Beam deflection boundary condition calculation

[xzibition8612]

Homework Statement

Find the deflection at x=L/4 and x=L/2 for the beam

Homework Equations

See attached pic.

The Attempt at a Solution

So I have the solution derived in class. Only 0<x<L/2 is derived because since the load on the beam is at L/2, the equation is valid for the entire beam since its symmetric (or something like that, if this isn't the correct explanation somebody tell me). My question concerns the boundary conditions. I know that at x=0, the displacement v=0. Hence you get C2=0. Now the second thing:

dv/dx=0 at x=L/2

I have no idea what dv/dx means, why its 0, and why its taken at x=L/2. Any help would be appreciated thanks.

 

[SteamKing]

You ever heard of the 'slope' of the beam being discussed in your class? The slope = dv/dx.

 

[xzibition8612]

ok, so what is the slope physically mean? Why is it 0 at x=L/2? It has something to do with the load P? I see that the derivative of moment is shear force, and the derivative of shear force is distributed force. But that has nothing to do with deflection right?

 

[SteamKing]

Start with deflection. What's the first derivative of a curve represent?

 

[xzibition8612]

the slope, y/x. Why would dv/dx=0 at x=L/2? That makes no physical sense, because at x=L/2 there is a load P upward and it must deflect. Hence dv/dx can't be 0.

 

[SteamKing]

What's the slope of a curve in calculus class?