1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Internal Force Diagram for Rigid Body and Distributed Load

  1. Mar 30, 2015 #1
    I'm working on a homework problem for Statics and I'm stuck. Could someone please help?


    Problem: Draw the internal force (N,V,M) diagrams and include all significant figures
    HW5-9_Book7-80.png


    Here is all of my work:

    Resulting F from W1:
    W(x) = W1
    ∴ F2=∫02b W1dx
    eq (1) ⇒ F2 = W1⋅2b​
    x1 = (F1)-102b W1dx = 1/(W1⋅2b)⋅(W1/2 ⋅ x2) |02b
    eq (2) ⇒ x1 = b;​

    SUPPORT REACTIONS DIAGRAM &FBD:​
    DIAGRAM: FBD.png
    FBD: FBD%2B(1).png

    SUPPORT REACTIONS VARIABLE SOLUTION:
    [+→]∑Fx = 0 : 0 = RAX
    eq (3) ⇒ RAX = 0 ;​
    [+∧]∑Fy = 0 : 0 = (RAY + RBY + RDY) - (F1 + F2)
    eq (4) ⇒ RAY + RBY + RDY = F1 + W1⋅2b;​
    [+CW]∑MA = 0 : 0 = (F2⋅b + F1(2b + a)) - (RBY⋅b + RDY(2b+2a))
    →RBY⋅b + RDY(2b+2a) = W1⋅2b⋅b + F1(2b + a)
    eq (5) ⇒ RBY⋅b + RDY(2b+2a) = 2⋅W1⋅b2 + F1(2b + a) ;​
    [+CW]∑MB = 0 : 0 = (RAY⋅b + F1(b+a)) - RDY(b+2a)
    eq (6) ⇒ RDY(b+2a) = RAY⋅b + F1(b+a) ; ​
    [+CCW]∑MB = 0 : 0 = (F1⋅a + F2(2a+b)) - (RBY(2a+b)+RAY(2b+2a))
    eq (7) ⇒ RBY(2a+b)+RAY(2b+2a) = F1⋅a + F2(2a+b) ;​

    SEPARATE FBDs:
    I: FBD%2B(2).png II: FBD2.png
    Separating the system at the hinge (C) will allow us to solve for the support reactions: In Figure 1, we can solve for the support reactions in the separate part, and therefore solve for the support reactions in the whole beam.

    [+CW]∑MB = 0 : 0 = RAY⋅b
    eq (8) ⇒ RAY = 0 ;​
    [+∧]ΣFy = 0 : 0 = RAY + RBY - F2
    eq (9) ⇒ RBY = F2

    So now, using eq's (8) & (9) in eq (4) we can determine RDY
    eq (4) : RAY + RBY + RDY = F1 + W1⋅2b
    →F2 + RDY = F1 + W1⋅2b
    → RDY = F1 + W1⋅2b - W1⋅2b
    eq (10) ⇒ RDY = F1

    (9) > (5) : F2⋅b + RDY(2b+2a) = 2⋅W1⋅b2 + F1(2b + a)
    → RDY(2b+2a) = 2⋅W1⋅b2 + F1(2b + a) - (W1⋅2b)⋅b
    → RDY= 1/(2b+2a)[F1(2b + a) + 2⋅W1⋅b2 - (W1⋅2b)⋅b]
    → RDY= F1(2b + a)/(2b+2a)


    I'm just super lost, I guess, on where to go and if I'm even starting in the right direction... could someone please help me?
     
  2. jcsd
  3. Mar 31, 2015 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Are you sure you are intpreting W1 correctly? You are taking it as the weight per unit length. From the info in the OP, it could just as easily be the total weight (your F2).
    Your first eqn for x1 has some typos, but it worked out ok.
    When you considered the hinged parts separately, you overlooked that there may be a force at C between the two parts. Your eqn 10 is clearly wrong if you consider moments about C.
     
  4. Apr 2, 2015 #3
    Yes W1's units are Force/Distance.
    what are the typos for x1, just so I know?
    and
    is supposed to be Sum of Moments at C, so considering, what do I do with this information?
     
  5. Apr 2, 2015 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Shouldn't that be ##x_1 = (F_2)^{-1}\int _0^{2b} W_1x.dx##?
    Looks like moments about D to me.
     
  6. Apr 2, 2015 #5
    The key to getting the reaction forces in this problem is recognizing that the bending moment at point C is zero. I think that you noted this, but you still got the wrong values for the reaction forces.

    If I do a moment balance on the portion of the beam to the right of C by taking moments about the hinge at point C, I get:
    $$2aR_D-aF_1 = 0$$
    This yields ##R_D=F_1/2##

    Now, what do you get if you do the same kind of thing for the portion of the beam to the left of point C?

    (I get ##R_A=-F_1/2## and ##R_B=2W_1b+F_1##)

    Chet
     
    Last edited: Apr 2, 2015
  7. Apr 5, 2015 #6
    okay so using your logic...
    [+CW]ΣMC = 0 : 0 = F1a - 2a⋅RDY
    →RDY = (2a)-1(F1a)
    ⇒ RDY = F1/2
    then plugging this solution for RDY into eq (5) I get:
    RBY = 1/b (2⋅W1⋅b2 + F1(2b+a)-RDY(2b+2a))
    →RBY = (2⋅W1⋅b2)/b + F1⋅(2b+a)/b - F1/2⋅(2b+2a)/b
    →RBY = 2⋅W1⋅b + F1⋅(2+a/b) - F1⋅(1+a/b)
    ⇒RBY = 2⋅W1⋅b + F1

    then plugging this solution for RBY into eq (4) I get:
    RAY = F1 + 2bW1 - RBY - RDY
    →RAY = F1 + 2bW1 - (2bW1 + F1) - (F1/2)
    ⇒RAY = - F1/2
    Now that I have the support reactions, I should be able to determine the Internal Force equations and create a diagram from that information.

    What is the best strategy to go about this?
     
  8. Apr 5, 2015 #7
    I don't know what the best strategy is. I would start at the left and work my to the right, first getting the V's, and then starting at the left again and getting the M's.

    Chet
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Internal Force Diagram for Rigid Body and Distributed Load
  1. Forces on a rigid body (Replies: 5)

Loading...