- #1
Bill Nye Tho
- 48
- 0
Homework Statement
I have a 6 meter long beam that is supported by a roller at 0m and a pin at 6m.
There is a triangular distributed load that begins at 0m to 3m which is 2kN/m. (Max load is at 0m)
There is an additional triangular distributed load that begin at 3m which mirrors the first triangular load (2kN/m, max load is at 6m)
And there is a counter-clockwise moment acting at 0m of 18 kNm
Homework Equations
ƩFx = 0
ƩFy = 0
ƩMr = 0
Vertical Reaction @ A (0m) = 6kN
Vertical Reaction @ C (6m) = 0kN
The Attempt at a Solution
So my shear functions are fine but my bending functions always run into a similar problem.
From 0m to 3m my bending function is f(x) = -(2/3)^2(x)^3(.5) + 6(x)
From 3m to 6m my bending function is f(x) = -(.5)(2)(3)(x-1) + 6(x) - ((.5)(2/3)(x-3)^3)/3
My method for drawing internal force diagrams involves checking shared values of each function. So in this case, I like to use 3m in each function to see if I get the same thing, since I know that the bending function is continuous.
When I plug 3 into my first equation I get: -12 kNm
When I plug 3 into my second equation I get: -12 kNm
Everything checks out!
Now, if I calculate my bending forces at 0 m, I get 0 kNm; If I calculate my bending forces at 6 m I get -18kNm.
So basically I now have a bending force diagram that begins at 0, hits -12 kNm at 3m, finally hits -18 kNm at 6m.
I'm looking at the solution and for some reason it's reversed. So I decided to see what I did wrong by including the external force into my functions and while I get the correct answer at 0m and 6m, my 3m answer is completely off.
Is my solution correct? Should it be reversed? Why/Why not? (I solve left to right)