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Chapter 17: Problem 13

Explain why it might be difficult to weld aluminum to steel or to weld any two unlike metals together.

Short Answer

Expert verified
Answer: Welding dissimilar metals like aluminum and steel is difficult due to differences in melting points, coefficients of thermal expansion, and metallurgical reactions between the metals. These differences can lead to poor joint formation or destructive effects on the metals. However, specialized techniques or processes can be employed to overcome these challenges.

Step by step solution

01

Introduction to welding

Welding is a process that uses heat to fuse two metals, often of the same type, together by melting the base metals along with a filler material. The filler material cools down and forms a strong joint between the base metals.
02

Melting point differences

When welding two dissimilar metals like aluminum and steel together, one major issue faced is the difference in melting points. Metals have specific melting points, which is the temperature at which they change from a solid to a liquid state. Aluminum has a melting point of around 660°C (1220°F), while steel has a much higher melting point that ranges from 1370°C to 1510°C (2500°F to 2750°F). Attempting to weld these metals together without specialized techniques may lead to poor joints or destructive effects on the metals.
03

Coefficient of thermal expansion

Another problem in welding unlike metals is that each metal has its coefficient of thermal expansion. This factor determines how much a material will expand or contract under thermal stress. When heating dissimilar metals with different thermal expansion rates, they may expand and contract at significantly different rates, leading to warping, deforming, or weak joint formation.
04

Metallurgical reactions

When welding dissimilar metals together, there may be metallurgical reactions between the different materials. These reactions can lead to brittle intermetallic phases or the formation of inclusions, which reduces the strength of the joint. For example, aluminum and steel may form intermetallic compounds that weaken the joint and make it susceptible to cracking.
05

Techniques to overcome welding difficulties

To successfully weld dissimilar metals together, specialized techniques or processes have been developed. These methods include brazing, where a lower melting point material is used to join the metals without melting the two base metals, or using specialized filler materials that provide a metallurgical transition between the two dissimilar metals. In conclusion, welding unlike metals like aluminum to steel or any other dissimilar metal combination is difficult due to differences in melting points, thermal expansion, and metallurgical reactions. However, by employing specialized techniques and understanding the unique properties of each metal, these challenges can be overcome.

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Most popular questions from this chapter

Some textbooks use the unit \(\mathrm{K}^{-1}\) rather than \({ }^{\circ} \mathrm{C}^{-1}\) for values of the linear expansion coefficient; see Table 17.2 How will the numerical values of the coefficient differ if expressed in \(\mathrm{K}^{-1}\) ?

Thermal expansion seems like a small effect, but it can engender tremendous, often damaging, forces. For example, steel has a linear expansion coefficient of \(\alpha=1.2 \cdot 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\) and a bulk modulus of \(B=160\) GPa. Calculate the pressure engendered in steel by a \(1.0^{\circ} \mathrm{C}\) temperature increase.

The lowest air temperature recorded on Earth is \(-129^{\circ} \mathrm{F}\) in Antarctica. Convert this temperature to the Celsius scale.

A plastic-epoxy sheet has uniform holes of radius \(1.99 \mathrm{~cm}\). The holes are intended to allow solid ball bear- ings with an outer radius of \(2.00 \mathrm{~cm}\) to just go through. Over what temperature rise must the plastic-epoxy sheet be heated so that the ball bearings will go through the holes? The linear expansion coefficient of plastic-epoxy is about \(1.3 \cdot 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\).

For food storage, what is the advantage of placing a metal lid on a glass jar? (Hint: Why does running the metal lid under hot water for a minute help you open such a jar?)
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