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Factor of Safety - Mechanics of Materials

  1. Feb 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Rigid beam ABC is supported as shown in the figure. The pin connections at B, C, and D are each double shear connections, and the ultimate shear strength of the pin material is 640 MPa. Tie rod (1) has a yield strength of 350 MPa. A uniformly distributed load of w = 11 kN/m is applied to the beam as shown. A factor of safety of 3.5 is required for all components. Assume a = 575 mm, b = 725 mm, c = 250 mm, and d = 525 mm.

    Determine

    (a) the minimum diameter required for tie rod (1).
    (b) the minimum diameter required for the double shear pins at B and D.
    (c) the minimum diameter required for the double shear pin at C.

    http://edugen.wileyplus.com/edugen/courses/crs6838/art/qb/qu/ch0/13425408508040_6521478752676759.jpg
    2. Relevant equations

    σ = F/A
    τ = F/A
    σallow = σfailure/FS
    τallow = τfailure/FS


    3. The attempt at a solution

    First I got the resultant force of w → 14.3 kN/m.
    I started by finding the x and y components of F1(force on the rod), which were F1cos35.9 and F1sin35.9, respectively.
    Then I summed the moments about C which gave me ΣM = (-14.3 kN/m)(13m) + (0.810F1)(0.25m) = (0.5864F1)(0.725)
    ⇒ F1 = 14.8 kN
    Using σ ≤ σallow, I plugged in my values → 14809.4 N/π(d/2)2 ≤ 350 N/mm2/3.5
    ⇒ drod = 188.6 mm

    14809.4 N/2π(d/2)2 ≤ 640 N/mm2/3.5
    ⇒dB,D = 51.6 mm

    Then I summed the x and y forces to get the force at C, which was C = 13.2 kN

    13242.3 N/2π(d/2)2 ≤ 640 N/mm2/3.5
    ⇒ d = 46.1 mm

    All of my d values were wrong. I have recalculated and changed the signs on my forces and moments just to see if I did that wrong but I still can't get the right answer. Am I going about the problem the wrong way? I'm just not sure where I'm going wrong and I've been working on this and a similar problem for a couple days now to no avail.
     
  2. jcsd
  3. Feb 25, 2015 #2

    PhanthomJay

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    You mean 14.3 kN
    You are getting a bit careless I think on math and signage, and using the wrong lever arm for the distributed load moment. The beam is 1.3 m long. Its resultant load acts at its cg when determining moments. The pins are in double shear.
     
  4. Feb 25, 2015 #3
    Oops I meant .65m not 1.3m. I calculated it using the 0.65 but I typed it wrong. I also thought I was calculating the stress in the pins correctly, by dividing the force by 2. Where else am I going wrong?
     
  5. Feb 25, 2015 #4

    PhanthomJay

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    I will have to check your numbers I haven't done that ....when you say your force at C is the sum of the x and y components I hope you meant vector sum not algebraic sum. You have the 2 in there for dbl shear so that's ok , your approach is good I'll check math later
     
  6. Feb 26, 2015 #5

    PhanthomJay

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    Your math is way off when calculating the diameters. Try again.
     
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