Plane-Stress Plate — Inverse Problem: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q2

Question 2: Plane-Stress Plate — Inverse Problem (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. Square plate, side $1.6$ m; biaxial plane stress ($\sigma_z=\tau=0$); $\delta_x=0.8$ mm, $\delta_y=0.25$ mm; $\sigma_y=60$ MPa; $E=200$ GPa.

Find. (a) $\sigma_x$ and Poisson’s ratio $v$; (b) the through-thickness strain $\varepsilon_z$.

Approach. Convert the measured elongations to strains, then invert the two-dimensional Hooke’s law: two equations in the two unknowns $\sigma_x$ and $v$ (eliminating $\sigma_x$ leaves a quadratic in $v$). Finally apply the out-of-plane Hooke relation for $\varepsilon_z$.

In-plane strains. Uniform stress over a side of $1600$ mm gives $$\varepsilon_x=\frac{0.8}{1600}=5.00\times10^{-4},\qquad \varepsilon_y=\frac{0.25}{1600}=1.5625\times10^{-4}.$$
Plane-stress Hooke’s law. With $\sigma_z=0$, $$E\varepsilon_x=\sigma_x-v\,\sigma_y=100\ \text{MPa},\qquad E\varepsilon_y=\sigma_y-v\,\sigma_x=31.25\ \text{MPa}.$$
Eliminate $\sigma_x$. From the first equation $\sigma_x=100+60v$; substituting into the second and rearranging, $$60\,v^{2}+100\,v-28.75=0\;\Rightarrow\;\boxed{v=0.25}$$ (the negative root $v=-1.92$ is inadmissible). This value is physically valid ($0\lt v\lt 0.5$).
Back-substitute for $\sigma_x$. $$\sigma_x=100+60(0.25)=\boxed{115\ \text{MPa (tension)}}.$$
Thickness strain. Out-of-plane, $\varepsilon_z=-\dfrac{v}{E}\left(\sigma_x+\sigma_y\right)$, $$\varepsilon_z=-\frac{0.25}{200\,000}\,(115+60)=-2.19\times10^{-4}.$$ The plate thins slightly, as expected under biaxial tension.

Quantity	Result
Poisson’s ratio, $v$	$0.25$
Normal stress $\sigma_x$	$115$ MPa (tension)
Thickness strain $\varepsilon_z$	$-2.19\times10^{-4}$ ($-218.8\ \mu\varepsilon$)

Topic: Inverting the plane-stress constitutive law from measured strains.

Key relations: $\varepsilon_x=(\sigma_x-v\sigma_y)/E$, $\varepsilon_y=(\sigma_y-v\sigma_x)/E$, $\varepsilon_z=-v(\sigma_x+\sigma_y)/E$.

Why this works: For a thin plate the faces are traction-free, so $\sigma_z=0$ and the state is plane stress. The two in-plane Hooke equations contain exactly the two unknowns here ($\sigma_x$ and $v$), so the problem is fully determined. Eliminating $\sigma_x$ produces a quadratic whose admissible root ($0\lt v\lt 0.5$) is the material constant; $\varepsilon_z$ then follows from the Poisson contraction driven by the sum of the in-plane stresses.

Common pitfall: Using the plane-strain relations (which carry extra $(1+v)$ factors) instead of plane stress, or accepting the negative/greater-than-0.5 root. Remember $\varepsilon_z\neq0$ even though $\sigma_z=0$—plane stress does not mean plane strain.

Source: Ugural & Fenster, Advanced Strength and Applied Elasticity 4th ed., Ch. 3 (generalized Hooke’s law); Hibbeler, Mechanics of Materials §10.6.

Question 2 of 8: Plane-Stress Plate — Inverse Problem

Question 2: Plane-Stress Plate — Inverse Problem (20 marks)