Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable

Question

Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable shearing force in each nail is 700 N, determine the largest permissible spacing s between the nails.

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Edana Edana 5 months 2021-08-03T00:55:37+00:00 2 Answers 23 views 0

Answers ( )

    0
    2021-08-03T00:57:00+00:00

    Answer:

    Using equation

    qs=F

    q=VQ/I   V=1200

    q=998.45 N

    qs=F

    s=F/q

    s=700/998.45

    spacing=s=0.701 m

    0
    2021-08-03T00:57:21+00:00

    Answer:

    The question is incomplete. The mass of the board is needed in order to calculate the moment of inertia of the beam. 91.67 * 10⁶mm⁴ is used as the moment of inertia in the calculation below. You can always substitute to get your answer.

    The spacing is  95.1mm

    Explanation:

    Given’

    vertical shear V = 1200N

    Shearing force = 700N

    Thickness of each board = 50mm

    The spacing can be calculated using the formula;

    F = q * s—————————1

    but, q = VQ/I———————————2

    Therefore,  F = VQ/I  * s ——————————3

    where,

    s = spacing

    v = vertical shear

    f = shearing force

    I = moment of inertia

    Making s subject formula in equation 3, we have

    s = FI/VQ——————————-4

    but Q = YA

    A = area = 50 *(150)

    Y = 75

    therefore, Q = 75 *50*150

    Q= 5.625* 10⁶mm⁴

    Substituting into equation 4, we have

    s = 700 * 91.67 *10⁶) / (1200*5.625* 10⁶)

       =  6.42*10¹⁰ /6.75*10⁸

      = 95.1mm

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