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Homework Help: Tractions on the surface of the block, and resultant forces and moment

  1. Sep 16, 2013 #1
    1. The problem statement, all variables and given/known data
    The plane strain solution for the stresses in the rectangular block 0 < x < a, 0 < y < b, 0 < z < c under certain loading conditions is given by.
    [itex] \sigma_{xx} = \frac{3Fxy}{2b^3}[/itex]
    [itex] \sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}[/itex]
    [itex] \sigma_{yy} = 0[/itex]
    [itex] \sigma_{zz} = \frac{3\nu Fxy}{2b^3}[/itex]
    Find the tractions on all the six surfaces of the block. Determine the resultants (Forces and Moments) on the surface x = a.

    2. Relevant equations
    [itex]F = \int\int \sigma dA[/itex]
    [itex]M_x = Fx[/itex]
    [itex]M_y = Fy[/itex]
    [itex]M_z = Fz[/itex]

    3. The attempt at a solution
    For the traction on each of the surfaces, I just plug the min and max values of the dimensions into the stress given.

    x faces:
    x = 0,
    [itex]\sigma_{xx} = 0[/itex]
    [itex]\sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}[/itex]
    [itex]\sigma_{xz} = 0[/itex]

    x = a,
    [itex]\sigma_{xx} = \frac{3Fay}{2b^3}[/itex]
    [itex]\sigma_{xy} = \frac{3F(b^2-y^2)}{4b^3}[/itex]
    [itex]\sigma_{xz} = 0[/itex]

    y faces:
    y = 0,
    [itex]\sigma_{yy} = 0[/itex]
    [itex]\sigma_{yx} = \frac{3F}{4b}[/itex]
    [itex]\sigma_{yz} = 0[/itex]

    y = b,
    [itex]\sigma_{xx} = 0[/itex]
    [itex]\sigma_{xy} = 0[/itex]
    [itex]\sigma_{xz} = 0[/itex]

    z faces:
    z = 0,
    [itex]\sigma_{zz} = \frac{3 \nu Fxy}{2b^3}[/itex]
    [itex]\sigma_{zy} = 0[/itex]
    [itex]\sigma_{xz} = 0[/itex]

    z = c,
    [itex]\sigma_{zz} = \frac{3 \nu Fxy}{2b^3}[/itex]
    [itex]\sigma_{zy} = 0[/itex]
    [itex]\sigma_{xz} = 0[/itex]

    The part I am confused about is finding the resultant forces on the face x = a.
    I believe that all I do is use the tractions that I found for x = a. Thus,

    [itex]F_x = \int \limits_0^b \int \limits_0^c \sigma_{xx}dzdy = \int \limits_0^b \int \limits_0^c \frac{3Fay}{2b^3}dzdy[/itex]

    [itex]F_y = \int \limits_0^b \int \limits_0^c \sigma_{xy}dzdy = \int \limits_0^b \int \limits_0^c \frac{3F(b^2-y^2)}{4b^3}dzdy[/itex]

    [itex]F_z = \int \limits_0^b \int \limits_0^c \sigma_{xz}dzdy = \int \limits_0^b \int \limits_0^c (0)dzdy[/itex]

    Then for the moments I just take these forces and multiply them by the distance to the x axis.

    Is my work correct so far or am I missing something.
  2. jcsd
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