Diagonal (vertical plane) trusses???

[dave_m]

Quick question, though I'm sure it won't be as quick to answer!

 

Say I wanted to hang a truss diagonally in the vertical plane, what would the calculations involved be? Keeping the maths simple for now, lets assume the following:

• hung at 45 deg to ground (0 deg)
  
• 5m between points in grid
  
• 100kg udl
  
• hung on two points
  
• Truss is litec QX30
  

 

So, I'm assuming the lower point, will have some kind of moment acting on it, and as such, that point will have more load on it?

 

I've attached an image to make my ramblings clear if they aren't already. http://i307.photobucket.com/albums/nn300/dave_mcc/truss.jpg

 

Usual stuff applies, I would get a competent person to make/check these calculations and install the system, I'm just interested!

[mumbles]

A moment would only be created if there was a fixed joint at the connection to the truss below B, which is unlikely, as the hook on the end of the motor would act as a pin.

 

Taking moments about point A, we get a vertical load acting down along the 5m cross section of 100kg plus truss spread as a UDL. This is balanced as the truss is in equilibrium by a reactant force upwards at point B. As you say, if the truss was horizontal, then this load would be half of the UDL, plus the weight of the motor, but as you only have vertical loads, and your supports are only acting vertically, the load will stay the same, with half the UDL plus the motor on each point.

 

As far as the supports see the situation, it is identical to having a 5m span of truss hangin horizontally, with a 2m drop, and the angle of the truss wouldn't have any effect.

 

HTH,

 

M

[dave_m]

Cheers for that mumbles, you've cleared something up that I've been thinking about for a while now! It was reading about forces acting on bridles that did it!

[Goran Mitic]

Ola,

 

I believe, that lower point (B) will bear more load.

 

Plus, making one point higher will make dead-hangs a bit angled (V shaped) since point A ON Truss is going to move greater towards point B ON Truss, and point B ON truss is going to move less towards point A ON Truss,

(since moving A up will produce B bearing more load. Moving truss totally vertical will end in point A ON truss being vertical above point B ON truss, and point A bearing nothing, point B all)

 

 

imo

[roryfm]

Mumbles is correct that the lines will bear identical loads (if the lines are vertical as shown in the diagram), however goran is also correct in stating that the drops will not be vertical if the hoists are 5m apart and the truss is at 45 degrees.

 

Assuming the hoists are 5m apart the pickups would have to be off the ends to keep the lines vertical (5m across, 5m drop as shown gives a hypotenuse of √50 = 7.07m)

 

If the lines were to be vertical and the points on the truss were to be one metre in from each end (ie 5m apart) the hoists would be 5*cos45 = 3.54 m apart. This would give equal loads on both points equivalent to 50kg.

 

 

 

This can be calculated by modelling the truss as a rigid body, with mass 100kg acting in the centre (2.5m from each point, 3.5m from the ends) As the truss is in equilibrium (its not moving) moments about any point will sum to zero

 

Letting the tension in line A being equal to Ta and tension in line B equal Tb

 

Taking clockwise moments about pickup A on the truss gives:

 

100*g*2.5*cos45 - Tb*5*cos45 = 0

 

therefore Tb = 100*g*2.5*cos45)/(5*cos45) = 50*g N

 

 

Taking clockwise moments about pickup B gives:

 

 

-100*g*2.5*cos45 + Ta*5*cos45 = 0

 

therefore

Ta = (100*g*2.5*cos45)/(5*cos45) = 50*g N

 

Which shows that Ta = Tb at 50*g Newtons

 

 

hope this clears things up a little, PM if you want me to email a diagram

 

ror

 

PS I'm not a rigger, just an (electrical) engineering student that should be revising for the first of my finals tomorrow...(yeah on a SATURDAY)

 

edit... I'm now away to think about what would happen if the points were 5m apart, and the lines weren't vertical, so I may prove that mumbles was right all along.

[Goran Mitic]

about non equal load bearings on A and B

 

if two people are taking wardrobe (case) from bottom floor to top floor(for funny reason let say ten store building), the one being on the lower side will bear much more weight.

 

Or have I always taken the upper side for equal share? :) Hell!!!

[mumbles]

yes... but the person taking the weight at the bottom isn't taking the weight simply in a vertical direction, they are also acting in the opposite direction to movement.

 

As I said before, and as rory has said again and demonstrated, if you take moments about either point, you must have a moment of zero or the truss will be rotating. From this you can caluclate the forces on the supports at either end, and you will find them to be equal, (or slightly out as the points on the truss aren't 7.07 m apart).

 

If the truss isn't symmetrical this wouldn't apply, and likewise if the loadings weren't symmetrical, so if you aren't sure about doing the calculations yourself make sure you get someone who can, as the OP said they would.