Answer:
[tex]e_{ab} = 4.18*10^(-3)[/tex]
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
[tex]AB = \sqrt{600^2 + 800^2} = 1000 mm[/tex]
We calculate distance A and B'
[tex]AB' = \sqrt{(600 - (-6.9815))^2 + (0-800)^2} = 1004.2 mm[/tex]
Normal Strain in AB is:
[tex]e_{ab} = \frac{AB' - AB}{AB} \\\\= \frac{1004.2 - 1000}{100}\\\\= 4.18 * 10^(-3)[/tex]
The solution is :
[tex]e_{ab} = 4.18 * 10^(-3)[/tex]
