Tricksy thermal stress question

[Vuldoraq]

Homework Statement

A steel rod 0.350 {\rm m} long and an aluminum rod 0.250 {\rm m} long, both with the same diameter, are placed end to end between rigid supports with no initial stress in the rods. The temperature of the rods is now raised by 60.0 {\rm ^\circ C}.

What is the stress in each rod?

Youngs moduli:$$Steel=20*10^{10}, Aluminium= 7*10^{10}$$

Coef. of linear expansion: $$Steel=1.2*10^{-5}, Aluminium=2.4*20^{-5}$$

Homework Equations

$$\frac{F}{A}= Y(\frac{\Delta L}{L_{0}}-\alpha \Delta T)$$

The Attempt at a Solution

I am completely flumuxed, I know that that the total length of the two rods doesn't change but how do I find out which rod will expand and which be compressed? Any help getting started would be great.

 

[anathema]

Firstly, in order to solve this problem, we need to make a few assumptions. We will assume that the rods are uniform in composition and that the temperature increase is the same for both rods. We will also assume that the rods are free to expand and contract without any external forces acting on them.

Now, let's start by looking at the equation you have provided. This is known as Hooke's Law, which states that the stress (F/A) in a material is directly proportional to its Young's modulus (Y) and the strain, which is the change in length (ΔL) divided by the original length (L0), minus the coefficient of thermal expansion (α) times the change in temperature (ΔT).

Based on this equation, we can see that the stress in each rod will depend on three factors: the Young's modulus, the coefficient of thermal expansion, and the change in temperature. Since the rods have the same diameter, we can assume that they have the same cross-sectional area (A). Therefore, the only variables that will affect the stress in each rod are the Young's modulus and the coefficient of thermal expansion.

Now, let's look at the values given for each material. The Young's modulus for steel is 20*10^10, which is significantly higher than the Young's modulus for aluminum, which is 7*10^10. This means that steel is much stiffer than aluminum and will require more force to produce the same amount of strain. Additionally, the coefficient of thermal expansion for steel is 1.2*10^-5, which is half of the coefficient of thermal expansion for aluminum, which is 2.4*10^-5. This means that steel will expand or contract less than aluminum for the same change in temperature.

Based on these values, we can conclude that the stress in the steel rod will be higher than the stress in the aluminum rod. This is because the steel rod is stiffer and will require more force to produce the same amount of strain, and it will expand or contract less for the same change in temperature.

To find the exact stress in each rod, we can plug in the given values into the equation and solve for F/A. Since we are not given the exact dimensions of the rods, we cannot calculate the actual stress in each rod. However, we can still determine which rod will experience a higher stress based on the material properties.

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[Moetasim]

Firstly, it is important to note that the expansion or contraction of a material is directly related to its coefficient of linear expansion. A higher coefficient of linear expansion means that the material will expand more for a given change in temperature.

To solve this problem, we can use the formula for thermal stress:

\frac{F}{A}= Y(\frac{\Delta L}{L_{0}}-\alpha \Delta T)

Where F is the stress, A is the cross-sectional area, Y is the Young's modulus, \Delta L is the change in length, L_{0} is the original length, \alpha is the coefficient of linear expansion, and \Delta T is the change in temperature.

We know that the total length of the rods remains constant, so we can set up the following equation:

L_{total} = L_{steel} + L_{aluminum}

L_{0, total} + \Delta L_{total} = L_{0, steel} + \Delta L_{steel} + L_{0, aluminum} + \Delta L_{aluminum}

Since the initial stress in the rods is zero, we can also say that the total stress on the rods is equal to zero. So, we can set up another equation:

F_{total} = F_{steel} + F_{aluminum} = 0

Now, we can plug in the given values for the coefficients of linear expansion and Young's moduli for steel and aluminum into the first equation and solve for F_{steel} and F_{aluminum}.

\frac{F_{steel}}{A} = (20*10^{10})(\frac{\Delta L_{steel}}{0.350}-1.2*10^{-5}*60)

\frac{F_{aluminum}}{A} = (7*10^{10})(\frac{\Delta L_{aluminum}}{0.250}-2.4*10^{-5}*60)

We also know that the cross-sectional area is the same for both rods, as stated in the problem. So, we can set \frac{F_{steel}}{A} = \frac{F_{aluminum}}{A} and solve for \Delta L_{steel} and \Delta L_{aluminum}.

\frac{\Delta L_{steel}}{0.350}-1.2*10^{-5}*60 = \frac{\Delta L_{al