5.2 Adams Cooling Rate, the t8/5 Window & HAZ Peak-Temperature Distribution

Adams Centerline Cooling Rate Formulations

C. M. Adams simplified Rosenthal's complex exponential equations to derive practical engineering expressions for the cooling rate ($R = \left| \frac{dT}{dt} \right|$) along the weld centerline at a specified critical temperature $T_c$.

3D Cooling Rate (Thick Plate, $\tau > 0.9$)

$$R_{3D} = \left| \frac{dT}{dt} \right|_{T_c} = \frac{2 \pi k (T_c - T_0)^2}{H_{\text{net}}} = \frac{2 \pi k v (T_c - T_0)^2}{\eta V I}$$

2D Cooling Rate (Thin Plate, $\tau < 0.6$)

$$R_{2D} = \left| \frac{dT}{dt} \right|_{T_c} = 2 \pi k \rho C_p \left( \frac{d}{H_{\text{net}}} \right)^2 (T_c - T_0)^3 = 2 \pi k \rho C_p \left( \frac{v d}{\eta V I} \right)^2 (T_c - T_0)^3$$

Comparison of Mathematical Sensitivities

Exam Trap Alert: In 3D heat flow, doubling the plate thickness has zero effect on the weld centerline cooling rate, because heat is already conducting hemispherically into an effectively semi-infinite heat sink. In 2D heat flow, doubling the plate thickness quadruples the cooling rate ($2^2 = 4$) for a fixed heat input per unit length! Do not confuse the two regimes on the CWEng exam.

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Critical Cooling Time $\Delta t_{8/5}$ and Metallurgical Kinetics

In the welding of structural, pressure vessel, and pipeline carbon and low-alloy steels, the single most critical thermal parameter is the cooling time from $800^\circ\text{C}$ to $500^\circ\text{C}$, designated as $\Delta t_{8/5}$.

   Temperature (°C)
     1000 ┬
          │         Austenite Phase Region
      800 ┼─────────────────┐ (Start of phase transformation window)
          │                 │
          │                 │ ◄─── Δt_8/5 (Critical Cooling Time Window)
          │                 │
      500 ┼─────────────────┴ (End of phase transformation window)
          │         Ferrite / Pearlite / Bainite / Martensite Formation
      200 ┴─────────────────────────────────────────────► Time (s)

Why the $800^\circ\text{C} \to 500^\circ\text{C}$ Window Dominates

During heating above the $Ac_3$ temperature, the base metal transforms to austenite ($\gamma$). Upon cooling:

• Above $800^\circ\text{C}$, austenite remains stable; diffusion is rapid, but no structural transformations occur.
• Between $800^\circ\text{C}$ and $500^\circ\text{C}$, the austenite decomposes into equilibrium or non-equilibrium microstructures depending on cooling rate (Continuous Cooling Transformation, or CCT, kinetics).
• If $\Delta t_{8/5}$ is too short (excessively fast cooling), carbon cannot diffuse out of the face-centered cubic (FCC) austenite lattice, triggering a diffusionless shear transformation into untempered martensite. Martensite exhibits high hardness, high residual lattice strain, and extreme vulnerability to Hydrogen-Induced Cracking (HIC) / cold cracking.
• If $\Delta t_{8/5}$ is too long (excessively slow cooling), excessive austenite grain growth occurs in the coarse-grained HAZ (CGHAZ), producing coarse upper bainite and ferrite with aligned second phases, leading to severe degradation of Charpy V-notch impact toughness.

Closed-Form Equations for $\Delta t_{8/5}$

Integrating Adams cooling formulations across the temperature interval from $800^\circ\text{C}$ to $500^\circ\text{C}$ yields:

For 3D Heat Flow (Thick Plate, $\tau > 0.9$):

$$\Delta t_{8/5}^{3D} = \frac{H_{\text{net}}}{2 \pi k} \left[ \frac{1}{500 - T_0} - \frac{1}{800 - T_0} \right]$$

For 2D Heat Flow (Thin Plate, $\tau < 0.6$):

$$\Delta t_{8/5}^{2D} = \frac{H_{\text{net}}^2}{4 \pi k \rho C_p d^2} \left[ \left( \frac{1}{500 - T_0} \right)^2 - \left( \frac{1}{800 - T_0} \right)^2 \right]$$

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Peak Temperature Distribution $T_p(y)$ across the Heat-Affected Zone (HAZ)

The microstructure and mechanical properties across a welded joint vary continuously because each point experiences a distinct thermal cycle characterized by a unique peak temperature ($T_p$). Adams formulated expressions relating peak temperature $T_p$ at a transverse distance $y$ from the fusion boundary:

For 2D Thin-Plate Conduction:

$$\frac{1}{T_p - T_0} = \frac{1}{T_m - T_0} + \frac{4.13 \, \rho C_p d \, y}{H_{\text{net}}}$$

For 3D Thick-Plate Conduction:

$$\frac{1}{T_p - T_0} = \frac{1}{T_m - T_0} + \frac{2 \pi e \, \rho C_p \, y^2}{H_{\text{net}}}$$

where:

• $T_m$ = Liquidus melting temperature of the alloy ($^\circ\text{C}$; for carbon steel, $\approx 1530^\circ\text{C}$)
• $y$ = Perpendicular distance from the fusion line into the base metal ($\text{mm}$)
• $e = 2.71828$ = Base of the natural logarithm

The Four Distinct Sub-Zones of the Steel HAZ

   Fusion Line  CGHAZ        FGHAZ        ICHAZ        SCHAZ         Unaffected Base Metal
       │       1100°C-Tm    Ac3-1100°C   Ac1-Ac3      < Ac1
       │◄─────►│◄──────────►│◄──────────►│◄──────────►│◄───────────►│
       │ Coarse│ Fine       │ Partially  │ Tempered   │ Original Base
       │ Grains│ Equiaxed   │ Transformed│ Spheroid-  │ Microstructure
       │ Hard/ │ Tough/     │ M-A Islands│ ized       │
       │ Brittle Ductile    │ Local Soft │ Softened   │

1. Coarse-Grained HAZ (CGHAZ, $1100^\circ\text{C} < T_p < T_m$): Extreme peak temperatures drive dissolution of grain-pinning precipitates (e.g., $\text{Nb(C,N)}$, $\text{TiN}$), leading to massive austenite grain growth ($>100,\mu\text{m}$). Upon rapid cooling, it transforms to martensite or coarse upper bainite, presenting the lowest toughness and highest cold cracking risk.
2. Fine-Grained HAZ (FGHAZ, $Ac_3 < T_p < 1100^\circ\text{C}$): Complete re-austenitization occurs, but lower peak temperatures prevent grain coarsening. Subsequent cooling produces ultra-fine, equiaxed ferrite-pearlite or fine lower bainite, exhibiting optimal impact toughness.
3. Intercritical HAZ (ICHAZ, $Ac_1 < T_p < Ac_3$): Partial austenitization occurs within the two-phase ferrite + austenite field. Carbon partitions preferentially into the austenite islands, which upon rapid cooling transform into hard, brittle Martensite-Austenite (M-A) constituents, creating localized brittle zones (LBZs).
4. Subcritical HAZ (SCHAZ, $T_p < Ac_1$): Peak temperatures remain below the lower transformation temperature. No phase changes occur, but existing base metal carbides or cold-worked grains undergo tempering, recovery, and localized softening.

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Comprehensive Worked Numerical Example: Complete Thermal Profile

Problem Statement

A heavy structural girder fabricated from ASTM A572 Grade 50 steel plate with thickness $d = 35.0\text{ mm}$ is joined using Submerged Arc Welding (SAW). The welding procedure operates under the following conditions:

• Arc Voltage $V = 30.0\text{ V}$
• Welding Current $I = 500.0\text{ A}$
• Travel Speed $v = 6.0\text{ mm/s}$ ($360\text{ mm/min}$)
• Initial Workpiece Preheat $T_0 = 50.0^\circ\text{C}$
• Arc Thermal Efficiency $\eta = 0.95$ (typical SAW process)
• Liquidus Melting Temperature $T_m = 1530.0^\circ\text{C}$
• Material Properties:
  • Thermal Conductivity $k = 40.0\text{ W/(m}\cdot\text{K)} = 0.040\text{ J/(mm}\cdot\text{s}\cdot^\circ\text{C)}$
  • Volumetric Heat Capacity $\rho C_p = 4.50 \times 10^6\text{ J/(m}^3\cdot\text{K)} = 0.00450\text{ J/(mm}^3\cdot^\circ\text{C)}$
  • Thermal Diffusivity $\alpha = 8.89\text{ mm}^2/\text{s}$

Calculate:

1. The gross and net heat inputs per unit length ($H_{\text{gross}}$ and $H_{\text{net}}$).
2. The relative plate thickness $\tau$ at critical temperature $T_c = 540^\circ\text{C}$ to establish the heat flow regime (2D vs. 3D).
3. The cooling rate $R$ at the weld centerline at $T_c = 540^\circ\text{C}$.
4. The critical cooling time $\Delta t_{8/5}$.
5. The peak temperature $T_p$ at a transverse distance $y = 5.0\text{ mm}$ from the fusion boundary.

Step-by-Step Solution

Step 1: Calculate Gross and Net Heat Input

$$H_{\text{gross}} = \frac{V \cdot I}{v} = \frac{30.0\text{ V} \times 500.0\text{ A}}{6.0\text{ mm/s}} = \frac{15,000\text{ W}}{6.0\text{ mm/s}} = 2500.0\text{ J/mm} = 2.50\text{ kJ/mm}$$ $$H_{\text{net}} = \eta \cdot H_{\text{gross}} = 0.95 \times 2500.0\text{ J/mm} = 2375.0\text{ J/mm} = 2.375\text{ kJ/mm} = 2,375,000\text{ J/m}$$

Step 2: Determine Relative Plate Thickness $\tau$ (2D vs. 3D Regime)

$$\tau = d \sqrt{\frac{\rho C_p (T_c - T_0)}{H_{\text{net}}}}$$

Using SI units: $d = 0.035\text{ m}$, $\rho C_p = 4.50 \times 10^6\text{ J/(m}^3\cdot\text{K)}$, $T_c - T_0 = 540 - 50 = 490^\circ\text{C}$, $H_{\text{net}} = 2.375 \times 10^6\text{ J/m}$.

$$\frac{\rho C_p (T_c - T_0)}{H_{\text{net}}} = \frac{4.50 \times 10^6 \times 490}{2.375 \times 10^6} = \frac{2.205 \times 10^9}{2.375 \times 10^6} = 928.42\text{ m}^{-2}$$ $$\sqrt{928.42\text{ m}^{-2}} = 30.47\text{ m}^{-1}$$ $$\tau = 0.035\text{ m} \times 30.47\text{ m}^{-1} = 1.066$$

Evaluation: Since $\tau = 1.066 > 0.9$, the weldment exhibits unambiguous three-dimensional (3D) thick-plate heat flow. Thermal conduction proceeds hemispherically into the thickness without experiencing thermal reflection from the plate bottom.

Step 3: Calculate 3D Weld Centerline Cooling Rate at $T_c = 540^\circ\text{C}$ Using Adams 3D cooling rate equation:

$$R_{3D} = \frac{2 \pi k (T_c - T_0)^2}{H_{\text{net}}}$$

Substitute values in consistent units ($\text{mm}, \text{s}, ^\circ\text{C}$):

• $k = 0.040\text{ J/(mm}\cdot\text{s}\cdot^\circ\text{C)}$
• $T_c - T_0 = 540 - 50 = 490^\circ\text{C}$
• $H_{\text{net}} = 2375.0\text{ J/mm}$

$$(T_c - T_0)^2 = (490)^2 = 240,100\text{ (}^\circ\text{C)}^2$$ $$R_{3D} = \frac{2 \times \pi \times 0.040 \times 240,100}{2375.0} = \frac{60,343.89}{2375.0} = 25.41^\circ\text{C/s}$$

Step 4: Calculate Critical Cooling Time $\Delta t_{8/5}$ Using the 3D thick-plate formulation:

$$\Delta t_{8/5}^{3D} = \frac{H_{\text{net}}}{2 \pi k} \left[ \frac{1}{500 - T_0} - \frac{1}{800 - T_0} \right]$$ $$\frac{H_{\text{net}}}{2 \pi k} = \frac{2375.0}{2 \times \pi \times 0.040} = \frac{2375.0}{0.251327} = 9449.83\text{ s}\cdot^\circ\text{C}$$ $$\frac{1}{500 - 50} - \frac{1}{800 - 50} = \frac{1}{450} - \frac{1}{750} = 0.0022222 - 0.0013333 = 0.0008889\text{ }^\circ\text{C}^{-1}$$ $$\Delta t_{8/5}^{3D} = 9449.83 \times 0.0008889 = 8.40\text{ seconds}$$

Step 5: Calculate Peak Temperature at $y = 5.0\text{ mm}$ from Fusion Boundary Using the Adams 3D peak temperature distribution equation:

$$\frac{1}{T_p - T_0} = \frac{1}{T_m - T_0} + \frac{2 \pi e \, \rho C_p \, y^2}{H_{\text{net}}}$$ $$\frac{1}{T_m - T_0} = \frac{1}{1530 - 50} = \frac{1}{1480} = 6.7568 \times 10^{-4}\text{ }^\circ\text{C}^{-1}$$ $$2 \pi e \, \rho C_p \, y^2 = 2 \times 3.14159 \times 2.71828 \times 0.00450 \times (5.0)^2 = 17.0795 \times 0.00450 \times 25.0 = 1.9214\text{ J/(mm}\cdot^\circ\text{C)}$$ $$\frac{1.9214}{H_{\text{net}}} = \frac{1.9214}{2375.0} = 8.0901 \times 10^{-4}\text{ }^\circ\text{C}^{-1}$$ $$\frac{1}{T_p - T_0} = 6.7568 \times 10^{-4} + 8.0901 \times 10^{-4} = 1.4847 \times 10^{-3}\text{ }^\circ\text{C}^{-1}$$ $$T_p - T_0 = \frac{1}{1.4847 \times 10^{-3}} = 673.54^\circ\text{C}$$ $$T_p = 673.54 + 50.0 = 723.5^\circ\text{C}$$

Metallurgical Conclusion: At $y = 5.0\text{ mm}$, $T_p = 723.5^\circ\text{C}$, which falls just below the lower transformation temperature $Ac_1 \approx 727^\circ\text{C}$. This location resides in the Subcritical Heat-Affected Zone (SCHAZ), undergoing carbide tempering without austenite phase re-transformation.

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Real-World Engineering Scenarios & Exam Pitfalls

Industrial Case: Hydrogen-Induced Cold Cracking in Offshore Monopiles

During the fabrication of thick-walled ($d = 65\text{ mm}$) S355ML structural steel offshore wind turbine monopiles, ultrasonic testing revealed extensive transverse cracking in the coarse-grained HAZ within 24 hours of welding. The welding engineer discovered that shop personnel omitted the mandated $125^\circ\text{C}$ preheat because the ambient shop temperature was $20^\circ\text{C}$. Calculating the thermal cycle revealed that without preheat, $\tau = 65 \times 10^{-3} \sqrt{4.5 \times 10^6 \times (540 - 20) / 2.0 \times 10^6} = 2.22$ (deep 3D conduction). The 3D cooling time $\Delta t_{8/5}$ dropped from $12.8\text{ s}$ (with preheat) to $4.9\text{ s}$ (without preheat). This ultra-fast quench transformed the CGHAZ into hard untempered martensite ($385\text{ HV10}$), which, combined with weld residual stress and diffusible hydrogen from flux moisture, caused catastrophic cold cracking. Reinstating preheat extended $\Delta t_{8/5}$ above $11.0\text{ s}$, capping hardness at $270\text{ HV10}$ and eliminating cracking.

Common Exam Traps

Exam Trap 1: Conflating Gross Heat Input with Net Heat Input Always check whether a question provides arc efficiency $\eta$. If given, $H_{\text{net}} = \eta V I / v$. Rosenthal and Adams cooling formulations rely strictly on net heat input ($H_{\text{net}}$). Substituting gross heat input ($V I / v$) will underestimate cooling rates by $10% - 35%$, leading to non-conservative engineering errors.

Exam Trap 2: Applying the 2D Equation to Thick Plates A classic examination error is automatically using the 2D cooling rate equation because it explicitly includes plate thickness $d$. Remember: for $\tau > 0.9$, the plate acts as a semi-infinite 3D heat sink, and cooling rate is completely independent of thickness! Only when $\tau < 0.6$ does thickness enter the equation ($R \propto d^2$).

Exam Trap 3: Mixing Thermal Conductivity and Thermal Diffusivity Units Thermal conductivity $k$ is expressed in $\text{W/(m}\cdot\text{K)}$ or $\text{J/(mm}\cdot\text{s}\cdot^\circ\text{C)}$, while thermal diffusivity $\alpha = k/(\rho C_p)$ is expressed in $\text{m}^2/\text{s}$ or $\text{mm}^2/\text{s}$. When evaluating the exponential terms in Rosenthal's solution ($v/2\alpha$), verify that travel speed $v$ and diffusivity $\alpha$ share identical length units (e.g., $\text{mm/s}$ and $\text{mm}^2/\text{s}$).