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

`e_(ab) = 4.18*10^(-3)`

Step-by-step 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 = √(600^2 + 800^2) = 1000 mm`

We calculate distance A and B'

`AB' = √((600 - (-6.9815))^2 + (0-800)^2) = 1004.2 mm`

Normal Strain in AB is:

`e_(ab) = (AB' - AB)/(AB) \\\\= (1004.2 - 1000)/(100)\\\\= 4.18 * 10^(-3)`

The solution is :

`e_(ab) = 4.18 * 10^(-3)`