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Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ = W / A
we substitute
σ = 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ - σ
)A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb