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Offset crank slider mechanism connecting rod analysis

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data

    A) Determine the value of the angle θ (measured from vertical) when the angular velocity of link AB a maximum.

    B) What is the maximum angular velocity of link AB.

    slider crank.JPG

    2. Relevant equations


    3. The attempt at a solution
    This is a part of a bigger question. I've managed to answer all other part apart of those two. Can anyone point me in the right direction. I'm not an expert in this subject. Just let me know if you require any more data.
     
  2. jcsd
  3. Aug 6, 2015 #2

    SammyS

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    It's likely that the other parts of the problem will help answer these.

    Also, as you should know by now, you need to make an attempt at a solution.
     
  4. Aug 6, 2015 #3
    No problem SammyS, I will include the full question and all my workings done so far.
     
  5. Aug 6, 2015 #4
    1) For the mechanism shown in the figure included within the #1 post determine for the angle θ=45°:

    a) the velocity of the piston relative to the fixed point O (VBO)
    b) the angular velocity of AB about point A (i.e. ωAB)
    c) the acceleration of point B relative to A (αBA)

    Note: Link AB is horizontal when θ=45°.

    2) Determine the value of the angle θ (measured from vertical) when:

    a) the velocity of point B=0
    b) the angular velocity of link AB a maximum

    3) What is the maximum angular velocity of link AB?

    ANSWER:
    ωOA=300 revsmin-1=10π rads-1
    lOA=0.05m
    lAB=0.2m

    1)
    a)
    VAO=lOAOA=0.05*10π=1.5708 ms-1
    using velocity triangle
    cosθ=VBO/VAO
    VBO=cos45°*VAO=(√2/2)*1.5708=1.1107 ms-1

    b)
    sinθ=VBA/VAO
    VBA=sin45°*VAO=(√2/2)*1.5708=1.1107 ms-1
    VBA=lABAB
    ωAB=VBA/lAB=1.1107/0.2=5.5535 rads-1

    c)
    αBA(r)=lABAB2=0.2*5.55352=6.1683 ms-2
    ωOA=VAO/lOA=1.5708/0.05=31.416 rads-1
    αAO(r)=lOAOA2=0.05*31.4162=49.3482 ms-2

    drawing acceleration diagram using following scale (see acceleration diagram attachment)
    1 cm=20 mm
    1 cm=5 ms-2

    measured from the graph
    αBA(t)=6.9cm=34.5 ms-2

    2)
    a)
    The velocity of point B=0 occurs at maximum and minimum piston displacement.

    calculating the offset height lOD (see the graph attachment)
    cosθ=lOD/lOA
    lOD=cos45°*lOA=(√2/2)*50=35.3553 mm

    calculating at max displacement.
    lOB=lOA+lAB=50+200=250 mm
    let the angle symmetry line through O and the crank-connection rod at max piston displacement be γ
    sinγ=lOD/lOB=35.3553/250=0.1414
    γ=arcsin(0.1414)=8.1289°
    so θ at max piston displacement (measured from vertical) is θ=90°=8.1289°=81.8711°

    calculating at min displacement.
    let the angle symmetry line through O and the crank-connection rod at min piston displacement be β
    sinβ=lOD/(lAB-lOA)=35.3553/150=0.2357
    β=arcsin(0.2357)=13.6329°
    so θ at min piston displacement (measured from vertical) is θ=270°-13.6329°=256.3671°

    b)
    To be able to determine the value of the angle θ (measured from vertical) when the angular velocity of link AB a maximum I need to work out at which position the link AB has the max angular velocity.

    3)
    Once I know the answer to 2b) I should be using either trigonometry or velocity diagram calculate that velocity. At least I think/hope so.
     

    Attached Files:

  6. Aug 10, 2015 #5
    Guys, can anyone help me with question 2b and 3 - see post #4. Much appreciated.
     
  7. Aug 10, 2015 #6

    SammyS

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    I'm sorry about not getting back to you sooner. I was off traveling.

    I will try to wade through that material.

    One thing I noticed on drawing for this problem, the angle that OA makes with the horizontal is not θ. It's (π/2) - θ , i.e. 90° - θ. It's only equal to θ for θ = 45° .

    graph-jpg.86968.jpg
     
  8. Aug 24, 2015 #7
    Good Morning SammyS,

    I believe you don't have any further information on my question? Thanks for trying anyway. I think the connection rod is at the max velocity when is perpendicular to the crank. I will use this assumption and try to work out the angle and the velocity.
     
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