# Distributed load with an offset?

I have a question on my work which I cannot figure out. I have three chains which support a beam. On the beam acts a distributed load. I want to calculate how much load will go to each chain.

I know the sum of each force/moment is equal to zero. The magnitude of the distributed load ($w$ Newton/meter). The length of the beam $L$ (Meter) and the sub-distances ($x_a$, $x_b$, $x_c$, $x_d$ Meter). $F_{vt}$ (newton) represents the concentrated force of the distributed load ($F_v = w*l*0.5$) and $F_{av}$ is the concentrated load in section $x_a$.

So if you go through all the point you get four formulas. The first one for point $a$ is:

$$\begin{gather} \sum M_a = 0\\ F_b(x_b) * F_c(x_b+x_c) - (w *L)* \dfrac{L}{2} = 0 \end{gather}$$

But in $x_a$ section forces are also action, which should a moment around point $a$. But how can I calculate the resultant force ($F_{av}$) for section $x_a$?

I have tried the following but I get a net force in this section of 0 N ($F_{av} = w\cdot\dfrac{0}{L} = 0$). So there is no force acting in section $x_a$?

Can somebody help me out with this part?

• This question can certainly be answered here, but some clarification is necessary. You first call the distributed load $w$, but the diagram shows $F_v$ instead. And then your moment calculation contains the multiplication of $w$ and $F_v$. Also, what are the red loads $F_{av}$ and $F_{vt}$? Is $F_{av}$ a concentrated force while $F_{vt}$ is the resultant force of the distributed load? Or is it also another concentrated force? – Wasabi Mar 7 '16 at 11:13
• @Wasabi thanks for your reply i have edited the questions to make it (hopefully) more clear – pwghost Mar 7 '16 at 12:59
• Something still seems strange. Both $w$ and $F_V$ are still appearing in your moment calculation but I do not see $w$ on the figure. – William S. Godfrey- S.E. Mar 7 '16 at 18:43
• @william s. godfrey i have edited it and hope you guys understand my question now :S – pwghost Mar 8 '16 at 8:43

The simplest way to solve this problem would be to hinge the beam in the middle, so that you have two statically determinate beams. Then there is no Bending Moment at B, so that

Fa = Fc = (w*2.5*2.5/2)/2 = 1.5625*w

Fb = w*5 - Fa – Fc = 1.875*w

Does the central chain serve any purpose?

There is a slight mistake in your formula – taking moments about A to the right-hand side, the moment due to the UDL is not w * L * L/2, but w*(Xb + Xc + Xd) * (Xb + Xc + Xd)/2

If you keep it as a continuous beam, then I hope you do not design it “on the limit”, especially if it is a safety-critical situation (And 50kg dropping onto someone could harm them.) I assume you will have a means of adjusting say the chain at B, before applying the UDL, until the three chains all feel to be carrying equal, or near-equal loads. Even 1mm may be enough to go from the chain at B being completely slack, to it carrying the total load on the beam

At a theoretical level, FEA is the way to go. It is so easy to change the assumptions about chain elasticity, beam bending elasticity, pre-tensioning of each chain.

Practically, on the basis of the limited information you have given us, I don't think you will be far out if you assume the beam is hinged in the centre, even if it is continuous there, provided you adjust the chains before applying the UDL.

A simple beam such as this one supported at three locations is statically indeterminate. You have three unknown reactions, one for each support, but there are only two equations of static equilibrium available, hence you have one degree of indeterminacy.

This type of beam is called a continuous beam, and there are different methods of calculating the reactions at each support.

• hi thanks for you comment, could you tell me how to approach this problem ? – pwghost Mar 8 '16 at 8:43

As has been said, this is a Statically Indeterminate Problem. The force in each chain will depend on the elasticity of each chain, the bending elasticity of the beam, and how much each chain is tensioned. One chain could be completely slack, and the other two could carry the whole load. Or, theoretically, the central chain could carry the whole load. In practice it would all depend on how much you choose to tension each chain. A Finite Element Analysis would give you a solution for whatever beam section, and therefore elasticity, chain elasticity, and initial chain tensions you choose.

• all chains are the same, could i do it then without an finite element analysis? – pwghost Mar 9 '16 at 12:56

I have looked at the other solution that you mentioned. That solution would be ok provided that the two assumptions hold. That is, the beam is not flexible, and the chains are adjusted before any load is applied, to be of equal length and carrying equal almost-zero loads. Depending on the Load/Extension value of the chains, this may involve very small adjustments to their initial lengths, and resulting tensions. Without the length and elastic modulus of the chains, and knowledge of the temperature stability of the support structure, it is all guesswork, but I expect that an adjustment in length of the centre chain of about 0.1 mm would be enough for it to go from carrying no load, to carrying the whole load.

The loads are so low, is an accurate answer really necessary? Could you change it to only two chains? Could you hinge the beam where the middle chain is attached?

Have a look at strand7.com, and their demo download. This is a fully working Finite Element package, limited to 20 beams, so ok for your problem. In my personal opinion it is the best FE package for user-friendliness. (Or - it is dead easy to understand and use.)

Sorry it took so long to come back. I am currently struggling to understand how emails work, and how ubuntu works.

• thank you very much for your follow up comment, it's really helping me. the beam has 3 to 8 chains attached to it depending on the load it as to carry in the desinged machine. the beam is not rigid it is very flexible L profile of 5m long 4mm steel with a load between the 5kN and 80 kN – pwghost Mar 17 '16 at 19:36