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I am reading a book on fundamental physics for mechanics. There is a truss shown in the figure. A train of mass $M=56$ ton is resting at the middle point of AC, ignore all the mass of truss and rails. All triangles are equilateral. I am trying to find the force exerted on strut AB.

enter image description here

I know this issue is all about static equilibrium on force. I am looking at point A, I have the force $F_{AB}\sin(60) = 0.5*Mg$. I use "half" of the mass because there are two strut AB and BC sharing the weight of the train. But this doesn't match the answer ($F_{AB}$ should be $7.9\times 10^4 $N). What's wrong with my calculation.

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closed as off-topic by Emilio Pisanty, Dan, Manishearth Aug 8 '13 at 4:13

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Doesn't strut DC, for example, also share some of the weight? I think the point here might be that you'll need to include most if not all of the struts in your analysis. You should then get a system of equations that can be solved for all of the forces. –  joshphysics Apr 3 '13 at 14:58
    
Thanks for your reply. I am thinking should we look at the joint at a time. I wonder for point A, along the vertical direction, how force BC and CD comes into the equation? –  user1285419 Apr 3 '13 at 21:54
    
What are the end conditions at A and E like? –  ja72 Aug 7 '13 at 15:46

2 Answers 2

You will need to clarify the support conditions for the truss because whther or not there is horizontal restraint at the supports will matter to the result. Assume pinned one side and a roller bearing on the other side then your approach should be:

  1. Resolve vertically for the whole bridge (eq.1)
  2. Take moments about one of the supports (eq.2)
  3. Using system of equations eq.1 and eq.2 you should be able to derive the vertical support reactions for the bridge.
  4. Resolve vertically at A using the support reaction you have just derived will give you the required force in AB.
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Don't forget that member AC can apply a vertical load on A and C whereas all other members can only apply tensile/compressive loads with zero lateral loads. –  ja72 Aug 7 '13 at 15:05

Yes you can split the load $W = M\,g$ between nodes A and C. Then you have 5 nodes and two equations for each node with a total of 10 equations. You have 7 members for 7 element forces and 3 reaction forces = 10 unknowns. So this sysytem is statically determinant (solvable).

Here is an example of the force balance on node A:

$$ A_x - F_{AC} - F_{AB} \cos 60° = 0 $$ $$ A_y - \frac{W}{2} - F_{AB} \sin 60° = 0$$

Using equal and opposite forces on AB the force balance on node B is:

$$ F_{AC} - F_{CE} + F_{BC} \cos 60° - F_{CD} \cos 60° = 0 $$ $$ -\frac{W}{2} - F_{BC} \sin 60° - F_{CD} \sin 60° = 0 $$

and on and on to the rest of the nodes. Note that at point E there can only be a vertical reaction force, with zero horizontal reaction. Otherwise the system cannot be solved withough considering the elasticity of each component.

In the end you get a 10x10 matrix system (which is solvable in Excel, or by hand)

$$ \begin{bmatrix} 1 & 0 & 0 & \mbox{-}\frac{1}{2} & -1 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \mbox{-}\frac{\sqrt{3}}{2} & 0 & \cdots & 0 & 0 \\ \vdots & & & & & \ddots & & \vdots \\ 0 & 0 & 0 & 0 & 0 & \cdots & 0 & \mbox{-}\frac{1}{2} \end{bmatrix} \begin{pmatrix} A_x \\ A_y \\ E_y \\ F_{AB} \\ \vdots \\ F_{CE} \\ F_{DE} \end{pmatrix} = \begin{pmatrix} 0 \\ \frac{W}{2} \\ 0 \\ \frac{W}{2} \\ \vdots \\ 0 \\ 0 \end{pmatrix} $$

Note that positive member force is considered compressive by me, but you are free to switch it around if you want. The standard convention is positive = tension though.

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