# Springs stacked on each other in series with a mass on top, is the deformation the same?

This is a practical question which I am trying to determine the life of 5 rubber pucks which act as suspensions/shock absorbers for my airplane.

Essentially 5 rubber pucks are stacked on each other with the weight of the front of the plane on top (in a static system) between the body of the plane and the front wheel.

Now for a physics question, if I have a mass M, and 5 springs (all the same with same constant), such as the diagram below, would the deformation/displacement of each spring be the same?

I am finding the formula to calculate the total spring constant of the system but having difficulty figuring out how to see which spring has more displacement or should they all be equal? Imagine one single coil of wire making a spring of length L and constant k. You can cut it in half to get two springs, each with length L/2 and constant k. If you stack the two springs, you have an apparatus that is indistinguishable from the original spring - two springs stuck together is just one spring. Assuming a uniform spring constant along a spring's length, we see that one spring compresses uniformly along its length, therefore a series of identical springs must as well.

All of the pucks experience the entire compressive force that is holding up the airplane and, since the puck are the same and thus have the same individual spring constants, the deformation of all the pucks should be about the same. This is assuming that the mass of the pucks is insignificant compared to the mass of the airplane that they are supporting. If that assumption doesn't hold then the pucks at the bottom of the stack would be compressed more as they have to support more mass than the ones near the top.

@MEnns is right. I have further explanation.

The purpose of stacking 5 springs on top of each other isn't to make life easier for each spring. It is to make a spring 5 times less stiff, which reduces the forces transmitted to the airplane by a factor of 5.

The important property of a puck is the spring constant, $$k$$. The important equation for a spring is $$F = kx$$.

Suppose you have a single puck suspension and you taxi down the runway. The entire weight of the plane, F, compresses the puck by distance $$x = F/k$$. The spring pushes up with this same force. This holds the plane up.

Suppose you hit a bump that raises the wheel a distance $$\Delta x$$. The plane doesn't go up instantly, so the spring is compressed to $$x + \Delta x$$. The force pushing up on the plane becomes $$k(x + \Delta x) = F + \Delta F$$.

Repeat this with your suspension. As you taxi smoothly, each spring has the weight of the plane pushing down on it. Each is compressed by $$\Delta x$$. The total compression is $$5 \Delta x$$. The spring constant for the stack is $$F/(5 \Delta k) = k/5$$.

Suppose you hit the same bump. The wheel goes up $$\Delta x$$. The force on the plane becomes

$$\frac{k}{5}(5x + \Delta x) = F + \frac{\Delta F}{5}$$.

The plane is jostled less.

To understand a stack of springs, consider how you compress a spring. You push both ends with the same force.

When the plane sits on the runway, gravity exerts a downward force. This is called the weight of the plane. I called this $$F$$ above.

Forces accelerate objects. If the runways wasn't there, the plane would fall.

If an object is not accelerating, the total force must be $$0$$. The runway pushes up on the plane with an equal force $$F$$.

If you put a spring between the runway and the plane, these two equal forces compress the spring.

The spring presses back on both ends with the same force. The plane is still held up with the same force. The load on the runway is the same.

If you have a stack of springs, each spring pushes on the next, and the next pushes back. The second spring from the top doesn't know if the plane is pushing down, or it is the plane plus a spring. All the springs are compressed with the same force $$F$$. All compress the same distace $$x$$. From the FBD you obtain

$$F_{i+1}=F_i\quad,i=1..5$$ where $$F_i=k\,(x_{i+1}-x_i)\quad ,i=1..5$$

and $$~F_1=m\,g=W~,x_6=0~,F_6=0$$

solve these 5 equations for $$~x_i~,i=1..5~$$ you obtain

$$x_i=-\left[ \begin {array}{c} 5\,{\frac {W}{k}}\\ 4\,{ \frac {W}{k}}\\ 3\,{\frac {W}{k}} \\ 2\,{\frac {W}{k}}\\ {\frac {W }{k}}\end {array} \right]$$

from here the spring deflections

$$\Delta x_i=x_{i+1}-x_i=\frac Wk=\text{constant}$$