Inverse plane-stress problem for a square plate: 22-Mec-A7 Advanced Strength of Materials, May 2016, Q4

Question 4: Inverse plane-stress problem for a square plate (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. Side L = 1.25 m = 1250 mm; elongations Δ_x = 1.95 mm, Δ_y = 0.20 mm, so ε_x = 1.56 × 10⁻³, ε_y = 1.6 × 10⁻⁴; σ_x = 200 MPa; E = 110 GPa; plane stress (σ_z = 0).

Find. (a) σ_y and ν; (b) the yield strength under Tresca.

Approach. The two plane-stress Hooke’s-law equations contain the two unknowns σ_y and ν. Eliminating σ_y gives a quadratic in ν; keep the admissible root (ν < 0.5). Then apply Tresca to the resulting principal stresses.

Plane-stress Hooke’s law. $$E\varepsilon_x=\sigma_x-\nu\sigma_y=171.6,\qquad E\varepsilon_y=\sigma_y-\nu\sigma_x=17.6$$ From the first, νσ_y = 28.4 MPa; from the second, σ_y = 17.6 + 200ν.
Quadratic in ν. Substitute: $$\nu(17.6+200\nu)=28.4 \;\Rightarrow\; 200\nu^2+17.6\nu-28.4=0$$ $$\nu=\frac{-17.6+\sqrt{17.6^2+4(200)(28.4)}}{400}=0.335$$ $$\boxed{\nu=0.335\ (\text{admissible, }\nu\lt 0.5)}$$
Back-substitute for σ_y. $$\sigma_y=17.6+200(0.335)=84.7\ \text{MPa}\quad(\text{check: }\nu\sigma_y=28.4\ \checkmark)$$ $$\boxed{\sigma_y=84.7\ \text{MPa}}$$
Tresca yield strength (part b). The principal stresses are σ₁ = 200, σ₂ = 84.7, σ₃ = 0 (out-of-plane). Maximum shear governs on the largest difference: $$\sigma_Y=\sigma_1-\sigma_3=200-0 \;\Rightarrow\; \boxed{\sigma_Y=200\ \text{MPa}}$$

Question 4 — results
Quantity	Value
σ_y	84.7 MPa
Poisson’s ratio ν	0.335
Yield strength (Tresca)	200 MPa

Topic. Back-calculating material constants and stress from measured biaxial strains (inverse plane-stress problem).

Key relations. ε_x = (σ_x−νσ_y)/E, ε_y = (σ_y−νσ_x)/E; Tresca σ_Y = σ_max−σ_min.

Why this works. Two independent strain measurements give two equations in the two remaining unknowns, so the state is fully determined. Because ν multiplies an unknown stress, elimination produces a quadratic rather than a linear equation; physical admissibility (0 < ν < 0.5 for isotropic materials) selects the correct root. In plane stress the third principal stress is zero, and because both in-plane stresses here are tensile, the maximum shear spans from the largest tensile stress down to that zero — so the Tresca yield strength equals σ_x.

Common pitfall. Dropping the out-of-plane zero stress when applying Tresca (using σ_x−σ_y instead of σ_x−0 gives the wrong, smaller value), or keeping the inadmissible negative/large root of the quadratic.

Source. Boresi & Schmidt Ch. 3 (generalised Hooke’s law); Hibbeler §10.6 (yield criteria).

Question 4 of 8: Inverse plane-stress problem for a square plate

Question 4: Inverse plane-stress problem for a square plate (equal value)