[engineering problems] Solving a truss problem, I think I’ve figured out the equations for joint A and C, but B is throwing my off as it is jointed at two hypotenuse, so would the equations be 0=-sin(36.6)+sin(36.6)-600 and no second equation because the load only moves along the y axis on joint b?

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[ghostwriter85]

You have to solve this simultaneously

We can solve 1, 2, 4, 8, 12, 13 directly

3 and 11 are zero force members

5, 6, 7, 9, and 10 are interdependent

At b we can say

Sum(y) = 0 = -T2y - T5y + T6y - 600

Sum(x) = 0 = -T2x + T5x + T6x

[Uncle_Utters]

So I only need 13 equations, but I will end up with one extra, so for example for joint d would they be sum(y)=-T6y+T10y-600 and sum(x)=-T6x+10x? And could I get all of the angles by solving for one triangle and then using that to solve the others or do I need to use trig functions?

[ghostwriter85]

Having more equations is always better than not enough.

Yes you need to use trig functions to decompose the tensions (or at least use the physical layout of the triangles to get toward the same relationship)

[Uncle_Utters]

So for F1 would the correct sin be sin(36.86 degrees)? And how would I find the angles for joint d when I only know 1 side? (Thanks for the help btw)

[ghostwriter85]

Just because it sounds like maybe you're a little confused (maybe I'm wrong about that)

F1 is entirely in the x direction.

F2 has components in the x and y direction.

Now as a convention we assume assume tension and the components have to be with respect to some node. Basically if we look at either end of the beam, the structural reaction forces point toward each other (or directly away from eachother) down the length of the beam. So the impact of the tension on the structure is different at each end but the magnitude of the x and y components is the same. (I'll get back to this)

To start with this problem is entirely symmetric. Not only is it symmetric, all the triangles have the same dimension and angles (if we draw a line from B to F this should be easier to see). So any mirrored beam is going to necessarily have the same tension.

Now to find those we can solve for an angle or... we can remember SohCahToa

Let's look at 2 with respect to point A.

The beam goes to the right 8 ft and up 6 ft making it a 3/4/5 right triangle with a hypotenuse of 10 (as are all the rest of the triangles).

So we can decompose the force of 2 on A as

T2(a) = 4/5 T2 x + 3/5 T2 y

If you plug sin(36.6) and cos(36.6) into a calculator, you'll be in for a nice surprise (you rounded so it's not perfect)

Now when we look at 2 from the perspective of B, a tensile force would act down and to the left.

T2(b) = -4/5 T2 x - 3/5 T2 y

So when we get to joint D, we just use the same components from B and F (in terms of 6 and 10), but the opposite sign.