Statically Determinate: Insight into this equation please?

[mishima]

Hi, I teach a unit on bridge building for American high schools. We use software to show which members are in compression and tension and adjust design prior to physically building (ie doubling up members in compression).

We use the equation:
2*Joints = Members + Reactions

as a way to enable the software to work (advanced classes do it by hand). Typically I say something like "The software won't be able to compute the forces unless the equality is true". I was looking for something more related to real world experience of an actual engineer...what is the significance of this equation in practice?

 

[PhanthomJay]

That equation is used in order to determine whether the truss is statically determinate , indeterminate, or unstable. T

 

[PhanthomJay]

Hi, I teach a unit on bridge building for American high schools. We use software to show which members are in compression and tension and adjust design prior to physically building (ie doubling up members in compression).

We use the equation:
2*Joints = Members + Reactions

as a way to enable the software to work (advanced classes do it by hand). Typically I say something like "The software won't be able to compute the forces unless the equality is true". I was looking for something more related to real world experience of an actual engineer...what is the significance of this equation in practice?

This equation allows you to determine whether or not the truss is statically determinate or indeterminate both externally (reactions) or internally (members). Statically determinate trusses can be solved by the equilibrium equations based on Newton's first law, either by hand or apparently your computer software. Statically indeterminate trusses must use other equations besides the equilibrium equations, and can be solved with some difficulty by hand, or by an applicable computer program more advanced than the one you are using. Essentially, there are 2 equilibrium equations at each joint, that's the 2j part, and the unknowns are the member forces and the x and y component of the reaction forces ( r + m). When those values are equal, there are as many unknowns as the equilibrium equations, so the problem is solvesble using basic statics alone.

 

[mishima]

So, a bridge not satisfying this equality is not necessarily a bad/unstable design, it just means analysis is more complex?

 

[PhanthomJay]

So, a bridge not satisfying this equality is not necessarily a bad/unstable design, it just means analysis is more complex?

If m+r is greater than 2j, the design is fine and stronger, but analysis requires not only the static equilibrium equations, but also the calculations for member deformations under loading in order to determine member forces. It becomes a more tedious calculation, but a good computer software program can handle it fairly easily (as long as you trust its results with a sanity check).
If m+r is less than 2j, then this is no good a design, because the truss is unstable and will likely collapse.

See here:
http://richardson.eng.ua.edu/Former...russes/Stability_&_Determinacy_of_Trusses.pdf

You should note on the first page Figure 1 that there are 4 members , 4 joints, and 3 reaction force components (2 at the pin and one at the roller). Thus 2j is 8, and m+r is 7, which is less than 8 , so the truss is unstable and will collapse.
In figure 2, we count 5 members, 4 joints, and 3 reactions: here, m + r = 8 and 2j = 8, the truss is stable and statically determinate.
in Figure 3, again 4 joints and 3 reactions, but this time 6 members, thus 2j = 8, and m + r = 9, the truss is stable but statically indeterminate, so it requires more equations to solve (one member of the X brace will be in tension and the other in compression, under the loading shown in the figures, and this is a very good design.