Plane-stress plate — back-out of σ x and E: 22-Mec-A7 Advanced Strength of Materials, December 2018, Q1

Question 1: Plane-stress plate — back-out of σ_x and E (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 square plate, side $L=1.5\text{ m}$, in plane stress ($\sigma_z=\tau=0$).

Given data
Side length $L$	1.5 m (both directions)
Elongation in x, $\delta_x$	0.75 mm
Elongation in y, $\delta_y$	0.25 mm
Transverse stress $\sigma_y$	60 MPa
Poisson's ratio $\nu$	0.32

Find. (a) the in-plane stress $\sigma_x$ and Young's modulus $E$; (b) the through-thickness strain $\varepsilon_z$.

Approach. Convert the elongations to strains, write the two plane-stress Hooke's-law equations, take their ratio to eliminate $E$ and solve for $\sigma_x$, then back out $E$; finally use the plane-stress relation for $\varepsilon_z$.

Strains from the elongations. Uniform stress gives uniform strain, so $$\varepsilon_x=\frac{\delta_x}{L}=\frac{0.75}{1500}=5.00\times10^{-4},\qquad \varepsilon_y=\frac{\delta_y}{L}=\frac{0.25}{1500}=1.667\times10^{-4}.$$
Plane-stress Hooke's law. With $\sigma_z=0$, $$\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E},\qquad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E}.$$
Eliminate E by taking the ratio. Dividing the two equations removes $E$ and leaves one equation in $\sigma_x$: $$\frac{\varepsilon_x}{\varepsilon_y}=\frac{\sigma_x-\nu\sigma_y}{\sigma_y-\nu\sigma_x}\;\Longrightarrow\; \sigma_x=\sigma_y\,\frac{\varepsilon_x+\nu\varepsilon_y}{\varepsilon_y+\nu\varepsilon_x}.$$ Substituting the numbers, $$\sigma_x=60\cdot\frac{5.00\times10^{-4}+0.32(1.667\times10^{-4})}{1.667\times10^{-4}+0.32(5.00\times10^{-4})}=\boxed{101.6\text{ MPa}}.$$
Modulus from the x-equation. $$E=\frac{\sigma_x-\nu\sigma_y}{\varepsilon_x}=\frac{101.6-0.32(60)}{5.00\times10^{-4}}=\boxed{164.9\text{ GPa}}.$$
Through-thickness strain. In plane stress the z-strain is driven by the two in-plane stresses: $$\varepsilon_z=-\frac{\nu}{E}\left(\sigma_x+\sigma_y\right)=-\frac{0.32}{164\,900}\,(101.6+60)=\boxed{-3.14\times10^{-4}}.$$ The plate thins, as expected under net in-plane tension.

Final results — Question 1
Quantity	Value
Stress $\sigma_x$	101.6 MPa (tension)
Young's modulus $E$	164.9 GPa
Thickness strain $\varepsilon_z$	$-3.14\times10^{-4}$ (contraction)

Topic. Inverse plane-stress elasticity — recovering material constants from measured strains.

Key relations. Plane-stress Hooke's law $\varepsilon_x=(\sigma_x-\nu\sigma_y)/E$, $\varepsilon_y=(\sigma_y-\nu\sigma_x)/E$, and $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$.

Why this works. Two independent strain measurements give two equations in the two unknowns $\sigma_x$ and $E$ (with $\sigma_y$, $\nu$ known). Taking the ratio of the equations cancels $E$, isolating $\sigma_x$; a single Hooke equation then yields $E$. Because $\sigma_z=0$ in a thin plate, the thickness strain is not independent — it is fixed by the two in-plane stresses through Poisson coupling, so no extra measurement is needed to obtain $\varepsilon_z$.

Common pitfall. Writing $\varepsilon_z=-\nu\varepsilon_x$ (a uniaxial shortcut). In a biaxial field $\varepsilon_z$ depends on both in-plane stresses; use $-\nu(\sigma_x+\sigma_y)/E$. Also watch the millimetre-to-metre conversion in the strains.

Source. Boresi & Schmidt, Advanced Mechanics of Materials, Ch. 3 (Linear stress–strain relations); Hibbeler, Mechanics of Materials, §10.6.

Question 1 of 8: Plane-stress plate — back-out of σ x and E

Question 1: Plane-stress plate — back-out of σx and E (20 marks)

Question 1: Plane-stress plate — back-out of σ_x and E (20 marks)