Question

The pin - connected rigid rods AB and BC are inclined at theta = 30 when they are unloaded. When the force P is applied theta becomes 30.2. Determine the average normal strain developed in wire AC.

Answer 1

Concepts and reason

Normal strain is a measure of longitudinal deformation that is elongation or contraction in the body. Normal strain is denoted by $\varepsilon$ .

Normal stress can be calculated using known values of Young’s modulus and normal strain.

Normal stress is used in various real time applications.

Fundamentals

Force is a vector quantity. It is represented by a magnitude, direction, and point of application.

The direction of force can be either compressive or tensile.

The normal strain is calculated as follows:

From the figure, initial length is ${L_1}$ and the final length is ${L_2}$ .

The normal strain is,

$\varepsilon = \frac{{{L_2} - {L_1}}}{{{L_1}}}$

Write the relation to calculate initial length of wire AC.

${L_1} = 2L\sin {\theta _1}$

Here, the length of the rod is L and the angle between the rods before load is applied is ${\theta _1}$ .

Substitute $0.6\,{\rm{m}}$ for L and $30^\circ$ for ${\theta _1}$ .

$\begin{array}{c}\\{L_1} = 2 \times 0.6 \times \sin 30^\circ \\\\ = 0.6\,{\rm{m}}\\\end{array}$

Write the relation to calculate final length of wire AC.

${L_2} = 2L\sin {\theta _2}$

Here, the angle between the rods after load is applied is ${\theta _2}$ .

Substitute $0.6\,{\rm{m}}$ for L and $30.2^\circ$ for ${\theta _1}$ .

$\begin{array}{c}\\{L_2} = 2 \times 0.6 \times \sin 30.2^\circ \\\\ = 0.60362\,{\rm{m}}\\\end{array}$

Write the relation to calculate the average normal strain developed in the wire AC.

$\varepsilon = \frac{{{L_2} - {L_1}}}{{{L_1}}}$

Substitute $0.60362\,{\rm{m}}$ for ${L_2}$ and $0.6\,{\rm{m}}$ for ${L_1}$ .

$\begin{array}{c}\\\varepsilon = \frac{{0.60362 - 0.6}}{{0.6}}\\\\ = 6.033 \times {10^{ - 3}}\,\\\end{array}$

Ans:

The average normal strain developed in the wire AC is $6.033 \times {10^{ - 3}}\,$ .