to help construct continuous‐cooling transformation diagrams useful for studying phase transformations in welding and heat treating of steels.
By performing high‐speed tensile testing during weld thermal simulation, the elevated‐temperature ductility and strength of metals can be evaluated. This is often called the hot‐ductility test. Nippes and Savage [64, 65], for instance, used this test to investigate the HAZ fissuring in austenitic stainless steels.
Charpy impact test specimens can also be prepared from specimens (1 cm by 1 cm in cross section) subjected to various thermal cycles. This synthetic‐specimen or simulated‐microstructure technique has been employed by numerous investigators to study the HAZ toughness.
2.5.3 Limitations
Weld thermal simulators, though very useful, have some limitations. First, extremely high cooling rates during electron and LBW cannot be reproduced, due to the limited cooling capacity of the simulators. Second, because of the surface heat losses, the temperature at the surface can be lower than that at the centerline of the specimen, especially if the peak temperature is high and the thermal conductivity of the specimen is low [66]. Third, the temperature gradient is much lower in the specimen than in the weld heat‐affected zone, for instance, 10 °C/mm, as opposed to 300 °C/mm near the fusion line of a stainless‐steel weld. This large difference in the temperature gradient tends to make the specimen microstructure differ from the real HAZ microstructure. For example, the grain size tends to be significantly larger in the specimen than in the heat‐affected zone, especially at high peak temperatures such as 1100 °C and above.
Examples
Example 2.1 Bead‐on‐plate welding of a thick steel plate is carried out using GTAW at 200 A, 10 V, and 2 mm/s. Based on Rosenthal's 3D equation, calculate the 500 °C cooling rates along the x‐axis of the workpiece for zero and 250 °C preheating. The arc efficiency is 70% and the thermal conductivity is 35 W/m°C.
Answer:
Along the x‐axis of the workpiece as shown in Figure 2.18,
(2.13)
Therefore, Eq. (2.9) becomes
(2.14)
Therefore, the temperature gradient is
From the above equation and
(2.16)
the cooling rate becomes
Without preheating the workpiece before welding,
(2.18)
With 250 °C preheating,
(2.19)
It is clear that the cooling rate is reduced significantly by preheating. Preheating is a common practice in welding high‐strength steels because it reduces the risk of HAZ cracking. In multiple‐pass welding the inter‐pass temperature is equivalent to the preheat temperature T 0 in single‐pass welding.
Thus, Eq. (2.17) shows that the cooling rate decreases with increasing Q/V, and Eq. (2.15) shows that the temperature gradient decreases with increasing Q.
Example 2.2 Consider 2D (x, y) heat flow and 3D (x, y, z) heat flow in the workpiece. (a) When heat flow in the workpiece is 2D, does the temperature distribution (including the weld width) change much in the depth direction of the workpiece? (b) What about 3D heat flow? (c) Which equation works better for the thick plate (25 mm thick) in Figure E2.2, and why? (d) How about the thin sheet (3.2 mm thick)? (e) At the same Q and V, how does preheating affect the weld width and cooling rate?
Figure E2.2 Transverse cross‐sections of welds.
Answer:
1 (a) No, with 2D heat flow the temperature distribution changes little in the depth direction.
2 (b) With 3D heat flow, the temperature distribution changes significantly in the depth direction.
3 (c) 2D equation works better because the weld width is essentially uniform in the depth direction, suggesting 2D heat flow.
4 (d) 3D equation works better because the weld width changes significantly in the depth direction, suggesting 3D heat flow.
5 (e) Increasing preheating temperature increases the weld width but decreases the cooling rate.
Example 2.3 Consider the transverse cross‐section of the weld pool based on Rosenthal's 3D heat flow equation. What is the shape of the transverse cross‐section of a weld based on Rosenthal's 3D equation?
Answer:
From Rosenthal's 3D equation
At a fixed value of x, x = c, where c is a constant. For a given material under a given welding condition, T o , k, Q, V, and α are all constant. Furthermore, T = T m (the melting point) at the fusion boundary. Therefore, everything in Eq. (2.20) is constant, and the radial distance R between the origin and a point at the fusion boundary must, therefore, also be constant.
(2.21)
