# The forces acting on the tensegrity structure

1. Are these the only forces acting on the body. I assumed the COM to be somewhere there in between. since the tension in main string doesnt perfectly coincide with COM theres a small torque being generated which is being balanced by the string on the other side.

2. If some other force was applied anywhere else say point A, a tension would be developed on the B, C strings to cancel the net torque but there would be a movement in the centre of mass downwards since translational equilibrium was not maintained?

Is my take right? I'd appreciate it if someone could show the real forces acting if this isnt right, I saw the FBDs of the structure with 2 strings and main string but I'm not being able to wrap my head around this one since it's more 3D. I get lost. Thank you.

• The point of the strings that connect points A, B, and C to the base is to maintain lateral stability, That is to prevent sideways movement. So to understand the role they play, you need to not only apply the downward force $W$ on the upper frame but also some side load $S$ away from one of the strings. Commented Feb 24 at 19:17

The free body diagram for the structure as a whole has only two external forces - the weight of the structure, acting downwards, and the normal force from the surface it is resting on, acting upwards. When we consider the structure as a whole, the tensions in the three strings and in the central spring are internal forces, so do not appear in the free body diagram.

A more informative point of view is to consider free body diagrams for the upper and lower components of the structure separately. The external forces acting on the upper component are its weight (downwards), the tension in the central spring (upwards) and the tensions in the three outer strings (downwards - and maybe only two if one of the strings is slack). The external forces acting on the lower component are its weight (downwards), the tension in the central spring (downwards), the tensions in the three outer strings (upwards - and maybe only two if one of the strings is slack), and the normal force (upwards).

I don't think we can say any more than this without more information about the masses and dimensions of the various parts.

• yes apologies i edited my post. my main intention was indeed to ask how all the forces are acting on this structure not on the system as a whole. Can you tell me what the 'compressive' forces refer to in this diagram? i keep seeing that the structure is a result of a balance between the tension forces of string and compression forces of stick. So I was hoping for a complete description of forces acting. Also one small clarification does the reaction force consider the upper portions mass or are we sticking with only mass of lower body*g. Commented Feb 23 at 18:25
• @ZoruaChan You can divide the structure into even more components if you like - you can consider the forces acting on each of the individual struts for example - but I think that is overcomplicating things. I see no mention of compressive forces in the diagram. The normal force (is that what you mean by the "reaction force" ?) must equal the total weight of the whole structure - both upper and lower parts - in order for the structure to be in equilibrium. Commented Feb 23 at 18:56
• yep the normal force thanks. I believe the 'compression' forces refer to the restoring forces in the stick. I'm trying to get expressions for tensions in the string, the maximum load capacity until the string breaks, net torque required to destabilize the structure etc. such info. i believe finding this much is possible from the forces you mentioned, and measuring the weights + positon of center of mass. p.s - im making the structure and noting down observations. Commented Feb 23 at 19:25
• However if possible i'd also like to see the stress on the sticks. though i know the string would most definitely yield first, i want to know at what load will the stick break, for which i think i'll require the modulus of elasticity + the force acting and maybe even the angle of inclination right? Commented Feb 23 at 19:27

The way 3D free-body diagrams work is that they need to work as a 2D free-body diagram on every plane possible.

Consider the section below of the upper frame

The weight or payload on the frame is designated $$W$$ and might act not in line with the support force $$P$$ where the middle spring extends to counteract the load applied. The offset between the two forces is designated $$d$$ above.

In addition, tension might develop on the outer strings but not both simultaneously. In the diagram above tension $$B=0$$, but $$A>0$$.

You do the balance of forces and torques about the support point to get

\begin{aligned} P - W - A & = 0 \\ a A- d W & = 0 \end{aligned}

with solution of $$P = \tfrac{d}{a} W$$ and $$P = \left( 1 + \tfrac{d}{a} \right) W$$

as you can see when $$d=0$$ and the applied load is in line with the support, you have also $$A=0$$ and $$P=W$$.