Question

Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB. Originally the cable is unstretched. If a force is applied to the end B of the member and causes it to rotate by ?=0.58?, determine the normal strain in the cable.

Answer 1

Concepts and reason

Strain: When the alloy is stressed in specific direction the response of the system for this stress is strain. The deformation of an alloy in specified direction divided by the original length gives the strain the alloy.

The deformation in a rigid body is represented by deformation diagram.

When load is applied in a specific direction the body deforms in that direction.

The change in position of a point under action of load is called as displacement.

Fundamentals

Calculate the normal strain in a body using following relation:

$\varepsilon = \frac{{\delta l}}{l}$

Here, the change in length is $\delta l$ and the original length is $l$ .

For small angles calculate the arc length using the following relation:

$\Delta = R\theta$

Here, the radius of the arc is $R$ .

Cosine rule:

${a^2} = {b^2} + {c^2} - 2bc\cos A$

Draw the deformation diagram for the rigid body AB.

Calculate the length of the cable AB when no load is acting.

${L_{AB}} = \sqrt {{{\left( {{L_{AC}}} \right)}^2} + {{\left( {{L_{BC}}} \right)}^2}}$

Here, the length of control link $AC$ is ${L_{AC}}$ and the length of control link $BC$ is

${L_{BC}}$ .

Substitute $600\,{\rm{mm}}$ for ${L_{AC}}$ and $800\,{\rm{mm}}$ for ${L_{BC}}$ .

$\begin{array}{c}\\{L_{AB}} = \sqrt {{{600}^2} + {{800}^2}} \\\\ = 1000\,{\rm{mm}}\\\end{array}$

Calculate the length of the cable ${L_{AB'}}$ , when there is load P acting at D using cosine rule.

${L_{AB'}} = \sqrt {{L_{AC}}^2 + {L_{CB}}^2 - 2\left( {{L_{AC}}} \right)\left( {{L_{CB}}} \right)\cos \left( {90 + \theta } \right)}$

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Substitute $600\,{\rm{mm}}$ for ${L_{AC}}$ , $800\,{\rm{mm}}$ for ${L_{BC}}$ , and ${\rm{0}}{\rm{.58}}^\circ {\rm{ for }}\theta$ .

$\begin{array}{c}\\{L_{AB'}} = \sqrt {{{600}^2} + {{800}^2} - 2\left( {600} \right)\left( {800} \right)\cos \left( {90 + 0.58} \right)} \\\\ = 1004.84\,{\rm{mm}}\\\end{array}$

From the strain relation, write the expression for strain in the cable AB.

$\begin{array}{c}\\{\varepsilon _{AB}} = \frac{{{\rm{change}}\,{\rm{in}}\,{\rm{length}}}}{{{\rm{original}}\,{\rm{length}}}}\\\\{\varepsilon _{AB}} = \frac{{{L_{AB'}} - {L_{AB}}}}{{{L_{AB}}}}\\\end{array}$

Substitute $1004.84\,{\rm{mm}}$ for ${L_{AB'}}$ and $1000\,{\rm{mm for }}{L_{AB}}$ .

$\begin{array}{c}\\{\varepsilon _{AB}} = \frac{{1004.84 - 1000}}{{1000}}\\\\ = 0.00484\,{\rm{mm/mm}}\\\end{array}$

Ans:

Therefore, the normal strain in the cable is $0.00484\,{\rm{mm/mm}}$ .