Static Forces physics problem / General Science and Electronics / Forums

Forums

4hv.org :: Forums :: General Science and Electronics

« Previous topic | Next topic »

Static Forces physics problem

Move Thread

LAN_403

IamSmooth

Fri Oct 10 2008, 01:30AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I want to solve a problem with three variables, but I can only come up with two unique equations. I have a 30' pole fixed at the ground with two guy wires attached to the same point at the ground 20' away. Guy wire A attaches at the top (56deg angle) and the other, B, 15' up (37deg angle). There is a 400lb force at the top (30' up) pulling against the guy wires. What are the tensions in guy wire A and B? Here is what I have so far:
Horizontal forces: Acos56 + Bcos37 = 400 + Fx (horizontal ground force acting on pole base)
Torque forces: 30Acos56 + 15Bcos37 = 400, using base as reference
Torque forces: 15BCos37 = 30Fx, using top of pole as reference
Vertical forces neglected
This gives:
1: 0.56A + 0.8B = 400 + Fx
2: 16.8A + 12B = 12000
3: 12B = 30Fx
However, subing 3 into 1 results in equation 2, so I still have two equations and three unknowns. Is there any way to solve this? What other equation am I missing?

G^3

Fri Oct 10 2008, 04:59AM

Registered Member #97 Joined: Thu Feb 09 2006, 05:40PM
Location:
Posts: 61

You have 3 equations and 3 unknowns so you should be good. Substitute 3 into 1 and then 1 into 2 and you should have everything in terms of one variable.

...

Fri Oct 10 2008, 06:41AM

Registered Member #56 Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445

No, his poin is that eq2 is the same as eq1 and 2
rewrite 3 to be Fx=.4B
rewrite 1 to be .56A+.8B=400+.4B -> .56A = 400 - .4B -> A = 714 - .714B
rewrite 2 to be 16.8(714-.714B) + 12B = 12000 -> 12000 - 12B + 12B = 12000 -> 0 = 0
.: no soln

I would say it can probably be done with only sum of torques
ie, ignore the x direction info, just to 0=T(400lb force)+T(guidewire1)+T(guidewire2)

Dr. Slack

Fri Oct 10 2008, 07:10AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Does this diagram represent the complete problem as given? If so, then the problem is overdetermined, what a civil engineer would call a hyperstatic structure, there is no unique solution in terms of a tensiin in each guy.

If the mast is assumed to be stiff, then either guy could be removed, and a tension could be found for the other. If you assume all components to be stiff, then an infinitessimal change in the relative length of the guys would throw all tension from one to the other.

However, if that *is* the complete problem, then you must assume a tension x in one guy, and find an equation in terms of x for the tension in the other.

In one common engineering solution to this type of problem, the peg at 20ft is a pulley, which forces the two tensions to be equal. In real life of course none of the components are stiff, in particular the mast can bend, and the middle guy is used to control the bend.

Steve Conner

Fri Oct 10 2008, 09:45AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Yeah, it's statically indeterminate, you need to leave out one of your guy wires to get a solution

IamSmooth

Fri Oct 10 2008, 04:17PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

Dr Slack and Steve,

Thank you for clarifying what I suspected. This is a simplification of a wind tower I am building for a wind turbine. I designed it so each guy wire can withstand the tension generated by the thrust load, but I was curious if there was a way to solve the system for the two guy wires simultaneously.

Dr. Slack

Sat Oct 11 2008, 09:19AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Each guy operates differently when supporting the wind load by itself.

The top throws only a compression load onto the mast, and so the mast only has to be of adequate geometry to avoid buckling (and withstand the compression load obviously, but that's easy), or be guyed in the middle to stabilise it against buckling.

The middle guy causes a bending load, so the mast would have to be designed as a beam to resist the bending moment, possibly making for a far heavier structure.

For redundancy against guy failure, I'd be inclined to attach multiple guys to the top of the mast, either 5 or 6 radially, or 3 paired guys. Then either make the mast a lattice with no greater length/diameter ratio than 20 to avoid buckling. Or make it thinner and guy in the middle to steady it, each section between guys having a local L/D of <20.

Steve Conner

Mon Oct 13 2008, 10:16AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

In that case, I'd suggest calculating as if the top guy wires were the only guy wires, and taking all the load.

Then add guy wires to the middle, and assume these take none of the thrust load, but only stabilize the mast against buckling. (But use the same kinds of wires and fittings as for the top guys, so they could withstand the same tension.)

In practice, the middle guys will take some of the thrust load, but the bending moment they generate on the mast in doing so, when combined with the forces from the top guys, is the very thing that stabilizes it against buckling. So this method should err on the safe side.

coillah

Thu Oct 16 2008, 08:42AM

Registered Member #1517 Joined: Wed Jun 04 2008, 06:55AM
Location: Chico CA
Posts: 304

There is a way to solve this problem.

To get a third equation you have to include deflection, that is, how much the pole bends under the stresses provided by the tension in the wires. If you do not know this value, or cannot assume one, then this problem will remain indeterminate.

By knowing this value you can use Young's Modulus to generate a third equation containing your deflection and your tensions. However this will require an analysis of a couple of parts of the beam, but is a problem that is solved within the first 4 weeks of a strength of materials course.

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob