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Stress at half of tensile strength

  1. Jan 8, 2007 #1
    Can you guys solve a debate for me?


    If you have two objects and apply a physical stress to them, let's say in the forum of pressure.
    First object: Tensile strength 400 lbs. and you apply 200 lbs
    Second object: Tensile strength 200 lbs and you apply 100 lbs.
    So each is under 1/2 it's tensile strength.

    The debate:
    Choice A: First object is under twice the stress since the load is double
    Choice B: Both are under the same stress since each are being subjected to 50% of their max.

    Thanks,
    Ron
     
    Last edited: Jan 8, 2007
  2. jcsd
  3. Jan 8, 2007 #2

    Hootenanny

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    Depends, do the two objects have the same cross-sectional area?
     
  4. Jan 8, 2007 #3
    The larger object has twice the cross sectional area.


    _____________________________________________________________

    Where this debate stems from:

    It's concerning weightlifting. The details are this:

    A person is just starting out. Say they are able to curl 50 lbs 10 times and this induces a certain amount of muscular stress. Then, years later, after their muscles have increased in CSA (cross sectional area) they can now curl 100 lbs 10 times. Would the stress to their muscles be the same since they are now larger and stronger and it's still the same percentage of strength. I'm saying yes it's the same stress since stress is relative to 'what is being stressed and it's present condition'.

    My example was that if you had two objects, one with tensile strength of 100 and you applied 100, and a second object with a tensile strength of 200 and you applied 150, that even though 150 is greater, it's less stressful since it's a smaller percentage of the tensile strength of the second object,
    Object one: tensile 100 applied 100
    Object two: tensile 200 applied 150

    __________________________________________________________________________

    I had found these,
    http://en.wikipedia.org/wiki/Stress_(physics)

    In the more general setting of continuum mechanics, stress is a measure of the internal distribution of force per unit area that balances and reacts to the external loads or boundary conditions applied to a body.

    http://www.answers.com/topic/stress-physics

    Britannica
    In the physical sciences and engineering, the force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation. Normal stress refers to the stress caused by forces that are perpendicular to a cross-section area of the material. Shear stress arises from forces that are parallel to the plane of the cross section. Stress is expressed as the quotient of a force divided by an area.

    Science
    In physics, the internal resistance of an object to an external force that tends to deform it.
     
    Last edited: Jan 8, 2007
  5. Jan 8, 2007 #4

    Q_Goest

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    Hi Ron.
    First, you're using the term "tensile strength" incorrectly. Tensile strenth is measured in force per unit area. It is the stress at which something is expected to break. Also called "ultimate tensile strength".

    To determine "stress" or "tensile stress" in something such as a bar, take the force being resisted and divide by the area. Units are in force per unit area. (ex: psi)
     
  6. Jan 8, 2007 #5
    Ah ok, oops :blushing: forget the term tensile then.Basically, I'm just trying to query on the term 'stress' and how it applies to physical objects.

    In more basic terms then, (assume all else is relatively equal, joints, leverages, ect etc.)

    Person A's bicep CSA of 25 with 100 lbs
    Person B's bicep CSA of 50 with 200 lbs

    In reference to the term 'stress' would both biceps feel the same stress?
     
  7. Jan 8, 2007 #6

    Hootenanny

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    Yes, stress([itex]\sigma[/itex]) is defined as force per unit area ([itex]\sigma = F/A[/itex]). So, although person B exerts twice the force, his biceps have a CSA twice that of person A; so both person's biceps are under the same stress.
     
  8. Jan 8, 2007 #7
    Thanks much, much appreciated :)
     
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