For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with diameter of 4

Question

For the material in the previous question that yields at 200 MPa, what is the maximum mass, in kg, that a cylindrical bar with diameter of 4.5 cm can support without yielding ? Recall that a mass of 1 kg has a weight of 9.81 N at sea-level. Answer Format: X (no decimal) Unit: kg Example: of you calculate 41321.78…. enter 41322

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Acacia 3 years 2021-08-24T22:15:17+00:00 1 Answers 47 views 0

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  1. thienthanh
    0
    2021-08-24T22:16:26+00:00
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    Answer:

    The maximum mass the bar can support without yielding = 32408.26 kg

    Explanation:

    Yield stress of the material (\sigma) = 200 M Pa

    Diameter of the bar = 4.5 cm = 45 mm

    We know that yield stress of the bar is given by the formula

                    Yield Stress = \frac{Maximum load}{Area of the bar}

    ⇒                                \sigma = \frac{P_{max} }{A}  —————- (1)

    ⇒ Area of the bar (A) = \frac{\pi}{4} ×D^{2}

    ⇒                            A  = \frac{\pi}{4} × 45^{2}

    ⇒                            A = 1589.625 mm^{2}

    Put all the values in equation (1) we get

    ⇒ P_{max} = 200 × 1589.625

    ⇒ P_{max} = 317925 N

    In this bar the P_{max} is equal to the weight of the bar.

    ⇒ P_{max} = M_{max} × g

    Where M_{max} is the maximum mass the bar can support.

    ⇒ M_{max} = \frac{P_{max} }{g}

    Put all the values in the above formula we get

    ⇒ M_{max} = \frac{317925}{9.81}

    ⇒ M_{max} = 32408.26 Kg

    There fore the maximum mass the bar can support without yielding = 32408.26 kg

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