Moment diagram with triangular load

[chickenalfredo]

Homework Statement

For the overhanging beam in the figure, A) draw the moment diagram indicating all critical values including the maximum moment (Value and location), and B) write the moment function, M(x), for B-C section in terms of x coordinate as shown in the figure.

Homework Equations

F = ma
M = F*d
(1/3x) <---- (Equation for the slope of the triangular load from B-C I think)
-1/6x^2 <---- (Equation for the slope of the shear load from B-C I think)
-1/18x^3 <---- (Equation for the slope of the moment diagram from B-C I think)

The Attempt at a Solution

Mc = -Ra(10) + (20)(6) + 1/2(6)(2)(1/3(6)) - (2)(4)(2) = 0 therefore, Ra = 11.6 kips
Fy = Ra - 20 - 1/2(6)(2) - 2(4) + Rc = 0 therefore, Rc = 22.4 kips


I'm trying to figure out how to find where the middle cubic function crosses the x-axis and I'm having a really hard time. Could I get a push in the right direction please?

 

[SteamKing]

Step 1: Figure out the reactions at A and C.
Step 2: Construct the shear force diagram for the beam with these reactions.
Step 3: Using the shear force diagram, construct the bending moment diagram.

You are trying to construct the moment diagram by jumping in the middle of the process without completing the basic steps (1 and 2 above) first.

 

[chickenalfredo]

Alrighty thanks for the reply SteamKing, I did as you described and did the shear diagram, and that helped me complete the moment diagram. Now I am having trouble finding where the cubic function crosses the x-axis. I tried using the formula -x^3/18 = 46.4 and solving for x which gives me 9.417. I need an answer of 8.795 and I have no idea at the moment on how to get that.

 

[nvn]

chickenalfredo: The book answer, x = 8.7948, is indeed correct. So no problem there.

Your moments at points B and C are correct. Nice work. I do not know why you currently think M2(x) = -(x^3)/18 might be the correct function for the moment from point B to C. I have no idea how you derived it. Using your expression for M2(x), notice M2(4) = -(4^3)/18 = -3.556, which is obviously wrong, because it is not the moment at point B (which is 46.40). Also, using your expression, M2(10) = -(10^3)/18 = -55.56, which is obviously wrong, because it is not the moment at point C (which is -16.00).

Therefore, derive a valid equation for M2(x), the moment from point B to C. It will be more complicated than your above expression, but not too complicated. And then, solving M2(x) by trial and error, iteratively, or numerically, you can solve for x when M2(x) = 0. Try again.

Aside: If there is a clever way to factor the M2(x) expression (which I did not research, but I currently assume there is not), then you alternately could solve M2(x) analytically for x when M2(x) = 0.

By the way, please do not post wide images on a forum. Reduce each image width to 640 pixels wide before posting.

 

[rezeew]

As a scientist, you are correct in your equations for the shear force and moment diagrams. To find where the middle cubic function crosses the x-axis, you can set the equation equal to zero and solve for x. In this case, the equation is -1/18x^3 = 0. This means that x = 0 is one point where the cubic function crosses the x-axis. To find the other points, you can use the quadratic formula on the remaining two terms (-1/6x^2 and -1/3x). This will give you two additional points where the cubic function crosses the x-axis. From there, you can plot these points on your moment diagram to complete your solution. Good luck!