A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length

1.49 mm

Explanation:

The modulus of elasticity, Y = stress/strain = σ/ε

σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.

ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,

So, Y = σ/ε

Y = F/A ÷ ΔL/L

Y = FL/AΔL

making the change in length ΔL subject of the formula, we have

ΔL = FL/AY

substituting the value of A into the equation, we have

So, ΔL = FL/(πd²/4)Y

ΔL = 4FL/πd²Y

Since Y = 200 GPa = 200 × 10⁹ Pa

Substituting the values of the variables into the equation, we have

ΔL = 4FL/πd²Y

ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]

ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]

ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N

ΔL = 0.149 × 10⁻² m

ΔL = 1.49 × 10⁻³ m

ΔL = 1.49 mm

The change in length of the circular rod is 1.49 mm

answer:

1.3 + 20

explanation:

20.1.3