Solving for Work in a Stressed Aluminum Wire: Figure 12-56 Example

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In summary, the conversation involves a problem with finding the work done by a machine on an aluminum wire stretched in opposite directions. The problem provides information about the wire's initial length, cross-sectional area, and the scale of the stress axis. The conversation also mentions a formula for calculating work and asks for help in finding the answer. The conversation ends with a question about the elasticity modulus and a similar problem being faced by another individual.
  • #1
slydg895
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hey guys i have tried this problem many times and can't figure out the answer. please help

Heres the problem:
Figure 12-56 shows the stress versus strain plot for an aluminum wire that is stretched by a machine pulling in opposite directions at the two ends of the wire. The scale of the stress axis is set by s = 5.50, in units of 10^7 N/m2. The wire has an initial length of 0.950 m and an initial cross-sectional area of 2.50 × 10^-6 m2. How much work does the force from the machine do on the wire to produce a strain of 1.90 × 10^-3?

This is my work:
W=.5(stress)(strain)(volume)
volume=(area)(lenght)
And i get .124094.

What am i doing wrong?
 
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Does anyone have any ideas?
 
  • #3
slydg895 said:
Does anyone have any ideas?
It is unclear what is the Elasticity modulus or the slope of the stress - strain curve or the value of the strain at the given stress? You didn't attach a figure.
 
  • #4
I have a similar problem, my elasticity modulus (stress/strain) = 7E10.
 
  • #5


As a scientist, it is important to approach problems with a systematic and analytical mindset. It seems that you have attempted to solve the problem using the equation W=.5(stress)(strain)(volume), which is a good starting point. However, it is important to make sure that all of your units are consistent and that you have the correct values for each variable.

In this case, it seems like you have the correct values for stress and strain, but the units for stress should be in N/m^2, not 10^7 N/m^2. Additionally, it is important to double check your calculations to ensure that you are using the correct values for the area and length of the wire.

In order to solve for work, you can use the equation W = FΔx, where F is the force and Δx is the displacement. In this case, the force can be calculated using the stress and area of the wire, and the displacement can be calculated using the initial length of the wire and the strain.

I recommend going back through your calculations and double checking your units and values. If you are still having trouble, it may be helpful to ask a classmate or teacher for assistance. Remember to approach problems with a logical and systematic approach, and don't be afraid to ask for help when needed.
 

1. What is the purpose of solving for work in a stressed aluminum wire?

The purpose of solving for work in a stressed aluminum wire is to determine the amount of energy required to deform the wire and the amount of energy that is stored in the wire due to the deformation. This information is important in understanding the behavior of the wire under stress and can help engineers design more efficient and durable structures.

2. How is work calculated in a stressed aluminum wire?

Work is calculated by multiplying the force applied to the wire by the distance the wire is stretched or compressed. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

3. What factors affect the amount of work done on a stressed aluminum wire?

The amount of work done on a stressed aluminum wire is affected by the force applied to the wire, the distance the wire is stretched or compressed, and the properties of the wire such as its elasticity and strength. The temperature and environment in which the wire is stressed can also affect the amount of work done.

4. How is work related to the deformation of a stressed aluminum wire?

Work and deformation are directly related in a stressed aluminum wire. As the wire is deformed, work is done on the wire, resulting in the storage of energy. The amount of work done is directly proportional to the amount of deformation, meaning the more the wire is stretched or compressed, the more work is done and the more energy is stored.

5. How is solving for work in a stressed aluminum wire useful in real-world applications?

Solving for work in a stressed aluminum wire is useful in real-world applications because it allows engineers to accurately predict the behavior of the wire under different levels of stress. This information can be used to design structures that are more efficient and can withstand higher levels of stress without failure. It can also help in troubleshooting and identifying potential issues in existing structures made of aluminum wire.

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