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# Plotting singularity functions watch

1. at x=6, the value shown is 0 but I get

5.75 -3(6) + 3(3) + 1/6(6^2) - 1/6(3^2)+ 2(3) + 1.75
=9

can someone explain where I'm going wrong?
Attached Images

2. On that graph, where X = 6 (horizontal axis) the value of V (vertical) is -1.75kN

If you are referring to the final equation on the bottom right, it's a quadratic and they are working out the value of X for which V= 0 not the value of V where x = 6
3. (Original post by Stonebridge)
On that graph, where X = 6 (horizontal axis) the value of V (vertical) is -1.75kN

If you are referring to the final equation on the bottom right, it's a quadratic and they are working out the value of X for which V= 0 not the value of V where x = 6
but how did they get that value at x=6?

I'm getting
5.75 -3(6) + 3(3) + 1/6(6^2) - 1/6(3^2)+ 2(3) = 7.25
4. (Original post by therealone7)
but how did they get that value at x=6?

I'm getting
5.75 -3(6) + 3(3) + 1/6(6^2) - 1/6(3^2)+ 2(3) = 7.25
You can't put the value of x=6 in that equation.

The equation given expresses what happens (the value of V) between x = 0 and x = 3

After x = 3 there is clearly another relationship.

It looks like the value of x then remains constant (due to whatever this is supposed to represent).
5. (Original post by Stonebridge)
You can't put the value of x=6 in that equation.

The equation given expresses what happens (the value of V) between x = 0 and x = 3

After x = 3 there is clearly another relationship.

It looks like the value of x then remains constant (due to whatever this is supposed to represent).

Using this table
http://postimg.org/image/3x0wywr4v/

could you explain why the terms in red represent the red distributed load in the attached diagram?
Attached Images

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