Question

Resultado: 4

Answer 1

Answer:

The magnitude of the maximum load that can be applied to the wire is approximately  381703.51 N

The elongation to a 1.2 m length of the wire at the maximum load is approximately 8.696 mm

Explanation:

Result: 4

Calculate the magnitude of the maximum load

that can be applied to a

wire

made of tempered steel 1.8 cm in diameter

so as not to exceed its elastic limit;

also determine the elongation that will suffer

If the calculated maximum load is applied and

It has an initial length of 1.2m.

The given information are;

The diameter, D, of the tempered steel wire = 1.8 cm = 0.018 m

The initial length of the wire = 1.2 m

The cross-sectional area of the wire = π × D²/4 = π × (0.018)²/4 = 0.000254469 m²

When we take the yield strength for tempered steel as 1500 MPa, we have;

$Yield \ strength = \dfrac{Yield \ force}{Original \ cross-sectional \ area} = \dfrac{F}{A}$

Therefore;

F = (Yield strength) × (Original cross-sectional area)

F = 1,500 MPa × 0.000254469 m² ≈ 381703.51 N.

The magnitude of the maximum load that can be applied to the wire =  381703.51 N

2. We have the modulus of elasticity, E = 207 GPa

∴ E = Yield strength/(Elastic strength)

Elastic strength = Yield strength/(E) = 1500 MPa/207 GPa ≈ 7.25×10⁻³

Elastic strength = Δl/l = (Change in length)/(Original length)

∴ 7.25×10⁻³ = Δl/1.2

Δl = 1.2 × 7.25×10⁻³ ≈ 8.696×10⁻³ m = 8.696 mm

The elongation to a 1.2 m length of the wire at the maximum load ≈ 8.696 mm.