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10 Guys and Anchors (10m)

Instructions:

  1. Cover the following content as a group (either reading out loud or independently) then give an opportunity to answer any questions.
  2. Have students do the review questions independently, then take up answers.
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Figure 21 PISA anchor

Power installed anchors (PISA) are used to carry the strain of overhead conductors attached to poles in conjunction with guy wires and fiber strain insulators. PISA anchors are screwed into the earth to anchor the structure’s load. Anchors come in a variety of sizes and either single or multiple flights are used dependent upon the soil conditions and the amount of strain to be carried.

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Figure 22 Large single PISA (power installed screw anchor) anchor attr. hubbell.com
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Figure 24 Double PISA, attr. uus.coop Hubbell Inc
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Figure 23Single PISA complete with anchor rod and triple eye nut
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Figure 25 PISA being installed

Applications in Line Construction (30m)

You have learned how to calculate the angles and lengths involved with right angle triangles. You can use this information and the parallelogram method to solve tensions for guys, conductors, and the stress on poles which we call compression.

Example 1

In the drawing below, a line 15m in the air has 800 lbs. of tension with an anchor 15m from the pole.

image
Figure 26 Attr. SaskPower Training Manual.

Find:

  1. The length of the guy
  2. The tension on the guy
  3. The compression on the pole

 

Solution

a) Length of Guy

c2=a2+b2

c2=152+152

c = 21.2m

b) Tension on guy

Since the two sides of the right angle triangle are of equal length, the angle is 45 degrees. Find the hypotenuse:

cosine theta equals adjacent over hypotenuse

the cosine of 45 degrees equals 800 over h of y p

0.707 equals 800 over h of y p

0.707 times h of y p equals 800

h of y p equals 800 over 0.707

h of y p equals 1131.5 l b s

c) Compression on the pole

800 lbs., because a 45 degree right angle triangle is equal on both sides.

Example 2:

image
Figure 27 Attr. SaskPower Training Manual

From the above example, what happens when the anchor is moved in 5m as shown in the diagram below?

Find:

  1. Length of guy
  2. Tension on guy
  3. Compression on pole

 

 

Solution

a) Length of Guy

c squared equals a squared plus b squared

c squared equals 10 squared plus 15 squared

c equals 18.03 m

b) Tension on Guy

cosine theta equals adjacent over hypotenuse

cosine theta equals 10 over 18.03 equals 56.3 degrees

Therefore,

cosine 56.3 degrees equals 800 over h of y p

0.555 times h of y p equals 800

h of y p equals 800 over 0.555 equals 1441.4 l b s

c) Compression on Pole

tangent theta equals o p p o s i t e over a d j a c e n t

1 lines Line 1: the tangent of 56.3 degrees equals o p p o s i t e over 800

1.499 equals o p p over 800

o p p equals 1199.5 l b s

We can conclude from the two examples that if the anchor is shortened, there will be more tension on the guy and more compression on the pole.

Example 3:

Weight with Slings

To calculate the tension on a sling, you must realize it would take two slings to raise the load. The load is effectively split 50% on one sling, and 50% on the other.

Figure 28 Attr. SaskPower Training Manual.

If you make a vector out of the sling angles, you can solve the tension easier.

Figure 29 Attr. SaskPower Training Manual.

You know that the slings can be represented as two straight lines as shown below.

Figure 30 Attr. SaskPower Training Manual.

These can be solved using vectors in much the same manner as the pole and guy wire examples.

Figure 31 Attr. SaskPower Training Manual.

sine theta equals opposite over hypotenuse

T e n s i o n S l i n g equals the sine of 30 degrees equals 50 over cap T

T equals 100 l b s period p e r s l i n g

cosine theta equals 50 over cap T

T e n s i o n S l i n g equals cosine 60 degrees equals 50 over T

T equals 100 l b s period p e r S l i n g

NOTE: If the sling angle was reduced from 30 degrees to 10 degrees, there would be 288 lbs. of tension on the slings. (Even though the load is only 100 lbs.)

 

Review Questions: Guys and Anchors (30m)
  1. Use the diagram below to find the following:
    1. Length of guy-wire
    2. Angle of the guy-wire to the ground
    3. Tension on guy-wire
    4. Compression on the pole
  2. What would be the solution to Question 1 if the guy was moved in 4 metres?
    1. Length of guy-wire
    2. Angle of the guy-wire to the ground
    3. Tension on guy-wire
    4. Compression on the pole
  3. By looking at the values in Questions #1 and #2, why do we put 45 degree angles on our guys?
  4. The job at hand requires you to rig a 50kVA transformer using two 1-metre slings. The transformer weighs 350 kg. The slings must be rigged so the tension does not exceed 200 kg per sling.
    1. What must the angle be where each sling meets the transformer?
    2. What would the tension on each sling be if the above angle was decreased to 15 degrees?
Answer Key

1 a. 16.97m, b. 45 degrees, c. 4243N, d. 3000N compression on pole.

2 a. 14.4m, b. 56.3 degrees, c. 5406N, d. 4497N

3. causes less strain on the poles and guys. You don’t want to stress your guys out.

4. a. 61 degrees, b. 676 kg

License

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Powerline Tech Prep Program Manual Copyright © by Saskatchewan Indian Institute of Technologies-Trades and Industrial is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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