Ultrasonic testing formulas turn travel time and echo amplitude into a flaw's position and size. The most used is depth = velocity × time ÷ 2: at 5,890 m/s, sound crosses 10 mm of steel and returns in 3.4 µs. The second is the decibel rule, where a 6 dB drop halves the amplitude.
This guide gives each formula with a worked number in steel, using velocities published by Evident (Olympus). It covers wavelength, decibels, acoustic impedance, near field, beam spread, Snell's law and critical angles, angle-beam trigonometry and skip distance, TOFD depth, and the AWS D1.1 attenuation factor. It ends with what the formulas can't tell you.
Contents: Velocities · Wavelength · Time of flight · Decibels · Impedance · Near field · Beam spread · Snell's law · Skip distance · TOFD · AWS D1.1 · Limits · FAQ
Which sound velocities should you use in UT calculations?
Use the measured velocity of the actual part when it matters, and published values only as a starting point. Every formula below depends on velocity, and Evident's velocity table calls its own figures "only a general guide". Alloy, heat treatment and temperature all shift the real value.
| Material | Longitudinal velocity | Shear velocity | Source |
|---|---|---|---|
| Carbon steel (1020) | 5,890 m/s (0.232 in/µs) | ≈3,240–3,250 m/s (0.128 in/µs) | Evident velocity table; Olympus Technical Notes |
| Alloy steel (4340) | 5,850 m/s | ≈3,240 m/s | Olympus Technical Notes |
| Acrylic (wedge material) | 2,730 m/s | 1,430 m/s | Evident; Olympus Technical Notes |
| Water (20 °C) | 1,480 m/s | none (liquids don't carry shear waves) | Evident |
The worked examples in this article use 5,890 m/s longitudinal and 3,250 m/s shear for steel, and 2,730 m/s for an acrylic wedge. TOFD procedures written to ISO 10863 assume 5,920 ± 50 m/s longitudinal and 3,255 ± 30 m/s shear in steel, so expect small differences between references. Calibrating on a block of the same material removes most of the error. Our guide to ultrasonic testing calibration covers how that is done on rail work.
Key takeaway: a 1% velocity error is a 1% depth error, so set velocity before you trust any number on the screen.
How are velocity, frequency and wavelength related?
Wavelength equals velocity divided by frequency: λ = v / f. Velocity belongs to the material, frequency to the transducer, and wavelength follows from the two.
| Formula | Meaning | Worked example in steel |
|---|---|---|
| λ = v / f | Wavelength | 5 MHz longitudinal: 5.89 mm/µs ÷ 5 MHz = 1.18 mm |
| λ = v / f | Wavelength (shear) | 2.25 MHz shear: 3.25 ÷ 2.25 = 1.44 mm |
| T = 1 / f | Period of one cycle | 5 MHz: 0.2 µs |
| PRF limit ≈ 1 / round-trip time | Pulse repetition frequency before echoes overlap | 500 mm steel bar: round trip 170 µs, so keep PRF well under 5.9 kHz |
A shorter wavelength resolves smaller and closer reflectors but is scattered and absorbed faster, so higher frequency trades penetration for resolution. That is why coarse-grained welds and castings are often tested at 1 to 2.25 MHz while thin-wall thickness gauging uses 5 to 10 MHz or more. The PRF figure is a floor, not a setting: multiple back-wall echoes in a low-attenuation part can still wrap into the next pulse, so leave margin.
How do you convert time of flight into depth?
In pulse-echo testing, depth d = v × t / 2, because the sound travels to the reflector and back. In through-transmission, with a separate transmitter and receiver on opposite faces, the factor of 2 drops out: d = v × t.
Worked example: pulse-echo time of flight in steel
- 10 mm steel, longitudinal wave: t = 2 × 10 mm ÷ 5.89 mm/µs = 3.40 µs round trip
- A measured round trip of 20 µs: d = 5.89 × 20 ÷ 2 = 58.9 mm
- The same 20 µs read with the shear velocity (3.25 mm/µs) would give 32.5 mm, which is why the instrument's velocity must match the wave mode in use
Olympus' technical notes give the same thickness relation for gauging. In practice the instrument also subtracts a zero offset (probe delay or wedge delay) before applying the formula. If the zero is wrong, every depth is wrong by the same amount, which a two-point calibration on a step block catches.
How do decibels relate to echo amplitude?
The decibel difference between two echoes is dB = 20 · log10(A1 / A2), where A1 and A2 are their amplitudes. The factor is 20, not 10, because UT instruments display amplitude (pressure), not energy. So a 6 dB drop halves the amplitude and quarters the energy.
| dB difference | Amplitude ratio A1 : A2 | Example on screen |
|---|---|---|
| 3 dB | 1.41 : 1 | 100% vs 71% full screen height (FSH) |
| 6 dB | 2 : 1 | 80% vs 40% FSH |
| 12 dB | 4 : 1 | 80% vs 20% FSH |
| 14 dB | 5 : 1 | 100% vs 20% FSH |
| 20 dB | 10 : 1 | 100% vs 10% FSH |
| 40 dB | 100 : 1 | 100% vs 1% FSH |
The 3, 6, 12, 20 and 40 dB rows match the table in Olympus' Ultrasonic Transducers Technical Notes. To reverse it, A1 / A2 = 10^(dB/20): adding 14 dB of gain multiplies amplitude by 5.01.
Worked example: comparing an indication with a reference
A side-drilled hole is set to 80% FSH. An indication peaks at 40% FSH at the same gain. It is 20 · log10(80/40) = 6.0 dB below reference. If you instead add gain until the indication reaches 80%, the added gain (6 dB) is the same answer. Attenuation in dB per unit length, signal-to-noise ratio and distance-amplitude correction all use this relationship. Our guide to the DAC curve in ultrasonic testing shows how the reference line is built.
Key takeaway: −6 dB means an amplitude ratio of about 0.5. It is not a beam-spread constant, a point the original version of this article blurred.
How much sound reflects at an interface?
Reflection depends on the acoustic impedance mismatch, Z = ρ × v. The amplitude (pressure) reflection coefficient is R = (Z2 − Z1) / (Z2 + Z1), and the reflected energy fraction is R².
| Formula | What it gives |
|---|---|
| Z = ρ × v | Acoustic impedance (often in 10⁶ kg/m²s, or "MRayl") |
| R = (Z2 − Z1) / (Z2 + Z1) | Reflected amplitude as a fraction; the sign shows phase reversal |
| Reflected energy (%) = R² × 100 = [(Z2 − Z1)² / (Z2 + Z1)²] × 100 | Reflected energy |
| Transmitted energy (%) = 100 − reflected energy (%) | Energy that crosses the boundary |
Worked example: steel–water and steel–air
Olympus lists Z ≈ 45.41 for 1020 steel and 1.48 for water (both in 10⁶ kg/m²s).
- Steel to water: R = (1.48 − 45.41) / (1.48 + 45.41) = −0.937. The back-wall echo is 20 · log10(0.937) = −0.57 dB, and 87.8% of the energy reflects.
- Transmitted energy is 12.2%, or 10 · log10(0.122) = −9.1 dB. Olympus gives the same −0.57 dB and −9.13 dB values.
- Steel to air: with Z for air near 0.0004, R is about −0.99998 and essentially all the energy reflects.
This is why an open crack gives a strong echo, and why couplant is needed at all: an air film between probe and part reflects nearly everything. A tight crack or a fusion-line flaw between similar steels can still give a weak echo, so impedance predicts the best case, not every case. Our pulse-echo ultrasonic inspection guide covers how the instrument uses these echoes.
How do you calculate near field length?
For a flat circular element, near field length N = D² f / (4 v), or N = D² / (4λ), where D is the element diameter, f the frequency and v the velocity in the material. Olympus describes the near field this way:
"Because of the variations within the near field it can be difficult to accurately evaluate flaws using amplitude based techniques." (Olympus NDT, Ultrasonic Transducers Technical Notes)
Evident adds that N "represents the natural focus of the transducer": the last on-axis amplitude maximum.
Worked example: near field in steel and water
| Probe | Medium | Calculation (D in mm, f in MHz, v in mm/µs) | Near field N |
|---|---|---|---|
| 5 MHz, 10 mm element | Steel, longitudinal | 10² × 5 ÷ (4 × 5.89) | 21.2 mm |
| 2.25 MHz, 12.7 mm (0.5 in) element | Steel, longitudinal | 12.7² × 2.25 ÷ (4 × 5.89) | 15.4 mm |
| 2.25 MHz, 25.4 mm (1.0 in) element | Water | 25.4² × 2.25 ÷ (4 × 1.48) | 245 mm (9.65 in) |
The water result is a useful check: Olympus' table gives 9.55 in for the same probe, slightly shorter because real elements are a little smaller than nominal. Doubling the diameter quadruples N, and raising the frequency lengthens it, so a large high-frequency probe can put a thin part entirely inside its near field. Amplitude comparisons with a DAC are most reliable beyond about one near-field length. For the difference between the zones, see near field vs far field in UT.
These are flat, circular, contact or immersion cases. Angle-beam probes, rectangular elements and focused probes need the manufacturer's effective dimensions or modeling, and the wedge path uses up part of the near field.
How do you calculate beam spread?
The half-angle of beam spread is sin(α/2) = K × v / (f × D). Olympus gives K = 0.514 for the −6 dB pulse-echo beam edge, the definition a flaw detector user normally needs. Larger elements and higher frequencies narrow the beam.
The constant K depends on how the beam edge is defined, and on whether you mean the one-way (transmitted) field or the two-way pulse-echo response. That is where the original version of this article went wrong: its K values were one-way values, and its "0 dB = 1.22" row was really the first null (zero amplitude).
| Beam edge | Amplitude ratio at edge | K, pulse-echo (two-way) | K, one-way (transmitted field) |
|---|---|---|---|
| −3 dB | 0.71 | 0.37 | 0.51 |
| −6 dB | 0.50 | 0.51 (Olympus: 0.514) | 0.70 |
| −10 dB | 0.32 | 0.65 | 0.87 |
| −12 dB | 0.25 | 0.70 | 0.93 |
| −20 dB | 0.10 | 0.87 | 1.09 |
| First null (no signal) | 0 | 1.22 | 1.22 |
The K values are calculated from the standard circular-piston directivity function, 2J₁(x)/x, and the pulse-echo −6 dB value reproduces Olympus' 0.514. A pulse-echo −6 dB edge is the one-way −3 dB edge, which is why 0.51 appears in both columns. Some study guides quote slightly different constants; state which definition your procedure uses.
Worked example: beam spread and beam width
- 5 MHz, 10 mm probe in steel: sin(α/2) = 0.514 × 5.89 ÷ (5 × 10) = 0.0605, so α/2 = 3.5°. At 100 mm depth the −6 dB beam is roughly 2 × 100 × tan 3.5° ≈ 12 mm wide.
- 2.25 MHz, 12.7 mm probe in steel: α/2 = 6.1°, about 21 mm wide at 100 mm. To the first null, the same probe spreads to 14.6°.
Beam width matters for length sizing: a 6 dB drop length measured with a 21 mm wide beam on a small reflector mostly measures the beam, not the flaw.
How does Snell's law set the refracted angle?
Snell's law is sin θ1 / sin θ2 = v1 / v2, where θ1 is the incident angle in the wedge and θ2 the refracted angle in the part. It is how a plastic wedge produces a 45°, 60° or 70° shear wave in steel.
| Result (acrylic wedge, 2,730 m/s, on steel) | Calculation | Angle |
|---|---|---|
| First critical angle (longitudinal refracts to 90°) | sin⁻¹(2,730 / 5,890) | 27.6° |
| Second critical angle (shear refracts to 90°) | sin⁻¹(2,730 / 3,250) | 57.1° |
| Wedge angle for 45° shear | sin⁻¹(2,730 / 3,250 × sin 45°) | 36.4° |
| Wedge angle for 60° shear | sin⁻¹(2,730 / 3,250 × sin 60°) | 46.7° |
| Wedge angle for 70° shear | sin⁻¹(2,730 / 3,250 × sin 70°) | 52.1° |
| Water on steel, first and second critical | sin⁻¹(1,480 / 5,890), sin⁻¹(1,480 / 3,250) | 14.6° and 27.1° |
Between the two critical angles only the shear wave enters the steel, which gives a clean single-mode beam. Using 3,230 m/s for shear moves the second critical angle to about 57.7°. Olympus notes that Snell's law "doesn't hold" well at low frequency and small element size because of beam spread, so check the actual angle on an IIW-type block. Shear-wave behavior and wedge selection are covered in our guide to shear wave ultrasonic testing.
How do you calculate skip distance, depth and surface distance?
For an angle beam of refracted angle θ in a plate of thickness T: depth = sound path × cos θ, surface distance = sound path × sin θ, and full skip distance = 2 × T × tan θ. Half skip (T × tan θ) is where the beam first reaches the back wall.
| Quantity | Formula |
|---|---|
| Depth (first leg) | d = SP × cos θ |
| Surface distance from the probe index | SD = SP × sin θ |
| Leg length (sound path to the back wall) | SP_leg = T / cos θ |
| Full skip distance | Skip = 2 × T × tan θ |
| Depth on the second leg | d = 2T − SP × cos θ |
Worked example: 70° shear wave in a 25 mm (1 in) plate
- Leg length: 25 ÷ cos 70° = 73.1 mm (2.92 in).
- Full skip distance: 2 × 25 × tan 70° = 137.4 mm (5.49 in). Half skip is 68.7 mm.
- Indication at 50 mm sound path (first leg): depth = 50 × cos 70° = 17.1 mm; surface distance = 50 × sin 70° = 47.0 mm.
- Indication at 100 mm sound path: that is past the 73.1 mm leg, so it is on the second leg. Depth = 2 × 25 − 100 × cos 70° = 15.8 mm, measured from the top surface, at a surface distance of 94.0 mm.
Step 4 is where reports go wrong: reading 100 × cos 70° = 34.2 mm as a depth puts the reflector below a 25 mm plate. Always compare the sound path with the leg length first. For comparison, a 60° beam in 20 mm plate has a 40.0 mm leg and a 69.3 mm skip, and a 45° beam in 25 mm plate has a 35.4 mm leg and a 50.0 mm skip.
The same trigonometry applies on rail. A 70° probe that sees a transverse defect at 40 mm sound path in the rail head puts it about 13.7 mm below the running surface and 37.6 mm ahead of the index point. Our guide to angle beam testing in UT explains scanning patterns.
Key takeaway: check the probe index and actual refracted angle on a calibration block before mapping, because a 2° angle error moves a 70° reflector several millimeters.
How is flaw depth calculated in TOFD?
TOFD does not use the pulse-echo formula. It uses the geometry of two probes facing each other at a fixed separation: the diffracted signal travels from the transmitter down to the flaw tip and back up to the receiver, so the path is two hypotenuses, not a straight down-and-back.
For half probe separation S (PCS = 2S), velocity c and diffracted-signal time t (after removing probe delay):
d = √[(c × t / 2)² − S²]
Equivalently, using the time Δt measured after the lateral wave (the first signal, which travels along the surface in 2S / c):
d = √[((c × Δt + PCS) / 2)² − (PCS / 2)²]
Both forms assume the flaw tip sits midway between the probes. Baker Hughes' TOFD guide uses the Δt form and notes that depth "has a tolerance due to the unknown transverse coordinate of the diffraction centre." For a tip centered between the probes, it gives a typical tolerance of about ±0.2 mm, depending on wall thickness and depth zone.
Worked example: TOFD tip depth in a 30 mm weld
- PCS = 60 mm (S = 30 mm), c = 5.92 mm/µs (the ISO 10863 steel value)
- Lateral wave arrives at 60 ÷ 5.92 = 10.14 µs. The back wall of a 30 mm plate arrives at 2 × √(30² + 30²) ÷ 5.92 = 14.33 µs.
- An upper-tip signal arrives at t = 13.0 µs: c × t / 2 = 38.48 mm, so d = √(38.48² − 30²) = 24.1 mm
- The pulse-echo formula would give 38.5 mm, deeper than the plate itself. That was the error in the original version of this article, which gave "2 × distance ÷ velocity" for TOFD.
TOFD has a near-surface dead zone hidden under the lateral wave. Baker Hughes' example puts it at 3.9 mm (10 MHz) to 5.5 mm (5 MHz) for a 60 mm PCS, and it also has a back-wall dead zone. TOFD is therefore normally paired with pulse-echo or PAUT. ASTM E2373/E2373M describes TOFD as "proven effective" on 9 to 300 mm (0.375 to 12 in.) thickness. See PAUT vs TOFD for rail weld inspection and TOFD rail inspection for where it does and doesn't fit.
How does the AWS D1.1 attenuation factor work?
In AWS D1.1 angle-beam UT, each indication is rated as d = a − b − c, where c is the attenuation factor: c = 2 × (sound path in inches − 1) dB. That is 2 dB per inch of sound path beyond the first inch. Evident's AWS weld inspection tutorial gives the same formula.
| Term | Meaning |
|---|---|
| a | Indication level: gain (dB) that brings the indication to the reference line |
| b | Reference level: gain for the reference reflector (a 1.5 mm side-drilled hole) |
| c | Attenuation factor: 2 × (sound path in inches − 1) |
| d | Indication rating, a − b − c |
Worked example: rating one indication
An indication peaks at 3.5 in. sound path. It reaches the reference line at 50 dB, and the reference hole needed 46 dB.
- c = 2 × (3.5 − 1) = 5 dB
- d = 50 − 46 − 5 = −1 dB
A lower (more negative) rating means a stronger reflector. Whether −1 dB is Class A, B, C or D depends on the weld thickness, probe angle and static or cyclic acceptance table in the edition named in the contract, so no fixed dB band applies. The current edition is AWS D1.1/D1.1M:2025, and secondary summaries say it changed how c is handled on instruments that show fractional dB, so use the rounding rule in your edition. The full rating procedure, with class tables, is in our guide to AWS D1.1 weld inspection criteria. How rejection levels differ between codes is in our guide to ultrasonic testing acceptance criteria standards.
The 2 dB/in. figure is a code convention, not a measured attenuation. Holloway's 2017 CINDE Journal modeling found it within about 2 dB at 45°, but nearly 6 dB off at 60° and 70° with sound paths around 150 mm (6 in.). On thick welds that can move an indication across a class boundary.
What ultrasonic testing formulas can't tell you
Every formula here describes an idealized beam in a uniform, isotropic material. Real inspections break those assumptions in predictable ways.
| Formula | Assumption | When it fails |
|---|---|---|
| d = v × t / 2 | Known, uniform velocity | Velocity varies with alloy, temperature and stress; austenitic and coarse-grained welds can bend and scatter the beam |
| dB = 20 · log10(A1/A2) | Same coupling and beam position for both echoes | Rough or curved surfaces, couplant changes and probe rocking alter amplitude more than the flaw does |
| R = (Z2 − Z1)/(Z2 + Z1) | Large, flat, perpendicular interface | Tight cracks, tilted planar flaws and small reflectors return far less than the formula predicts |
| N = D² f / 4v and sin(α/2) = K v / fD | Flat circular element, far field | Rectangular, focused and angle-beam probes need effective dimensions; near-field amplitude is unreliable |
| Snell's law | Single, flat interface | Low frequency, small elements and curved surfaces shift the true angle; verify on a block |
| Skip and depth trigonometry | Parallel surfaces, known thickness | Weld crowns, misalignment, tapered sections and mode-converted echoes put indications in false positions |
| TOFD depth | Tip midway between probes | Off-axis tips add depth error; near-surface and back-wall dead zones hide flaws |
| AWS c = 2 × (SP − 1) | Code-defined attenuation | Not the real attenuation; error grows at 60° and 70° on long paths |
Formulas also say nothing about acceptance. A correctly located and rated indication still has to be judged against the code and edition named in the contract, by personnel qualified under the employer's written practice. In the US that is usually SNT-TC-1A, a recommended practice under which the employer, not ASNT, certifies the technician. ASTM E164-24, the contact-weld UT practice for 0.250 to 8 in. thickness, likewise leaves reference standards and acceptance limits to agreement between the parties.
Near-surface resolution is another gap. The dead zone under a contact probe depends on pulse length and ring-down, so there is no single formula. A rough estimate is velocity × ring-down time ÷ 2: a 1 µs ring-down in steel hides about the top 3 mm. Our dead zone in ultrasonic testing guide covers how to measure it.
Ultrasonic testing formulas: quick reference
| Calculation | Formula | Check value (steel) |
|---|---|---|
| Wavelength | λ = v / f | 5 MHz L: 1.18 mm |
| Period | T = 1 / f | 5 MHz: 0.2 µs |
| Pulse-echo depth | d = v × t / 2 | 10 mm ↔ 3.40 µs |
| Decibels | dB = 20 · log10(A1 / A2) | 2:1 = 6 dB |
| Acoustic impedance | Z = ρ × v | Steel ≈ 45.4 × 10⁶ kg/m²s |
| Reflection coefficient | R = (Z2 − Z1) / (Z2 + Z1) | Steel–water: −0.937 (−0.57 dB) |
| Near field | N = D² f / (4 v) | 5 MHz, 10 mm: 21.2 mm |
| Beam spread (−6 dB, pulse-echo) | sin(α/2) = 0.514 v / (f D) | 5 MHz, 10 mm: 3.5° |
| Snell's law | sin θ1 / sin θ2 = v1 / v2 | 70° shear needs a 52.1° acrylic wedge |
| Angle-beam depth | d = SP × cos θ (first leg) | 70°, 50 mm SP: 17.1 mm |
| Surface distance | SD = SP × sin θ | 70°, 50 mm SP: 47.0 mm |
| Skip distance | 2 × T × tan θ | 70°, 25 mm: 137.4 mm |
| TOFD depth | d = √[(c t / 2)² − S²] | S = 30 mm, t = 13.0 µs: 24.1 mm |
| AWS D1.1 attenuation | c = 2 × (SP in inches − 1) | 3.5 in.: 5 dB |
How EMA Industries applies this
EMA Industries provides AWS D1.1 structural weld UT, MT and PT, ultrasonic rail flaw testing and NDT Level III services from Southern California, and mobilizes nationwide. On a weld or rail job, these formulas belong in the written procedure and the report: the velocity and zero used, the probe angle verified on a block, the reference level, the attenuation factor and the sound path, depth and surface distance of each indication. A Level III review of a procedure checks exactly these numbers. To scope an inspection or a procedure review, see our ultrasonic testing services or contact EMA Industries.
Related reading: Shear wave ultrasonic testing · AWS D1.1 weld inspection criteria · Ultrasonic testing acceptance criteria standards · Near field vs far field in UT · DAC curve in ultrasonic testing
Frequently asked questions
What is the basic formula for ultrasonic waves?
The core relation is λ = v / f: wavelength equals sound velocity divided by frequency. In steel, a 5 MHz longitudinal wave at 5,890 m/s has a wavelength of about 1.18 mm. Wavelength sets resolution and how quickly the beam is scattered.
How do you calculate range or depth in ultrasonic testing?
In pulse-echo, depth = velocity × time of flight ÷ 2, because the sound goes out and back. A 20 µs round trip in steel at 5,890 m/s is 58.9 mm. Through-transmission drops the factor of 2.
How do you calculate skip distance in ultrasonic testing?
Full skip distance = 2 × thickness × tan θ, where θ is the refracted angle in the part. A 70° beam in 25 mm plate skips every 137.4 mm, with the first back-wall bounce at 68.7 mm.
How many dB is half the amplitude?
About 6 dB. From dB = 20 · log10(A1/A2), a 2:1 amplitude ratio is 6.02 dB, 4:1 is 12 dB and 10:1 is 20 dB. Because the factor is 20, 6 dB halves the amplitude, not the energy.
What K constant should be used for beam spread?
For the −6 dB pulse-echo beam edge, K ≈ 0.514, as given by Olympus. K = 0.70 is the one-way −6 dB value, and 1.22 is the edge at the first null. State which edge definition your procedure uses.
How do you calculate the dead zone in ultrasonic testing?
There is no exact formula, because the dead zone depends on pulse length, ring-down and instrument recovery. A rough estimate is velocity × ring-down time ÷ 2, about 3 mm in steel for 1 µs. Measure the real value on a reference block with near-surface holes.
Sources
- Ultrasonic Transducers Technical Notes, Olympus NDT, 2006 (dB table, impedance example, near field, −6 dB pulse-echo beam spread, Snell's law, material velocities)
- Beam Characteristics, Evident (Olympus IMS) flaw detection tutorial, section 2.4
- Material Sound Velocities, Evident ultrasonic tutorial, accessed October 2026
- Ultrasonic flaw detection tutorial: AWS weld inspection, Evident
- What is time of flight diffraction (TOFD)?, Baker Hughes (Waygate Technologies) blog
- ASTM E2373/E2373M, Standard Practice for Use of the Ultrasonic Time of Flight Diffraction (TOFD) Technique, ASTM International
- ASTM E164-24, Standard Practice for Contact Ultrasonic Testing of Weldments, ASTM International
- AWS Announces Release of D1.1/D1.1M:2025, Structural Welding Code – Steel, American Welding Society, April 17, 2025
- Structural UT: Variables Affecting Attenuation and Review of the 2 dB per Inch Model, P. Holloway, CINDE Journal Vol. 38 No. 3, May/June 2017

