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I have a four-bar linkage of known geometry (all angles, lengths, etc.) and I'm trying to solve for the required force to balance an input force.

Per this diagram, the input force (Fc, perpendicular to ABC) is known and the required force to counteract it (Freq, only in Y-axis) must be found. A and E are able to rotate, but not translate. I've tried different methods to solve for the force, but my results don't make sense once I plug-in some example values.

Diagram

This plot shows the calculated ratio of Freq to Fc across a range of phi values. I wouldn't expect the ratio to suddenly change by several orders of magnitude, so I believe my solution is wrong. Force Plot

Edit: I've added plots of the three positions that result in the extreme points on the force plot, with forces shown as vector and X-Y components:

(AX=37, AY=50, EX=0, EY=30, LDE=50, LBD=24, LAB=13, LAC=120, phi=[-0.8, 0.8, 1.25]) Angles

I've evaluated the FBDs of each link: FBDs

Then tried solving the equations directly and via matrix multiplication, but get the same result. I think the FBDs are where I must have gone wrong then. Can someone confirm, please?

My attempted matrix solution, for reference (blank cells are zeros): Matrix

Update: using SAM 8.4 I simulated the linkage and found the Freq for a 1N FC. Intuitively the curve makes sense because the spike on the left side is where AB approaches parallel with BD and the force is otherwise near LAC/LAB (~10):

Parameters: (AX=37, AY=45, EX=0, EY=30, LDE=50, LBD=25, LAB=13, LAC=120) SAM

I'll assume that this represents an accurate curve to compare against.

Mandias
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3 Answers3

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The strange angles might occur when points B, D, and E are on one line or BD is perpendicular to $F_{req}$. In such cases, $F_{req}$ will hardly produce any $F_c$, i.e., $\left|F_{req}\right| \ggg \left|F_c\right|$ and consequently:

$$\frac{F_{req}}{F_c} \rightarrow \pm \infty$$

So, your calculation might actually be right.

Edit: One problem I see is that the force $F_{req}$ is considered only in member ED, but this force acts on point D, not on a specific member. So you should take it into account also in $\sum F_Y$ equilibrium of member BD: $$\sum F_Y = F_{DY}-F_{BY}+F_{req}$$

Tomáš Létal
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If we use the method of virtual work, all the forces need not be computed.

Let's start by assuming that the bar ED is at an angle $\theta$ with the horizontal.

The position of the point D is $\{\text{ED} \cos (\theta ),\text{ED} \sin (\theta )\}$.

The position of the point B is $\{\text{EA}_x+\text{AB} \cos (\phi ),\text{EA}_y+\text{AB} \sin (\phi )\}$.

Since BD is of constant length we have the constraint equation

$$\fbox{$(\text{ED} \cos (\theta )-\text{EA}_x-\text{AB} \cos (\phi )){}^2+(\text{ED} \sin (\theta )-\text{EA}_y -\text{AB} \sin (\phi)){}^2=\text{BD}^2$}$$

We can find the virtual displacements from the above equation as

$$(2 \text{ED} \cos (\theta ) \left(-\text{AB} \sin (\phi )-\text{EA}_y+\text{ED} \sin (\theta )\right)\\-2 \text{ED} \sin(\theta ) \left(-\text{AB} \cos (\phi )-\text{EA}_x+\text{ED} \cos (\theta )\right))\delta \theta \ + \\ (2 \text{AB} \sin (\phi ) \left(-\text{AB} \cos (\phi )-\text{EA}_x+\text{ED} \cos (\theta )\right)\\-2 \text{AB} \cos (\phi ) \left(-\text{AB} \sin (\phi )-\text{EA}_y+\text{ED} \sin (\theta )\right)) \delta \phi = 0 $$

This simplifies to $$\text{EA}_x (\text{ED}\ \delta \theta \sin (\theta )-\text{AB} \ \delta \phi \sin (\phi ))\\+\text{EA}_y (\text{AB} \ \delta \phi \cos (\phi )-\text{ED} \ \delta \theta\cos (\theta ))\\+\text{AB}\ \text{ED} (\delta \theta -\delta \phi ) \sin (\theta -\phi )=0$$

Which gives

$$\delta \theta =\frac{\text{AB}(\text{EA}_x \sin (\phi )- \text{EA}_y \cos (\phi )+\ \text{ED} \sin (\theta -\phi ))}{\text{ED}(\text{EA}_x \sin (\theta )-\text{EA}_y \cos (\theta )+\text{AB} \sin (\theta -\phi ))} \delta \phi$$.

Now applying the principle of virtual work we get

$$ \text{ED}\ \delta \theta\ F_{\text{req}} \cos (\theta )-\text{AC} \ \delta \phi \ F_c = 0$$

Substitute for $\delta\theta$ and simplify to get $$\fbox{$F_{req}=\frac{AC(\text{EA}_x \sin (\theta )-\text{EA}_y \cos (\theta )+\text{AB} \sin (\theta -\phi ))}{\text{AB} \left(\text{EA}_x \sin (\phi )-\text{EA}_y \cos (\phi )+\text{ED} \sin (\theta -\phi )\right)} \ F_c\ \sec (\theta )$}$$

For a given value of $\phi$, you need to solve for $\theta$ from the constraint equation and then substitute that into the force equation.

Suba Thomas
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The solution I found was to translate forces into normal and parallel along each link:

Where:

  • θ is the angle between ABC's normal and BD
  • ω is the angle between BD and DE's normal
  • γ is the angle between BD and the X-axis

Equations

This resulted in a force curve that matches the SAM software: Force Curve

Mandias
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