Bending Moment in bar 1

[leisure17]

Am I right in thinking the BM in the bar shown is the resultant load R multiplied by the perpendicular distance to the opposite pin?. I suppose it's similar to a spreader bar but the few calcs I've seen for spreader bars haven't taken bending into account and I'm sure there must be some.

 

[electricpete]

First glance - my vote is bending moment in the bar is zero.

=====================================
(2B)+(2B)' ?

 

[desertfox]

Hi leisure17

Moment in the bar is zero, the wires act as ties and the centre bar is acting as a srut.

desertfox

 

[hydtools]

The bar is a two-force pinned member, no moment. Just compression.

Ted

 

[rb1957]

assuming pinned ends ... (which, of course, they are in practice)

 

[rb1957]

you're picture is correct, the wires carry tension load, the spreader bar reacts the (compression) component of the diagonal wire

P horizontal wire = P/2
P diagonal wire > P/2 = P/2/sin(theta)
(theta measured to the vertical)
Pspreader = (P/2/sin(theta))*cos(theta)

 

[rb1957]

to clarify (?) the side loads on the spreader (from the diagonal wire and from teh horizontal wire) balance out, and there is only axial compression in the spreader.

 

[lisa247]

The horizontal wires react the horizontal component of the applied force, so for each side the tension in the wire will be: P/2 x cos O
The bar reacts the vertical component of the applied force, so the compression in the bar will be:
P/2 x sin O
Where O is the angle in degrees. Beacuse both angles are the same (45 degrees) then both values should be the same.
If you imagine taking the bar away, you will hopefully be able to picture the horizontal wires wanting to come together - this is caused by the forces which are putting the bar into compression.

 

[leisure17]

Thanks Guys, yes I understand all that but for some reason can't shake the idea that there will be a resultant load (R in diagram) of the horizonal wire and sloping wire trying to act on it also.
I will try and wipe it from my memory banks.

 

[Ussuri]

The way you have drawn it, with a resultant acting at angle/2 you assumed the load in both wires (the horizontal one and the one at 45 degrees) is the same. I.e if you were running a wire over a sheave for instance and the wire was a single wire.

In the case you have above the load in the horizontal wire P/2 is different to the load in the 45 degree wire P/(2 cos 45).

 

[lisa247]

Try to think of the horizontal and angled wires as completely separate things than arent attached together and can move independently of one another.

 

[leisure17]

lisa247 that's made it clear now.

 

[rb1957]

i understand your problem ... you're assuming the tension in both wires (horizontal and diagonal) is the same.

for me the diagonal wire is balanced by compresison in the spreader and tension in the hoizontal wire ... the two react components of the diagonal load. The spreader has one wire for the load, and another for the support.

if it is a continuous wire, and there's a roller at the ends, then i think your picture holds. i think there's a physical problem keeping the spreader in place, it's perfectly neutrally balanced ...

try this.
imagine you have two diagonal wires holding the load, both in tension. now you come along with a prop and push it into the wire, bending it. the prop supplies the force R to the wire, balanced by the tenisons in the wire ... the tensions are the same (as the wire is continuous). rotate the prop away from the resultant and yes there's bending in it, but there's also a problem with it slipping along the wire.

 

[btrueblood]

"bending moment in the bar is zero"

Until it buckles.

 

[hydtools]

The bar has no fixed ends and therefore cannot have a bending, moment reaction. Pinned ends cannot have a moment reaction.

Ted

 

[rb1957]

ok, we understand a conventional spreader arrangement.

but, assume the load is supported by two wires, so that the wires have constant tension.

now take a prop, which has a roller (or a nicely radiused end) and bend the wire. The tension in the wire is still constant, the prop bisects the angle between the deflected wire portions (as shown in the pic above).

now assume you've done this to both wires. it seems to me that you could have a bowed spreader, the spreader would take the shape of an arc. it is clearly a very "ropey" sort of structure, the ends of the spreader are just begging to slip out, it is a severely bent column ('cause it'll be in compression), but there would be bending at the mid-point.

 

[btrueblood]

"The bar has no fixed ends and therefore cannot have a bending, moment reaction"

Until it buckles.

 

[btrueblood]

Or, as Rb notes, if there is any eccentricity in the bar.

 

[dhengr]

Generally speaking, there is no moment in the vertical bar in your sketch. Certainly, you could do something with the end fittings and the eccentricities btwn. the lines of force in the cables to cause a moment in that bar, that is top and/or bottom fixed end moments, but you would try to minimize these. There will be no moment in the vertical bar if the lines of force in the cables meet on the center line (C.G.) of the vertical bar. Any eccentricities must be resolve right there at the fittings and connection detail, and then they are reacted by the moment cap’y. of the vertical bar. My biggest issue with your sketch is that you have an unstable structural system unless you put at least one diagonal from one of the wall supports to the opposite (furthest, diagonally opposite) end of the vertical bar. Top at wall to bot. of bar diagonal is in tension; bot. at wall to top of bar must be a compression strut.

 

[leisure17]

Thanks dhengr, it was a diagramatic sketch only so I could illustrate what I was thinking regarding just the forces on the end of the bar. Thank's to everyone it has been fully explained.
Mick
Roll on Saturday