## 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**

**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**

**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