The elastic energy stored in a wire

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SUMMARY

The discussion focuses on calculating the elastic energy stored in a steel wire and a lifting cable under specific loads. A steel wire with a diameter of 0.50mm and a length of 1.84m supports a 60N weight, resulting in an extension of 2.81mm. The elastic energy stored in the wire is calculated using the formula W = ½ force x extension, yielding a result of 84mJ. Additionally, a steel lifting cable with a diameter of 40mm lifting a 10kN weight extends by 2.98mm, confirming the calculations provided in the discussion.

PREREQUISITES
  • Understanding of Young's modulus and its application in material science.
  • Familiarity with the formula for calculating extension in materials: Δl=(lo F)/EA.
  • Knowledge of elastic potential energy calculations in physics.
  • Basic arithmetic skills for unit conversions and calculations.
NEXT STEPS
  • Study the implications of Young's modulus in different materials, particularly in construction and engineering.
  • Learn about the relationship between strain and stress in elastic materials.
  • Explore more complex calculations involving elastic potential energy in various contexts.
  • Investigate the significance of elastic limits and how they affect material performance under load.
USEFUL FOR

Students in physics or engineering, material scientists, and professionals involved in structural design and analysis will benefit from this discussion.

Littlemonkey
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A steel wire diameter 0.50mm and length 1.84m is suspended vertically from a fixed point and used to support a 60N weight hung from the lower end of the wire. The Young modulus for steel is 2.0 x 10^11 N m-2

a) The extension of the wire

b) The elastic energy stored in a wire given the elastic limit for steel is at a strain of 0.0025

Formula:
Δl=(lo F)/EA

a)Δl=(1.84m×60N)/((2.0×10^11 Nm-2) )×(Π×0.00025^2 ) ) = 2.81×10-3m or 2.81mm

I know the answer to a) is correct I found the book that the original question was taken from I also know the answer to b) is 84mJ but I have no clue where to even start any help would be greatly appreciated.

Also could someone check my calculations on the problem below as again I have the answer from the book but I cannot get the equation to work:

A crane used to lift iron girders on a building site has a steel lifting cable diameter 40 mm. When the crane lifts an iron girder of weight 10 kN, the cable length with the girder just off the ground is 75.0 m. Calculate
a) The extension of the cable for this length and load

Formula:
Δl=(lo F)/EA

Answer:
Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

The answer in the book is 2.98×10-3 m or 2.98 mm I just can't work out where I am going wrong.
 
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Littlemonkey said:
I also know the answer to b) is 84mJ but I have no clue where to even start any help would be greatly appreciated.
How do you calculate elastic potential energy? (Compare to a stretched spring.)

Answer:
Δl=(75m×10kN)/((2.0×10^11 Nm-2) )×(Π×0.02^2 ) ) = 2.98×10-4 m or .298 mm

The answer in the book is 2.98×10-3 m or 2.98 mm I just can't work out where I am going wrong.
Double check your arithmetic.
 
Oh I am a balloon I was conviced that 75*10kN was only 75000N not 750000N. One down one to to go.

Thanks
 
Ok I have the answer:

Formula:
W = ½ force x extension
Answer:
W = ½ × 60 N × 2.81×10-3m = 8.43×10-2J or 84 mJ

One last question and its why it took me so long to work it out. Why is the strain of 0.0025 mentioned when it does not seem to be relevant when working out this answer.
 
Littlemonkey said:
Why is the strain of 0.0025 mentioned when it does not seem to be relevant when working out this answer.
To calculate the elastic potential energy you need to be assured that the elastic limit is not exceeded. It doesn't factor into the calculation.
 
That makes sense thanks Doc.
 

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