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).
  1. Calculate the strain if the sample is elongated to 10.5 cm.
  2. With a Poisson’s ratio of 0.27, calculate the width of the sample if the bar is elongated to 10.5 cm under applied stress.
  3. Calculate the stress required to elongate the sample to 10.5 cm if the elastic modulus is 2.10 GPa.

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
$$$\sigma =E\varepsilon$$$
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.
  1. 0.05
  2. 2.072 cm
  3. 105 MPa