Posted by *Curbie* on February 24, 2009, 4:24 pm

wrote:

*>> If four guy anchor/wires are properly placed at 0, 90, 180, & 270*

*>> degrees apart and the wind coming in from 45 degrees is the horizontal*

*>> wind load evenly dispersed between anchors/guys at 0 and 90 degrees*

*>> and the base?*

*>>*

*>> Or maybe more properly asked: given the same arrangement, if the wind*

*>> coming in from 0 degrees is the horizontal wind load evenly dispersed*

*>> between anchor/guy at 0 and the base?*

*>>*

*>In the case where the wind is coming 'from' the same direction as a guy, *

*>that guy will have to carry all the wind force and the other guys will have *

*>no force at all (except for any 'pre-load' tension in them).*

*>The tension in the guy will be *more* than the wind force alone. If the guy *

*>is at a 45 degree angle, the tension in the guy will be about 1.414 ( *

*>1/sin(45)).*

Where does (1/sin(45)) come into play? (1/sin(45)) = 1.175221363

*>The force on the base will be equal to all the weight plus an amount equal *

*>to the wind load (if the guy is at 45 degrees).*

*>So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload *

*>will have a guy wire tention of 141 lbf and the tower base will be *

*>supporting 150 lbf.*

In other words, a horizontal 100 lb force wind load on a tower/turbine

will present an additional 50 lb diagonal force.

Force on guy is horizontal 100 lb force wind load x 1.414

Force on base is weight of tower/turbine + horizontal 100 lb force

wind load + additional 50 lb diagonal force

*>hope this helps*

*>daestrom*

Thanks.

Curbie

Posted by *Curbie* on February 24, 2009, 5:26 pm

Ok, moron alert 1,414 = 1 / SIN(RADIANS(45))

I was so busy with the overall concept that I didn't convert degrees

to radians. (moron)

For those who are trying to follow this thread either now or in the

future.

Curbie

wrote:

*>wrote:*

*>>*

*>>> If four guy anchor/wires are properly placed at 0, 90, 180, & 270*

*>>> degrees apart and the wind coming in from 45 degrees is the horizontal*

*>>> wind load evenly dispersed between anchors/guys at 0 and 90 degrees*

*>>> and the base?*

*>>>*

*>>> Or maybe more properly asked: given the same arrangement, if the wind*

*>>> coming in from 0 degrees is the horizontal wind load evenly dispersed*

*>>> between anchor/guy at 0 and the base?*

*>>>*

*>>*

*>>In the case where the wind is coming 'from' the same direction as a guy, *

*>>that guy will have to carry all the wind force and the other guys will have *

*>>no force at all (except for any 'pre-load' tension in them).*

*>>*

*>>The tension in the guy will be *more* than the wind force alone. If the guy *

*>>is at a 45 degree angle, the tension in the guy will be about 1.414 ( *

*>>1/sin(45)).*

*>Where does (1/sin(45)) come into play? (1/sin(45)) = 1.175221363*

*>>*

*>>The force on the base will be equal to all the weight plus an amount equal *

*>>to the wind load (if the guy is at 45 degrees).*

*>>*

*>>So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload *

*>>will have a guy wire tention of 141 lbf and the tower base will be *

*>>supporting 150 lbf.*

*>In other words, a horizontal 100 lb force wind load on a tower/turbine*

*>will present an additional 50 lb diagonal force.*

*>Force on guy is horizontal 100 lb force wind load x 1.414*

*>Force on base is weight of tower/turbine + horizontal 100 lb force*

*>wind load + additional 50 lb diagonal force*

*>>*

*>>hope this helps*

*>>*

*>>daestrom*

*>Thanks.*

*>Curbie*

Posted by *daestrom* on February 25, 2009, 1:04 am

*> wrote:*

<snip>

*>>The tension in the guy will be *more* than the wind force alone. If the *

*>>guy*

*>>is at a 45 degree angle, the tension in the guy will be about 1.414 (*

*>>1/sin(45)).*

*> Where does (1/sin(45)) come into play? (1/sin(45)) = 1.175221363*

45 degrees converted to radians...

*>>*

*>>The force on the base will be equal to all the weight plus an amount equal*

*>>to the wind load (if the guy is at 45 degrees).*

*>>*

*>>So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload*

*>>will have a guy wire tention of 141 lbf and the tower base will be*

*>>supporting 150 lbf.*

*> In other words, a horizontal 100 lb force wind load on a tower/turbine*

*> will present an additional 50 lb diagonal force.*

*> Force on guy is horizontal 100 lb force wind load x 1.414*

*> Force on base is weight of tower/turbine + horizontal 100 lb force*

*> wind load + additional 50 lb diagonal force*

Uh... no, not quite right.

A 100 lbf horizontal load must be countered by the guy wire. So the guy

wire, because its at 45 degrees, has 141 lbf tension in it. That 141 lbf in

the guy at 45 degrees breaks down into a 100 lbf vertical component and 100

lbf horizontal component. Of course the horizontal component is just what

we wanted to counteract the wind force (also horizontal).

But the 100 lbf vertical component of the guy wire *adds* to whatever the

weight of the windmill and tower is. I misspoke before in saying:

*>>So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload*

*>>will have a guy wire tention of 141 lbf and the tower base will be*

*>>supporting 150 lbf*

It was supposed to read 50 lbf *gravity* load. (I know a 50 lb windmill is

tiny). But the point I was trying to make is that this vertical component

of the guy force is pulling down, adding to the load exerted on the base.

At a 45 degree guy wire angle, it just happens to work out that whatever the

side load of the wind is, the additional downward force caused by acting

through the guy wire is the same amount. 100 lbf wind load horizontal gets

turned into 100 lbf *additional* down force on the base.

And the wire tie down has a similar force. The 141 lbf diagonal force in

the guy wire will pull *up* on the base with 100 lbf and *sideways* on the

guy (toward the tower of course) of 100 lbf.

daestrom

Posted by *vaughn* on February 24, 2009, 12:37 am

*> So a 100 lbf wind load on a tower/turbine that presents a 50 lbf windload *

*> will have a guy wire tention of 141 lbf and the tower base will be *

*> supporting 150 lbf.*

Plus the reaction from the combined preload on the guys (depends on angle

of guys with ground) plus the total weight of mast and equipment.

Vaughn

Posted by *vaughn* on February 24, 2009, 4:35 pm

Just a point of personal experience here: Don't undersize your base

foundation on a guyed tower on the theory that everything else is more

important to holding up the tower. As we have established in this thread,

most of your stresses somehow end up combining right there!

Because of the recent spate of hurricanes here in Florida, I was

involved in the re-engineering and subsequent upgrading of a couple

communications towers so that they could survive a higher wind speed. In

each case, the limiting factor turned out to be the foundation at the base

of the tower. There is a point where adding or strengthening guys will not

help, if there is no way of beefing up the base foundation.

Vaughn

>> If four guy anchor/wires are properly placed at 0, 90, 180, & 270>> degrees apart and the wind coming in from 45 degrees is the horizontal>> wind load evenly dispersed between anchors/guys at 0 and 90 degrees>> and the base?>>>> Or maybe more properly asked: given the same arrangement, if the wind>> coming in from 0 degrees is the horizontal wind load evenly dispersed>> between anchor/guy at 0 and the base?>>>In the case where the wind is coming 'from' the same direction as a guy,>that guy will have to carry all the wind force and the other guys will have>no force at all (except for any 'pre-load' tension in them).>The tension in the guy will be *more* than the wind force alone. If the guy>is at a 45 degree angle, the tension in the guy will be about 1.414 (>1/sin(45)).