility, displacement integration); Boresi & Schmidt Ch. 2.

Question 5 of 8: Strain Compatibility and Displacement Field

Question 5: Strain Compatibility and Displacement Field (20 marks)