Shear and moments

  • Thread starter rsq_a
  • Start date
  • #1
107
1

Main Question or Discussion Point

I can't seem to wrap my head around the signs of shears and moments when applied to beams. Consider a cantilever beam that goes from x = 0 to x = L (with positive deflection, or y(x), corresponding to a deflection upwards).

The standard equations tell us that

[tex]
\text{Moment} = EI \frac{d^2 y}{dx^2}
[/tex]

[tex]
\text{Shear} = EI \frac{d^3 y}{dx^3}
[/tex]

[tex]
\text{Load} = EI \frac{d^4 y}{dx^4}
[/tex]


Now consider what happens when we change [tex]x = -x[/tex] (that is, we put our coordinate system so that the beam begins at x = 0 and goes to x = -L).

Why does that change the shear to negative, but keep the sign of the moments and loads the same?
 

Answers and Replies

  • #2
nvn
Science Advisor
Homework Helper
2,128
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rsq_a: In post 1, it is somewhat unclear what "begins" means. Does "begins" mean the cantilever clamped support, in both cases you mentioned? Or does "begins" mean the cantilever free tip, in both cases?
 
  • #3
86
1
Apply a uniform load (UDL) across the span of a beam. Then apply the three equations you quoted. By the way, you could also add a fourth equation, first derivative which conveys curvature of the beam. Then you can note that the 1st and 3rd derivatives will yield opposite signed values if you reverse your coordinate, but remains equal for 2nd and 4th derivatives.

[tex]
\text{Curvature} = EI \frac{d^1 y}{dx^1}
[/tex]

The reason is that the loading (UDL) is symmetric across the coordinate span, by the choice of your loading example.

Now choose a non-symmetric loading pattern, say one-sided triangular loading which rises from zero at one end to max at other end. Then all four derivatives will give values that are not symmetric with regard to a change of the coordinate system.
 

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