Plate under biaxial stress — plane stress vs. plane strain: 22-Mec-A7 Advanced Strength of Materials, December 2019, Q1

Question 1: Plate under biaxial stress — plane stress vs. plane strain

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. An in-plane stress state on a thin plate, with the elastic constants from which Poisson's ratio follows.

Given data
Quantity	Symbol	Value
Normal stress, x	$\sigma_{xx}$	65 MPa
Normal stress, y	$\sigma_{yy}$	100 MPa
Shear stress	$\tau_{xy}$	75 MPa
Young's modulus	$E$	200 GPa
Shear modulus	$G$	77 GPa

Find. $\varepsilon_{xx},\ \varepsilon_{yy},\ \varepsilon_{zz},\ \sigma_{zz},\ \gamma_{xy}$ under (A) plane stress ($\sigma_{zz}=0$) and (B) plane strain ($\varepsilon_{zz}=0$).

Approach. Recover Poisson's ratio from $E$ and $G$, then apply the 3-D generalized Hooke's law, imposing the defining constraint of each condition.

Poisson's ratio from the two moduli. The isotropic relation $G=\dfrac{E}{2(1+\nu)}$ inverts to $$\nu=\frac{E}{2G}-1=\frac{200}{2(77)}-1=\boxed{0.299}$$
Plane-stress condition (A): set $\sigma_{zz}=0$. Hooke's law for each normal strain, $\varepsilon_{ii}=\tfrac{1}{E}\left[\sigma_{ii}-\nu(\sigma_{jj}+\sigma_{kk})\right]$, gives $$\varepsilon_{xx}=\frac{65-0.299(100)}{200\,000}=1.757\times10^{-4}$$ $$\varepsilon_{yy}=\frac{100-0.299(65)}{200\,000}=4.029\times10^{-4}$$ $$\varepsilon_{zz}=\frac{-\,0.299(65+100)}{200\,000}=-2.464\times10^{-4}$$ The through-thickness strain is non-zero (the plate is free to thin), while $\sigma_{zz}=0$ by definition.
Plane-strain condition (B): set $\varepsilon_{zz}=0$. The constraint forces an out-of-plane normal stress $$\sigma_{zz}=\nu(\sigma_{xx}+\sigma_{yy})=0.299(165)=49.3\ \text{MPa}$$ Re-evaluating the in-plane strains with this $\sigma_{zz}$, $$\varepsilon_{xx}=\frac{65-0.299(100+49.3)}{200\,000}=1.020\times10^{-4}$$ $$\varepsilon_{yy}=\frac{100-0.299(65+49.3)}{200\,000}=3.293\times10^{-4}$$ with $\varepsilon_{zz}=0$ by definition.
Shear strain is identical for both cases. Shear couples only to $\tau_{xy}$: $$\gamma_{xy}=\frac{\tau_{xy}}{G}=\frac{75}{77\,000}=9.740\times10^{-4}$$

Results — Question 1
Quantity	(A) Plane stress	(B) Plane strain
$\varepsilon_{xx}$	$1.757\times10^{-4}$	$1.020\times10^{-4}$
$\varepsilon_{yy}$	$4.029\times10^{-4}$	$3.293\times10^{-4}$
$\varepsilon_{zz}$	$-2.464\times10^{-4}$	$0$ (imposed)
$\sigma_{zz}$	$0$ (imposed)	$49.3$ MPa
$\gamma_{xy}$	$9.740\times10^{-4}$	$9.740\times10^{-4}$

Topic. Generalized (3-D) Hooke's law and the two 2-D idealizations.

Key relations. $\nu=\tfrac{E}{2G}-1$; $\varepsilon_{ii}=\tfrac{1}{E}[\sigma_{ii}-\nu(\sigma_{jj}+\sigma_{kk})]$; plane stress $\Rightarrow\sigma_{zz}=0$; plane strain $\Rightarrow\varepsilon_{zz}=0,\ \sigma_{zz}=\nu(\sigma_{xx}+\sigma_{yy})$.

Why this works. A thin plate loaded in its plane carries no traction on its faces, so $\sigma_{zz}=0$ (plane stress). A long prism or a slice fully restrained axially cannot strain out of plane, so $\varepsilon_{zz}=0$ (plane strain) and a reactive $\sigma_{zz}$ appears.

Common pitfall. Assuming the two conditions give the same in-plane strains — the reactive $\sigma_{zz}$ in plane strain measurably reduces $\varepsilon_{xx}$ and $\varepsilon_{yy}$.

Source. Ugural & Fenster, Advanced Strength and Applied Elasticity, Ch. 3 (Hooke's law; plane stress and plane strain).

Question 1 of 7: Plate under biaxial stress — plane stress vs. plane strain

Question 1: Plate under biaxial stress — plane stress vs. plane strain