Statics: Shear and bending moment, distributed load

[yaro99]

Homework Statement

Homework Equations

ƩF_y=0
ƩM=0

The Attempt at a Solution

Finding the reaction at A:

ƩF_y=0: A + (L*w₀)/6 - (L*w₀)/2 = 0
A = (L*w₀)/3

ƩM_A=0: M_A - [(L*w₀)/2]*(L/3) + [(L*w₀)/6]*((2L)/3) = 0
M_A = (w₀*L²)/18

For the shear and bending moment:

ƩF_y=0: (L*w₀)/3 - (x*w₀)/2 + (x*w₀)/6 - V = 0
V = (L*w₀)/3 - (x*w₀)/3

ƩM=0: (w₀*L²)/18 + [(x*w₀)/2]*[(2*x)/3] - [(x*w₀)/6]*(x/3) + M = 0
M = -(w₀*L²)/18 - (5/18)*w₀*x²

 

[SteamKing]

What happens to your equations for V(x) and M(x) when x = 0? Would these values of V and M match your statics equilibrium calculations?

 

[yaro99]

What happens to your equations for V(x) and M(x) when x = 0? Would these values of V and M match your statics equilibrium calculations?

It looks to me like they do:
V = (L*w₀)/3 which balances out the force at A (V is pointing down)

M = -(w₀*L²)/18 = (w₀*L²)/18 clockwise, which balances the equivalent counterclockwise moment at A.

 

[haruspex]

ƩF_y=0: (L*w₀)/3 - (x*w₀)/2 + (x*w₀)/6 - V = 0
V = (L*w₀)/3 - (x*w₀)/3

x, I assume, is a distance from the wall.
I don't understand how you obtained these equations. Can you spell that out more clearly?

 

[yaro99]

x, I assume, is a distance from the wall.
I don't understand how you obtained these equations. Can you spell that out more clearly?

Yes, that is correct.
I summed up the vertical forces in my diagram.

(L*w₀)/3 is the value I obtained for A in the beginning of the problem.
(x*w₀)/2 is from the triangle area formula (Bh)/2, where the base is distance x, and w₀, the magnitude of the force, is the height
(x*w₀)/6 : also using area of a triangle, where the base is again x, and the height is (1/3)*w₀ as indicated in the problem.


V = (L*w₀)/3 - (x*w₀)/3 is what I get when I solve the ƩF_y equation for V

 

[haruspex]

I summed up the vertical forces in my diagram.

I still don't get the reasoning behind it.
Since the system is static, the total moment about any point on the beam is zero. The bending moment at X is the contribution from one side of the point X only. I would expect to see something like $\int_{y=x}^L (y-x)F(y).dy$. Maybe you have a neat way of avoiding the integration, but if so I'm not seeing it.