18 Materials & Heat Treatment Practice Questions & Answers

1. 1. A part is quenched to maximum hardness and put straight into service without tempering. What has the gain in hardness cost that part?

   • A.Toughness — the part is now brittle and can chip or crack under shock.✓ Answer
   • B.Density — the part weighs measurably less than it did before quenching.
   • C.Its modulus of elasticity, which drops sharply as soon as the steel is hardened.
   • D.Nothing at all; a harder part is stronger and tougher than a soft one in every respect.

   Hardness is resistance to penetration and abrasion, while toughness is the ability to absorb shock without fracturing, and the two move in opposite directions. Quenching forms martensite, which is hard but brittle, so a hardened part is tempered afterward to trade a little hardness back for toughness.

   Source: NIMS/ANSI 101-2001 Machining Level I, KSAO 7.3 'Material Properties' (Standards, p.90); ASM heat-treating fundamentals — as-quenched martensite must be temperedReport a problem with this question

2. 2. A 6061 aluminum plate has to be surface ground. Why will it not hold on the magnetic chuck, and what should the machinist do about it?

   • A.Nothing prevents it; any metal will hold on a magnetic chuck as long as the seating face is clean and flat.
   • B.Aluminum is non-ferrous and non-magnetic, so it gives the flux no path and develops no holding force; the plate must be held mechanically in a precision vise, with clamps, or in a fixture.✓ Answer
   • C.Aluminum is too soft for a magnet to grip, so the chuck has to be demagnetized and then re-energized at full power to build up enough holding force.
   • D.The magnetic field would case harden the aluminum surface, so the plate must be annealed first and then set on magnetic parallels.

   A magnetic chuck holds work only by passing magnetic flux through a ferrous part, so a non-ferrous, non-magnetic metal such as aluminum gives the flux no path and develops no holding force. Non-ferrous work must be clamped in a precision vise, held with clamps or a fixture, or stuck down with double-sided adhesive.

   Source: NIMS Grinding Level I task list — workholding: magnetic chuck versus non-ferrous work; KSAO 7.3 ferrous/non-ferrous recognitionReport a problem with this question

3. 3. A machinist holds a magnet against a bar of 304 stainless steel and it does not stick. What is the correct conclusion?

   • A.The bar is in the annealed condition; 304 becomes magnetic again only after it is quenched and tempered to full hardness.
   • B.The magnet is too weak — every stainless grade, including the 400 series, is strongly magnetic when the surface is clean.
   • C.304 is still a ferrous, iron-based alloy; its austenitic structure simply is not magnetic.✓ Answer
   • D.The bar is non-ferrous, because every alloy that contains iron is attracted to a magnet without exception.

   Ferrous simply means iron-based. The 300-series stainless steels are ferrous but austenitic, and austenite is essentially non-magnetic, so a magnet alone cannot sort ferrous from non-ferrous. The 400-series martensitic and ferritic stainless steels, by contrast, are magnetic.

   Source: NIMS/ANSI 101-2001 Machining Level I, KSAO 7.3 — recognizes differences between ferrous, non-ferrous and magnetic materials; AISI austenitic (300-series) stainless classificationReport a problem with this question

4. 4. A print calls out the stock as SAE 1070. What do the last two digits tell the machinist?

   • A.The minimum tensile strength of the bar in thousands of psi, so roughly 70,000 psi.
   • B.The percentage of the principal alloying element other than carbon in the steel.
   • C.The Rockwell C hardness the material will reach once it is quenched and tempered.
   • D.The carbon content in hundredths of one percent — about 0.70% carbon.✓ Answer

   In the SAE/AISI four-digit system the last two digits state carbon content in points — hundredths of one percent — so 1070 is about 0.70% carbon. Carbon level is what tells the machinist whether the steel will through-harden and roughly how hard it will be to cut.

   Source: SAE/AISI steel numbering system (Machinery's Handbook); NIMS Job Planning, Benchwork & Layout Level I — 'Materials' task: SAE steel identification systemReport a problem with this question

5. 5. A print specifies 4140 for a shaft. What do the digits '41' tell the machinist?

   • A.They identify the alloy family — a chromium-molybdenum alloy steel.✓ Answer
   • B.They state the Rockwell C hardness the supplier guarantees on delivery of the bar.
   • C.They give the carbon content, so the steel contains about 4.1% carbon.
   • D.They are a mill heat-lot number and say nothing about composition or machinability.

   In the SAE/AISI system the first digit names the major alloy class and the second the approximate amount of that alloying element, so the 41xx series is chromium-molybdenum alloy steel; only the last two digits carry carbon. The machinist looks the number up in the Machinery's Handbook to get the composition and the recommended cutting speed.

   Source: SAE/AISI steel numbering system (Machinery's Handbook); NIMS Job Planning, Benchwork & Layout Level I — 'Materials' task: SAE steel identification systemReport a problem with this question

6. 6. Bar stock is marked 6061-T6. What does that designation tell the machinist?

   • A.It is a steel containing 0.61% carbon that has been through-hardened by an oil quench and then drawn back by tempering.
   • B.It is an aluminum alloy whose principal alloying elements are magnesium and silicon, strengthened by solution heat treating and aging.✓ Answer
   • C.It is an aluminum alloy that gains its strength exactly the way carbon steel does, by heating above the critical temperature and quenching, and it can be re-hardened after machining.
   • D.It is a copper alloy whose suffix reports the Brinell hardness measured after the final cold-drawing pass.

   Aluminum uses its own four-digit system in which the first digit names the principal alloying element — the 6xxx series is magnesium and silicon — and the letter suffix names the temper: -O annealed, -H cold worked, -T solution heat treated and aged. Aluminum is strengthened by precipitation (age) hardening, not by the quench-and-temper cycle used on steel, and at the machine it runs at the highest cutting speeds with sharp, polished, high-rake tooling and plenty of coolant to keep a built-up edge from forming.

   Source: Aluminum Association alloy and temper designation system (Machinery's Handbook); NIMS/ANSI 101-2001 KSAO 7.3 material recognitionReport a problem with this question

7. 7. A print calls for a shaft in 0.18% carbon steel that must have a hard, wear-resistant surface over a tough core. Which process meets that requirement?

   • A.Through hardening by heating above the critical temperature and quenching in brine, which is all that low-carbon steel needs to reach full hardness.
   • B.Annealing followed by a slow furnace cool, which raises surface hardness while leaving the core soft.
   • C.Carburizing — carbon is diffused into the surface at high temperature and the part is then quenched, hardening the high-carbon case while the low-carbon core stays tough.✓ Answer
   • D.Stress relieving below the critical temperature, which hardens the outer skin without changing the core.

   Steel needs roughly 0.30% carbon or more to harden by quenching, so a 0.18% carbon steel cannot be through-hardened and must be case hardened. Carburizing diffuses carbon into the surface at high temperature; the following quench hardens that high-carbon case while the low-carbon core stays tough and shock absorbing.

   Source: NIMS/ANSI 101-2001 KSAO 7.3 — 'Understands the changes which heat-treat impart to materials'; case hardening (carburizing) of low-carbon steelReport a problem with this question

8. 8. A forged alloy-steel blank is too hard to rough machine efficiently. Which heat treatment leaves it in the softest, most machinable condition?

   • A.Tempering — reheat the blank below the lower critical temperature and hold it there before cooling in air.
   • B.Hardening — heat above critical and quench, then machine the blank in the untempered condition.
   • C.Normalizing — heat above critical, soak, then cool in still air outside the furnace, which leaves it somewhat harder.
   • D.Annealing — heat above the critical temperature, soak, then cool very slowly in the closed furnace.✓ Answer

   Annealing heats the part above the critical temperature and then cools it as slowly as possible, normally in the closed furnace. That slow cool produces the softest, most ductile, lowest-stress structure and therefore the best machinability, which is why annealing is the treatment used before heavy roughing or before re-hardening.

   Source: NIMS/ANSI 101-2001 KSAO 7.3 heat-treat effects; ASM heat treating — full annealing (furnace cool)Report a problem with this question

9. 9. A weldment has coarse, uneven grain and locked-in stress from welding. The shop wants a refined, uniform grain structure before machining, with slightly more strength than the fully annealed condition. Which process is correct?

   • A.Normalizing — heat above the critical temperature, soak, and cool in still air, which refines the grain and relieves the stress left by welding.✓ Answer
   • B.Annealing, cooling the weldment slowly inside the closed furnace, which gives the softest and coarsest structure of the four.
   • C.Tempering below the lower critical temperature, which refines grain without ever heating above critical.
   • D.Quenching in brine straight from welding heat, which refines the grain by rapid transformation.

   Normalizing heats above the critical temperature and then cools in still air. Cooling faster than a furnace but far slower than a quench refines and homogenizes the grain and relieves stresses left by welding, forging or casting, leaving the part slightly harder and stronger than annealed but still easy to machine.

   Source: NIMS/ANSI 101-2001 KSAO 7.3 heat-treat effects; ASM heat treating — normalizing (still-air cool)Report a problem with this question

10. 10. A long cold-drawn steel bar bows as soon as one side is roughed off. What is the correct machining sequence?

    • A.Finish the whole part in one setup, then harden and temper it to pull it back straight.
    • B.Rough machine, then stress relieve below the critical temperature and slow cool to release the locked-in cold-drawing stress, then finish to size.✓ Answer
    • C.Anneal the bar above the critical temperature before roughing, since that is the only way to remove residual stress from cold drawing.
    • D.Increase the depth of cut so the bar is roughed and finished in one heavy pass before the stress has time to release.

    Cold drawing locks residual stress into the bar, and removing metal from one side unbalances that stress so the bar moves. Stress relieving heats below the critical temperature and slow cools, which releases the locked-in stress without changing hardness or grain structure, so the part stays dimensionally stable while it is finished.

    Source: NIMS/ANSI 101-2001 KSAO 7.3 heat-treat effects; ASM heat treating — stress relieving below the lower critical temperatureReport a problem with this question

11. 11. Why is a quench-hardened steel part reheated below the lower critical temperature before it goes into service?

    • A.To dissolve the carbon back into solution so the part can be machined with ordinary high-speed steel tooling.
    • B.To add carbon to the surface, since reheating in air lets carbon diffuse into the outer skin of the part.
    • C.To reduce brittleness and restore toughness, giving up a little hardness in exchange.✓ Answer
    • D.To raise hardness still further, because the second heating completes the transformation to martensite.

    Tempering, also called drawing, reheats the quench-hardened part below the lower critical temperature so that some of the brittle as-quenched martensite breaks down. Hardness drops slightly while toughness and ductility come back, and the higher the tempering temperature the softer and tougher the finished part.

    Source: NIMS/ANSI 101-2001 KSAO 7.3 heat-treat effects; ASM heat treating — tempering below the lower critical temperatureReport a problem with this question

12. 12. An alloy-steel part is already ground to size and now needs a very hard, thin, wear-resistant surface with the least possible distortion. Which surface treatment fits?

    • A.Through hardening in a furnace above critical, followed by an oil quench and a full tempering cycle.
    • B.Carburizing, which raises the surface carbon at high temperature and then requires a quench to harden the case.
    • C.Flame hardening, which heats the surface above critical with a torch and immediately quenches it with a water spray.
    • D.Nitriding — nitrogen diffuses in at a comparatively low temperature and no quench is required.✓ Answer

    Nitriding diffuses nitrogen into alloy steel at a comparatively low temperature and needs no quench, so there is no rapid transformation and almost no distortion or scale. That is why it is the surface treatment chosen for parts that are already machined and ground to final size.

    Source: NIMS/ANSI 101-2001 KSAO 7.3 heat-treat effects; ASM heat treating — nitriding (no quench, minimum distortion)Report a problem with this question

13. 13. A hardened tool-steel block must be checked for hardness. Which Rockwell scale and indenter are correct?

    • A.The C scale — a diamond brale cone under a 150 kgf major load, the combination meant for hardened and tool steel, where a steel ball would simply flatten.✓ Answer
    • B.The A scale — a brale under a 60 kgf major load, intended mainly for cemented carbide and very thin stock.
    • C.Any scale will do, because every Rockwell scale reads the same number on the same piece of material.
    • D.The B scale — a 1/16 in. steel ball under a 100 kgf major load, the scale meant for soft or annealed steel and for brass.

    Rockwell C pairs a diamond brale cone with the heavy 150 kgf major load, the combination intended for hardened steel and tool steel. A steel-ball indenter would flatten against a hardened surface and give no usable reading, which is why the ball scales are reserved for softer materials such as annealed steel, brass and aluminum.

    Source: ASTM E18 Rockwell hardness testing — scale and indenter selection; NIMS Measurement, Materials & Safety Level IReport a problem with this question

14. 14. A carburized part has a case only a few thousandths of an inch deep. Why should a superficial Rockwell scale such as 15N or 30N be used instead of the regular C scale?

    • A.Superficial scales use a ball indenter, which cannot damage a hardened surface the way a diamond can.
    • B.The lighter loads keep the indenter inside the thin case instead of punching through to the soft core.✓ Answer
    • C.The regular C scale may never be used on any ferrous material once it has been given any kind of surface-hardening treatment.
    • D.Superficial scales measure the diameter of the impression under a microscope, which is more accurate on thin work.

    Superficial Rockwell scales use a 3 kgf minor load and light major loads, so the indenter stays within a shallow case and reports the true surface hardness. A regular C-scale test drives the brale through the thin case and the reading averages case and soft core, understating the hardness that was actually achieved.

    Source: ASTM E18 — superficial Rockwell scales (15N/30N/45N) for thin material and case-hardened surfaces; NIMS Measurement, Materials & Safety Level IReport a problem with this question

15. 15. How does a Brinell test differ in procedure from a Rockwell test, and where is the Brinell number most useful in the shop?

    • A.Brinell is a rebound test in which a small hammer is dropped on the work and the height of the bounce is read off a scale.
    • B.Brinell reads penetration depth straight off the dial the instant the load is removed, which makes it the quickest and handiest check for finished precision parts.
    • C.Brinell presses in a ball and the impression diameter is measured and converted; the number suits rough castings and indexes the speed-and-feed tables.✓ Answer
    • D.Brinell uses a diamond brale under a 150 kgf load and is read directly, so it is used only on fully hardened tool steel and carbide tooling.

    Rockwell reads depth of penetration directly on the machine, while Brinell presses a ball under a heavy load and the operator must measure the impression diameter with a microscope and convert it to a number. The large impression averages out coarse structure, which suits castings and forgings, and the Machinery's Handbook cutting-speed tables are indexed by Brinell hardness.

    Source: ASTM E10 Brinell hardness test; Machinery's Handbook cutting-speed tables indexed by Brinell hardnessReport a problem with this question

16. 16. Two identical parts are turned with the same high-speed steel tool: one is 150 BHN, the other 300 BHN. How should cutting speed be handled on the harder part?

    • A.Raise the cutting speed, because harder material resists deflection and can take a higher surface speed.
    • B.Keep the sfm the same and double the feed rate instead, since only feed affects tool wear in hard material.
    • C.Cutting speed depends only on the diameter of the part, so hardness has no bearing on the rpm chosen.
    • D.Lower the cutting speed in sfm, which lowers the rpm for the same diameter.✓ Answer

    The harder the material, the more heat and pressure it puts on the cutting edge, so the recommended surface speed goes down as hardness goes up. Because rpm = (CS x 4) / D, a lower sfm automatically gives a lower rpm at the same diameter; running the harder part at the softer part's speed simply burns the tool.

    Source: NIMS Job Planning, Benchwork & Layout Level I — 'Materials and cutting speeds (relating to sfm and rpm)': harder materials require lower cutting speeds and lower rpmReport a problem with this question

17. 17. Which abrasive is the correct choice for grinding aluminum and other non-ferrous metals?

    • A.Silicon carbide, the abrasive matched to non-ferrous metals, which keeps the wheel cutting instead of loading up and rubbing.✓ Answer
    • B.Cubic boron nitride, the abrasive reserved for hardened ferrous alloys and nickel-based super alloys.
    • C.Diamond, which is used on cemented carbide, ceramics and glass rather than on soft metals.
    • D.Aluminum oxide, the general-purpose abrasive normally chosen for steel and other ferrous metals.

    Silicon carbide is the abrasive matched to non-ferrous metals such as aluminum, brass and copper, and also to cast iron and cemented carbide, while aluminum oxide is the general-purpose abrasive for steel and other ferrous metals. Matching the abrasive to the workpiece keeps the wheel from loading up and keeps it cutting instead of rubbing.

    Source: NIMS Grinding Level I preparation guide — 'Types of Abrasives': silicon carbide for non-ferrous metals, cast iron and carbide; aluminum oxide for ferrous metalsReport a problem with this question

18. 18. While turning 304 stainless, a machinist takes a light spring pass, the tool rubs instead of cutting, and the next pass will not cut at all. What happened and what is the fix?

    • A.The coating came off the carbide insert; simply dropping the speed with the same worn edge will restore the cut.
    • B.The surface work hardened under the rubbing tool; use a sharp edge and a firm, uninterrupted feed.✓ Answer
    • C.304 is magnetic and is being pulled toward the tool; demagnetizing the bar in the chuck will let the tool bite again.
    • D.The steel air hardened from the cutting heat the way a tool steel does, so the bar has to be annealed in a furnace and then machined again from the beginning.

    Austenitic stainless steels such as 304 work harden rapidly: cold plastic deformation from a rubbing or dwelling tool raises the surface hardness above what the edge can cut. The cure is a keen, sharp tool and a positive, continuous feed heavy enough to keep the edge under the hardened skin — the tool must never dwell or rub.

    Source: NIMS/ANSI 101-2001 KSAO 7.3 applied in Job Execution — respond to cutting conditions imposed by material properties; work hardening of austenitic stainless steelReport a problem with this question

Practice questions based on NIMS/ANSI 101-2001, Duties and Standards for Machining Skills Level I, and the published content of the NIMS Machining Level I theory exams, together with standard precision-machining practice. NIMS is a mark of the National Institute for Metalworking Skills; this site is not affiliated with or endorsed by NIMS. Machining Level I is a set of separate credentials, and most of them also require a hands-on performance test that this bank does not cover. The NIMS theory exams are open-reference, but questions here never depend on recalling a handbook table value, a citation number or a machine rating — always work from the print in front of you, your employer's written procedures, and the machine's own documentation, and confirm current requirements before testing. About the NIMS machining credentials →