Plane-Stress Square Plate — Inverse Determination: 22-Mec-A7 Advanced Strength of Materials, December 2013, Q2

Question 2: Plane-Stress Square Plate — Inverse Determination (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. A 2 m square plate in biaxial plane stress with measured edge elongations and one known stress.

Given data
Plate size	2 m × 2 m (L = 2000 mm each side)
Elongations	δ_x = 0.8 mm ⇒ ε_x = 4.0×10⁻⁴; δ_y = 0.2 mm ⇒ ε_y = 1.0×10⁻⁴
Known stress / modulus	σ_y = 200 MPa, E = 80 GPa
Stress state	σ_z = τ = 0 (plane stress)

Find. (a) σ_x; (b) ν; (c) ε_z.

Biaxial plane stress; σ_x tensile (horizontal), σ_y = 200 MPa tensile (vertical).

Approach. Two biaxial Hooke’s-law equations relate (ε_x, ε_y) to (σ_x, ν) with σ_y and E known. Eliminating σ_x leaves a quadratic in ν; the thickness strain then follows from the plane-stress form of ε_z.

Biaxial Hooke’s law. With Eε written out, $$E\varepsilon_x=\sigma_x-\nu\sigma_y=32\ \text{MPa},\qquad E\varepsilon_y=\sigma_y-\nu\sigma_x=8\ \text{MPa}.$$
Eliminate σ_x. From the first equation σ_x = 32 + 200ν; substituting into the second, $$\nu(32+200\nu)=192\ \Longrightarrow\ 25\nu^{2}+4\nu-24=0.$$
Solve for ν (part b). The positive root is $$\boxed{\nu=\frac{-4+\sqrt{16+2400}}{50}=0.903.}$$
Back-substitute for σ_x (part a). $$\boxed{\sigma_x=32+200(0.903)=212.6\ \text{MPa}.}$$
Thickness strain (part c). In plane stress the out-of-plane strain is $$\varepsilon_z=-\frac{\nu}{E}\,(\sigma_x+\sigma_y)=-\frac{0.903}{80\,000}(212.6+200)=\boxed{-4.66\times10^{-3}.}$$

Check — data inconsistency. The measured elongations force ν = 0.903, which exceeds the isotropic upper bound ν < 0.5 (an isotropic material with ν → 0.5 is incompressible; ν > 0.5 gives a negative bulk modulus). The given δ_x, δ_y, σ_y and E are therefore not mutually consistent for a real isotropic plate — a data flaw in the printed exam. Following the exam’s “state your assumptions” rubric, the algebra is carried through as posed (the method is what is examined); the numerical ν and ε_z should be reported with the note that they are non-physical. Had δ_y been, say, 0.35 mm, the data would return an admissible ν ≈ 0.28.

Results — Question 2 (as-given data)
Quantity	Value
σ_x	212.6 MPa (tension)
Poisson’s ratio ν	0.903 (inadmissible — see note)
Thickness strain ε_z	−4.66×10⁻³

Topic: Generalized Hooke’s law in plane stress and the inverse problem — recovering stresses and elastic constants from measured strains.

Key relations: ε_x = (σ_x − νσ_y)/E, ε_y = (σ_y − νσ_x)/E, and ε_z = −ν(σ_x+σ_y)/E. Two in-plane measurements are enough to solve for two unknowns.

Why this works: for a biaxial stress state the three strains are linear in the two non-zero stresses through ν and E; with σ_y and E supplied, the system closes. The thickness strain is non-zero even though σ_z = 0 — the Poisson contraction of the in-plane tensions.

Common pitfall: assuming ε_z = 0 (that is plane strain, not plane stress), or accepting an out-of-range ν without checking ν < 0.5. Physically-impossible constants signal inconsistent input data, not an arithmetic error.

Source: Hibbeler, Mechanics of Materials 10th ed., §10.6 (generalized Hooke’s law; dilatation and bulk modulus bounds on ν).

Question 2 of 7: Plane-Stress Square Plate — Inverse Determination

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