chilled76 said » Having a look at it, isn't aracers solution for a fixed bridge? Does your solution take into account it's fixed at one end and free roaming at the other?
No - fixed ends would complicate things as you'd have to consider material effects (ie how much the members would expand and compress under load and the resultant horizontal loads on the support points). Which is why things like this are usually done with one unconstrained support - as indeed is the case for real bridges, though in the real world you also have other issues like thermal expansion to worry about. My working is done assuming no horizontal loads at the end supports, which is the effect of having one end freely supported and makes things a lot simpler.
mrodgers said » You can also solve it by considering half of the truss and taking moments about point C. This is a bit simpler as it means you don't have to work out all of the forces in the other members
Depends what you mean by simpler! Certainly less total calcs in your method and it's very elegant, but I'd argue that each of the individual calcs for my method is simpler as you're just adding and subtracting forces at each point, then it's just a case of repeating at each joint (and actually I don't need to do 2 of my calcs to get the load in HG). The other advantage being that it confirms the initial assumption about the member with the highest load. I'd probably give your method higher marks though (if I'd tried to do moments I'd have been busy trying to do them around point A!) The question has to be whether you could do your method in your head as I did with mine (just writing down the forces in each member).
chilled76 said » Also, any chance of having a look at your workings? Seems a touch simple for 22 marks?
First of all I should probably add details of where I'm resolving for each load:
C, vertical: CH = 200.root2 compression
H, vertical: BH = 200.root2 tension
B, vertical: BJ = 600.root2 compression
J, vertical: AJ = 600.root2 tension
A, horizontal: AB = 600 compression
B, horizontal: BC = 1400 compression
J, horizontal: JH = 1200 tension
H, horizontal: HG = 1600 tension
For point C where I start from, clearly you can split the truss into two as everybody else has also done, hence half of the 400N load is taken by each half. Resolving vertically, the only vertical forces at C are the 400N load and the force in CH. Clearly the vertical component of the force in CH is 200N compressive in order to support half the 400N load. As it is a 45 degree member then force in the member is 200 / sin 45 = 200.root2 (alternatively you can consider that horizontal and vertical components are equal in a 45 degree member, so by Pythagorus the total load is root(200^2 + 200^2)).
Vertically at H the only forces are CH and BH, so the force in BH must be the same as the force in CH but in the opposite direction.
when we get into resolving horizontally having determined the forces in all members with vertical components it's just more of the same, but with more members involved and you just have to keep track of the direction of each force. As mentioned above it is assumed there is no horizontal force at the support. The most complex calc is probably horizontally at point B - for this I'll use shorthand of tension -ve compression +ve (no idea if that's the usual convention) so I can do it with formulae not words:
Fbc = Fab + Fbj sin45 - Fbh sin45
= 600 + (600.root2 / root2) - (-200.root2 / root2)
= 600 + 600 + 200