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Tensile strength/ force question help Watch

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    • Thread Starter
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    Can someone help me with this question:


    So stress= force/cross-sectional area

    We are given the stress?
    We want the force

    I need to work out the cross-sectional area.


    Is that pi*(15mm-0.7mm)^2?

    Also I don't get what is meant by a factor of safety of 3.Name:  15857361_1334707363255477_1725193876_o.jpg
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    • PS Reviewer
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    (Original post by Super199)
    I need to work out the cross-sectional area.

    Is that pi*(15mm-0.7mm)^2?
    Close.

    A = \pi \left(\dfrac{15 \times 10^{-3}}{2}\right)^2 - \pi \left(\dfrac{15 \times 10^{-3}}{2} - 0.7 \times 10^{-3}\right)^2

    (Original post by Super199)
    Also I don't get what is meant by a factor of safety of 3.
    The factor of safety is used to account for many uncertainties e.g. the accuracy of your model. To apply it, you simply multiplied your applied loads by the safety factor i.e. you are designing for greater applied loads than you would expect to apply.
 
 
 
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