Angled beam support reaction loads

[Rigga81]

Hi,
Im trying to find the method for working out the support reaction loads on a beam at given angle.
example: a 6m beam with a point load 600kg or UDl of 100kg/m perpendicular to the beam. Suspended at each end by a chain hoist, "a" and "b". "b" is 2m higher than "a". I believe using angle is the key but I can't find many examples I understand to check my results against.

Thanks

James

 

[Master1022]

Hi,
Im trying to find the method for working out the support reaction loads on a beam at given angle.
example: a 6m beam with a point load 600kg or UDl of 100kg/m perpendicular to the beam. Suspended at each end by a chain hoist, "a" and "b". "b" is 2m higher than "a". I believe using angle is the key but I can't find many examples I understand to check my results against.

Thanks

James

Yes, the angle will be very helpful here in determining certain angles. So I believe this will require us to take moments (don't see much point in resolving as there aren't many unknowns...). Have you tried taking moments about each end of the beam, knowing that the net moment = 0? If you do that, you should be able to find the tensions.

Are the chains vertical? I have written my response assuming that they are? It becomes a bit more tricky if they aren't, but then you can use 3 coplanar forces ideas...

 

[Rigga81]

Hi, Thanks for you reply. Yes in this instance the chains are vertical. Could you explain further how you would solve this please?

 

[Master1022]

Hi, Thanks for you reply. Yes in this instance the chains are vertical. Could you explain further how you would solve this please?

Most important thing when doing moment questions is to draw a nice, clear, and big diagram so you can familiarise yourself with the geometry of the problem.

We also know that Moment = Force x perpendicular distance to pivot from line of action:
\textbf{Moment} = \textbf{r} \times \textbf{F}

If we are comfortable with this idea, then there is no need for us to use the angles... If you are not clear about taking moments, I would go and look at some videos on youtube as there are many sources out there.

Taking moments about A (the lower end) couter-clockwise, we know that the net moment is equal to 0 in order for the beam to be in rotational equilibrium:
0 = T_{b}(4\sqrt2) - W(\frac{4\sqrt2}{2})

We can calculate W from the problem: if beam has uniform distribution of mass of 100 kg/m, then the total mass of 6m of the beam is... (and we can treat this as acting through the center of the beam as it is uniform distribution of mass)

Then: we can do the same thing for Ta or we can resolve vertically and simply solve from there for Ta

Hope that is of some help.