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

Answer:

$e_{ab} = 4.18*10^(-3)$

Explanation:

This question will be solved with the help of diagram (see attachment)

Given:

Δ∅ = 0.50 degrees (correction)

BC = 800 mm

AC = 600 mm

Solution:

We use coordinate system with point C as origin (0,0)

Hence,

Point A = (600,0)

Point B = (0,800)

The change in length or displacement can be calculated BB' :

BB' = BC * tan (Δ∅) = (800 mm) * tan (0.5) = 6.9815 mm

Hence, Point B' = (-6.9815,800) and we calculate distance A and B

$AB = \sqrt{600^2 + 800^2} = 1000 mm$

We calculate distance A and B'

$AB' = \sqrt{(600 - (-6.9815))^2 + (0-800)^2} = 1004.2 mm$

Normal Strain in AB is:

$e_{ab} = \frac{AB' - AB}{AB} \\\\= \frac{1004.2 - 1000}{100}\\\\= 4.18 * 10^(-3)$

The solution is :

$e_{ab} = 4.18 * 10^(-3)$