Science:Math Exam Resources/Courses/MATH110/April 2013/Question 08

MATH110 April 2013

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[hide] Question 08
In mechanics, a force exerted on a steel rod is called stress, and denoted by σ. The deformation of the rod due to this stress is called strain, and denoted ε. Stress and strain are related by the Ramberg-Osgood equation $\displaystyle \epsilon =a\sigma +b\sigma ^{n}$ where a, b and n are constants. Suppose the stress applied to a steel rod increases at a constant rate of R. Determine the rate at which the strain is increasing when the stress is equal to σ₀. Hint: your answer will be an expression involving several constants.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

[show] Hint
Note the mention of rates of change - this will be a related rates problem. Before you differentiate, make sure you know which values in the given equation are constants and which are variables.

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[show] Solution
The question is asking us for the rate at which strain is increasing. Since strain is indicated by the variable $\epsilon$ , this means we are seeking ${\frac {d\epsilon }{dt}}$ . To find ${\frac {d\epsilon }{dt}}$ we differentiate the given formula with respect to t. Remember that a, b and n are constants, while $\epsilon$ and $\sigma$ are variables that depend on t. So $\displaystyle \epsilon (t)=a\sigma (t)+b(\sigma (t))^{n}$ becomes ${\frac {d\epsilon }{dt}}(t)=a{\frac {d\sigma }{dt}}(t)+bn(\sigma (t))^{n-1}{\frac {d\sigma }{dt}}(t)$ We were asked to find ${\frac {d\epsilon }{dt}}$ where $\sigma (t)=\sigma _{0}$ , so we can plug that value into the expression above. Finally, recall we are also told earlier in the problem that the stress is increasing at a constant rate R. In the expression above, the rate of stress increasing is given by ${\frac {d\sigma }{dt}}(t)$ , so we replace ${\frac {d\sigma }{dt}}(t)$ with R. This gives our final answer of: ${\frac {d\epsilon }{dt}}=aR+bn\sigma _{0}^{n-1}R$