# Statically determinate system

The exercise is taken from linear algebra and its applications, prof. G. Strang

Exercise 2.4.8 Truss

I have already found that: m=n=8 → system is statically determinate, therefore solvable by equilibrium equation.

A is a 8 by 8 square matrix.

Also we know, there aren’t any rigid movements since the truss is fixed. Also, because 8-8 is 0, we're expecting 0 solutions to Au=0

I wrote the forces eq. equilibrium:

$$f_{H_1 }=-y_2-y_4 cos⁡30$$

$$f_{H_2 }=y_2+y_5 cos⁡30$$

$$f_{H_3 }=-y_7-y_5 cos⁡30$$

$$f_{H_4 }=y_7+y_4 cos⁡30$$

$$f_{V_1 }=-y_2-y_4 sin⁡30$$

$$f_{V_2 }=y_2+y_5 sin⁡30$$

$$f_{V_3 }=-y_7-y_5 sin⁡30$$

$$f_{V_4 }=y_7+y_4 sin⁡30$$

How should I continue and say something regarding the stability of the system?

• IMHO, any lateral force would make this collapse. Nodes 1234 indeed create a statically determinate body. However, the rod 3-5 and 4-6 are parallel and of equal length. Therefore body 1234 can freely rotate. Jan 23 at 8:44
• yes, but how can i show it mathematically? Jan 23 at 8:54
• @Efrat Brayer Too far away from my school years, so I can't tell you how to write the mathematic expressions required to demonstrate the stability of this system. But, if you provide a horizontal force at node 1, then you shall find horizontal forces on nodes 3 and 4, which are acting on member 7, and force it to go side way (horizontal displacement). Now you can see why the system is unstable -: the pin supports (joints 5 and 6) can't resist rotation due to the horizontal force on member 7, nor joints 3 and 4. From this observation, you shall be able to write the equilibrium equations.
– r13
Mar 13 at 14:07

IMHO, any lateral force would make this collapse. Nodes 1234 indeed create a statically determinate body.

However, the rod 3-5 and 4-6 are parallel and of equal length. Therefore body 1234 can freely rotate.

Regarding how you show that mathematically, (without knowing the context of your assignment):

• 1234 is a rigid body
• you can substitute 1234 with a single rod 34 .
• for the new system you have nodes 3,4,5,6 and rods 53, 34, 56.

Therefore in the new equivalent system you have 4 nodes and 3 rods. Therefore, there are 8 unknowns and only 6 equations.

The frame is externally unstable.

ED = R - 3 - S = 4 - 3 - 4 = -3 < 0, unstable.

R = number of support reaction S = number of internal joint releases (4 moment releases at the hinges)

You got the horizontal equilibrium equations correct. Now add them all up and you'll get $$f_{H_1} + f_{H_2} + f_{H_3} + f_{H_4} = 0$$. That equation says that the sum of all external loads equals zero. So any combination of these that doesn't fulfill that requirement will result in collapse of the structure, mathematically represented by the equation's LHS not being equal to the RHS. For example, if $$f_{H_1} = 1$$ and all of the others = 0, you'll get 1 = 0, so mechanically speaking the system is not in equilibrium.

You do not need the vertical direction to solve the task, but it can demonstrate the difference. You got the vertical equations wrong, though, because you kept the hotizontal trusses in there. You need to leave those out and account for the vertical ones instead (apart from 1 and 3, 6 and 8 will also appear). Now if you get them right and add them up just like the horizontal ones, you'll see that there will actually be unknowns left, not just external forces, which means that strictly in the vertical direction, the structure is not unstable.