To find the area of the cross-section of the bar for a factor of safety of 2, we need to consider the Soderberg principle. The Soderberg principle combines the effects of static and fatigue loading on a material.

First, let's find the maximum static stress on the bar. The maximum static stress is given by the maximum tensile load divided by the area of the cross-section of the bar. The maximum tensile load is 100 kN, which is equivalent to 100,000 N. To convert this to MPa, we divide by 1,000,000. Therefore, the maximum static stress is 100,000 N / Area (mm^2).

According to the Soderberg principle, the maximum static stress should be less than or equal to the yield strength of the material divided by the factor of safety. In this case, the yield strength is 240 MPa and the factor of safety is 2. Therefore, we have the following inequality:

100,000 N / Area (mm^2) ≤ (240 MPa / 2)

To find the minimum area of the cross-section, we rearrange the inequality:

Area (mm^2) ≥ (100,000 N) / (240 MPa / 2)

Now, let's consider the endurance limit in reversed bending. The endurance limit is the maximum alternating stress that a material can withstand for an infinite number of cycles without failure. In this case, the endurance limit is 160 MPa.

According to the Soderberg principle, the maximum alternating stress should be less than or equal to the endurance limit of the material divided by the factor of safety. In this case, the maximum alternating stress is the difference between the maximum and minimum tensile loads divided by the area of the cross-section of the bar. The difference between the maximum and minimum tensile loads is 100 kN - 20 kN = 80 kN, which is equivalent to 80,000 N. Therefore, we have the following inequality:

80,000 N / Area (mm^2) ≤ (160 MPa / 2)

To find the minimum area of the cross-section, we rearrange the inequality:

Area (mm^2) ≥ (80,000 N) / (160 MPa / 2)

Now, let's compare the two inequalities we derived:

(100,000 N) / (240 MPa / 2) ≥ (80,000 N) / (160 MPa / 2)

Simplifying the inequality:

(100,000 N) / (240 MPa / 2) ≥ (80,000 N) / (160 MPa / 2)
(100,000 N) / (240 MPa / 2) ≥ (80,000 N) / (80 MPa)

Cancelling out the common terms:

2 * (100,000 N) / 240 MPa ≥ (80,000 N) / 80 MPa
(100,000 N) / 240 MPa ≥ (80,000 N) / 80 MPa

Simplifying further:

(100,000 N) / 240 MPa ≥ (100,000 N) / 100 MPa

Since the left side of the inequality is greater than the right side, the inequality holds true. Therefore, the minimum area of the cross-section of the bar is given