## Shear Stress & Strain

Consider a copper test sample with a modulus of elasticity of 117 GPa is under a shear stress of 10 GPa. If the sample experiences a 0.2 shear strain, what is its Poisson’s ratio?

Hint
$$G=\frac{E}{2(1+\nu )}$$$where $$G$$ is the shear modulus or modulus of rigidity, $$E$$ is the modulus of elasticity, and $$\nu$$ is Poisson’s ratio. Hint 2 The shear modulus (modulus of rigidity) is a shearing force’s coefficient of elasticity. It is the ratio of shear stress to the displacement per unit length (shear strain). $$G=\frac{\tau }{\gamma }$$$
where $$\tau$$ is shear stress and $$\gamma$$ is shear strain.
The shear modulus (modulus of rigidity) is a shearing force’s coefficient of elasticity. It is the ratio of shear stress to the displacement per unit length (shear strain).
$$G=\frac{\tau }{\gamma }$$$where $$\tau$$ is shear stress and $$\gamma$$ is shear strain. $$G=\frac{10GPa}{0.2 }=50GPa$$$
Recall that Poisson’s ratio measures a material’s deformation in directions perpendicular to the loading/pressure direction:
$$\upsilon=-\frac{\varepsilon_{lateral}}{\varepsilon_{longitudinal}}$$$where $$\varepsilon_{lateral}$$ is the lateral strain, and $$\varepsilon_{longitudinal}$$ is the longitudinal strain. The relationship between the shear modulus and Poisson’s ratio is: $$G=\frac{E}{2(1+\nu )}$$$
where $$G$$ is the shear modulus or modulus of rigidity, $$E$$ is the modulus of elasticity, and $$\nu$$ is Poisson’s ratio.
$$50GPa=\frac{117GPa}{2(1+\nu )}$$$Solving for Poisson’s ratio: $$(1+\nu )=\frac{117}{(2)50}=1.17$$$
$$\nu=1.17-1=0.17$$\$
0.17