25-Nav-A5 Ship Design · May 2016

Question 4 of 8: Thin Plate in Biaxial Stress — Elastic Constants and Yield

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 4: Thin Plate in Biaxial Stress — Elastic Constants and Yield (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. Square plate, plane stress ($\sigma_z=0$).

Given data
Quantity	Value
Plate side	1.25 m × 1.25 m
Elongation, $x$	1.95 mm
Elongation, $y$	0.20 mm
$\sigma_x$	200 MPa
$E$	110 GPa

Find. (a) $\sigma_y$, $\nu$; (b) Tresca yield strength.

Square plate under uniform biaxial normal stress (plane stress, $\sigma_z=0$).

Approach. Convert elongations to strains, write the two plane-stress Hooke's-law equations for $\varepsilon_x,\varepsilon_y$, solve the resulting pair simultaneously for $\sigma_y$ and $\nu$, then apply Tresca to the three principal stresses ($\sigma_x,\sigma_y,\sigma_z=0$).

Strains from the given elongations. $$\varepsilon_x=\frac{1.95}{1250}=1.560\times10^{-3}\qquad \varepsilon_y=\frac{0.20}{1250}=1.600\times10^{-4}$$
Plane-stress Hooke's law (two equations, two unknowns $\sigma_y,\nu$). $$\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E}\qquad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E}$$ Eliminating $\sigma_y=E\varepsilon_y+\nu\sigma_x$ from the second equation and substituting into the first gives a quadratic in $\nu$: $200\nu^2+17.6\nu-28.4=0$ (coefficients in MPa), whose physically valid root ($0<\nu<0.5$) is $$\boxed{\nu=0.3354}$$ $$\boxed{\sigma_y=E\varepsilon_y+\nu\sigma_x=17.60+0.3354(200)=84.68\text{ MPa}}$$
Part (b) — Tresca yield strength. Plane stress gives a third principal stress $\sigma_z=0$. The three principal stresses are $\sigma_x=200$, $\sigma_y=84.68$, $\sigma_z=0$ MPa, so the extremes are $\sigma_x$ (max) and $\sigma_z$ (min): $$\boxed{\sigma_{\text{yield}}=\sigma_{\max}-\sigma_{\min}=200-0=200\text{ MPa}}$$

Question 4 — final results
Quantity	Value
$\sigma_y$	84.68 MPa
Poisson's ratio $\nu$	0.3354
Tresca yield strength	200 MPa

Topic: Plane-stress Hooke's law and the maximum-shear-stress (Tresca) yield criterion.

Key relations: $\varepsilon_x=(\sigma_x-\nu\sigma_y)/E$, $\varepsilon_y=(\sigma_y-\nu\sigma_x)/E$; Tresca $\sigma_y^{\text{yield}}=\sigma_{\max}-\sigma_{\min}$ over all three principal stresses including the through-thickness $\sigma_z=0$.

Why this works: Two measured strains and one known stress leave exactly two unknowns ($\sigma_y,\nu$) in the two Hooke's-law equations, which is a well-posed (if quadratic) simultaneous system; once the in-plane stresses are known, Tresca requires ranking ALL THREE principal stresses — not just the two in-plane ones — because the unloaded through-thickness direction in plane stress is still a principal stress and, being zero here, sits between $\sigma_y$ and $\sigma_x$ rather than being the true minimum in a signed sense.

Common pitfall: using only $\sigma_x-\sigma_y$ for the Tresca criterion in plane stress and forgetting $\sigma_z=0$ is itself a principal stress that can control the maximum shear (it does not change the answer here only because $\sigma_z=0$ is already the minimum, but must always be checked).

Source: Hibbeler, Mechanics of Materials, Ch.10 (plane stress, generalized Hooke's law); Ch.10 App. (failure theories).

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