If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of end A when th

Question

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of end A when the 60-kN force is applied. Take E = 200 GPa.

in progress 0
Vodka 3 years 2021-08-02T15:25:11+00:00 1 Answers 87 views 0

Answers ( )

  1. minhkhoi
    0
    2021-08-02T15:26:52+00:00
    Reply

    Answer:

    σ = 1.09 mm

    Explanation:

    Step 1: Identify the given parameters

    rod diameter = 20 mm

    stiffness constant (k) = 55 MN/m = 55X10⁶N/m

    applied force (f) = 60 KN = 60 X 10³N

    young modulus (E) = 200 Gpa = 200 X 10⁹pa

    Step 2: calculate length of the rod, L

    K = \frac{A*E}{L}K=

    L

    A∗E

    L = \frac{A*E}{K}L=

    K

    A∗E

    A=\frac{\pi d^{2}}{4}A=

    4

    πd

    2

    d = 20-mm = 0.02 m

    A=\frac{\pi (0.02)^{2}}{4}A=

    4

    π(0.02)

    2

    A = 0.0003 m²

    L = \frac{A*E}{K}L=

    K

    A∗E

    L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

    55X10

    6

    (0.0003142)∗(200X10

    9

    )

    L = 1.14 m

    Step 3: calculate the displacement of the rod, σ

    \sigma = \frac{F*L}{A*E}σ=

    A∗E

    F∗L

    \sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

    (0.0003142)∗(200X10

    9

    )

    (60X10

    3

    )∗(1.14)

    σ = 0.00109 m

    σ = 1.09 mm

    Therefore, the displacement at the end of A is 1.09 mm

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )