AWS CWEng Part 1 - Certified Welding
Engineer Fundamentals of Science and
Mathematics Exam QUESTIONS AND
1. A carbon steel weld undergoes a thermal cycle with a peak temperature of 1350°C
and a cooling time from 800°C to 500°C (t8/5) of 12 seconds. Using the Rosenthal
thick-plate solution, which change would most effectively reduce the risk of cold
cracking without altering the heat input?
A. Increase the welding current while decreasing travel speed proportionally.
B. Preheat the base metal to 150°C prior to welding.
C. Increase the electrode extension (stick-out) to raise resistance heating.
D. Switch from DCEN to DCEP to increase heat input.
Answer: B. Preheat the base metal to 150°C prior to welding.
Rationale: Preheating reduces the cooling rate (increases t8/5), allowing more time for
hydrogen diffusion and reducing the likelihood of martensite formation and cold cracking,
without changing the heat input. Increasing current or changing polarity alters heat input,
while stick-out affects deposition rate and heat input indirectly.
2. In a shielded metal arc welding (SMAW) operation, the arc voltage is 25 V and the
current is 150 A. The arc length is doubled while maintaining the same current.
Assuming a linear arc characteristic, what is the approximate new arc voltage?
A. 12.5 V
B. 25 V
C. 50 V
D. 37.5 V
Answer: C. 50 V
Rationale: For a linear arc characteristic, voltage is proportional to arc length (V = a +
bL). Doubling the arc length while keeping current constant doubles the voltage component
related to length, so the voltage increases. If the original voltage was 25 V at length L, at
2L it becomes approximately 50 V (assuming the constant 'a' is negligible).
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,3. A welded joint in a pressure vessel is inspected using ultrasonic testing. The sound
velocity in steel is 5900 m/s. If the time of flight for a longitudinal wave is 20 s, what is
the approximate path length?
A. 59 mm
B. 118 mm
C. 29.5 mm
D. 236 mm
Answer: B. 118 mm
Rationale: Path length = velocity × time = 5900 m/s × 20×10^-6 s = 0.118 m = 118 mm.
This is a direct application of the wave equation.
4. Which of the following best explains why the heat-affected zone (HAZ) in a
high-strength low-alloy (HSLA) steel weldment can exhibit lower toughness than the
weld metal, even when matching filler metal is used?
A. The HAZ experiences rapid heating and cooling that can produce untempered
martensite and coarse grains.
B. The weld metal has higher nickel content that refines its microstructure.
C. The HAZ is shielded from atmospheric contamination by the slag, while the weld
metal is not.
D. The weld metal undergoes post-weld heat treatment that the HAZ does not.
Answer: A. The HAZ experiences rapid heating and cooling that can produce
untempered martensite and coarse grains.
Rationale: The HAZ is subjected to steep thermal gradients that can form brittle phases like
untempered martensite and cause grain coarsening, reducing toughness. The weld metal
may have a more favorable microstructure due to controlled solidification and alloying.
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,5. In a statistical process control (SPC) chart for weld tensile strength, the process
mean is 500 MPa and the standard deviation is 25 MPa. If the specification limits are
450 MPa and 550 MPa, what is the process capability index (Cpk)?
A. 0.67
B. 1.00
C. 1.33
D. 2.00
Answer: A. 0.67
Rationale: Cpk = min[(USL - mean)/(3), (mean - LSL)/(3)] = min[(550-500)/(75),
(500-450)/(75)] = min[0.67, 0.67] = 0.67. This indicates the process is not capable of
meeting specifications consistently.
6. A weld is made on a restrained butt joint. After cooling, the residual stress
distribution is measured. Which statement best describes the typical longitudinal
residual stress pattern?
A. Tensile stresses in the weld center, balanced by compressive stresses away from the
weld.
B. Compressive stresses in the weld center, balanced by tensile stresses away from the
weld.
C. Uniform tensile stress across the entire plate.
D. Zero stress everywhere due to stress relief from phase transformations.
Answer: A. Tensile stresses in the weld center, balanced by compressive stresses away
from the weld.
Rationale: During cooling, the weld metal contracts but is restrained by the cooler base
metal, resulting in tensile residual stresses in the weld and HAZ, balanced by compressive
stresses in the adjacent base metal to maintain equilibrium.
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, 7. For a given welding process, the heat input is calculated as (V × I × 60) / (1000 × v)
kJ/mm, where V is volts, I is amperes, and v is travel speed in mm/min. If V = 25 V, I =
200 A, and v = 300 mm/min, what is the heat input?
A. 0.5 kJ/mm
B. 1.0 kJ/mm
C. 1.5 kJ/mm
D. 2.0 kJ/mm
Answer: B. 1.0 kJ/mm
Rationale: Heat input = (25 × 200 × 60) / (1000 × 300) = 300,,000 = 1.0 kJ/mm.
This is a straightforward calculation using the standard formula.
8. A weldment is to be made from a quenched and tempered steel with a carbon
equivalent (CE) of 0.45. Which combination of preheat and interpass temperature is
most appropriate to avoid hydrogen-assisted cracking?
A. No preheat, interpass below 100°C
B. Preheat to 100°C, interpass 100-150°C
C. Preheat to 200°C, interpass 200-250°C
D. Preheat to 300°C, interpass 300-350°C
Answer: C. Preheat to 200°C, interpass 200-250°C
Rationale: For a CE of 0.45, typical guidelines recommend a preheat of at least 200°C to
slow cooling and allow hydrogen to escape, reducing cracking risk. Interpass should be
maintained within a similar range to avoid excessive cooling or overheating.
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Engineer Fundamentals of Science and
Mathematics Exam QUESTIONS AND
1. A carbon steel weld undergoes a thermal cycle with a peak temperature of 1350°C
and a cooling time from 800°C to 500°C (t8/5) of 12 seconds. Using the Rosenthal
thick-plate solution, which change would most effectively reduce the risk of cold
cracking without altering the heat input?
A. Increase the welding current while decreasing travel speed proportionally.
B. Preheat the base metal to 150°C prior to welding.
C. Increase the electrode extension (stick-out) to raise resistance heating.
D. Switch from DCEN to DCEP to increase heat input.
Answer: B. Preheat the base metal to 150°C prior to welding.
Rationale: Preheating reduces the cooling rate (increases t8/5), allowing more time for
hydrogen diffusion and reducing the likelihood of martensite formation and cold cracking,
without changing the heat input. Increasing current or changing polarity alters heat input,
while stick-out affects deposition rate and heat input indirectly.
2. In a shielded metal arc welding (SMAW) operation, the arc voltage is 25 V and the
current is 150 A. The arc length is doubled while maintaining the same current.
Assuming a linear arc characteristic, what is the approximate new arc voltage?
A. 12.5 V
B. 25 V
C. 50 V
D. 37.5 V
Answer: C. 50 V
Rationale: For a linear arc characteristic, voltage is proportional to arc length (V = a +
bL). Doubling the arc length while keeping current constant doubles the voltage component
related to length, so the voltage increases. If the original voltage was 25 V at length L, at
2L it becomes approximately 50 V (assuming the constant 'a' is negligible).
Page 1 of 26
,3. A welded joint in a pressure vessel is inspected using ultrasonic testing. The sound
velocity in steel is 5900 m/s. If the time of flight for a longitudinal wave is 20 s, what is
the approximate path length?
A. 59 mm
B. 118 mm
C. 29.5 mm
D. 236 mm
Answer: B. 118 mm
Rationale: Path length = velocity × time = 5900 m/s × 20×10^-6 s = 0.118 m = 118 mm.
This is a direct application of the wave equation.
4. Which of the following best explains why the heat-affected zone (HAZ) in a
high-strength low-alloy (HSLA) steel weldment can exhibit lower toughness than the
weld metal, even when matching filler metal is used?
A. The HAZ experiences rapid heating and cooling that can produce untempered
martensite and coarse grains.
B. The weld metal has higher nickel content that refines its microstructure.
C. The HAZ is shielded from atmospheric contamination by the slag, while the weld
metal is not.
D. The weld metal undergoes post-weld heat treatment that the HAZ does not.
Answer: A. The HAZ experiences rapid heating and cooling that can produce
untempered martensite and coarse grains.
Rationale: The HAZ is subjected to steep thermal gradients that can form brittle phases like
untempered martensite and cause grain coarsening, reducing toughness. The weld metal
may have a more favorable microstructure due to controlled solidification and alloying.
Page 2 of 26
,5. In a statistical process control (SPC) chart for weld tensile strength, the process
mean is 500 MPa and the standard deviation is 25 MPa. If the specification limits are
450 MPa and 550 MPa, what is the process capability index (Cpk)?
A. 0.67
B. 1.00
C. 1.33
D. 2.00
Answer: A. 0.67
Rationale: Cpk = min[(USL - mean)/(3), (mean - LSL)/(3)] = min[(550-500)/(75),
(500-450)/(75)] = min[0.67, 0.67] = 0.67. This indicates the process is not capable of
meeting specifications consistently.
6. A weld is made on a restrained butt joint. After cooling, the residual stress
distribution is measured. Which statement best describes the typical longitudinal
residual stress pattern?
A. Tensile stresses in the weld center, balanced by compressive stresses away from the
weld.
B. Compressive stresses in the weld center, balanced by tensile stresses away from the
weld.
C. Uniform tensile stress across the entire plate.
D. Zero stress everywhere due to stress relief from phase transformations.
Answer: A. Tensile stresses in the weld center, balanced by compressive stresses away
from the weld.
Rationale: During cooling, the weld metal contracts but is restrained by the cooler base
metal, resulting in tensile residual stresses in the weld and HAZ, balanced by compressive
stresses in the adjacent base metal to maintain equilibrium.
Page 3 of 26
, 7. For a given welding process, the heat input is calculated as (V × I × 60) / (1000 × v)
kJ/mm, where V is volts, I is amperes, and v is travel speed in mm/min. If V = 25 V, I =
200 A, and v = 300 mm/min, what is the heat input?
A. 0.5 kJ/mm
B. 1.0 kJ/mm
C. 1.5 kJ/mm
D. 2.0 kJ/mm
Answer: B. 1.0 kJ/mm
Rationale: Heat input = (25 × 200 × 60) / (1000 × 300) = 300,,000 = 1.0 kJ/mm.
This is a straightforward calculation using the standard formula.
8. A weldment is to be made from a quenched and tempered steel with a carbon
equivalent (CE) of 0.45. Which combination of preheat and interpass temperature is
most appropriate to avoid hydrogen-assisted cracking?
A. No preheat, interpass below 100°C
B. Preheat to 100°C, interpass 100-150°C
C. Preheat to 200°C, interpass 200-250°C
D. Preheat to 300°C, interpass 300-350°C
Answer: C. Preheat to 200°C, interpass 200-250°C
Rationale: For a CE of 0.45, typical guidelines recommend a preheat of at least 200°C to
slow cooling and allow hydrogen to escape, reducing cracking risk. Interpass should be
maintained within a similar range to avoid excessive cooling or overheating.
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