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Can someone explain why the M is assigned to be anticlockwise here ?

1 When i assign it as clockwise , i will get -P(δ -v) , which is different from the author ... If i do so , then , all the sign will be different Can i do so ? Why ?

or the second example here , i can understand that M = -Pv , since M+Pv = 0 at either end when it's in equlibrium.

P/s : I know the the sign convention of the bending moment of beam is positive when the beam upwards as shown ...

For the first example , i gt M+P(∂-v) = 0 , so M = - P(∂-v) . I am not sure whether is my concept correct or not . 2

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You choose your coordinate system with a FBD, however the different pin supports have to counteract the exerted force to be static. Typically bending moments are defined using the rights hand rule. In example 1, M is negative by convention only because of the orientation typically used. Notice the author has x as the y axis. You can also ignore all that as the anti clockwise moment is opposing the bending moment caused by P (negative, aka clockwise). It's a cut section.

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Simple. The bending moment caused by $P\delta$ is clockwise, so the bending moment reaction must be counter-clockwise to resist it. The bending equilibrium equation should be

$$\begin{gather} \sum M_{support} = M + (-P\delta) = 0 \\ \therefore M = P\delta \end{gather}$$

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  • $\begingroup$ Do you mean the case2 ? buckling of column with pin support ? We can see that M = -Pv , which is M+Pv = 0 , not M-P𝛿 = 0 , M =- P𝛿 as you showed . Why you said that M = P𝛿 ? $\endgroup$ – kelvinmacks Dec 25 '16 at 12:18
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    $\begingroup$ @kelvinmacks: Your question, as I understand it, is for the first example (cantilevered). My answer is therefore for the first example. $\endgroup$ – Wasabi Dec 25 '16 at 12:39
  • $\begingroup$ OK, for the second example, why it's M = - Pv? $\endgroup$ – kelvinmacks Dec 25 '16 at 12:41
  • $\begingroup$ why the M has same direction with Pv? $\endgroup$ – kelvinmacks Dec 25 '16 at 12:42
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    $\begingroup$ @kelvinmacks: Oh, woops, sorry. Brain fart. Counter-clockwise bending moment is positive by definition. I'll add another comment later, no time now. $\endgroup$ – Wasabi Dec 27 '16 at 11:52

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