A commonly used method of fastening one part to another part is called "shrink fitting." A steel rod has a diameter of 2.0046 cm, and a flat plate contains a hole whose diameter is 2.0000 cm. The rod is cooled so that it just fits into the hole. When the rod warms up, the enormous thermal stress exerted by the plate holds the rod securely to the plate. By how many Celsius degrees should the rod be cooled?
delta T = delta L / (coefficient of thermal expansion) * (original length)
delta T = 0.000046 / (0.02) * (0.000012) = 191.67 degrees Celsius
Well, we can't let that poor steel rod become too "cool" for school, can we? Let's keep it chill with some calculations!
To find out how many Celsius degrees the rod needs to be cooled, we'll use the formula for thermal expansion:
Change in length = original length * coefficient of linear expansion * change in temperature
However, in this case, we want the difference in diameter, so we'll use the same formula, but with the coefficient of linear expansion doubled.
Doubling the coefficient of linear expansion (α) for the rod, we get:
ΔD = 2 * D * α * ΔT
Given:
D = 2.0046 cm
ΔD = 0.0046 cm (since it needs to shrink from 2.0046 to 2.0000)
α = coefficient of linear expansion of steel (approx. 12 x 10^-6 cm/°C)
Now, let's solve for ΔT:
ΔT = ΔD / (2 * D * α)
= 0.0046 cm / (2 * 2.0046 cm * 12 x 10^-6 cm/°C)
Calculating this, we find that the rod needs to be cooled by approximately 192.3°C.
So, you could say the rod needs to "chill out" by about 192.3 degrees Celsius to perfectly fit into the hole. Stay cool, Mr. Rod!
To determine how many Celsius degrees the rod should be cooled, we can use the principle of thermal expansion and the concept of shrink fitting.
The basic idea is that the rod will contract when cooled, allowing it to fit into the hole in the plate. Then, when the rod warms up, it will expand and create a tight fit with the plate.
The difference in diameter between the rod and the hole is given as 2.0046 cm - 2.0000 cm = 0.0046 cm.
Now, let's calculate the change in diameter of the rod due to cooling. The coefficient of linear expansion for steel is typically around 12 x 10^-6 per °C.
change in diameter = (coefficient of linear expansion) * (original diameter) * (change in temperature)
0.0046 cm = (12 x 10^-6 per °C) * (2.0046 cm) * (change in temperature)
Simplifying the equation, we have:
change in temperature = 0.0046 cm / ((12 x 10^-6 per °C) * (2.0046 cm))
change in temperature ≈ 0.0046 cm / (2.4052 x 10^-5 °C^-1)
change in temperature ≈ 190.993 °C
Therefore, the rod should be cooled by approximately 190.993 degrees Celsius in order to fit into the hole in the plate using the shrink fitting method.
To determine by how many Celsius degrees the rod should be cooled, we need to calculate the thermal expansion of the rod and the plate.
The thermal expansion of a material can be calculated using the equation:
ΔL = αLΔT
Where:
ΔL is the change in length or diameter
α is the coefficient of linear expansion
L is the initial length or diameter
ΔT is the change in temperature
In this case, the rod has a diameter of 2.0046 cm and the plate has a hole with a diameter of 2.0000 cm. We can assume the coefficient of linear expansion (α) for steel to be approximately 12 x 10^-6 1/°C.
Let's calculate the change in diameter for both the rod and the plate to determine the required change in temperature to achieve shrink fitting.
For the rod:
ΔL_rod = α_rod * L_rod * ΔT
ΔL_rod = (12 x 10^-6 1/°C) * (2.0046 cm) * ΔT
For the plate:
ΔL_plate = α_plate * L_plate * ΔT
ΔL_plate = (12 x 10^-6 1/°C) * (2.0000 cm) * ΔT
To achieve shrink fitting, the change in diameter of the rod should match the change in diameter of the plate. Therefore:
ΔL_rod = ΔL_plate
(12 x 10^-6 1/°C) * (2.0046 cm) * ΔT = (12 x 10^-6 1/°C) * (2.0000 cm) * ΔT
ΔT = (2.0000 cm - 2.0046 cm) / [(12 x 10^-6 1/°C) * (2.0046 cm - 2.0000 cm)]
ΔT ≈ 0.0046 cm / (12 x 10^-6 1/°C)
Calculating this, we find that ΔT ≈ 383.3 °C.
Therefore, the rod should be cooled by approximately 383.3 °C to achieve shrink fitting with the plate.