Why use bearing reaction force instead of applied force for shaft stress

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I've got this shaft you can see in my drawing below, and I'm supposed to find the tension at the locations C and D. At point C I use the following equation:

[tex]\sigma=\frac{M}{W}=\frac{2500\cdot350}{\frac{50^3\cdot\pi}{32}}=71,3 N/mm^2[/tex]

This is apparently the right thing to do. 2500 comes from the reaction forces from the bearings at the ends. What I don't really see is why can't I use the force of 5 kN to calculate this. Putting in 5000*150 instead and exchanging 50 for 60 in the denominator. Could someone explain this to me...?
 
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If you are referring to the normal stresses due to bending, then it looks like you are using the section modulus in the denominator which is right. The reason you can not do what you are thinking is becasue a force has to be created by a reaction to create the stress. The 5kN load on it's own will do nothing to induce the stress. It is the reactions that create the stress. If you have done beam deflections, you always start at one end where you know the reaction and calculated the moment and shear diagrams moving from end to end. The same thing applys here.
 
What I don't really see is why can't I use the force of 5 kN to calculate this. Putting in 5000*150 instead ...
Just to clarify what Fred is saying, actually you CAN put the 5000*150 in, but that's not the ONLY moment if you take moments around point C to the right of point C. You also have the 2500 N load at a distance of 650. The total moment then at point C, if you take moments created on the right side, is:
-5000*150+2500*650 which is the same as 2500*350.

... and exchanging 50 for 60 in the denominator. Could someone explain this to me...?
The location of point C is implied. It is implied that it is at the corner where the shaft is at 50 mm and about to step up to 60 mm. There is NO stress at the corner where it is at 60 mm, there is nothing at that corner to react against. If the diagram was a bit more specific, it would have shown point C with an arrow pointing at the inside corner at the 50 mm diameter.

The stress on the OD of the shaft where it steps up to a larger diameter is zero at the step and will gradually increase to the value calculated as you show here. There is no easy way of calculating that stress distribution by hand, though you would see it if you performed an FEA analysis.