Bending Book rev2, Study notes of Trigonometry

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REVISION IIREVISION IIREVISION IIREVISION II

Created By: David L Hensley

Local Union # 903 IBEW

Acknowledgment:

This little book was created from many different books, but mainly from the one that

has helped me to bend conduit for many years. Thank you to the man that wrote

“Professor Brown’s Guide to Conduit Bending”

30º

9/16” 6 9/16”

6”

2” (C–

2” (C–

2” (C–

adj

opp

Parallel Offsets Progression of Bends

When bending two or more offsets it is necessary to advance the centers of the bends for the progressive conduits in order to maintain an equal center to center spacing.

Multiply the (C-C) measurement of the conduits by the tangent of 1/2 the bend angle. Add this figure to the center of bend measurement of 1st conduit. This will be center of bend measure- ment of the 2nd conduit. Advance the center of bend measurement of each succeeding conduit by this figure.

Example:

tanӨ = opp adj

opp = tan 15º x adj (C-C)

opp = .2679 x 2”

opp =.5358’ or 9/16”

To Bend Kicks to Any Given Angle

After determining angle to use, bend this angle in a piece of scrap conduit. Measure from the front of the shoe to the center of the bend. This is the shoe factor (SF). Multiply the cosecant of the bend angle by the amount of kick. Add 1/2 O.D. of the conduit. This is the center of the bend measured from the back of the 90°. Deduct the (SF) figure and place the front of bending shoe on this mark. Pull through proper amount of travel for desired angle.

Example:

30° Kick 2” amount of kick 2” (amt of kick) x 2 (csc 30°) 4”

• 1/2 (1/2 O.D.) 4 1/2” (Center of Bend)

• SF Front of Shoe

4 1/2”

4” SF

1/2”

4”

2”

Center of Bend

30 ° bend^ Amount of Kick

Back of 90

Center of Bend (^) Front of Shoe

Kicks with Conduits running Perpendicular to Cabinet

2 7/8” C – C of KO’s hyp

2 1/2” C – C

2 1/2 “ C – C

cos Ө = adj hyp

Ө = 30º

cos (30º) = adj hyp

hyp = adj cos(30º)

hyp = 2.5” .

hyp = 2.887”

hyp = 2 7/8”

To find centers of KO’s in the cabinet and maintain centers (2 1/2”) of conduits, divide the (C – C) measurement by the cosine bend angle.

adj

3”

6”

A

B

C

D

E

F

1/4”

1/4”

3 7/8” Measure from corner (A) to center of conduit (B). Multiply this figure by 2. This is (C–D) measurement. Add the O.D. of the conduit and the depth of supports on each side and multiply this figure by 1.414. Subtract this figure from the (C–D) measurement. This figure is the (E–F) measure- ment, or center to center measurement.

Example:

(1) 3” (A–B) (2) 1.0” O.D. x 2 .25 support 6” (C–D) + .25 support 1.50”

(3) 1.414 (4) 6.00” (C–D) x 1.5” - 2.12” 2.121” 3.88” (E–F) (center to center)

1”

To Find Centers of 45° bends Having Conduit Supports

4 1/4” (C)

3 3/4” (C-C)

1”

1 1/2” (A)

1 1/2” (B)

To Find Centers of 45* bends with Square Obstruction

Add (A) and (B) measurements and multiply by 1. This is the back of the conduit (C). For the center to center measurement (C-C) deduct 1/2 O.D. of the conduit. Example:

(1) 1.5” (A) (2) 1.

• 1.5” (B) x 3” 3” 4.242” 4 1/4” (C) (3) 4 1/4” (C)

• 1/2” (1/2 O.D.) 3 3/4” (C-C)

Note: - If the conduit is on supports add the depth of the supports on each side and multiply by 1.414. Subtract this figure from (C-C) measurement.

- To obtain more clearance increase the (C) measurement 2” for each 1” of clearance.

Measure from (A) to (B). Mul- tiply by 2.4. This is to outside radius of the bend. Deduct 1/ O.D. of the conduit for the centerline radius. To obtain more clearance, increase the distance from (A) to (B).

Example:

1 1/2”

To Find a Radius Required to Clear an Obstruction

1”

3 5/8”

3 1/8”

1.5” (A-B) x 2. 3.6” 3 5/8” (outside radius)

• .5” 1/2 O.D. 3.1” 3 1/8” (centerline radius)

A

B

1 1/2”

1 1/2”

To Find Centers of 45° bends with a Square Obstruction

Multiply (A) measurement by 3. This is the back of the conduit (B). For center to center measurement (C-C) deduct 1/2 O.D. of the conduit.

Example:

(1) 1.5” (A) (2) 4.5” (B) x 3 - .5 (1/2 O.D.) 4.5” (B) 4.0” (C-C)

Note: - If the conduit is on supports add the depth of the supports on each side and multiply by 1.414. Subtract this figure from (C-C) measurement.

- To obtain more clearance increase the (B) measurement 2” for each 1” of clearance. .

4 1/2” (

B ) 4” (C-C)

1”

1 1/2” (A)

2”

1 1/2”

1

A

B

C 4”

3 3/4”

30°

30°

30°

3”

2 3/4”

2

A

B

C

To bend a 90° bend using 3 - 30° bends as shown:

Multiply side ‘A’ of by the cosecant of 30° , that equals side ‘C’ 2” X 2 = 4” Multiply side ‘A’ of by the cosecant of 30° , that equals side ‘C’ 1 1/2” X 2 = 3”

Multiply the tangent of 1/2 the bend angle by the O.D. of the conduit. Subtract this from the length of side ‘C’ in each for the center - center measurement. Tan 15° =. .2679 X 1” = 1/4” 4” - 1/4” = 3 3/4” 3” - 1/4” = 2 3/4”

1

2

1”

Bending a 90° Bend Using 3 - 30° Bends

Example:

Pythagorean theorem A² + B² = C²

C² = 4² + 3² C² = 25 C = √ 25 C = 5 (amount of offset)

To Figure Amount of Offset Needed for A Rolling Offset

1 2 3 4 5 6 7 8 9 10 11 12 13

13

12

11

10

9

8

7

6

5

4

3

2

1 5” (C) 6 3/8 (C) 7 3/4” (C)

4” (A)

3” (B)

5” (C)

Measure the distance the conduit has to offset up or down and the distance it has to offset right or left. Mark these two figures on any square and measure between them as shown. This will be amount of offset required.

C² = A² + B² C² = A² + B² C² = A² + B²

C² = 4² + 3² C² = 5² + 4² C² = 5.5² + 5.5²

C² = 25 C² = 41 C² = 60.

C = √ 25 C = √ 41 C = √ 60.

C = 5 C = 6.403 (6 3/8”) C = 7.778 (7 3/4”)

When laying out a slab to get a square wall use the carpenter’s rule called The 3, 4, 5 rule. Just like the triangle above, take any number multiply it by 3, 4, and 5 use these numbers to lay out a square wall.

5’ x 3 = 15’ C² = A² + B² 5’ x 4 = 20’ 25² = 15² + 20² 5’ x 5 = 25’ 25² = 625 25 = √ 625 25 = 25

Gain for 90º Bends

36”

(C)

7 3/4”

31 1/2”

• 36 ” (C actual amount of conduit) 3 1/4” (amount of gain)

Gain is the difference between the sum of right angle measurements of both legs of the 90º bend and the actual amount of conduit required to make the 90º bend. For standard shoes on Chicago benders the gain is 3 times the outside diameter of the conduit (+ or – a fraction of an inch).

To determine the gain of your bender take a scrap piece of conduit of the size needed. Use 3/4” for the example, and the length is 36”. Put the end of the conduit flush with the front of the shoe and bend a 90º. Measure both legs of the bend very carefully. Say that the two sides measured (7 3/4”) and (31 1/2”), add the two sides (7 3/4” + 31 1/2” = 39 1/4”) Subtract the original length of the conduit form this figure (39 1/4” - 36” = 3 3/4”). This is the gain for a 3/4” 90º.

This is used to cut and thread conduit before bending. Another use is to obtain a figure for back to back 90º bending. If you have back to back 90ºs that are 5 foot, 6 foot, or longer, it is more practical to reverse the conduit in the bender and bend up the short end of the conduit. By putting the original conduit even with the front of the shoe, you now have the stub-up length of the bender (7 3/4”). To obtain back to back figure subtract the gain from the stub up. ((7 3/4” - 3 3/4”) = 4”) Add this figure to the length of the back to back bends.

76”^ 4”

Front of Shoe

Circumference of a circle (360°) = D π = 2 R π

π = 3.

Circumference of 1/4 of a circle (90°) = Developed length

Dev. Length = R 2 π 4

Dev. Length = R 2 (3.1416) = R x 1. 4

Multiply the radius by 1.57 for the Dev. Length of a 90° bend. This is the amount of straight conduit required to make the bend.

4”D

2” R

Example:

To make a 90° bend with a 4” center line radius: Multiply the radius (4”) by 1.57 for the Dev. Length (6.28). Divide by one less than the amount of bends, for example 9 – 1 = 8 spaces. Bend 10° at each line. .785 or 3/4” Dev Length = 1.57 x 4” = 6.28” 8 spaces 6.28”

3/4 3/4 3/4 3/4 3/4 3/4 3/4 3/

1 2 3 4 5 6 7 8

6.28”

4” R

Concentric Bends

33 1/8” R

28 3/4” R

24 3/8” R

20” R

4 3/8”4 3/8” 4 3/8”

2” 2” 2” O.D. of 2” conduit is 2 3/8”

20” R

28 3/4” R

24 3/8” R

33 1/8” R

Example:

Bend four concentric 90° bends in 2” conduit with 2” spacing between con- duits. Use a 20” centerline radius for the smallest radius. The next radius will be 20” + the cen.– cen. measurement of the 2 conduits, which is 4 3/8”. The O.D. of the 2” conduit is 2 3/8” so 1/ O.D. (1 3/16”) + space (2”) + 1/2 O.D. (1 3/16”) = 4 3/8” center to center. The 2nd radius will be 24 3/8” The 3rd radius will be 28 3/4” The 4th radius will be 33 1/8”

Multiply each radius by 1.57 for Developed length 1 20” X 1.57 = 31.40 = 31 3/8” 2 24.375 X 1.57 = 38.269 = 38 1/4” 3 28.75 X 1.57 = 45.138 = 45 1/8” 4 33.125 X 1.57 = 52.006= 52” Divide each D.L. by 17 (# of spaces) 1 31.40 / 17 = 1.847 = 1 13/16” 2 38.269 / 17 = 2.251 = 2 1/4” 3 45.138 / 17 = 2.655 = 2 5/8” 4 52.00 / 17 = 3.058 = 3 1/ 16” Mark each conduit from the same starting point 17 spaces 18 marks. Bend 5°at each mark.

The chord is found by laying out 2 points on the circle. These points must be less than the diameter. Make these points (X and Y) an even measurement such as 6’, 8’, 10’ ect. From the center point of XY (O) measure at a right angle to the circle. This figure (A) is the height of the arc. To find the cord of 1/2 the arc use the Pythagorean theorem. To find the radius of the circle use the formula R= C² / 2(A).

C = 6240.25”

C = 78.9”

8’ (R)

6’ (B) 6’ (B)

32 1/2”

(C) (A)

(O)

To Find Radius of a Tank or Arc

XY — Chord of arc O — Center of Cord of arc A — Height of arc C — Cord of 1/2 the arc R — radius

A² + B² = C²

C = A² + B²

C = 32.5²” + 72²”

R = 6240.25 / 2(32.5)

R = 96.0” (8’)

R = C² / 2(A)

X (^) Y