## Stress and Strain

A rectangular sample of an unknown material has an initial length of 10 cm, and a cross-section of 2.1 cm X 2.1 cm dimensions. The sample is pulled in tension along the length (with the deformation described below all in the elastic region).

- Calculate the strain if the sample is elongated to 10.5 cm.
- With a Poissonâ€™s ratio of 0.27, calculate the width of the sample if the bar is elongated to 10.5 cm under applied stress.
- Calculate the stress required to elongate the sample to 10.5 cm if the elastic modulus is 2.10 GPa.

##
__
__**Hint**

**Hint**

Strain is defined as change in length per unit length; for pure tension the following apply:

$$$\varepsilon_z=\frac{l-l_0}{l_0}$$$

where
$$\varepsilon$$
is engineering strain,
$$l$$
is the new length, and
$$l_0$$
is the initial length.

##
__
__**Hint 2**

**Hint 2**

$$$\sigma =E\varepsilon$$$

where
$$\sigma$$
is the stress on the cross section,
$$E$$
is the modulus of elasticity, and
$$\varepsilon$$
is the engineering strain.

Strain is defined as change in length per unit length; for pure tension the following apply:

$$$\varepsilon_z=\frac{l-l_0}{l_0}=\frac{10.5cm-10cm}{10cm}=\frac{0.5cm}{10cm}=0.05$$$

where
$$\varepsilon$$
is engineering strain,
$$l$$
is the new length, and
$$l_0$$
is the initial length. Calculating the sample width if elongated:

$$$v=0.27=\frac{-\varepsilon_x}{\varepsilon_z}\rightarrow \varepsilon_x=-v\varepsilon_z=-0.27(0.05)=-0.0135$$$

$$$\varepsilon_x=\frac{w-w_0}{w_0}\rightarrow w=\varepsilon_xw_0+w_0$$$

$$$w=w_0(\varepsilon_x+1)=2.1cm(-0.0135+1)=2.072\: cm$$$

Finding the stress for sample elongation,

$$$\sigma =E\varepsilon =2.10\cdot 10^3MPa(0.05)=105\:MPa$$$

where
$$\sigma$$
is the stress on the cross section,
$$E$$
is the modulus of elasticity, and
$$\varepsilon$$
is the engineering strain.

- 0.05
- 2.072 cm
- 105 MPa