Answer:
change in length = 0.0913 in
Explanation:
given data
length = 6 ft
diameter = 0.2 in
load of w = 200 lb/ft
solution
we apply here first equilibrium moment about C that is express as
∑M(c) = 0 .............1
so it can express as to get force in AB
10× 200 × ( 5) - (T cos(30)) × 10 = 0
solve it we get
Tension is wire T(AB) = 1154.7 lb
and we know modulus of elasticity will be here for A992
E = 29000 ksi
and area will be
Area = [tex]\frac{\pi }{4}\times 0.2^2[/tex]
so change in length will be express as
change in length = [tex]\frac{PL}{AE}[/tex] ................2
put here value and we get
change in length = [tex]\frac{1154.7 \times 6 \times 12}{\frac{\pi }{4}\times 0.2^2 \times 29000 \times 1000}[/tex]
change in length = 0.0913 in
The length by which the guy wire stretches is; δ = 0.091 in
We are given;
Length of guy wire AB = 6ft = 72 in
Diameter of wire = 0.2 in
UDL = 200 lb/ft
From the free body diagram of the rigid beam system and the guy wire attached, we can say that;
Taking moments about point C gives;
∑Mc = 0; -(2000 * 5) + (F_BA * cos 30 * 10) = 0
-10000 + 8.66F_BA = 0
8.66F_BA = 10000
F_BA = 10000/8.66
F_BA = 1154.73 lb
Now, the length which it stretches is gotten from the formula;
δ = PL/AE
Where;
δ is change in length
P is Force
L is length of wire
A is area
E is modulus of elasticity of wire(steel) = 29000 ksi = 29000000 psi
A = πd²/4
A = π * 0.2²/4
A = 0.0314 in²
Thus;
δ = (1154.73 * 72)/(0.0314 * 29000000)
δ = 0.091 in
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