AWS Certified Welding Inspector Exam
Welding Metallurgy Practice Questions
40 practice questions with detailed explanations — aligned to the AWS Certified Welding Inspector Exam.
Q1.In the Iron-Carbon phase diagram, the temperature at which liquid steel begins to solidify upon cooling is called the:
A.Solidus lineB.Liquidus lineC.Eutectoid pointD.Transformation lineB. Liquidus lineExplanation: The Liquidus line represents the temperature above which the alloy is completely liquid. Below this line, solidification begins.
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Q2.The region of the base metal that has not been melted but whose mechanical properties or microstructure have been altered by the heat of welding is called the:
A.Fusion ZoneB.Weld InterfaceC.Heat Affected Zone (HAZ)D.Unaffected Base MetalC. Heat Affected Zone (HAZ)Explanation: The Heat Affected Zone (HAZ) is the area of base metal adjacent to the weld that was not melted but was heated high enough to cause microstructural changes.
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Q3.Hydrogen-induced cracking (cold cracking) in carbon steels is most likely to occur in the:
A.Weld metal centerlineB.Heat Affected Zone (HAZ)C.Unaffected base metalD.Fusion boundaryB. Heat Affected Zone (HAZ)Explanation: Hydrogen cracking typically occurs in the coarse-grained region of the HAZ where the microstructure is hardest and most brittle (martensite).
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Q4.Which of the following elements is the primary hardening agent in steel?
A.ManganeseB.SiliconC.CarbonD.SulfurC. CarbonExplanation: Carbon is the most significant alloying element controlling the hardness and strength of steel. Higher carbon content increases hardness but decreases weldability.
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Q5.Preheating is primarily used to:
A.Increase the cooling rateB.Decrease the cooling rateC.Increase the hardness of the HAZD.Burn off impuritiesB. Decrease the cooling rateExplanation: Preheating slows down the cooling rate of the weld and HAZ, allowing hydrogen to diffuse out and preventing the formation of brittle martensite.
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Q6.A steel with a Carbon Equivalent (CE) greater than ___ is generally considered difficult to weld without special precautions.
A.0.20B.0.30C.0.40D.0.15C. 0.40Explanation: A CE of 0.40 or higher generally indicates a higher risk of cracking and requires preheat and low-hydrogen practices.
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Q7.The crystalline structure of austenite is:
A.Body Centered Cubic (BCC)B.Face Centered Cubic (FCC)C.Body Centered Tetragonal (BCT)D.Hexagonal Close Packed (HCP)B. Face Centered Cubic (FCC)Explanation: Austenite (Gamma iron) has a Face Centered Cubic (FCC) crystal structure, which can hold more carbon in solution than Ferrite.
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Q8.Which type of stainless steel is non-magnetic and cannot be hardened by heat treatment?
A.FerriticB.MartensiticC.AusteniticD.Precipitation HardeningC. AusteniticExplanation: Austenitic stainless steels (e.g., 300 series) are non-magnetic and hardenable only by cold working, not by heat treatment.
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Q9.The rapid cooling of austenite in carbon steel results in the formation of:
A.PearliteB.FerriteC.MartensiteD.CementiteC. MartensiteExplanation: Rapid quenching prevents the diffusion of carbon needed to form pearlite, instead shearing the lattice into a hard, brittle structure called Martensite.
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Q10.Sensitization in austenitic stainless steels (carbide precipitation) leads to:
A.Hydrogen crackingB.Intergranular corrosionC.Lamellar tearingD.PorosityB. Intergranular corrosionExplanation: Heating stainless steel in the 800°F-1600°F range causes chromium carbides to form at grain boundaries, depleting chromium and reducing corrosion resistance (intergranular corrosion).
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Q11.Low hydrogen electrodes are used to prevent:
A.PorosityB.Underbead cracking (Hydrogen cracking)C.Lack of fusionD.Slag inclusionsB. Underbead cracking (Hydrogen cracking)Explanation: The primary purpose of low hydrogen electrodes is to minimize hydrogen introduction, reducing the risk of underbead or cold cracking.
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Q12.Lamellar tearing typically occurs in:
A.The weld metalB.The base metal, parallel to the rolling directionC.The HAZD.Cast ironB. The base metal, parallel to the rolling directionExplanation: Lamellar tearing is a separation in the base metal caused by through-thickness strains (shrinkage) pulling apart non-metallic inclusions flattened by rolling.
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Q13.Which element is considered an impurity in steel that can cause 'hot cracking' or shortness?
A.ManganeseB.SulfurC.ChromiumD.NickelB. SulfurExplanation: Sulfur forms iron sulfide films at grain boundaries which have a low melting point, causing cracking during solidification (hot cracking). Manganese is added to combine with sulfur (MnS) to prevent this.
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Q14.Stress relief heat treatment is performed:
A.Above the upper transformation temperatureB.Below the lower transformation temperatureC.At the melting pointD.At room temperatureB. Below the lower transformation temperatureExplanation: Stress relief is done below the transformation temperature (typically 1100-1200°F for steel) to relieve residual stresses without altering the metallurgical structure.
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Q15.In the designation '6061-T6' aluminum, the 'T6' indicates:
A.The chemical compositionB.The temper or heat treatment conditionC.The ultimate tensile strengthD.The filler metal to useB. The temper or heat treatment conditionExplanation: T6 indicates the aluminum has been solution heat treated and artificially aged to achieve specific mechanical properties.
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Q16.What effect does a high heat input have on the grain size of the HAZ in carbon steel?
A.Refines the grain sizeB.Promotes grain growth (coarse grains)C.Has no effectD.Causes graphitizationB. Promotes grain growth (coarse grains)Explanation: High heat input keeps the metal at high temperature longer, promoting grain growth. Coarse grains generally result in lower toughness.
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Q17.The term 'pass' in welding refers to:
A.A single progression of welding along a jointB.The layer of weld metalC.The complete weldD.The welding speedA. A single progression of welding along a jointExplanation: A pass is a single progression of the welding operation along a joint. A layer may consist of one or more passes.
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Q18.Which test is used to determine the notch toughness of a material?
A.Tensile testB.Charpy V-Notch testC.Guided bend testD.Brinell hardness testB. Charpy V-Notch testExplanation: The Charpy V-Notch test measures the energy absorbed by a specimen during fracture, indicating its toughness and transition temperature.
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Q19.Which of the following is NOT a form of heat treatment?
A.AnnealingB.NormalizingC.QuenchingD.PeeningD. PeeningExplanation: Peening is a mechanical working process used to induce compressive stresses, not a thermal heat treatment process.
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Q20.When welding Quenched and Tempered (Q&T) steels (e.g., A514), heat input must be:
A.Maximized to prevent crackingB.Controlled/Limited to prevent softening of the HAZC.IgnoredD.Kept extremely low to prevent hardeningB. Controlled/Limited to prevent softening of the HAZExplanation: Q&T steels derive strength from heat treatment. Excessive heat input acts like an over-tempering or annealing cycle, reducing the strength (softening) of the HAZ.
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Q21.The property of a metal that allows it to deform plastically without breaking is:
A.HardnessB.DuctilityC.StrengthD.StiffnessB. DuctilityExplanation: Ductility is the ability of a material to deform plastically (stretch/elongate) under tensile stress before rupture.
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Q22.What is the primary alloying element in '300 Series' stainless steel?
A.CarbonB.Chromium and NickelC.MolybdenumD.TitaniumB. Chromium and NickelExplanation: 300 series stainless (Austenitic) contains Iron, Chromium (typically 18%), and Nickel (typically 8%).
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Q23.Calculate the Carbon Equivalent (CE) for a steel with: C=0.18%, Mn=0.60%, Cr=0.10%, Mo=0.15%, V=0.05%, Ni=0.15%, Cu=0.15%. Using IIW Formula.
A.0.30B.0.36C.0.42D.0.25B. 0.36Explanation: Formula: CE = C + Mn/6 + (Cr+Mo+V)/5 + (Ni+Cu)/15. CE = 0.18 + (0.60/6) + (0.10+0.15+0.05)/5 + (0.15+0.15)/15. CE = 0.18 + 0.10 + 0.06 + 0.02 = 0.36.
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Q24.During the solidification of a weld, impurities tend to collect:
A.At the grain boundaries and centerlineB.At the surfaceC.In the HAZD.Evenly throughout the grainA. At the grain boundaries and centerlineExplanation: Impurities with lower melting points are pushed ahead of the solidifying front and concentrate at the center of the weld or grain boundaries, causing centerline cracking.
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Q25.The lowest temperature at which a steel exhibits ductile fracture behavior is the:
A.Melting pointB.Upper transformation temperatureC.Ductile-to-Brittle Transition Temperature (DBTT)D.Curie pointC. Ductile-to-Brittle Transition Temperature (DBTT)Explanation: DBTT is the temperature below which the material fracture mode changes from ductile to brittle. Operating below this temp increases risk of catastrophic failure.
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Q26.When welding gray cast iron, the primary difficulty is:
A.Excessive ductilityB.Formation of brittle white cast iron and martensite in the HAZC.Low melting pointD.Excessive porosityB. Formation of brittle white cast iron and martensite in the HAZExplanation: Gray cast iron has high carbon. Rapid cooling during welding forms hard, brittle white iron (cementite) and martensite, leading to cracking.
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Q27.A 'Schaeffler Diagram' is used to predict:
A.Preheat temperaturesB.Microstructure of stainless steel weldsC.Tensile strength of carbon steelD.Cooling ratesB. Microstructure of stainless steel weldsExplanation: The Schaeffler diagram uses Nickel and Chromium equivalents to predict the phases (Austenite, Ferrite, Martensite) present in stainless steel welds.
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Q28.Which of the following reduces the cooling rate of a weld?
A.Decreasing heat inputB.Increasing plate thicknessC.Increasing preheat temperatureD.Decreasing interpass temperatureC. Increasing preheat temperatureExplanation: Increasing preheat reduces the temperature gradient between the weld and the base metal, thereby slowing down the cooling rate.
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Q29.Sigma phase embrittlement can occur in stainless steels when held for long periods at:
A.Room temperatureB.1000°F - 1700°FC.Cryogenic temperaturesD.Near melting pointB. 1000°F - 1700°FExplanation: Sigma phase is a hard, brittle intermetallic phase that forms in high-chromium steels exposed to temperatures between 1000°F and 1700°F.
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Q30.The 'Yield Strength' of a material is defined as:
A.The point at which the material breaksB.The stress at which plastic deformation beginsC.The maximum stress the material can endureD.The hardness of the materialB. The stress at which plastic deformation beginsExplanation: Yield strength is the stress level where the material transitions from elastic (recoverable) deformation to plastic (permanent) deformation.
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Q31.Which gas is soluble in molten aluminum but precipitates upon freezing to cause porosity?
A.NitrogenB.OxygenC.HydrogenD.ArgonC. HydrogenExplanation: Hydrogen has high solubility in molten aluminum but very low solubility in solid aluminum. Upon freezing, trapped hydrogen forms gas pores.
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Q32.Ferrite number (FN) is a measurement used to quantify:
A.Carbon content in steelB.Delta ferrite content in austenitic stainless steel weldsC.Hardness of cast ironD.Hydrogen contentB. Delta ferrite content in austenitic stainless steel weldsExplanation: A small amount of Delta Ferrite (FN 3-10) is desired in austenitic welds to prevent hot cracking. FN is measured magnetically.
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Q33.Weld decay (intergranular corrosion) can be prevented by using:
A.High carbon electrodesB.Stabilized grades (e.g., 321, 347) or Low Carbon (L-grade) base metalsC.Higher heat inputD.Slower travel speedsB. Stabilized grades (e.g., 321, 347) or Low Carbon (L-grade) base metalsExplanation: Stabilized grades contain Titanium or Niobium to bind carbon, preventing chromium carbide formation. L-grades have <0.03% carbon.
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Q34.Which property is determined by the 'Percent Elongation' in a tensile test?
A.HardnessB.Tensile strengthC.DuctilityD.ToughnessC. DuctilityExplanation: Percent Elongation measures how much the specimen stretched before breaking, which is a direct measure of ductility.
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Q35.Heat input (J/in) is calculated by:
A.(Amps x Volts x 60) / Travel Speed (in/min)B.Amps x Volts x Travel SpeedC.(Amps x Travel Speed) / VoltsD.Amps + VoltsA. (Amps x Volts x 60) / Travel Speed (in/min)Explanation: Heat Input formula: H = (Voltage × Amperage × 60) / Travel Speed. This formula calculates the energy per unit length of weld, measured in joules per inch (J/in).
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Q36.The term 'Passivation' regarding stainless steel refers to:
A.Heat treating to anneal the metalB.Chemical treatment to remove surface iron and restore the passive oxide layerC.Hardening the surfaceD.Painting the surfaceB. Chemical treatment to remove surface iron and restore the passive oxide layerExplanation: Passivation (usually with nitric or citric acid) removes free iron from the surface and enhances the formation of the chromium-oxide layer that resists corrosion.
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Q37.When welding titanium, shielding is required:
A.On the face of the weld onlyB.On the face, root, and trailing HAZ until cooled below 800°FC.Titanium does not require shieldingD.Only for the first passB. On the face, root, and trailing HAZ until cooled below 800°FExplanation: Titanium is extremely reactive to atmospheric gases at high temperatures. It requires inert gas shielding on the face, root, and a trailing shield for the cooling metal.
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Q38.Martensite is best described as:
A.Soft and ductileB.Hard and brittleC.Tough and elasticD.LiquidB. Hard and brittleExplanation: Martensite is a supersaturated solid solution of carbon in iron, formed by rapid quenching. It is extremely hard but very brittle.
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Q39.Dilution is defined as:
A.The change in chemical composition of a welding filler metal caused by the admixture of the base metalB.Adding water to the fluxC.The width of the HAZD.The amount of hydrogen in the weldA. The change in chemical composition of a welding filler metal caused by the admixture of the base metalExplanation: Dilution is the percentage of base metal that melts and mixes with the filler metal to form the final weld metal composition.
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Q40.Which factor promotes hydrogen cracking?
A.High hydrogen levelB.Susceptible microstructure (Martensite)C.Tensile stressD.All of the aboveD. All of the aboveExplanation: Hydrogen cracking requires three conditions: Hydrogen, susceptible microstructure, and stress (plus temperature < 400°F).
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