- #1

TSN79

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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 cant 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...?

[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 cant 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...?