Bending Exam Prep
Conduit Bending Exam Questions and Answers
The conduit bending questions that show up on the apprentice exam, worked the way the test expects: the direct answer, then the one step that gets you there. The calculated answers are computed from the bend geometry by the same locked engine as the calculator, so you learn the method and not just the letter. Check any of them against your own numbers in the interactive 3D bender below.
Answer Key
| Question | Answer |
|---|---|
| 90 turn in 12 concentric bends, each bend | 7.5° |
| Multiplier for a 30° offset | 2.00 |
| Shrink per inch at 30° | 1/4" |
| Distance saved by a 90’s arc | gain |
| Developed length, 90° at 10" radius | 15.71" |
| Travel of a 45° vs a 90° | half |
| Cut length, 9 ft, two 90s, 3.5" gain each | 101" |
| Angle rising 14" over 34" run | 22.38° |
Worked Answers
Concentric bends: the angle of each bend
Q: What is the bend angle of each bend for a 90° turn made with 12 concentric bends?
A: 7.5°. A concentric 90 sweeps the turn through a run of equal small bends so a bank of parallel conduits keeps its spacing. Each bend is the total turn divided by the number of bends: 90 ÷ 12 = 7.5°. The same division answers any count: 90 in 8 bends is 11.25°, 90 in 6 bends is 15°.
The 30 degree offset multiplier
Q: What is the multiplier for a 30° offset?
A: 2.00. The multiplier is 1 ÷ sin(angle), the cosecant, and 1 ÷ sin(30°) = 2.00 exactly, which is why 30° is the angle everyone learns first. Multiply the offset height by it for the mark spacing. See the full offset multiplier chart for every angle.
Shrink per inch at 30 degrees
Q: The conduit shrink is how much per inch of offset when using 30° bends?
A: 1/4" per inch. Multiply the offset height by 1/4" to get total shrink, then subtract it from the distance to your first mark. A 4" offset at 30° shrinks 1". It never changes with conduit size; the full table is on the conduit shrink calculator.
Gain: the distance the arc saves
Q: What is the distance saved by the arc of a 90° bend called?
A: Gain. Because a 90 follows a curved arc instead of a sharp corner, the finished run is shorter than the two legs measured square to the corner, and that saved length is the gain. It comes in two references, so name the one you mean: the pure centerline arc saving is 1.93" for 3/4" EMT, and the field (outside-corner) gain you actually subtract from a tape leg is 2.85", exactly one pipe diameter larger. You take one gain off per 90 when cutting a bent piece. The conduit gain chart lists both by size.
Developed length of a 90 at a given radius
Q: What is the developed length of a 90° bend if the centerline radius is 10"?
A: 15.71". Developed length is the arc the centerline travels through the bend: L = radius × angle in radians = 10 × (π ÷ 2) = 15.71". It scales straight with radius and angle, so a 45 at the same radius is exactly half. The bend radius page works it by EMT size.
Travel of a 45 versus a 90
Q: The amount of travel of a 45° bend is half the amount of travel of what bend?
A: A 90° bend. The arc a single bend consumes is proportional to its angle at a fixed radius, and 45° is exactly half of 90°, so it travels half the arc (0.5 of it). This is the arc the bend itself sweeps, not the mark spacing of an offset.
Cut length with gain deducted
Q: What is the cut length for a 9 ft straight length with two 90° bends if the gain is 3.5" each?
A: 101". Take the straight length, 9 ft = 108", and subtract the gain for each 90: 108 − (2 × 3.5) = 101". One gain comes off per 90 because each bend saves that much versus a square corner.
Bend angle from rise over run
Q: What is the angle of a bend if the conduit rises 14" over a horizontal distance of 34"?
A: about 22.38°, the standard 22.5° bend. The sloped leg is the hypotenuse of a right triangle rising 14" while advancing 34", so its angle above horizontal is arctangent(14 ÷ 34) = 22.38°. Benders stop at standard angles, so you set the 22.5° mark.
How many bends to clear an obstruction
Q: How many bends are required if a conduit run must pass over an obstruction like a 12" I-beam?
A: four bends. A 12" I-beam is a wide, flat obstruction, so it calls for a four point saddle: two bends up and over to a flat top that spans the beam, then two bends back down, for 4 bends. A three point saddle (3 bends) would touch the beam only at its peak and would not clear the flanges, so it is the answer for a narrow, round obstruction like a single pipe, not a beam. Match the saddle to the obstruction: round and small takes a three point, wide and flat takes a four point. The saddle bend calculator lays out the marks for both.
Check Your Answer in 3D
Enter the numbers from your question and see the exact bend as an interactive 3D pipe you can spin, with the marks, travel, shrink, and developed length the exam is asking for. The picture is built from the same geometry as the answers above, so what you see is what you would bend.
Frequently Asked Questions
What is the bend angle of each bend for a 90 degree turn made with 12 concentric bends?
7.5 degrees. Spreading a 90 degree turn across a run of equal small bends (the exam calls this concentric; a bender calls it a segmented large-radius sweep) means each bend equals the total turn divided by the number of bends: 90 / 12 = 7.5 degrees. The same rule gives any count: 90 in 8 bends is 11.25 degrees each, 90 in 6 bends is 15 degrees each.
The amount of travel of a 45 degree bend is half the amount of travel of what bend?
A 90 degree bend. The developed arc a single bend consumes is proportional to its angle at a given radius, and 45 degrees is exactly half of 90 degrees, so a 45 travels half the arc of a 90. This is the arc the bend itself sweeps, not the mark spacing of an offset.
What is the cut length for a 9 foot straight length with two 90 degree bends if the gain is 3.5 inches each?
101 inches. Start with the straight length, 9 feet = 108 inches, then subtract the gain for each 90: 108 - (2 x 3.5) = 101 inches. Gain is subtracted once per 90 because each bend saves that much versus a square corner.
What is the angle of a bend if the conduit rises 14 inches over a horizontal distance of 34 inches?
About 22.38 degrees, which you round to the standard 22.5 degree bend. The sloped leg is the hypotenuse of a right triangle that climbs 14 inches while advancing 34 inches, so its angle above horizontal is arctangent(14 / 34) = 22.38 degrees. Benders are cut to standard angles, so you would use the 22.5 degree stop.
Study the Full References
Printable Cheat Sheet
Every multiplier, shrink, take-up, and gain constant on one page. Print it for the test bench.
Offset Multiplier Chart
Multipliers and shrink for 10° through 60° offsets, with worked examples.
Conduit Shrink Calculator
Shrink per inch of offset and kick by angle, and why pipe size does not change it.
Bend Radius and Developed Length
Standard centerline radius by EMT size and the developed length of a bend.
Conduit Gain Chart
Gain by EMT size and the cut-to-length formula for bent pieces.
Conduit Bending Calculator
All 7 bend types as an interactive 3D pipe with verified bender profiles.