A plastic dowel has a Young’s Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. (a) What is

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A plastic dowel has a Young’s Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted. (a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.70 cm? kN (b) If a force of this magnitude is applied compressively, by how much (in mm) does the 28.0 cm long dowel shorten? (Enter the magnitude.) mm

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Mộc Miên 3 years 2021-08-06T17:55:27+00:00 1 Answers 18 views 0

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  1. thanhcong
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    2021-08-06T17:57:05+00:00
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    Answer:

    (A) The maximum force is 85.8 kN.

    (B) Dowel shorten by 2.79 mm

    Explanation:

    Given:

    Young modulus E = 1.50 \times 10^{10} \frac{N}{m^{2} }

    Pressure fracture P = 1.50 \times 10^{8} \frac{N}{m^{2} }

    Diameter D = 2.70 \times 10^{-2} m

    Radius r = 1.35 \times 10^{-2} m

    (A)

    From the formula of force in terms of pressure,

    F=PA

    Where A = area of dowel

    F = 1.50 \times 10^{8} \times \pi (1.35 \times 10^{-2} ) ^{2}

    F = 85.8 \times 10^{3}

    F = 85.8 kN

    (B)

    From young modulus formula,

     E = \frac{FL}{A\Delta L}

    Where L = 28 \times 10^{-2}m

    \Delta L = \frac{FL}{AE}

    \Delta L = \frac{85.8 \times 10^{3} \times 28 \times 10^{-2} }{\pi (1.35 \times 10^{-2} )^{2} \times 1.50 \times 10^{10} }

    \Delta L = 2.79 \times 10^{-3}

    \Delta L = 2.79 mm

    Therefore, the maximum force is 85.8 kN and dowel shorten by 2.79 mm

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