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 first thing you should do is develop a <u>budget</u> to determine what vehicle you can afford.
<h3>What is an automobile?</h3>
An automobile is also referred to as a vehicle, car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.
<h3>What is a budget?</h3>
A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one year.
In this context, we can reasonably infer and logically deduce that the first thing anyone should do is to develop a <u>budget</u> in order to determine what vehicle they can afford.
Read more on budget here: brainly.com/question/13964173
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