Materials and Processes Flashcards
- Magnetic particle inspection is used primarily to detect
A— distortion.
B— deep subsurface flaws.
C— flaws on or near the surface.
C— flaws on or near the surface.
Magnetic particle inspection is used to detect flaws in ferromagnetic material on or near the surface. These flaws form north and south magnetic poles when the part is magnetized. Iron oxide suspended in a f luid pumped over the part is attracted to and held by the magnetism and it outlines the flaw.
- Liquid penetrant inspection methods may be used on which of the following?
A— Ferrous metals and nonporous plastics.
B— Porous and nonporous plastics and nonferrous metals.
C— Ferrous and nonferrous metals and nonporous plastics.
C— Ferrous and nonferrous metals and nonporous plastics.
Liquid penetrant inspection methods may be used to detect faults that extend to the surface on both ferrous and nonferrous metals and nonporous plastics.
- Which of these nondestructive testing methods is suitable for the inspection of most metals, plastics, and ceramics for surface and subsurface defects?
A— Eddy current inspection.
B— Magnetic particle inspection.
C— Ultrasonic inspection.
C— Ultrasonic inspection.
Ultrasonic inspection uses high-frequency sound waves to detect faults in a material. It can be used on a wide variety of materials such as ferrous and nonferrous metals, plastics and ceramics. It can detect subsurface as well as surface defects.
8221-1.
Which of the following defects are not acceptable for metal lines?
A— Scratches or nicks on the inside of a bend less than 10% of wall thickness.
B— Dents in straight section that are 10% of tube diameter.
C— Dents in straight section that are 20% of tube diameter.
C— Dents in straight section that are 20% of tube diameter.
Scratches or nicks less than 10% of the wall thickness of the tube are repairable if they are not in the heel of the bend, as are dents that are 10% of tube diameter. A dent of more than 20% of the tube diameter is not acceptable.
- What nondestructive testing method requires little or no part preparation, is used to detect surface or near-surface defects in most metals, and may also be used to separate metals or alloys and their heat-treated conditions?
A— Eddy current inspection.
B— Ultrasonic inspection.
C— Magnetic particle inspection.
A— Eddy current inspection.
Eddy current inspection requires relatively little preparation of the part being inspected. It induces a magnetic field into the part which causes eddy currents to flow. Variations in the magnitude of the eddy currents affect this magnetic field, and when it is analyzed electronically, it gives information regarding such structural characteristics as flaws, discontinuities, thickness, and alloy or heat-treated condition of the material. Eddy current inspection is used to locate defects both on the surface and below the surface.
- What method of magnetic particle inspection is used most often to inspect aircraft parts for invisible cracks and other defects?
A— Residual.
B— Inductance.
C— Continuous.
C— Continuous.
The continuous method of magnetic particle inspection is used for most aircraft parts because it provides the strongest magnetic field to attract the oxide from the fluid. In the continuous method of magnetic particle inspection, the part is either placed between the heads of the magnetizing machine or held inside the solenoid (coil). Magnetizing current flows while the fluid is pumped over the part. In the residual method of magnetic particle inspection, used for some smaller parts, the parts are magnetized and the magnetizing current is shut off. Only residual magnetism is left in the part to attract the oxide.
- Which of the following factors are considered essential knowledge for X-ray exposure?
A— Processing of the film and its characteristics.
B— Material thickness and density and the type of defect to be detected.
C— Processing of the film, characteristics of X-ray machine used, and film characteristics.
B— Material thickness and density and the type of defect to be detected.
The factors of radiographic exposure are so interdependent that it is necessary to consider all factors for any particular radiographic exposure. These factors include, but are not limited to, the following:
1. Material thickness and density.
2. Shape and size of the object.
3. Type of defect to be detected.
4. Characteristics of X-ray machine used.
5. Exposure distance.
6. Exposure angle.
7. Film characteristics.
8. Type of intensifying screen, if used.
- The testing medium that is generally used in magnetic particle inspection utilizes a ferromagnetic material that has
A— high permeability and low retentivity.
B— low permeability and high retentivity.
C— high permeability and high retentivity.
A— high permeability and low retentivity.
The testing medium used to indicate the presence of a fault in magnetic particle inspection is a finely ground iron oxide that has a high permeability and low retentivity, and is nontoxic. It is usually suspended in a light oil such as kerosine.
- Which statement relating to the residual magnetizing inspection method is true?
A— Subsurface discontinuities are made readily apparent.
B— It is used in practically all circular and longitudinal magnetizing procedures.
C— It may be used with steels which have been heat treated for stressed applications.
C— It may be used with steels which have been heat treated for stressed applications.
In the residual method of magnetic particle inspection, the part is magnetized and removed from the magnetic field before the oxide-carrying fluid is pumped over it. Steel that has a high retentivity (retains its magnetism after the magnetizing force has been removed) can be inspected by the residual method. Steel that has been heat-treated for stressed applications has a high retentivity and it can be inspected by the residual method.
- A mechanic has completed a bonded honeycomb repair using the potted compound repair technique. What nondestructive testing method is used to determine the soundness of the repair after the repair has cured?
A— Eddy current test.
B— Metallic ring test.
C— Ultrasonic test.
B— Metallic ring test.
After a bonded honeycomb repair has been made using the potted-compound repair technique, the soundness of the repair can be tested by using the metallic ring test. The repaired surface is tested by tapping it with the edge of a coin. If the repair is sound, the tapping will produce a metallic ringing sound. If there is any void in the material, the tapping will produce a dull, thudding sound.
- What two types of indicating mediums are available for magnetic particle inspection?
A— Iron and ferric oxides.
B— Wet and dry process materials.
C— High retentivity and low permeability material.
B— Wet and dry process materials.
The magnetic medium used for magnetic particle inspection can be applied either as a dry oxide powder dusted over the surface or (as is more commonly done) suspended in a light oil such as kerosine and pumped over the surface.
The iron oxide used as the indicating medium is often treated with a fluorescent dye that causes it to glow with a green light when an ultraviolet light (black light) is shone on it.
- Which of the following materials may be inspected using the magnetic particle inspection method?
A— Copper alloys.
B— Aluminum alloys.
C— Iron alloys.
C— Iron alloys.
In order for a part to be inspected by the magnetic particle method, it must be magnetizable. The only magnetizable metals listed are iron alloys.
- One way a part may be demagnetized after magnetic particle inspection is by
A— subjecting the part to high voltage, low amperage AC.
B— slowly moving the part out of an AC magnetic field of sufficient strength.
C— slowly moving the part into an AC magnetic field of sufficient strength.
B— slowly moving the part out of an AC magnetic field of sufficient strength.
A steel part is magnetized by holding it in a strong, steady magnetic field that aligns all of the magnetic domains in the material.
It is demagnetized by placing it in an AC magnetic field that continually reverses its polarity. This causes the domains to continually reverse their direction. As the domains are reversing, the part is slowly moved from the field so the domains remain in a disoriented state when the demagnetizing force is removed.
- Which type crack can be detected by magnetic particle inspection using either circular or longitudinal magnetization?
A— 45°.
B— Longitudinal.
C— Transverse.
A— 45°.
Longitudinal magnetization produces a magnetic field that extends lengthwise in the material. It is used to detect faults that extend across the part, perpendicular to the lines of magnetic flux.
Circular magnetization produces a magnetic field that extends across the material. It can detect faults that are oriented along the length of the part.
Either type of magnetization can detect a fault that runs at 45° to the length of the part.
- Which of the following methods may be suitable to use to detect cracks open to the surface in aluminum forgings and castings?
A— Dye penetrant inspection and eddy current inspection.
B— Dye penetrant inspection and magnetic particle inspection.
C— Magnetic particle inspection and metallic ring inspection.
A— Dye penetrant inspection and eddy current inspection.
Dye penetrant, eddy current, ultrasonic, and visual inspections may be used on aluminum forgings and castings. Magnetic particle inspection can only be used on ferrous metals, and the metallic ring inspection is used to check for delamination in bonded composite structural materials.
- To detect a minute crack using dye penetrant inspection usually requires
A— that the developer be applied to a flat surface.
B— a longer-than-normal penetrating time.
C— the surface to be highly polished.
B— a longer-than-normal penetrating time.
The amount of penetrant that can enter a small crack is determined by both the length of time the penetrant is allowed to remain on the surface and the temperature of the part.
When looking for very small cracks, the part can be heated (but not enough to cause the penetrant to evaporate from the surface), and the penetrant can be allowed to stay on the surface for a longer than normal time before it is washed off.
8233-1.
Which of the following is a main determinant of the dwell time to use when conducting a dye or fluorescent penetrant inspection?
A— The size and shape of the discontinuities being looked for.
B— The size and shape of the part being inspected.
C— The type and/or density of the part material.
A— The size and shape of the discontinuities being looked for.
The dwell time (the time the penetrant is allowed to remain on the surface) is determined by the size and shape of the discontinuity being looked for.
- When checking an item with the magnetic particle inspection method, circular and longitudinal magnetization should be used to
A— reveal all possible defects.
B— evenly magnetize the entire part.
C— ensure uniform current flow.
A— reveal all possible defects.
Since longitudinal magnetization detects faults that lie across a part, and circular magnetization detects faults that lie parallel to its length, a complete inspection that will show up all possible defects requires that the part be magnetized twice, longitudinally and circularly, and given two separate inspections.
- In magnetic particle inspection, a flaw that is perpendicular to the magnetic field flux lines generally causes
A— a large disruption in the magnetic field.
B— a minimal disruption in the magnetic field.
C— no disruption in the magnetic field.
A— a large disruption in the magnetic field.
In order to locate a defect in a part by the magnetic particle inspection method, it is essential that the magnetic lines of force pass approximately perpendicular to the defect. This causes the maximum disruption of the magnetic field and forms magnetic poles which attract the indicating medium across the defect.
- If dye penetrant inspection indications are not sharp and clear, the most probable cause is that the part
A— was not correctly degaussed before the developer was applied.
B— has no appreciable damage.
C— was not thoroughly washed before the developer was applied.
C— was not thoroughly washed before the developer was applied.
After the penetrant has been on the surface of a part for the correct dwell time, the surface must be thoroughly washed to remove all traces of the penetrant. When the surface is clean and dry, the developer is sprayed or dusted on. Any penetrant left on the surface or in the pores of the material will stain the developer and faults will not show up as sharp and clear marks.
(1) An aircraft part may be demagnetized by subjecting it to a magnetizing force from alternating current that is gradually reduced in strength.
(2) An aircraft part may be demagnetized by subjecting it to a magnetizing force from direct current that is alternately reversed in direction and gradually reduced in strength.
Regarding the above statements,
A— both 1 and 2 are true.
B— only 1 is true.
C— only 2 is true.
A— both 1 and 2 are true.
Statement 1 is true. A part is demagnetized by placing it in an AC magnetic field whose strength is gradually reduced while it continually reverses its polarity. This leaves the domains in a disoriented state when the demagnetizing force is removed.
Statement 2 is also true. A DC magnetic field whose direction is continually reversed and the strength is gradually reduced may be used to demagnetize an aircraft part that has been inspected by the magnetic particle inspection method.
- The pattern for an inclusion is a magnetic particle buildup forming
A— a fernlike pattern.
B— a single line.
C— parallel lines.
C— parallel lines
Inclusions are impurities trapped inside a piece of metal when it was cast.
When the part is inspected by magnetic particle inspection, the inclusion does not show up as a clearly defined fault but the indication is fuzzy.
Rather than sharply defined poles, there are several sets of poles that cause the oxide to form in a series of parallel lines.
- A part which is being prepared for dye penetrant inspection should be cleaned with
A— a volatile petroleum-base solvent.
B— the penetrant developer.
C— water-base solvents only.
A— a volatile petroleum-base solvent.
It is important when performing a dye penetrant inspection that the surface of the part be as clean as possible.
Volatile petroleum-based solvents such as Stoddard solvent and naphtha are widely used for cleaning parts to be inspected.
- Under magnetic particle inspection, a part will be identified as having a fatigue crack under which condition?
A— The discontinuity pattern is straight.
B— The discontinuity is found in a nonstressed area of the part.
C— The discontinuity is found in a highly stressed area of the part.
C— The discontinuity is found in a highly stressed area of the part.
Fatigue cracks usually show up in areas that have been subjected to high concentrations of stresses. They are likely to form where the cross-sectional area of the part changes sharply.
- In performing a dye penetrant inspection, the developer
A— seeps into a surface crack to indicate the presence of a defect.
B— acts as a blotter to produce a visible indication.
C— thoroughly cleans the surface prior to inspection.
B— acts as a blotter to produce a visible indication.
To perform a dye penetrant inspection, the part to be inspected is thoroughly cleaned and soaked in a liquid penetrant which seeps into any cracks or defects that extend to the surface. After the part is soaked for the required dwell time, the penetrant is washed from the surface, and the surface is covered with a developer which, acting as a blotter, pulls the penetrant from the fault. The penetrant pulled out by the developer shows up as a visible indication.
- What defects will be detected by magnetizing a part using continuous longitudinal magnetization with a cable?
A— Defects perpendicular to the long axis of the part.
B— Defects parallel to the long axis of the part.
C— Defects parallel to the concentric circles of magnetic force within the part.
A— Defects perpendicular to the long axis of the part.
A part magnetized longitudinally by current flowing through a cable wrapped around it will show up defects that are perpendicular (at right angles) to the long axis of the part.
- Circular magnetization of a part can be used to detect which defects?
A— Defects parallel to the long axis of the part.
B— Defects perpendicular to the long axis of the part.
C— Defects perpendicular to the concentric circles of magnetic force within the part.
A— Defects parallel to the long axis of the part.
A part magnetized circularly by the magnetizing current f lowing lengthwise through it, will show up defects parallel to the long axis of the part.
(1) In nondestructive testing, a discontinuity may be defined as an interruption in the normal physical structure or configuration of a part.
(2) A discontinuity may or may not affect the usefulness of a part.
Regarding the above statements,
A— only 1 is true.
B— only 2 is true.
C— both 1 and 2 are true.
C— both 1 and 2 are true.
Statement 1 is true. In nondestructive testing, a discontinuity may be defined as an interruption in the normal physical structure or configuration of a part.
Statement 2 is also true. A discontinuity may or may not affect the usefulness of a part.
- What type of corrosion may attack the grain boundaries of aluminum alloys when the heat treatment process has been improperly accomplished?
A— Concentration cell.
B— Intergranular.
C— Fretting.
B— Intergranular.
An aluminum alloy part is heat-treated by being heated in an oven and then removed and immediately quenched in cold water.
If there is a delay between the time the part is removed from the oven and the time it is quenched, the grains in the metal will grow. Because of this, there is a good probability that intergranular corrosion will develop along the boundaries of the grains within the metal.
- Which of the following describe the effects of annealing steel and aluminum alloys?
A— Decreasing internal stresses and softening of the metal.
B— Softening of the metals and improved corrosion resistance.
C— Improved corrosion resistance.
A— Decreasing internal stresses and softening of the metal.
Steel and aluminum alloys may be annealed to decrease internal stresses and soften the metal. Annealing does not improve corrosion resistance.
- Which heat-treating process of metal produces a hard, wear-resistant surface over a strong, tough core?
A— Case hardening.
B— Annealing.
C— Tempering.
A— Case hardening.
Case hardening is a heat treatment process for steel in which the surface is hardened to make it wear resistant, but the inside of the metal remains strong and tough.
Annealing is a heat treatment process for either ferrous or nonferrous metal that makes the metal softer.
Tempering is a method of heat treatment in which some of the hardness is removed from a hardened metal. Removing some of the hardness makes the metal less brittle.
- Which heat-treating operation would be performed when the surface of the metal is changed chemically by introducing a high carbide or nitride content?
A— Tempering.
B— Normalizing.
C— Case hardening.
C— Case hardening.
In case hardening, the surface of the metal is changed chemically by introducing a high carbide or nitride content. The core is unaffected chemically. When heat-treated, the surface responds to hardening while the core remains tough.
- Normalizing is a process of heat treating
A— aluminum alloys only.
B— iron-base metals only.
C— both aluminum alloys and iron-base metals.
B— iron-base metals only.
Normalizing is a heat treating process in which an iron base metal is heated to a temperature above its critical temperature and allowed to cool in still air. Normalizing reduces the stresses in the metal that were put there by the fabrication process.
- Which of the following occurs when a mechanical force such as rolling, hammering, or bending is repeatedly applied to most metals at room temperature?
A— The metals become artificially aged.
B— The metals become stress corrosion cracked.
C— The metals become strain or work hardened or cold worked.
C— The metals become strain or work hardened or cold worked.
When a mechanical force such as rolling, hammering, or bending is repeatedly applied to most metals at room temperature, the metals become strain or work hardened or cold worked. They may become so hard and brittle that they break.
- The reheating of a heat treated metal, such as with a welding torch
A— has little or no effect on a metal’s heat treated characteristics.
B— has a cumulative enhancement effect on the original heat treatment.
C— can significantly alter a metal’s properties in the reheated area.
C— can significantly alter a metal’s properties in the reheated area.
When a heat-treated metal is reheated with a welding torch there is no close control of the temperature and the metal’s properties in the reheated area may be significantly altered.
- Why is steel tempered after being hardened?
A— To increase its hardness and ductility.
B— To increase its strength and decrease its internal stresses.
C— To relieve its internal stresses and reduce its brittleness.
C— To relieve its internal stresses and reduce its brittleness.
Steel is tempered after it is hardened to remove some of the internal stresses and make it less brittle. Tempering is done by heating it to a temperature quite a way below its critical temperature and allowing it to cool in still air.
- What aluminum alloy designations indicate that the metal has received no hardening or tempering treatment?
A— 3003-F.
B— 5052-H36.
C— 6061-O.
A— 3003-F.
In the temper designations used with aluminum alloy, these letters have the following meanings:
F means “as fabricated.” There has been no control over its temper.
H36 means the metal is non heat-treatable, but it has been strain-hardened and stabilized to its 3/4 hard state.
O means the metal has been annealed.
- Which material cannot be heat treated repeatedly without harmful effects?
A— Unclad aluminum alloy in sheet form.
B— 6061-T9 stainless steel.
C— Clad aluminum alloy.
C— Clad aluminum alloy.
Clad aluminum alloy sheets have a core of high-strength aluminum alloy onto whose surface have been rolled a thin layer of pure aluminum.
When clad sheets are heated in the process of heat treatment, some of the pure aluminum diffuses into the core alloy and weakens the sheet.
The manufacturer of the aluminum specifies the number of times clad sheets can be heat-treated. Typically, they allow the sheet to be heat-treated only one to three times.
- (Refer to Figure 42.) Which of the bolthead code markings shown identifies an AN corrosion resistant steel bolt?
A— 1.
B— 2.
C— 3.
C— 3.
The single dash on the head of a bolt identifies it as a standard bolt made of corrosion-resistant steel.
The cross on the head of a bolt identifies it as a standard AN bolt made of nickel alloy steel.
The cross inside a triangle identifies the bolt as an NAS close-tolerance bolt.
- (Refer to Figure 43.)
Identify the clevis bolt illustrated.
A— 1.
B— 3.
C— 2.
B— 3.
The bolt shown in View 3 is a clevis bolt.
The bolt shown in View 1 is a drilled-head, hex-head bolt.
The bolt shown in View 2 is an eyebolt.
- (Refer to Figure 44.) Identify the weld caused by an excessive amount of acetylene.
A— 4.
B — 1.
C— 3.
C— 3.
The weld in view 3 has been made with a flame that had an excess of acetylene. There are bumps along the center of the bead and craters at the edge of the weld. Cross checks are apparent where the body of the weld is sound. If this weld were cross-sectioned, it would show pockets and porosity.
The weld in view 1 was made too rapidly. The long and pointed appearance of the ripples was caused by an excessive amount of heat or by an oxidizing flame. If this weld were cross-sectioned, it would probably show gas pockets, porosity, and slag inclusions.
The weld in view 2 has improper penetration and cold laps caused by insufficient heat. It appears rough and irregular and its edges are not feathered into the base metal.
The weld in view 4 has considerable variations in depth of penetration. It often has the appearance of a cold weld.
- (Refer to Figure 44.) Select the illustration which depicts a cold weld.
A— 3.
B— 2.
C— 4.
B— 2.
The weld in view 2 is a cold weld. It has improper penetration and cold laps caused by insufficient heat. It appears rough and irregular and its edges are not feathered into the base metal.
- (Refer to Figure 45.) What type weld is shown at A?
A— Fillet.
B— Butt.
C— Lap.
B— Butt.
Weld A is a single butt weld.
Weld B is a double butt weld.
Welds C are both butt welds.
Weld D is a rosette weld.
Weld E is a fillet weld.
Weld F is an edge weld.
Welds G are lap welds.
- (Refer to Figure 45.) What type weld is shown at B?
A— Butt.
B— Double butt.
C— Fillet.
B— Double butt.
Weld B is a double butt weld. Both sides of the material have been ground in a V and a bead is formed on both sides of the sheet.
- (Refer to Figure 45.)
What type weld is shown at G?
A— Lap.
B— Butt.
C— Joint.
A— Lap.
The two welds shown at G are lap welds. When both edges are welded as is done here, the joint is called a double lap joint.
- (Refer to Figure 46.) The measurement reading on the illustrated micrometer is
A— 0.2851
B — 0.2911
C— 0.2901
A— 0.2851
The thimble has been screwed out more than 0.275 inch (the third mark beyond the 2 is visible).
The thimble has rotated just a bit more than ten hundredths of the way around, as is shown by the line on the sleeve. It is just beyond the 10 on the thimble. This means that 0.010 is added to the 0.275 on the sleeve.
The vernier line on the barrel for 1 is lined up with one of the marks on the thimble, and this means that we add 0.0001 inch to the two other numbers we have.
The total reading of this vernier micrometer is 0.275 + 0.010 + 0.0001 = 0.2851 inch.
- (Refer to Figure 47.) What is the measurement reading on the vernier caliper scale?
A— 1.411 inches.
B— 1.436 inches.
C— 1.700 inches.
B— 1.436 inches.
The zero on the vernier scale is beyond the 1. This shows the measurement is more than 1 inch.
It is beyond the 4. This shows it is more than 1.4 inch. It is beyond the first 0.025 mark on the bar. This shows that it is more than 1.425 inch.
The 11 line on the vernier scale lines up with one of the marks on the bar. This 0.011 is added to the reading we have.
The total reading is 1.425 + 0.011 = 1.436 inches.
- (Refer to Figure 48.) What does the micrometer read?
A— .2974
B— .3004
C— .3108
B— .3004
The thimble has been screwed out slightly more than 0.300 inch (the line by the 3 is just barely visible).
The thimble has rotated just a bit beyond the zero line on the sleeve. The vernier line on the barrel for 4 is lined up with one of the marks on the thimble, and this means that we add 0.0004 inch to the other numbers we have.
The total reading of this vernier micrometer is 0.300 + 0.0004 = 0.3004 inch.
- (Refer to Figure 49.) The measurement reading on the micrometer is
A— .2758
B— .2702
C— .2792
C— .2792
The thimble has moved out beyond two tenths of an inch (0.200 inch).
It has moved out to a point that is beyond the third twenty-five thousandths of an inch mark. (0.075 inch.)
The 4 mark on the thimble has just passed the scale on the sleeve (0.075 + 0.004 inch = 0.079). The 2 line on the vernier scale lines up with a number on the thimble.
This adds 0.0002 inch. The total measurement is 0.2792 inch.
