Homework Help: Statics: Shear and bending moment, distributed load

1. Jul 28, 2013

yaro99

1. The problem statement, all variables and given/known data

2. Relevant equations
ƩFy=0
ƩM=0

3. The attempt at a solution

Finding the reaction at A:

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

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

For the shear and bending moment:

ƩFy=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

Last edited: Jul 28, 2013
2. Jul 28, 2013

SteamKing

Staff Emeritus
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?

3. Jul 28, 2013

yaro99

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.

4. Jul 28, 2013

haruspex

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?

5. Jul 28, 2013

yaro99

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 ƩFy equation for V

6. Jul 28, 2013

haruspex

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.

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