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Question about double shear pin connections

  1. Feb 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Double Shear Pin Connection

    2. Relevant equations

    ∑Fx = -2V + P = 0

    3. The attempt at a solution

    When calculating the shear force along the planes of the pin, where I have a reaction at Ax and Ay, why do I have to use the magnitude of the reaction force as opposed to the axial force that is acting colinear to the shear plane?

    For example, if I have a horizontal reaction of 34.64 kN and a vertical reaction of 20 kN, why am I wrong for using 34.64 kN, instead of √34.642 + 202, to determine my shear forces?

    Wouldn't the magnitude of the force have to be projected along the same shear plane thus returning me to my original horizontal reaction?
  2. jcsd
  3. Feb 23, 2014 #2


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    I'm having a hard time visualizing what you are asking. Can you provide a sketch?
  4. Feb 24, 2014 #3

    Pin A is the one I'm talking about.

    For some reason the solutions manual uses the magnitude of the vertical and horizontal reaction but I can't see why since that would not be along the surface of the shear plane that I'm looking at.

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  5. Feb 24, 2014 #4


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    The shear experienced by the pin is the resultant of the axial and vertical components of the reaction at pin A. The pin can't distinguish a separate horizontal force and vertical force; component forces are mathematical conveniences used to solve problems.

    Just like forces can be decomposed into different components, stresses can as well, by using techniques like the Mohr's circle:


    Once you have the basic stress components, you can use Mohr's circle to find the normal stress and the shear stress components for an arbitrary stress plane.
  6. Feb 24, 2014 #5


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    Well Bill, you seem to have no problem at all in correctly calculating the shear force and stress at the pin at C, using the 40 kN link force which is directed 30 degrees below the horizontal. Did you know that the 40 kN force can be resolved into its vector components as 40 sin 30 = 20 kN in the y direction and 40 cos 30 = 34.6 kN in the x direction? Which is precisely the same as the x and y force components at A. That is, the resultant of the force components at A is sq rt (34.64^2 + 20^2) = 40 kN directed 30 degrees above the horizontal. Same magnitude as at C. That is why you must use the 40 kN force when calculating shear forces on the pin at A. As SteamKing has pointed out, the pin doesn't much care about component shear forces, only resultant shear force.
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