Engineering problem

[CatNap3]

Compression load calculation
The problem: imagine a right triangle
The vertical leg is 10 ft long point a to b
The horizontal leg to the right is 4ft long point b to c
Imagine a mast with a spreader
The angle at a is 22digrees (b to c)
Angle at b is 90 degrees
A
|
|
|
B———-C
The vertical load at c is 37,500 lbs (down)
What is the compression load c to d ??
Thanks if you can help !!
Nick

 

[Rick486]

Compression load calculation
The problem: imagine a right triangle
The vertical leg is 10 ft long point a to b
The horizontal leg to the right is 4ft long point b to c
Imagine a mast with a spreader
The angle at a is 22digrees (b to c)
Angle at b is 90 degrees
A
|
|
|
B———-C
The vertical load at c is 37,500 lbs (down)
What is the compression load c to d ??
Thanks if you can help !!
Nick

Nick:
1. You need a more detailed drawing to allow someone to help.
2. Where is point d?

 

[shemandr]

Shouldn't the 22 degree angle be from a to c? The angle at b should be a few degrees less than 90 because spreaders are mounted slightly raised on the tip - say 88 degrees. Otherwise I think it's a trig problem which I happily left back in 10th grade. I'm almost bored enough to give it a go.

 

[dlochner]

Nick:
1. You need a more detailed drawing to allow someone to help.
2. Where is point d?

I think there are some typos.

He wants the compression load on the spreader which is the side bc of the triangle.

Angle A is 22°, Angle B is 90°, C = 90-22= 78°

The spreader is BC which is 4 feet.

I don't think there is enough information here. The solution is going to involve trig, which was not my strong suit.

 

[Hayden Watson]

Assuming that your geometry is correct and the tension in the shroud is 37,500 lbs and the shroud below the spreader is parallel to the mast you get the following.
Tension in shroud = 37,500 lbs
Compression in mast = 34,770 lbs (assuming that your uppers are just going slack at max load)
compression on spreader = 14,050 lbs.

22º seems like a pretty big angle for masthead to upper shroud.

Formula = c*SIN-A = 37,500 lbs. * SIN-22º = 37,500 lbs. * .3746 = 14,050 lbs.

 

[CatNap3]

Nick:
1. You need a more detailed drawing to allow someone to help.
2. Where is point d?

Point d should be point b

Shouldn't the 22 degree angle be from a to c? The angle at b should be a few degrees less than 90 because spreaders are mounted slightly raised on the tip - say 88 degrees. Otherwise I think it's a trig problem which I happily left back in 10th grade. I'm almost bored enough to give it a go.

forget the word spreaders it’s not a mast question
angel at a is 22deg
Angel at b ois 90deg
Point d should be point b

Assuming that your geometry is correct and the tension in the shroud is 37,500 lbs and the shroud below the spreader is parallel to the mast you get the following.
Tension in shroud = 37,500 lbs
Compression in mast = 34,770 lbs (assuming that your uppers are just going slack at max load)
compression on spreader = 14,050 lbs.

22º seems like a pretty big angle for masthead to upper shroud.

Formula = c*SIN-A = 37,500 lbs. * SIN-22º = 37,500 lbs. * .3746 = 14,050 lbs.

thanks for the help
its not a mast problem
the mast/spreader was just to give a visualtion of th

 

[Will Gilmore]

its not a mast problem

Since you asked for a

Compression load calculation

The compression load on the masthead will be as Hayden, suggests.

Compression in mast = 34,770 lbs

However, the compression on the mast itself will gradually increase as you get farther down the mast and the weight of the mast is added in. Include the spreader weight, but no need to include the cable weights since the tension already includes that. However, if you really want to get precise: assume the tension is measured at just above deck level. That means the tension at the top of the mast WILL include the cable weight as an added factor to the tension at the mast head.

Are you all following all that?

We can't speak numbers because we don't know how much the mast and cables weigh.

If you want to know what size compression post to go with below deck. What material are we talking? Yellow pine, for example, has a longitudinal compression resistance in the neighborhood of 2000 lbs per square inch. You would want something like 20 square inches of cross section (4.75" x 4.75"). That's a pretty big timber.

-Will (Dragonfly)

 

[Tedd]

Strictly speaking, if the compression load was purely vertical and could only be purely vertical then there would be no load in the angled member (a to c). If the compression load is, say, a mass that's not constrained horizontally, then you need to be looking at dynamic interactions, which gets a lot more complicated.

It would also help to know what the degrees of freedom are at each apex. Are these pinned or socket joints that have rotational freedom, or are they clamped in some way?

 

[Will Gilmore]

in re-reading your original question, you are actually asking how much compression force is the on the member running between B and C.
If you consider this a mast spreader illustration with upper shroud running freely from masthead at A to shroud end at C and then down to turnbuckles, you're saying the shroud tension is 37,500 pounds. Without accounting for any resistance to the downward pull on the cable, or the weight of the cable itself, the tension should be the same on either side of C, both from A to C and from C downward to the unlabeled turnbuckle.
A force diagram should look something like this

Although, this is a simplistic diagram that does not show the angle of the spreaders. Ideally, the spreaders are angled up to bisect the angle of the shroud above and below the spreader end. By setting the spreaders horizontal, the forces change slightly. The compression force is calculated for the bisecting angle only.

-Will (Dragonfly)

 

[CatNap3]

in re-reading your original question, you are actually asking how much compression force is the on the member running between B and C.
If you consider this a mast spreader illustration with upper shroud running freely from masthead at A to shroud end at C and then down to turnbuckles, you're saying the shroud tension is 37,500 pounds. Without accounting for any resistance to the downward pull on the cable, or the weight of the cable itself, the tension should be the same on either side of C, both from A to C and from C downward to the unlabeled turnbuckle.
A force diagram should look something like this

View attachment 176644
Although, this is a simplistic diagram that does not show the angle of the spreaders. Ideally, the spreaders are angled up to bisect the angle of the shroud above and below the spreader end. By setting the spreaders horizontal, the forces change slightly. The compression force is calculated for the bisecting angle only.

-Will (Dragonfly)

This actually wasn’t a mast question. Unfortunately I used it as a reference but it was a bad one.
The question was actually about the compression load on a spreader bar.
The total load is 75000lbs, 37500lbs on each side of the spreader bar.
The angel is 42deg.
What I couldn’t figure out is the compression force on the spreader bar.
/\ A. Load on crane hook=75,000#. (This = a 5:1 safety factor)
/ \ Angel at A = 44dig.
/ \ Distance A to B/C = 10’-4”
/ \ Distance B toC = 8’-0”
/ \ Compression load B to C = ?? (Spreader bar)
————————
B. C. Load at B/C = 37500# down
Any help would be appreciated!! Nick

 

[CatNap3]

This actually wasn’t a mast question. Unfortunately I used it as a reference but it was a bad one.
The question was actually about the compression load on a spreader bar.
The total load is 75000lbs, 37500lbs on each side of the spreader bar.
The angel is 42deg.
What I couldn’t figure out is the compression force on the spreader bar.
/\ A. Load on crane hook=75,000#. (This = a 5:1 safety factor)
/ \ Angel at A = 44dig.
/ \ Distance A to B/C = 10’-4”
/ \ Distance B toC = 8’-0”
/ \ Compression load B to C = ?? (Spreader bar)
————————
B. C. Load at B/C = 37500# down
Any help would be appreciated!! Nick

My attempt at a dwg didn’t work.

 

[Will Gilmore]

As per diagram

For T equals the tension in the cable,
ang A equals the angle between stay and mast head,
ang l equals the angle of the cable below the spreader,
the formula is;
T (sin (ang A) - sin(ang l)).
(For angle l = 0, ang A = 44 deg).
(sin 44 deg) = 0.6947).
37,5000 (0.6947 - 0) = 26051.25
Then
Multiply it by 2, since there is equal tension from both sides.

-Will (Dragonfly)

 

[Davidasailor26]

I think Will’s got it, except as I read the question 44 degrees is the total angle between the two cables, so you’d need to use sin(22) instead of sin(44). Assuming that’s correct, total load is 37500*sin(22)*2 = 28,095 lbs

 

[Hayden Watson]

This is different than want you asked for in the first question. the load on the bottom of the spreader bar is not equal to the cable above going to the hook.
Load = 37,500 lbs
cable @ 22º = 40,445 lbs each cable
spreader bar = (2) x 15,150 lbs = 30,302 lbs.

 

[Davidasailor26]

This is different than want you asked for in the first question. the load on the bottom of the spreader bar is not equal to the cable above going to the hook.
Load = 37,500 lbs
cable @ 22º = 40,445 lbs each cable
spreader bar = (2) x 15,150 lbs = 30,302 lbs.

Yep, upon closer inspection I think you’re right. The load on the cable below the spreader is 37,500, but above it must be more than that because of the angle.

This is not like a spreader that is fixed to a mast where the shroud can run continuously through it and have a uniform load. In this case the spreader must be fixed to the cable, or else it would be forced down the cable.

 

[Apex]

ahhh, MATH problems! Fun for some, and frustrating at times.
Cable tension provides the same load along it's entire length. When the "spreader" re-directs force direction, it then begs for vector math:
Here is a quick google search for cable force vectors and something that looks like your question:

Cable AB exerts a force of 80 N - Question Solutions

Cable AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom.

www.questionsolutions.com

 

[CatNap3]

As per diagram

For T equals the tension in the cable,
ang A equals the angle between stay and mast head,
ang l equals the angle of the cable below the spreader,
the formula is;
T (sin (ang A) - sin(ang l)).
(For angle l = 0, ang A = 44 deg).
(sin 44 deg) = 0.6947).
37,5000 (0.6947 - 0) = 26051.25
Then
Multiply it by 2, since there is equal tension from both sides.

-Will (Dragonfly)

So I presume that the total compression load on the spreader bar is aprox 52,000#
At present I’m taking care of my neighbors chickens, only 12, a couple of goats and a pig.
Thanks for your help !! Nick

 

[Hayden Watson]

ahhh, MATH problems! Fun for some, and frustrating at times.
Cable tension provides the same load along it's entire length. When the "spreader" re-directs force direction, it then begs for vector math:
Here is a quick google search for cable force vectors and something that looks like your question:

Cable AB exerts a force of 80 N - Question Solutions

Cable AB exerts a force of 80 N on the end of the 3-m-long boom OA. Determine the magnitude of the projection of this force along the boom.

www.questionsolutions.com

No spreader bar that i have ever used has a single cable that connects from the hook to the load. if it did, the spreader would slide down the cable until it hit the load. With a spread bar as the OP described in post #10 a cable will go from the crane hook to a padeye on the top of each end of the spreader bar. Due to the 22º incline of this cable (hypotenuse) it will have tension greater than the actual load.

 

[Davidasailor26]

So I presume that the total compression load on the spreader bar is aprox 52,000#
At present I’m taking care of my neighbors chickens, only 12, a couple of goats and a pig.
Thanks for your help !! Nick

Keep reading down the thread. Closer to 30,000#.

 

[Will Gilmore]

No spreader bar that i have ever used has a single cable that connects from the hook to the load. if it did, the spreader would slide down the cable until it hit the load. With a spread bar as the OP described in post #10 a cable will go from the crane hook to a padeye on the top of each end of the spreader bar. Due to the 22º incline of this cable (hypotenuse) it will have tension greater than the actual load.

My thoughts are, No. This is a simple redirection. If the spreader put more tension on the cable above, it would do so below, as well. The tension above would be greater than at the bottom only because of the weight of the cable.
If the angle were 90 deg. You're suggesting the force above the spreader would be nearly infinite. Torque, may come into play, but that has more to do with the angle of the spreader, as a lever arm, to the direction of force.

-Will (Dragonfly)