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Question: 

A sign (mass 1700 kg) hangs from the bottom end of a vertical steel girder with a cross-sectional area of $0.012 \textrm{ m}^2$.

  1. What is the stress within the girder?
  2. What is the strain on the girder?
  3. If the girder is 9.50 m long, how much is it lengthened? (Ignore the mass of the girder itself.)

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 
  1. $1.4 \times 10^6 \textrm{ N/m}^2$
  2. $6.9 \times 10^{-6}$
  3. $6.6 \times 10^{-5} \textrm{ m}$

Giancoli 7th Edition, Chapter 9, Problem 43

(1:16)

Chapter 9, Problem 43 is solved.

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Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The stress on this steel girder is the force applied to it divided by its cross-sectional area. So the thing that's hanging from it is 1700 kilograms, we multiply that by 9.8 newtons per kilogram to get the weight and divide by the cross-sectional area of the steel girder of 0.12 meters squared and that gives 1.4 times 10 to the 6 newtons per square meter. Strain is stress divided by elastic modulus and the elastic modulus for steel is 200 times 10 to the 9 newtons per meter squared. So we take the stress that we calculated in part (a) and divide that by the elastic modulus to get about 6.9 times 10 to the minus 6. And since strain is the change in length of something divided by its original length, we can rearrange that for Δl by multiplying both sides by l naught and we get Δl is strain times original length. And so the change in length will be this strain that we found before times 9.5 meters original length and it's gonna stretch by 6.6 times 10 to the minus 5 meters.