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    Anybody got an idea of the final part to this question? I've got the other 2 things it asks for fine. This last bit though......
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    (Original post by Dav1dJG)
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    From the first part of the question, you know the method from which you can calculate the force required to be applied to your cutting instrument (whether that be the guillotine or the punch) in order to shear the material.

    You can then use the value of this force and the area of the hollow punch in order to calculate the internal stress of the hollow punch. This value for the stress is hence the minimum compressive strength of the material from which the punch should be fabricated from.
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    (Original post by pleasedtobeatyou)
    From the first part of the question, you know the method from which you can calculate the force required to be applied to your cutting instrument (whether that be the guillotine or the punch) in order to shear the material.

    You can then use the value of this force and the area of the hollow punch in order to calculate the internal stress of the hollow punch. This value for the stress is hence the minimum compressive strength of the material from which the punch should be fabricated from.
    Thanks for getting back to me. I did try this, answer for first part is 1.254MN, answer for part 3 is 58.84MN/m^2. The area of the hollow punch I calculated as 3.1416x10^-4 m^2
    The force and this new area don't get to the answer for part 3.... 😢
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    (Original post by Dav1dJG)
    Thanks for getting back to me. I did try this, answer for first part is 1.254MN, answer for part 3 is 58.84MN/m^2. The area of the hollow punch I calculated as 3.1416x10^-4 m^2
    The force and this new area don't get to the answer for part 3.... 😢
    The first thing is that you value for the area of the hollow punch is incorrect. Make sure you are using this formula:

    

A = A_{\mathrm{external}} - A_{\mathrm{internal}} = \pi \left[ \left( \dfrac{D_{\mathrm{external}}}{2} - \dfrac{D_{\mathrm{internal}}}{2} \right) \right].

    Let * denote the strength of a material.

    By equivalence of externally applied forces and internal stresses (making a plane stress assumption):

    \sigma_{\mathrm{min}}^* A = \tau^* \pi D_{\mathrm{external}} t.

    I have run through the calculations and obtained the answer provided. If you are still stuck, feel free to ask for clarification.
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    (Original post by Dav1dJG)
    Thanks for getting back to me. I did try this, answer for first part is 1.254MN, answer for part 3 is 58.84MN/m^2. The area of the hollow punch I calculated as 3.1416x10^-4 m^2
    The force and this new area don't get to the answer for part 3.... 😢
    Carrying on from your PM here:

    Your area calculation is still wrong but that may be the formula I provided was slightly incorrect (sorry).

    Here is a corrected version:

    

A = A_{\mathrm{external}} - A_{\mathrm{internal}} = \pi \left[ \left( \dfrac{D_{\mathrm{external}}}{2} \right)^2 -  \left( \dfrac{D_{\mathrm{internal}}}{2} \right)^2 \right].

    The answer should be A = \dfrac{31}{10000} \pi = 9.74 \times 10^{-3} \ [\mathrm{m^2}].
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    (Original post by pleasedtobeatyou)
    Carrying on from your PM here:

    Your area calculation is still wrong but that may be the formula I provided was slightly incorrect (sorry).

    Here is a corrected version:

    

A = A_{\mathrm{external}} - A_{\mathrm{internal}} = \pi \left[ \left( \dfrac{D_{\mathrm{external}}}{2} \right)^2 -  \left( \dfrac{D_{\mathrm{internal}}}{2} \right)^2 \right].

    The answer should be A = \dfrac{31}{10000} \pi = 9.74 \times 10^{-3} \ [\mathrm{m^2}].
    Ok thank you. Wow I should've spotted that myself but I'm doubting everything I did with that question lol. Would you believe me if I said I got 98% on my unit exam for Statics and Strength of materials!? Only dropped 1 mark, doesn't look like it now though hahahahah
    Cheers again
 
 
 
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