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Advanced Rigger Certification Exam 2026/2027
Rigging Safety and Equipment Handling Study Guide.
Comprehensive 50-Question Assessment
Prepared for: NCCCO Advanced Rigger Certification Candidates
Total Questions: 50 (30 Multiple-Choice | 10 Calculation-Based | 5 Select-All-That-Apply | 5
True/False)
Time Allotment: 120 Minutes
Passing Score: 90%
DOMAIN 1: SLING ANGLES, TENSION & LOAD CALCULATIONS (14 Questions)
Sub-Topic 1.1: Load Angle Factor (LAF) Calculations (4 Questions)
Question 1 (Calculation-Based)
A rigger is using a two-leg bridle sling to lift a 12,000-pound steel beam. The horizontal angle
between each sling leg and the load is 30 degrees. Calculate the tension in EACH individual sling
leg using the Load Angle Factor (LAF) method.
Formula: Tension per leg = (Total Load / Number of Legs) × LAF
LAF at 30 degrees = 2.0
Answer: __________ pounds
[CORRECT: 12,000 pounds] Rationale:
Step 1: Determine the load per leg before angle factor: 12,000 lb / 2 legs = 6,000 lb per leg.
Step 2: Apply the Load Angle Factor for 30 degrees: LAF = 2.0.
Step 3: Calculate final tension per leg: 6,000 lb × 2.0 = 12,000 pounds.
Safety Logic: At a 30-degree horizontal angle, the tension in each sling leg equals the TOTAL load
weight, not half the load. This is why OSHA and ASME B30.9 strongly discourage sling angles
below 30 degrees. The horizontal force component creates a multiplier effect that doubles the
tension compared to a vertical lift.
Question 2 (Multiple-Choice)
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A rigger must lift a 20,000-pound generator using a two-leg sling assembly. The available rigging
configuration allows for a maximum horizontal sling angle of 45 degrees. The rigger has two
options: (1) use the 45-degree configuration with a standard LAF of 1.414, or (2) add a spreader
bar to increase the angle to 60 degrees with an LAF of 1.155. What is the tension reduction PER
LEG achieved by using the spreader bar?
A) 1,414 pounds
B) 2,590 pounds
C) 5,180 pounds
D) 10,360 pounds
[CORRECT: B] Rationale:
Step 1: Tension at 45 degrees: (20,000 lb / 2) × 1.414 = 10,000 × 1.414 = 14,140 lb per leg.
Step 2: Tension at 60 degrees: (20,000 lb / 2) × 1.155 = 10,000 × 1.155 = 11,550 lb per leg.
Step 3: Tension reduction: 14,140 lb − 11,550 lb = 2,590 pounds per leg.
Safety Logic: Increasing the sling angle from 45 degrees to 60 degrees reduces the tension in
each leg by 2,590 pounds (18.3% reduction). This demonstrates why riggers should always
maximize sling angles using spreader bars or wider anchor points when possible.
Question 3 (Calculation-Based)
A four-leg sling assembly is rigged to lift a 32,000-pound modular building section. The
horizontal angle of each sling leg is 45 degrees (LAF = 1.414). During the lift, one sling leg is
accidentally severed by a sharp edge, leaving only three legs to support the load. Assuming the
load remains evenly distributed among the remaining three legs, what is the new tension in
EACH remaining sling leg?
Answer: __________ pounds
[CORRECT: 15,082 pounds] Rationale:
Step 1: Original tension per leg (4-leg): (32,000 lb / 4) × 1.414 = 8,000 × 1.414 = 11,312 lb per
leg.
Step 2: After one leg fails, total load redistributes to 3 legs: 32,000 lb / 3 = 10,667 lb per leg
(vertical component).
Step 3: Apply LAF for 45 degrees: 10,667 lb × 1.414 = 15,082 pounds per leg.
Safety Logic: The loss of one leg in a four-leg sling increases the tension in each remaining leg by
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33.3%. This scenario illustrates why NCCCO standards require that multi-leg sling assemblies be
designed so that the failure of any single leg does not cause catastrophic load loss.
Question 4 (Multiple-Choice)
A rigger is planning a lift using a two-leg sling with a horizontal angle of 60 degrees (LAF =
1.155). The rigger's supervisor suggests lowering the angle to 30 degrees (LAF = 2.0) to clear an
overhead obstruction. The sling legs are rated for 15,000 pounds each in a vertical
configuration. What is the maximum safe load that can be lifted at EACH angle without
exceeding the sling WLL?
A) At 60 degrees: 25,970 lb total; At 30 degrees: 15,000 lb total
B) At 60 degrees: 12,985 lb per leg; At 30 degrees: 7,500 lb per leg
C) At 60 degrees: 25,970 lb total; At 30 degrees: 7,500 lb per leg
D) At 60 degrees: 12,985 lb total; At 30 degrees: 15,000 lb total
[CORRECT: A] Rationale:
At 60 degrees: Maximum load per leg = 15,000 lb / 1.155 = 12,985 lb per leg. Total load = 12,985
× 2 = 25,970 pounds.
At 30 degrees: Maximum load per leg = 15,000 lb / 2.0 = 7,500 lb per leg. Total load = 7,500 × 2
= 15,000 pounds.
Safety Logic: Reducing the sling angle from 60 degrees to 30 degrees decreases the total safe
working load by 42.3% (from 25,970 lb to 15,000 lb).
Sub-Topic 1.2: WLL Determination for Damaged Wire Rope Slings (3 Questions)
Question 5 (Multiple-Choice)
A 1-inch diameter 6×37 IWRC wire rope sling with a rated vertical WLL of 26,000 pounds is
found to have a crushed section approximately 6 inches from the eye fitting. The crushed area
shows visible flattening of the outer strands, but no broken wires are present. According to
ASME B30.9 and OSHA 1926.251, what is the proper action?
A) Reduce the WLL by 25% and continue using the sling with increased inspection frequency.
B) Reduce the WLL by 50% and mark the sling with the new capacity.
C) Remove the sling from service immediately; crushing damage permanently compromises the
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internal wire structure.
D) Cut out the damaged section and re-splice the sling; the remaining rope is undamaged.
[CORRECT: C] Rationale:
Regulatory Reference: OSHA 1926.251(c)(4) and ASME B30.9-2021, Section 9-1.10.2.
Safety Logic: Crushing damage to a wire rope sling causes permanent deformation of the
internal wire structure, including the core and inner strands that are not visible during external
inspection. ASME B30.9 explicitly requires removal from service for any sling with kinking,
crushing, or bird-caging.
Question 6 (True/False)
A wire rope sling with a single protruding broken strand may continue in service if the broken
strand is trimmed flush with the rope surface and the sling is de-rated by 10% of its original
WLL.
[CORRECT: FALSE] Rationale:
Regulatory Reference: OSHA 1926.251(c)(4)(iii) and ASME B30.9-2021, Section 9-1.10.2(c).
Safety Logic: A single broken wire or protruding strand is an indication of internal fatigue or
external abrasion damage. OSHA and ASME standards require that any wire rope sling with
broken wires be removed from service immediately. Trimming the broken strand does not
address the underlying cause of the failure.
Question 7 (Calculation-Based)
A 3/4-inch diameter wire rope sling with a rated vertical WLL of 14,800 pounds is discovered to
have severe kinking in one leg of a two-leg bridle assembly. The rigger must determine if the
remaining undamaged leg can safely support the entire load in an emergency lowering
operation. If the load weighs 18,000 pounds and the undamaged leg is rigged at a 45-degree
horizontal angle (LAF = 1.414), what is the tension in the undamaged leg, and does it exceed the
sling's WLL?
Answer: Tension = __________ pounds. Exceeds WLL? __________ (Yes/No)
[CORRECT: 25,452 pounds; Yes] Rationale:
Step 1: In an emergency single-leg support scenario, the entire 18,000-pound load is supported
Advanced Rigger Certification Exam 2026/2027
Rigging Safety and Equipment Handling Study Guide.
Comprehensive 50-Question Assessment
Prepared for: NCCCO Advanced Rigger Certification Candidates
Total Questions: 50 (30 Multiple-Choice | 10 Calculation-Based | 5 Select-All-That-Apply | 5
True/False)
Time Allotment: 120 Minutes
Passing Score: 90%
DOMAIN 1: SLING ANGLES, TENSION & LOAD CALCULATIONS (14 Questions)
Sub-Topic 1.1: Load Angle Factor (LAF) Calculations (4 Questions)
Question 1 (Calculation-Based)
A rigger is using a two-leg bridle sling to lift a 12,000-pound steel beam. The horizontal angle
between each sling leg and the load is 30 degrees. Calculate the tension in EACH individual sling
leg using the Load Angle Factor (LAF) method.
Formula: Tension per leg = (Total Load / Number of Legs) × LAF
LAF at 30 degrees = 2.0
Answer: __________ pounds
[CORRECT: 12,000 pounds] Rationale:
Step 1: Determine the load per leg before angle factor: 12,000 lb / 2 legs = 6,000 lb per leg.
Step 2: Apply the Load Angle Factor for 30 degrees: LAF = 2.0.
Step 3: Calculate final tension per leg: 6,000 lb × 2.0 = 12,000 pounds.
Safety Logic: At a 30-degree horizontal angle, the tension in each sling leg equals the TOTAL load
weight, not half the load. This is why OSHA and ASME B30.9 strongly discourage sling angles
below 30 degrees. The horizontal force component creates a multiplier effect that doubles the
tension compared to a vertical lift.
Question 2 (Multiple-Choice)
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A rigger must lift a 20,000-pound generator using a two-leg sling assembly. The available rigging
configuration allows for a maximum horizontal sling angle of 45 degrees. The rigger has two
options: (1) use the 45-degree configuration with a standard LAF of 1.414, or (2) add a spreader
bar to increase the angle to 60 degrees with an LAF of 1.155. What is the tension reduction PER
LEG achieved by using the spreader bar?
A) 1,414 pounds
B) 2,590 pounds
C) 5,180 pounds
D) 10,360 pounds
[CORRECT: B] Rationale:
Step 1: Tension at 45 degrees: (20,000 lb / 2) × 1.414 = 10,000 × 1.414 = 14,140 lb per leg.
Step 2: Tension at 60 degrees: (20,000 lb / 2) × 1.155 = 10,000 × 1.155 = 11,550 lb per leg.
Step 3: Tension reduction: 14,140 lb − 11,550 lb = 2,590 pounds per leg.
Safety Logic: Increasing the sling angle from 45 degrees to 60 degrees reduces the tension in
each leg by 2,590 pounds (18.3% reduction). This demonstrates why riggers should always
maximize sling angles using spreader bars or wider anchor points when possible.
Question 3 (Calculation-Based)
A four-leg sling assembly is rigged to lift a 32,000-pound modular building section. The
horizontal angle of each sling leg is 45 degrees (LAF = 1.414). During the lift, one sling leg is
accidentally severed by a sharp edge, leaving only three legs to support the load. Assuming the
load remains evenly distributed among the remaining three legs, what is the new tension in
EACH remaining sling leg?
Answer: __________ pounds
[CORRECT: 15,082 pounds] Rationale:
Step 1: Original tension per leg (4-leg): (32,000 lb / 4) × 1.414 = 8,000 × 1.414 = 11,312 lb per
leg.
Step 2: After one leg fails, total load redistributes to 3 legs: 32,000 lb / 3 = 10,667 lb per leg
(vertical component).
Step 3: Apply LAF for 45 degrees: 10,667 lb × 1.414 = 15,082 pounds per leg.
Safety Logic: The loss of one leg in a four-leg sling increases the tension in each remaining leg by
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33.3%. This scenario illustrates why NCCCO standards require that multi-leg sling assemblies be
designed so that the failure of any single leg does not cause catastrophic load loss.
Question 4 (Multiple-Choice)
A rigger is planning a lift using a two-leg sling with a horizontal angle of 60 degrees (LAF =
1.155). The rigger's supervisor suggests lowering the angle to 30 degrees (LAF = 2.0) to clear an
overhead obstruction. The sling legs are rated for 15,000 pounds each in a vertical
configuration. What is the maximum safe load that can be lifted at EACH angle without
exceeding the sling WLL?
A) At 60 degrees: 25,970 lb total; At 30 degrees: 15,000 lb total
B) At 60 degrees: 12,985 lb per leg; At 30 degrees: 7,500 lb per leg
C) At 60 degrees: 25,970 lb total; At 30 degrees: 7,500 lb per leg
D) At 60 degrees: 12,985 lb total; At 30 degrees: 15,000 lb total
[CORRECT: A] Rationale:
At 60 degrees: Maximum load per leg = 15,000 lb / 1.155 = 12,985 lb per leg. Total load = 12,985
× 2 = 25,970 pounds.
At 30 degrees: Maximum load per leg = 15,000 lb / 2.0 = 7,500 lb per leg. Total load = 7,500 × 2
= 15,000 pounds.
Safety Logic: Reducing the sling angle from 60 degrees to 30 degrees decreases the total safe
working load by 42.3% (from 25,970 lb to 15,000 lb).
Sub-Topic 1.2: WLL Determination for Damaged Wire Rope Slings (3 Questions)
Question 5 (Multiple-Choice)
A 1-inch diameter 6×37 IWRC wire rope sling with a rated vertical WLL of 26,000 pounds is
found to have a crushed section approximately 6 inches from the eye fitting. The crushed area
shows visible flattening of the outer strands, but no broken wires are present. According to
ASME B30.9 and OSHA 1926.251, what is the proper action?
A) Reduce the WLL by 25% and continue using the sling with increased inspection frequency.
B) Reduce the WLL by 50% and mark the sling with the new capacity.
C) Remove the sling from service immediately; crushing damage permanently compromises the
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internal wire structure.
D) Cut out the damaged section and re-splice the sling; the remaining rope is undamaged.
[CORRECT: C] Rationale:
Regulatory Reference: OSHA 1926.251(c)(4) and ASME B30.9-2021, Section 9-1.10.2.
Safety Logic: Crushing damage to a wire rope sling causes permanent deformation of the
internal wire structure, including the core and inner strands that are not visible during external
inspection. ASME B30.9 explicitly requires removal from service for any sling with kinking,
crushing, or bird-caging.
Question 6 (True/False)
A wire rope sling with a single protruding broken strand may continue in service if the broken
strand is trimmed flush with the rope surface and the sling is de-rated by 10% of its original
WLL.
[CORRECT: FALSE] Rationale:
Regulatory Reference: OSHA 1926.251(c)(4)(iii) and ASME B30.9-2021, Section 9-1.10.2(c).
Safety Logic: A single broken wire or protruding strand is an indication of internal fatigue or
external abrasion damage. OSHA and ASME standards require that any wire rope sling with
broken wires be removed from service immediately. Trimming the broken strand does not
address the underlying cause of the failure.
Question 7 (Calculation-Based)
A 3/4-inch diameter wire rope sling with a rated vertical WLL of 14,800 pounds is discovered to
have severe kinking in one leg of a two-leg bridle assembly. The rigger must determine if the
remaining undamaged leg can safely support the entire load in an emergency lowering
operation. If the load weighs 18,000 pounds and the undamaged leg is rigged at a 45-degree
horizontal angle (LAF = 1.414), what is the tension in the undamaged leg, and does it exceed the
sling's WLL?
Answer: Tension = __________ pounds. Exceeds WLL? __________ (Yes/No)
[CORRECT: 25,452 pounds; Yes] Rationale:
Step 1: In an emergency single-leg support scenario, the entire 18,000-pound load is supported
