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100+ Free CAAS SAR-66 Module 6 Materials & Hardware Practice Questions
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Key Facts: CAAS SAR-66 Module 6 Materials & Hardware Exam
72 Qs
Multiple-choice questions per Category B1/B2 Module 6 exam.
CAAS AC 66-13 / SAR-66 Appendix 1
90 Mins
Total allotted exam time.
CAAS Examination Matrix
75%
Passing mark required per module.
CAAS SAR-66 Subpart B
S$87.20
Exam fee per subject (ANO 12th Schedule from 1 Jan 2026).
CAAS Air Navigation Order
11 Sections
Sub-modules 6.1 through 6.11 covering metals, composites, hardware & EWIS.
CAAS SAR-66 Appendix 1 Syllabus
CAAS SAR-66 Module 6 Materials & Hardware is a 90-minute exam consisting of 72 multiple-choice questions with a required passing score of 75% per module.
Sample CAAS SAR-66 Module 6 Materials & Hardware Practice Questions
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1Under the SAE steel numbering system, what are the primary alloying elements and nominal carbon content of SAE 4130 steel?
A.Chromium and molybdenum with 0.30% carbon
B.Nickel and chromium with 0.40% carbon
C.Manganese and silicon with 0.13% carbon
D.Vanadium and tungsten with 3.00% carbon
Explanation: In the SAE four-digit index system, the first digit '4' designates molybdenum steel, specifically chromium-molybdenum (chromoly) steel when paired with '1'. The final two digits '30' represent the nominal carbon content in hundredths of a percent, indicating 0.30% carbon.
2Which crystal structure transition occurs when carbon steel is heated above its upper critical point into the face-centered cubic phase prior to rapid quenching?
A.Ferrite transforms into Austenite
B.Austenite transforms into Cementite
C.Martensite transforms into Ferrite
D.Pearlite transforms into Bainite
Explanation: Heating carbon steel above its upper critical temperature causes the room-temperature body-centered cubic (BCC) Ferrite phase to transform into face-centered cubic (FCC) Austenite. Upon rapid quenching, this Austenite phase transforms into a body-centered tetragonal (BCT) Martensite structure, trapping carbon atoms and achieving maximum hardness.
3What is a distinct operational advantage of gas nitriding over carburizing for case hardening high-precision aircraft steel parts?
A.It takes place below the critical temperature and requires no quenching, minimizing thermal distortion
B.It produces a much thicker case depth in a fraction of the processing time
C.It eliminates the need for alloy steels containing aluminum or chromium
D.It allows the part to be easily welded or cold-formed post-treatment
Explanation: Nitriding is conducted at relatively low temperatures (around 500°C to 538°C / 935°F to 1000°F), which is below the critical transformation range of steel. Because nitrogen diffuses into special alloy steel (containing aluminum, chromium, or molybdenum) to form hard nitrides without requiring liquid quenching, dimensional distortion and scaling are virtually eliminated.
4Which indentor and major load combination is specified for testing heat-treated aircraft alloy steels on the Rockwell C hardness scale?
A.120-degree Brale diamond cone indentor with a 150 kg major load
B.1/16-inch steel ball indentor with a 100 kg major load
C.1/8-inch steel ball indentor with a 60 kg major load
D.10 mm tungsten carbide ball indentor with a 3000 kg major load
Explanation: The Rockwell C scale (HRC) utilizes a 120-degree diamond spheroconical penetrator (Brale indentor) subjected to a 10 kg minor load followed by a 150 kg major load. This scale is standard for evaluating hard materials such as quenched and tempered aircraft alloy steels.
5What are the primary chemical constituents and magnetic characteristics of 18-8 austenitic stainless steel (AISI 302/304) in the annealed condition?
A.18% Chromium and 8% Nickel; non-magnetic in the annealed state
B.18% Nickel and 8% Chromium; strongly magnetic in all heat-treated states
C.18% Carbon and 8% Molybdenum; magnetic only when heated above 700°C
D.18% Manganese and 8% Silicon; non-magnetic only after severe cold working
Explanation: 18-8 corrosion-resistant steel contains approximately 18% chromium and 8% nickel. The high nickel content stabilizes the face-centered cubic Austenite phase down to room temperature, rendering the material non-magnetic in the fully annealed condition.
6How does normalizing differ from full annealing when treating tubular aircraft steel structures after welding?
A.Normalizing involves cooling the part in still air, producing higher strength and finer grain structure than annealing
B.Normalizing requires extremely slow cooling inside a shut-down furnace over 24 hours
C.Normalizing heats steel to a much lower temperature than annealing to preserve cold-work hardness
D.Normalizing rapidly quenches the steel in ice water to maximize brittleness and wear resistance
Explanation: Normalizing involves heating the steel approximately 55°C (100°F) above its upper critical point and cooling it in still room-temperature air. This air cooling rate is faster than furnace cooling (annealing), resulting in a finer pearlite grain structure with higher tensile strength and yield strength while effectively relieving internal welding stresses.
7Which ultra-high-strength steel alloy is widely used for heavy-duty commercial aircraft landing gear forgings due to its deep hardenability?
A.SAE 4340 steel
B.SAE 1020 steel
C.SAE 1112 steel
D.SAE 30302 steel
Explanation: SAE 4340 is a nickel-chromium-molybdenum steel renowned for its exceptional strength, toughness, and deep hardenability throughout heavy cross-sections. When heat-treated, it achieves tensile strengths exceeding 200,000 psi (1380 MPa), making it the primary choice for main landing gear cylinders, axles, and structural pins.
8What is the primary purpose of tempering steel immediately after it has been fully hardened by quenching?
A.To reduce extreme brittleness and internal stress while restoring acceptable toughness and ductility
B.To increase maximum hardness and ultimate tensile strength beyond the quenched level
C.To transform remaining ferrite into stable austenitic crystals at room temperature
D.To produce a heavy black oxide protective scale across the external surface
Explanation: As-quenched martensitic steel is extremely hard, brittle, and heavily stressed internally. Tempering involves reheating the hardened steel to a temperature below its lower critical limit (typically between 200°C and 650°C) and holding it, which relieves internal quenching stresses and trades a small amount of hardness for substantial ductility and impact toughness.
9In a standard Brinell hardness test on a steel forging using a 10 mm ball under a 3000 kg load, an indentation diameter of 4.0 mm is measured. What is the calculated Brinell Hardness Number (HB)?
A.229 HB
B.149 HB
C.350 HB
D.450 HB
Explanation: The Brinell formula is HB = 2P/π D (D - √(D^2 - d^2)). Substituting load P = 3000 kg, ball diameter D = 10 mm, and indentation diameter d = 4.0 mm: HB = 6000/π × 10 × (10 - √(100 - 16)) = 6000/31.416 × (10 - 9.165) = 6000/26.23 ≈ 228.7 HB (rounded to 229 HB).
10What thermal stress-relieving parameters are recommended for welded SAE 4130 tubular engine mount assemblies prior to final machining?
A.Heating to 600°C–650°C (1100°F–1200°F), holding for 1 hour per inch of thickness, and air cooling
B.Soaking at 900°C (1650°F) for 4 hours followed by an immediate cold brine quench
C.Heating to 300°C (570°F) for 10 minutes followed by rapid agitation in forced air
D.Cryogenic freezing at -70°C (-94°F) for 24 hours to stabilize residual austenite
Explanation: Stress relieving welded 4130 steel requires heating the assembly below the critical temperature range to approximately 600°C to 650°C (1100°F to 1200°F), holding for sufficient time (typically 1 hour per inch of maximum wall thickness), and cooling uniformly in still air to relieve residual welding stresses without altering phase balance.
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