Q) Part of a control linkage for an airplane consists of a rigid member CB and a flexible cable AB ' 800 mm 600 mm Originally the cable is unstretched. If a force is applied to the end of the member and causes it to rotate by θ = 0.45°, determine the normal strain in the cable. Express your answer to three significant figures. (I believe this is the complete question)
Answer:
Normal Strain = 0.00376 mm/mm
Explanation:
Here we are given with the length AC which is L_AB= 600 mm and the length BC which is L_BC = 800mm and the angle of rotation ∅ = 0.45°
First we will find the unstretched length AB by applying Pythagoras theorem:
L_AB = sqrt( (L_AC)^2 + (L_BC)^2)
L_AB = sqrt((600^2) + (800^2)
L_AB = 1000 mm
Now in order to find the stretched length we will use the cosine formula:
L_AB' = sqrt ( (L_AC)^2 + (L_AB)^2) - 2(L_AC)(L_AB)cos∅)
L_AB' = sqrt( (600)^2 + (800)^2 - 2(600)(800)cos(90+0.45))
L_AB' = 1003.762 mm
Now we will use the strain formula:
∈ = (L_AB' - L_AB)/L_AB
∈ = (1003.762 - 1000)/1000
∈ = 0.00376 mm/mm
