The rigid bar AD is supported by two steel wires of 1 16-in. diameter (E 5 29 3 106 psi) and a pin and bracket at A. Knowing that the wires were initially taut, determine (a) the additional tension in each wire when a 220-lb load P is applied at D, (b) the corresponding deflection of point D.

Rhe image is missing and i have attached it

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 (π/1024))] = 2[(Pc x 8)/(29 x 10^(6) x (π/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

d is the answer i came up with