A metal bar of 40 mm x 40 mm section is subjected to a tensile load of 320 kN. The extension of a 200 mm gauge length is found to be 0.2 mm and the decrease in thickness of 0.012 mm. The lateral and longitudinal strains are _________ respectively.
3 x 10-4, 10 x 10-4
Understanding how materials deform under load is fundamental in mechanics. When a tensile load is applied to a metal bar, it stretches in the direction of the load (longitudinally) and shrinks perpendicular to the load (laterally).
Strain is a measure of deformation. It is defined as the change in dimension divided by the original dimension. Strain is a dimensionless quantity, although sometimes it is expressed with units like mm/mm or as a percentage or in scientific notation like $10^{-4}$.
The longitudinal strain ($\varepsilon_L$) is calculated using the change in length ($\Delta L$) and the original gauge length ($L_0$).
Given:
The formula for longitudinal strain is:
$\varepsilon_L = \frac{\Delta L}{L_0}$
Substituting the given values:
$\varepsilon_L = \frac{0.2 \text{ mm}}{200 \text{ mm}}$
$\varepsilon_L = \frac{0.2}{200} = \frac{2 \times 10^{-1}}{2 \times 10^2} = 1 \times 10^{-1} \times 10^{-2} = 1 \times 10^{-3}$
In scientific notation asked in options, $1 \times 10^{-3}$ can also be written as $10 \times 10^{-4}$.
So, longitudinal strain is $10 \times 10^{-4}$.
The lateral strain ($\varepsilon_{Lat}$) is calculated using the change in thickness or width ($\Delta t$) and the original thickness or width ($t_0$).
Given:
The formula for lateral strain is:
$\varepsilon_{Lat} = \frac{\Delta t}{t_0}$
Substituting the given values:
$\varepsilon_{Lat} = \frac{0.012 \text{ mm}}{40 \text{ mm}}$
$\varepsilon_{Lat} = \frac{0.012}{40} = \frac{12 \times 10^{-3}}{40} = \frac{12}{40} \times 10^{-3} = 0.3 \times 10^{-3}$
In scientific notation, $0.3 \times 10^{-3}$ can also be written as $3 \times 10^{-4}$.
So, lateral strain is $3 \times 10^{-4}$.
The question asks for the lateral and longitudinal strains respectively.
Lateral strain $= 3 \times 10^{-4}$
Longitudinal strain $= 10 \times 10^{-4}$
Therefore, the lateral and longitudinal strains are $3 \times 10^{-4}$ and $10 \times 10^{-4}$ respectively.
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