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Direct Stress - Strain Help for Sons Homework Watch

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    • Thread Starter
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    Good Afternoon,
    Bare with me as I'm not the best at physics/maths. My son has asked me for some help on his physics work but I'm a bit rusty.

    The question is;
    A circular metal bar is to carry an axial load of 50kN. The modulus of elasticity is 175Gpa. The metal bar has a length of 2m and it's diameter is 30mm. Find:

    The tensile stress
    The Strain and hence the change in length.

    I've worked out Tensile stress as:

    TS=FORCE/AREA
    50X10^3/706.85m^2
    =0.10Pa

    Is this correct? Also any help on stain and change in length would be great!

    Many thanks!
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    (Original post by vxdturbo12)
    Good Afternoon,
    Bare with me as I'm not the best at physics/maths. My son has asked me for some help on his physics work but I'm a bit rusty.

    The question is;
    A circular metal bar is to carry an axial load of 50kN. The modulus of elasticity is 175Gpa. The metal bar has a length of 2m and it's diameter is 30mm. Find:

    The tensile stress
    The Strain and hence the change in length.

    I've worked out Tensile stress as:

    TS=FORCE/AREA
    50X10^3/706.85m^2
    =0.10Pa

    Is this correct? Also any help on stain and change in length would be great!

    Many thanks!
    I think your area is incorrect. The question give the diameter in mm, so make sure the conversion to metre is done.

    10 mm = 1 cm = 0.01 m

    To find the extension, you need to work from modulus of elasticity which is given by stress/strain, where strain is extension/length.

    Being a father is not easy this day.
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    (Original post by Eimmanuel)
    I think your area is incorrect. The question give the diameter in mm, so make sure the conversion to metre is done.

    10 mm = 1 cm = 0.01 m

    To find the extension, you need to work from modulus of elasticity which is given by stress/strain, where strain is extension/length.

    Being a father is not easy this day.
    Thanks for replying;

    So area should be; 50x10^3/0.706 = 70.82Kpa?

    Is that correct? or am I still making a fool of myself ha

    Many thanks!
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    (Original post by vxdturbo12)
    Thanks for replying;

    So area should be; 50x10^3/0.706 = 70.82Kpa?

    Is that correct? or am I still making a fool of myself ha

    Many thanks!
    Errrr...

    Area = \pi \left( \dfrac{15}{1000} \right)^2 m2.

    I think your conversion is wrong.
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    You're using the right equation to calculate the stress, but you are making mistakes in your conversions/units.

    It's simpler if you calculate the area separately.

    To calculate the change in length you need to know the stress (which you have calculated), the modulus of elasticity (which you are given), and the strain, which can be worked out.

    Do you know the stress/strain relationship? For the elastic/linear region, which I am assuming you have to use unless told otherwise, and there is nothing to suggest that is the case here, we can say that the modulus of elasticity (E) is proportional to stress divided by strain.

    Once you know the strain, you can then work out the change in length, as you are given the original length, and strain is effectively the change in length is a proportion of the original length.
 
 
 
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