NivaarExam PrepOfficial exam papers ↗

22-Mec-A7 Advanced Strength of Materials · May 2013

Question 6 of 8: Plane-Stress Square Plate — Back-Calculating σ x , v, ε z

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

Notes on this paper

Paper format: National Exams, May 2013 — 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours; any five of the eight problems constitute a complete paper and all problems are of equal value. All eight problems are solved as a study resource.

Reference texts: A. C. Ugural & S. K. Fenster, Advanced Strength and Applied Elasticity, 4th ed.; A. P. Boresi & R. J. Schmidt, Advanced Mechanics of Materials, 6th ed.; R. C. Hibbeler, Mechanics of Materials, 10th ed.; S. P. Timoshenko & J. M. Gere, Theory of Elastic Stability, 2nd ed.

Question 6: Plane-Stress Square Plate — Back-Calculating σx, v, εz (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 1 m × 1 m plate in biaxial plane stress.

Given data
ElongationsΔx = 0.4 mm, Δy = 0.1 mm ⇒ εx = 400×10−6, εy = 100×10−6
Known stressσy = 100 MPa
ModulusE = 80 GPa

Find. (a) σx; (b) v; (c) εz.

σx σy
Thin square plate under biaxial tension σx, σy (plane stress).

Approach. Write the two in-plane Hooke’s-law equations. With σy and both strains known, the two equations contain the two unknowns σx and v; solve simultaneously, then obtain the through-thickness strain.

  1. Plane-stress Hooke’s law. $$E\varepsilon_x=\sigma_x-v\sigma_y=32\ \text{MPa},\qquad E\varepsilon_y=\sigma_y-v\sigma_x=8\ \text{MPa}.$$
  2. Eliminate σx. From the first, $\sigma_x=32+100v$; substitute into the second, $v(32+100v)=92$, i.e. $$25v^2+8v-23=0.$$
  3. Solve. The positive root is $$\boxed{v=0.812,\qquad \sigma_x=32+100v=113.2\ \text{MPa}.}$$
  4. Thickness strain. In plane stress $\varepsilon_z=-\dfrac{v}{E}(\sigma_x+\sigma_y)$: $$\varepsilon_z=-\frac{0.812}{80\,000}(113.2+100)=\boxed{-2.17\times10^{-3}.}$$
Question 6 — results (as given data yield)
QuantityValue
σx113.2 MPa
Poisson’s ratio v0.812 (see note)
εz−2.17×10−3

Check: the data as printed give v = 0.812, which exceeds the thermodynamic limit v = 0.5 for an isotropic material (it would imply a negative bulk modulus). The exam numbers are therefore internally inconsistent; a physically admissible v ≈ 0.3 would require a y-elongation near 1.0 mm rather than 0.1 mm. The solution above follows the printed data exactly; the flagged inconsistency should be stated in the answer per the exam’s “state your assumptions” instruction.