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The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress in the aluminum and the steel due to thisloading.The pipe has an outer diameter of 80 mm and aninner diameter of 70mm.

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klefenz

Answer:

In the steel: 815 kPa

In the aluminum: 270 kPa

Explanation:

The steel pipe will have a section of:

A1 = π/4 * (D^2 - d^2)

A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core:

A2 = π/4 * d^2

A2 = π/4 * 0.7^2 = 0.3848 m^2

The parts will have a certain stiffness:

k = E * A/l

We don't know their length, so we can consider this as stiffness per unit of length

k = E * A

For the steel pipe:

E = 210 GPa (for steel)

k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum:

E = 70 GPa

k2 = 70*10^9 * 0.3848 = 2.69*10^10 N

Hooke's law:

Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:

ε = f / k

When the force is distributed between both materials will stretch the same length:

f = f1 + f2

f1 / k1 = f2/ k2

Replacing:

f1 = f - f2

(f - f2) / k1 = f2 / k2

f/k1 - f2/k1 = f2/k2

f/k1 = f2 * (1/k2 + 1/k1)

f2 = (f/k1) / (1/k2 + 1/k1)

f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN

f1 = 200 - 104 = 96 kN

Then we calculate the stresses:

σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa

σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

The answer is: within the steel: 815 kPa

When within the aluminum: 270 kPa

The aluminum

When The steel pipe will have a piece of:Then A1 = π/4 * (D^2 - d^2)After that A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2

The aluminum core is: Now A2 = π/4 * d^2Then A2 = π/4 * 0.7^2 = 0.3848 m^2

After that The parts will have a particular stiffness:k = E * A/l

We don't know their length, so we are able to consider this as stiffness per unit of length k = E * A

For the steel pipe: E = 210 GPa (for steel) k1 = 210*10^9 * 0.1178 = 2.47*10^10 N

For the aluminum: E = 70 GPak2 = 70*10^9 * 0.3848 = 2.69*10^10 NHooke's law:Δd = f / k

Since we are using stiffness per unit of length we use stretching per unit of length:ε = f / k

When the force is distributed between both materials will stretch the identical length:f = f1 + f2f1 / k1 = f2/ k2

Replacing: f1 = f - f2(f - f2) / k1 = f2 / k2f/k1 - f2/k1 = f2/k2f/k1 = f2 * (1/k2 + 1/k1)f2 = (f/k1) / (1/k2 + 1/k1)f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KNf1 = 200 - 104 = 96 kN

Then we calculate the stresses:σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPaσ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa

Find out more information about Aluminum here:

brainly.com/question/4229590

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