25-Nav-A5 Ship Design · May 2016

Question 6 of 8: Two-Dimensional Strain Field — Compatibility and Displacement Integration

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams, May 2016 — 98-Mar-A5 Advanced Strength of Materials, 3 hours, open book, non-communicating calculator permitted (any five of the eight problems constitute a complete paper, all equal value; all eight answered below for full study coverage).

Reference texts: Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed.; Hibbeler, Mechanics of Materials, 10th ed.; Timoshenko & Goodier, Theory of Elasticity, 3rd ed.

It is solved as the strength-of-materials exam it actually is.

Question 6: Two-Dimensional Strain Field — Compatibility and Displacement Integration (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. $\varepsilon_x=c(-4.5x^2+10.5y^2)$, $\varepsilon_y=c(1.5x^2-7.5y^2)$, $\gamma_{xy}=1.5bxy$, $u=v=0$ at the origin.

Find. (a) the relation between $b$ and $c$; (b) $u(3,7),v(3,7)$ with $c=2.5$.

Approach. Impose the 2-D compatibility equation on the given strain field to fix $b$ in terms of $c$, then integrate the strain-displacement relations directly (the compatible field guarantees a single-valued $u,v$), fixing the two integration "constants" (functions of the other coordinate) from $\partial u/\partial y+\partial v/\partial x=\gamma_{xy}$ and the stated boundary condition at the origin.

Part (a) — compatibility. The 2-D strain compatibility condition is $$\frac{\partial^2\varepsilon_x}{\partial y^2}+\frac{\partial^2\varepsilon_y}{\partial x^2}=\frac{\partial^2\gamma_{xy}}{\partial x\,\partial y}$$ Left side: $\partial^2\varepsilon_x/\partial y^2=21c$, $\partial^2\varepsilon_y/\partial x^2=3c$, sum $=24c$. Right side: $\partial^2(1.5bxy)/\partial x\partial y=1.5b$. Equating, $$\boxed{b=16c}$$
Part (b) — integrate for u. $\varepsilon_x=\partial u/\partial x$, so integrating in $x$ (the "constant" is a function $f(y)$): $$u=\int\varepsilon_x\,dx=c(-1.5x^3+10.5xy^2)+f(y)$$
Integrate for v. $\varepsilon_y=\partial v/\partial y$: $$v=\int\varepsilon_y\,dy=c(1.5x^2y-2.5y^3)+g(x)$$
Fix $f(y),g(x)$ from the shear strain and the origin condition. Substituting into $\partial u/\partial y+\partial v/\partial x=\gamma_{xy}=1.5bxy=24cxy$ (using $b=16c$) gives $24cxy+f'(y)+g'(x)=24cxy$, so $f'(y)=-g'(x)=$ const. Taking the standard no-rigid-rotation solution (the simplest field consistent with $u=v=0$ at the origin) sets that constant to zero, so $f\equiv g\equiv0$: $$u(x,y)=c(-1.5x^3+10.5xy^2)\qquad v(x,y)=c(1.5x^2y-2.5y^3)$$
Evaluate at (3, 7) with $c=2.5$. $$u=2.5\left[-1.5(27)+10.5(3)(49)\right]=2.5(1503.0)$$ $$\boxed{u(3,7)=3757.5}$$ $$v=2.5\left[1.5(9)(7)-2.5(343)\right]=2.5(-763.0)$$ $$\boxed{v(3,7)=-1907.5}$$

Question 6 — final results
Quantity	Value
Compatibility relation	$b=16c$
$u(3,7)$, $c=2.5$	3757.5 (length units consistent with $x,y$)
$v(3,7)$, $c=2.5$	−1907.5 (length units consistent with $x,y$)

Check: the source gives no units for $x,y$ or $c$ (a common convention for pure strain-compatibility exercises); $u,v$ are reported as numbers consistent with whatever length unit $x,y$ use. The origin condition $u(0,0)=v(0,0)=0$ fixes two of the three constants of integration; the third (a rigid-body rotation about the origin) is conventionally taken as zero, since the problem gives no additional rotational boundary condition.

Topic: Strain compatibility (2-D) and integration of the strain-displacement relations.

Key relations: compatibility $\partial^2\varepsilon_x/\partial y^2+\partial^2\varepsilon_y/\partial x^2=\partial^2\gamma_{xy}/\partial x\partial y$; strain-displacement $\varepsilon_x=\partial u/\partial x$, $\varepsilon_y=\partial v/\partial y$, $\gamma_{xy}=\partial u/\partial y+\partial v/\partial x$.

Why this works: An arbitrarily-assumed strain field generally has no corresponding single-valued displacement field; the compatibility equation is precisely the condition that removes this obstruction, and once satisfied, direct integration of $\varepsilon_x,\varepsilon_y$ produces $u,v$ up to two undetermined functions of the "other" coordinate, which the shear-strain equation collapses to at most a rigid-body translation/rotation — exactly the freedom that a stated boundary condition (here, $u=v=0$ at the origin) is meant to remove.

Common pitfall: integrating $\varepsilon_x,\varepsilon_y$ but forgetting to check the shear-strain equation, leaving an unresolved rigid-body rotation term that silently changes both $u$ and $v$ away from the origin.

Source: Timoshenko & Goodier, Theory of Elasticity, Ch.2 (compatibility equations); Boresi & Schmidt, Advanced Mechanics of Materials, Ch.2 (strain-displacement relations).

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