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Maximum Allowable Loads in 2-force beams

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data
    The pin-connected structure is subjected to a load P as shown. Inclined member (1) has a cross-sectional area of 250mm2 and a yield strength of 255MPa. It is connected to rigid member ABC with a 16-mm-diameter pin in a double shear connection at B. The ultimate shear strength of the pin material is 300MPa. For inclined member (1), the minimum factor of safety with respect to the yield strength is FSmin=1.5. For the pin connections, the minimum factor of safety with respect to the ultimate strength is FSmin=3.0.

    Question: Based on the capacity of member (1) and pin B, determine the maximum allowable load P that may be applied to the structure.

    Ignore the angle written on the page, it's not the work that I used to solve the problem
    http://img593.imageshack.us/img593/6229/photobp.jpg [Broken]


    2. Relevant equations
    [itex]\Sigma[/itex]Fx=0
    [itex]\Sigma[/itex]Fy=0
    [itex]\sigma[/itex]allow=[itex]\sigma[/itex]failure/(Safety Factor)
    [itex]\tau[/itex]allow=[itex]\tau[/itex]failure/(Safety factor)
    Safety Factor=[itex]\sigma[/itex]failure/([itex]\sigma[/itex]actual)
    Safety Factor=[itex]\tau[/itex]failure/([itex]\tau[/itex]actual
    Safety Factor=Pfailure/(Pactual)
    Safety Factor=[itex]\nu[/itex]failure/([itex]\nu[/itex]actual)


    3. The attempt at a solution

    http://img513.imageshack.us/img513/4791/photo1yi.jpg [Broken]

    I know this is not the correct answer, and I know what the correct answer is
    Pmax=15.03kN

    Where did I go wrong?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 20, 2012 #2

    PhanthomJay

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    Gold Member

    First, you have shown that the pin in double shear can take 40.22 kN based on its allowable shear stress, and that the diagonal member can take 42.5 kN based on its allowable tensile stress. These values are correct, well done. So it shows that the max load is controlled by the pin......the pin and the member BD see the same load.

    Secondly, you now need to find the force in the member BD (which is the same as the force on the pin) as a function of P, and set it equal to 40.22 KN to solve for the allowable applied force P. You should calculate the y-component of the reaction at D (in terms of P) by summing moments about A, and note that since BD is a 2 force member, the x and y components and resultant force in BD are trigonometrically related.
     
  4. Jan 20, 2012 #3
    Thanks for the help!
     
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