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Pin joint Trusses and Equilibrium

  1. Mar 31, 2008 #1
    I have a general question about solving these types of problems. I had two different problems for calculating the tensions and compressions of a pin-jointed truss. both problems were of a trapezoidal shape with 5 pins in total.

    in one problem, a weight was applied to the center truss on the center of the bridge.

    on another problem, the weight was in the middle of the left beam.


    here is where i get confused. both problems involve solving the normal forces on the far left and far right pin which is easy enough knowing the dimensions of the truss bridge and the weight by taking a net torque equaling zero equation.

    my problem is when calculating the sum forces on the far left and far right pins. in the first problem, there is only the normal force and two beams creating equilibrium. in the second problem ( do to issues with sheer stress???) there is an extra force included in the free body diagram to find the results.

    i dont get why the first calculation involves just three forces in the equilbrium on the far pin, while when in the second problem the weight between two pins creates a need to add a partial force. if you could explain this, it would be much appreciated.

    i should mention that in the second problem, the first thing asked is whether the weight is equally distributed between the two pins (since the weght of the object is equidistant form both).
     
  2. jcsd
  3. Mar 31, 2008 #2

    PhanthomJay

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    If I understand correctly, it is often standard practice to apply all loads at joints rather than in between (to simplify the analysis and avoid shear stresses, as you noted, in order to analyze the truss as a pure truss with axial loads only). So if a weight is applied to a center of a beam, you may distribute it half and half as a load on the adjacent joints. You may then need to later analyze the beam with the load at its center for bending stresses.
     
  4. Mar 31, 2008 #3
    here are two problems i'm talking about. would solving for all the forces in both of them require extra steps in 69?

    [​IMG]
     
  5. Apr 1, 2008 #4

    PhanthomJay

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    For problem 69, the 'extra step' is to divide the weight of the 1500 kg car (15,000N) by 2, and apply that value as a vertical load acting down at joints A and C (7500N at A and 7500N at C). Then solve for the truss reactions and forces ignoring the car at the center.
    (The truss reactions at A and E can be solved either with the car positioned as shown, or by dividing its weight up as a load at A and C.....the results will be the same for either case....the placement of the load at the joints greatly simplifies the force analysis of the members).
     
  6. Apr 1, 2008 #5
    so, i take it for pin-jointed trusses, the trick to solving is if the weight is directly over a pin joint, just leave the force there. but if the weight acts in between joints, you resolve the force into sheer forces on the the adjacent pins and use a net-torque equation to to calculate those sheer forces.

    am i correct in my assesment?
     
  7. Apr 1, 2008 #6

    PhanthomJay

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    Correct
    if the weight acts in between joints, you divide it up appropriately and apply the calculated values as an applied load at the adjacent joints. For example, in the example , it's 7500N at A and 7500N at C. If the car were located 3/4ths of the way toward C, it'd be 3750N applied at at A and 11,250N applied at C, if that's what you mean by the net torque equation.
     
  8. Apr 1, 2008 #7
    well, in the solution manual, the way they solved for the sheer forces was by taking the segment AC by itself, and then saying that there are two forces on the joints going straight up:

    net forces on y for strut = SheerForce1 + SheerForce2 - Weight.

    and then solved for one sheer force using a 0 torque:

    net torque at A = 0 = (Weight of Car) * 25m - SheerForce2 * 50m

    your way is faster though :)

    the thing i didn't like about the manual's way of solving was that it wasn't easy to follow for me. in the first FBD of just the segment AC, they drew the two sheer forces point up, but then in the FBD of joint A, they drew it down. this doesn't seem right...
     
  9. Apr 2, 2008 #8

    PhanthomJay

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    well, that is right, you wouldn't expect the car's weight to act up on the joint, but you would expect the reaction at that joint to act up. When you take a FBD of member AC, the shear force at the left (and right) end points up ; when you take a FBD of joint A, the shear force points down. This is per Newton's 3rd law. The shear force then acts down on the joint A, but the normal reaction force at A acts up. The algebraic difference beween those two 'y' forces goes to the diagonal member, such that the joint is in equilibrium in the y direction (and of course also in the x direction). Rather than talking shear forces, I prefer to treat these shear forces as "applied" forces acting down at joints A and C. Otherwise, you quickly get lost with the plus and minus signs, and joint equiilibrium equations.
     
    Last edited: Apr 2, 2008
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