As per the laws of friction, contact area does not have any impact on the frictional force. I welded Al 2021 alloy by FSW with two tools, such that they differ by their shoulder diameter (18 mm & 25 mm). Other process parameters remain the same. The temperature measured at the weld pool for the 25 mm tool is significantly higher. Why is it so?
Indeed classical definition of friction is that it does not depend on the contact area. However, this definition does not take into account adhesion between two metals welded. That is where dependence on the contact area arises.
Shoulder diameter/area is directly proportional to frictional heat generation in FSW. That's why the diameter is increased to join high melting point materials compared to aluminum and magnesium for a given plate thickness. Please refer the following paper where a detailed heat generation model for FSW is discussed.
Progress in Materials Science 57 (2012) 95–183
In the case of FSW, the adiabatic heat resulting from plastic deformation should be taken into account. As the shoulder diameter increases, the stirring area becomes larger. This suggests more strain and strain rate applied on the material beneath the shoulder, and then more adiabatic heat. Of course, the amount of heat resulting from the plastic deformation during FSW is appreciably lower than that of caused by friction. The frictional heat during FSW definitely depends on the shoulder diameter together with the FSW parameters. This is mainly due to the adhesion between the tool and stirring material.
I agree with Mahmoud's comment. Rotational welding processes generate more heating with increasing diameter. The rotational speed of the tool generates a velocity gradient on the tool's surface, increasing the velocity towards the border of the shoulder. A larger diameter generates more velocity, which is directly proportional to energy input in rotational welding such as FSW.
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