Rhe image is missing and i have attached it
Answer:
A) Tension in A (Pa) = 121.38N and Tension in B (Pb) = 75.86N
B) Deflection at B = 13.64 x 10^(-3) inches
Explanation:
A) Diameter = 1/16 inches.
Area = π(1/16)²/4 = π/1024
Deflection is given by a formula ;
δi = PLi/EA
A = π/1024
E = 29 x 10^(6) psi
Li is length AE on the image = 15 inches.
Thus,
δA = (Pa x 15)/(29 x 10^(6) x (π/1024))
Similarly,
δC = (Pc x 8)/(29 x 10^(6) x (π/1024))
δA = (24/16)δB = (3/2)δB
δC = 8/16δB = (1/2)δB
Thus,
(3/2)δB = (Pa x 15)/(29 x 10^(6) x (π/1024))
δB = 2/3[(Pa x 15)/(29 x 10^(6) x (π/1024))]
Also,
(1/2)δB = (Pc x 8)/(29 x 10^(6) x (π/1024))
δB = 2[(Pc x 8)/(29 x 10^(6) x (π/1024))]
Equating both δB, we have,
2/3[(Pa x 15)/(29 x 10^(6) x (Ï€/1024))] = 2[(Pc x 8)/(29 x 10^(6) x (Ï€/1024))]
Thus,
(2/3)Pa x 15 = 2(8Pc)
So,
Pa = (8/5)Pc
Thus,
(8/5)Pc(3/2) + (1/2)Pc = 220
Multiply through by 10;
24Pc + 5Pc = 2200
29Pc = 2200
Pc = 2200/29 = 75.86N
Pa = (8/5)Pc
So, Pa = (8/5) x 75.86 = 121.38N
B) from earlier,
δC = (Pc x 8)/(29 x 10^(6) x (π/1024))
So, δC = (75.86 x 8)/(29 x 10^(6) x (π/1024)) = 6.82 x 10^(-3)
Also,we saw that δC = (1/2)δB
Thus,
6.82 x 10^(-3) = (1/2)δB
δB = 2 x 6.82 x 10^(-3) = 13.64 x 10^(-3) inches
