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Choosing The Best Type of Roller Chain

I often get the question of what type of roller chain should be used in an application. Let’s unpack that question in this article.

A standard duty roller chain should be specified unless spacial constraints prohibit the use of a larger or double chain. Selecting a standard duty chain allows for the opportunity to increase in strength if needed by converting to a heavy duty or double row chain later.

If you know me at all, you know that I always like to have a Plan B in mind. If something isn’t working right, I will work on it to a point, but I always want to have a backup design ready.

I am no different with roller chain. If I design to standard duty roller chain, I have three different options immediately available when (er…I mean if) it isn’t working correctly. They are

  • Heavy duty chain
  • Multiple strand chain
  • The next pitch size up

Before we dive deeper, we need to gain a little knowledge in the components of a roller chain link. Click here to see a video on how roller chain is made.

Roller Chain Components

The main parts of a roller chain are the inner and outer links, pins, bushings, and rollers. The bushings are rolled plates that are pressed into the inner links which are the dogbone shaped plates. The roller is inserted over the bushing before the second inner link is pressed on. Next, the pins are pressed into an outer plate. Then two inner link assemblies are assembled over the pins and the second outer plate is added. The ends of the pins are mushroomed in the pressing step so that the link will stay together. That was all in the video, you should have watched it.

Please note that smaller chains, line the one shown below, do not have a bushing under the roller. This is because the bushing would be rather small and difficult to handle.

Designation of Roller Chains

Roller chain has an interesting designation system. The last number in the designation can be 0, 1 or 5. A 0 designation is the standard roller chain as described above. A 1 designation refers to a narrower variation of the standard, 0, configuration. A 5 designation refers to a chain without a roller.

The first one or two numbers of a roller chain refer directly to the chain’s pitch. To get the pitch in inches, divide the number by 8. If you desire millimeters, multiply by 3.175. For example, a 50 chain has a pitch of 5 / 8 or 0.625″. For metric, 5 times 3.175 is 15.9mm.

Chain is usually sold in feet, so it may not be critical to know exactly how many links are needed. However, in critical applications, the tensioning device may not be able to handle too many or too few links. If you want to know the number of links, you simply divide the length needed by the pitch.

Standard chain sizes are 25, 35, 40, 41, 50, 60, 80, 100, 120, 140, 160, 180, 200 and 240. For single rows of chain, you will see the chain specified as 50 or 50-1. If two or more rows of chain are needed, the chain will be specified as 50-2 or 50-3 etc. If a heavy duty chain is required, a chain designator of ‘h’ is added to the end (i.e. 50-1h).

A heavy duty chain will use stronger components made of better grades of steel than a standard duty chain. A different heat treatment is also applied to increase strength. Finally, the link plates are usually slightly thicker. All this is to say that you will only get an incremental improvement in performance.

The pin is usually the weakest component in the link and there is only so much that one can do to increase its strength. A pins greatest weakness is bending loads and chains are designed to minimize that, but as the shear load goes up, the bending load goes up exponentially. In fact, just having thicker plates makes the bending load on the pin go up.

Chain Strength

Manufacturer’s literature may have the chain’s strength listed as the tensile or breaking strength.

Do not design to the tensile or breaking strength!

The breaking strength is the amount of load that the chain can handle before breaking. That’s failure! It does not account for manufacturing tolerance, chain wear, misalignment, shock loading or twisting. We should design to the working load!

As engineers, we design to the working load, which for roller chain is a design factor of 8:1. You may want to increase or even possibly decrease this factor based on your application. Shock loading is probably the largest driver in choosing a design factor.

Below is a table of typical chain breaking strength values and the relative cost to a 25-1 standard chain. While this table is not thorough and exhaustive, it will give you the tools to make informed decisions for your application.

Chain SizeStandard
Breaking Strength (lb)
Relative Cost to 25-1 Heavy-Duty
Breaking Strength (lb)
Relative Cost to 25-1
25-19301.0010102.04
35-123200.7639202.04
40-139700.8851812.42
50-162601.1679402.46
50-2132302.64
60-192701.51121302.05
60-2185303.56
80-1165402.67198503.43
80-2330805.73
100-1253604.32308705.76
120-1326406.32363908.32
140-1452107.784851010.23
160-1577809.746063013.13

One thing that is almost immediately visible is the 35-1 and 40-1 chains are cheaper than a 25-1 chain. This probably has to do with two things. First, the 25-1 chain isn’t very strong so most people probably don’t use it. Second, the parts used in the 25-1 chain are small and require more handling and that leads to less production. If you are already considering moving away from a 25-1 chain, just go to the 35-1. You won’t regret it.

In all but a few cases, you will get more bang for your buck by moving up one chain size. This move dramatically improves your strength, with a minimal impact to cost. The most notable improvement of 149% increase is going from 25 to 35 chain at a cost savings and the next would be going from 35 to 40 chain for a 71% increase in strength.

In fact, the only cases where going from a standard chain to a heavy duty chain is for 60, 80 and 100. On the 60 chain you get 24% more strength for 36% more money. With 80 chain you get 17% more strength for 29% more cost. Finally, with 100 chain there is a 18% increase in strength at a 33% cost increase.

While these numbers are representative, you might be able to find better deals than I can. The last major observation is that there wasn’t a case where it made sense to switch to a double or even a triple chain.

Why upgrade to a double or triple strand roller chain?

Good question. There are a lot of good reasons to upgrade to a multiple strand roller chain. Let’s look at a few.

  • It doubles (or triples) the working load easily. There isn’t a need to fully redesign an existing system. Sprockets and chain guides are usually available in single and double strand configurations
  • Your design has height constraints, but can be widened.
  • Your sprocket needs to have a certain number of teeth. Increasing the tooth count may throw off running speeds unless all the other sprockets are increased accordingly. Also, a sprocket for the larger chain size may not be available in the same bore.
  • Since there are two or more strands, there should be better wear and less friction between the chain and sprocket. This assumes that the chain is not misaligned.
  • The chain will have a longer life

How to upgrade

So you’ve made the decision to upgrade your chain from a standard chain to heavy duty or a multiple strand chain. It is very important before any parts are purchased to check the design calculations. If your calculations are good, but you are having issues with chain breakage, something else is in play. My guess is there is misalignment, interference or a shaft that isn’t rotating freely. Please see my guide to preventing roller chain failures.

The next step is to examine every part of the design in 3D CAD (if possible) and in real life to identify the possibilities of rubbing, pinching or even if extra chain support is needed.

If you do nothing else, make sure that you can assemble the machine with the wider chain and sprocket. This is one thing that most young engineers will forget about. This causes headaches all around.

You will also need to investigate increasing the tensioning devices for your chain. The weight increase of a heavy duty chain is minimal, but going from single strand to double or triple strand will double or triple your chain weight. You need to be prepared for this

Pay special attention for misalignment. When going to multiple strand roller chain, it is often the case that alignment tolerances have been eaten up with the wider sprockets.

Finally, when upgrading to heavy duty roller chain, you need to make sure that all the chain components are heavy duty rated. The most overlooked components are the connecting links.

Conclusion

From our investigation into what roller chain type we should be using to design to, we have found that the standard single strand roller chain should be used. In almost all cases, increasing the chain to the next pitch size gives us the most improvement in strength for the least cost. We also looked into reasons why we would use a multiple strand chain.

Finally, be sure that when you change your roller chain for an existing application that there is plenty of room to assemble the larger sprockets.

Roller chains are great ways to transmit power. Like all power transmission, there are certain rules that need to be followed. This article should give you insight into selecting the right roller chain.

A Quick, Easy Guide to Extension Spring Design

Springs are essential to life.  In fact everything on the planet acts like a spring, which follow Hooke’s law.  When force is applied to an object, it will deflect.  If more force is applied, it will deflect more. 

When designing or selecting a extension spring you need to consider: the spring shape and end type, the number of coils, the preload and allowable stress, the spring index. You can then calculate the stress on the spring using one of five basic approaches.

One crucial thing to understand when designing extension springs is to be sure to operate them in their intended range.  All springs work well in their linear range, but as soon as they are loaded past the yield strength, the spring stops behaving as intended and the spring life is severely shortened. 

Furthermore, the consequences with an extension spring are much higher. When a compression spring fails, the pieces are still in a bore or around a shaft and the components remain in contact with each other. If an extension spring fails, the components will be separated from each other and the result could be disastrous.

To illustrate this, if a spring in your cars suspension fails, your car will still be in one piece and you could at least make it to the side of the road before too much damage was done. On the other hand, if your trampoline is missing too may springs, you are falling through to the ground.

One of the most fascinating things about springs is that the method they are used in is opposite of the load they see.  A compression or extension spring is loaded in torsion and a torsion spring is loaded in bending.  I liken this to tightening a hex head screw with a socket wrench and tightening a socket head screw with a hex wrench. It’s all backward!

1. Spring Shapes

Cylindrical extension springs with hooks on the end are by far the most common type available, but there are other shapes to consider.  The most popular are:

  • Swivel Hook
  • Straight Cut
  • Drawbar

Cylinderical Hooked Springs

Cylinderical springs are very common and come in a variety of sizes as an off the shelf part.  The OD and ID of the spring is consistent for the length of the spring coils they are generally intended to be hooked on each end.  The spring rate is constant once preload is overcome (more to come on that). Adding extension springs in series will enable more travel, while adding them in parallel allows for more force.

One important thing to note with hooked springs is that the highest stress is at the ends of the spring. You can see that the two spots where the spring is first bent to start the hook is where the stresses are the highest. Typically the hook is formed in at least two moves so that the increase in stress is minimal.

Cross Section of Extension Spring FEA
This image has an empty alt attribute; its file name is barrel-1.png

Swivel Hook

            Swivel hook extension springs are rarely used. This spring is a barrel spring coiled around a hook on each end. This design is more expensive but it does deal with the higher stresses at the end of the spring. The main benefit is that the ends can pivot so that if the end mounts aren’t in line additional stress on the spring is not added.

Straight Cut

Straight cut extension springs are like regular compression springs, but wound so that it is an extension spring. The main benefit of this is that springs can be ordered in long stock lengths and cut to the desired length. This allows the engineer to get the spring characteristics desired quickly and without purchasing multiple springs.

Terminating these springs is very unique. The spring comes in diameters that match available threaded hardware. These springs are terminated by threading a fastener into the spring. This allows you to have a complete control over terminating with a capscrew, eyebolt or hook. The sky is the limit. Just make sure that you subtract the engagement between screw and thread in your calculations.

Draw Bar Spring

Drawbar springs are a way to use a compressive spring in an entension spring fashion. The main benefit is that the drawbar sits inside the spring and prevents the possibility of the spring failure separating the components at each end. (However, the drawbar could still fail is loads are high enough.)

Drawbar springs are used in winter pool cover attachment, porch swings, porch doors, trampolines and any application where if the spring fails, the components need to stay attached.

Many people use these to tie off a boat so that if the spring breaks, they don’t have to go searching for their boat.

2. Extension Spring End Types

There are three basic types of extension spring ends, they are:

  • Machine Hook
  • Machine Loop
  • Double Full Loop

These spring ends are easy to identify when shown next to each other. A hook end will allow an object to reach around the end of the hook to engage. Generally speaking this allows the hook to engage and disengage without disassembly.

The loop configurations are closed so assembly usually requires it so stretch over an object and another part to secure it. Imagine that the spring stretches over the end of a threaded rod and a lock nut is added to hold it on.

A double loop end is used when strengthening of the loop is needed. If the spring was loaded laterally, there might be enough force to open the loop on a single machine loop end. The double loop prevents this.

Other end types that are less popular are

  • Full loop on the side
  • Offset hook on the side
  • V-Hook
  • Elongated Round Hook
  • Small loop on the side
  • Small extended loop

If none of these fit your fancy, you can create your own custom spring. They can pretty much make any end design to your specifications.

One other consideration for the end condition is the angle of the loops, hooks, or whatever you thought of is one end’s relative position to the other end. An off the shelf spring will have the ends in line. They assume that you are linking things together like you would a chain. You can specify if your spring should be inline, perpendicular or random. Random, of course, gives you the lowest cost, but the least control of your design. If you are willing to foot the bill for a custom spring, you can specify any orientation you would like.

3. Number of Coils

Unlike a compression spring, counting the number of coils on a spring is straight forward.  The only exception is the drawbar spring so you should consult the compression spring design guide. Most extension springs are designed with a preload and that will manifest itself with the coils touching each other.

Start counting each coil from one end to the other starting where the first overlap is. Keep counting until you get to the other end of the spring. Generally speaking, coils are counted as full turns, but half and quarter turns are also widely used. Many spring manufacturers will allow two digits of precision on the number of coils.

Loop spring has 8.5 coils
Hook Spring has 22.5 coils

4. Spring Preload

Extension springs are able to have a preload in the design. This allows a spring to have some load at no deflection. This idea has a parallel in fastener design because they both rely on Hooke’s Law. A fastener is torqued to a load higher than the design anticipates so that it will not be subject to fatigue.

The preload on a coiled spring comes from several, if not all of the coils being twisted so they want to be on top of each other. It is basically adding a torque in the opposite direction of the spring.

Theoretically, the spring will have no deflection under load until the preset tension is reached. After that it will follow the linear spring rate. However, when the load is right at the preload, the load-deflection curve will make a smooth transition instead of a sharp curve. This generally isn’t the part of the curve that we are working in so it usually is ignored.

This phenonemon is caused by unequal twist in the spring coils. The theoretical case is that all of the coils separate at the same time. In reality, some coils with separate first and others won’t.

The above chart is zoomed in so that the theoretical vs actual curves can be distinguished. When the chart is zoomed out, the magnitude of the preload effects can be seen in proportion to the rest of the springs working envelope.

5. Allowable Spring Stress

All springs function according to their dimensions.  The dimensions will specify the stresses seen in the spring.  Based on general stress calculations, we know that the largest stresses in a extension spring will be on the outside surface of the wire as load is applied. 

When designing a spring, the dimensions are critical to spring characteristics.  Often we are confined by the form factor where the spring is applied, i.e., the spring needs to fit in a ¾” hole or over a ½” rod.  As the designer, if you find that there is only one or two choices for springs available, you may want to ask yourself if you are being too strict with the design.  I have often found that in my design I needed to go up in capacity, but the form factor was already set in stone.  Now it is really hard to get the performance I need.  Had I adjusted my design to have four or five choices of springs, I could simply change out a spring.  In practice, I desire to have five choices; my ideal spring plus two stronger and two weaker spring alternatives.

The life of the spring is dictated by fatigue.  As the spring is cycled, grains in the steel rub against each other.  As they repeatedly rub, small cracks will get larger and larger until failure occurs.  There are two main components that dictate the life of a spring; the number of cycles and the stress intensity.

The number of cycles is pretty easy to determine, it is something that is easily counted or estimated.  We find that most materials have an endurance limit, where if the spring makes it to that number of cycles, it will last forever.  For steel, this occurs around 1-2 million cycles.  Unless your spring is custom, vendors only make springs with infinite life.  Even then, vendors will fight you on a finite life spring.

Stress intensity is much more complicated.  A lot of it has to do with the mean (average) and alternating stresses.  The average stress, alternating stress and stress range are defined by the following equations.  We will use a maximum stress of 50 ksi and minimum stress of 10 ksi.

Since we are only dealing with extension springs, the minimum stress is usually zero unless the spring has a preload. In this case, the preload would be a negative stress.  In the case above and after the spring is installed, we will always have a preload on our spring.  This is usually the case in applications like a gas motor where the carbeurator needs a minimal amount of force to keep a valve open or shut.  We want preload on it so that the valve doesn’t vibrate when at rest and it also requires some force to move. 

In any case, our stress range is 40 ksi.  To extend our spring life, we will want to decrease this as much as possible.  If an extension spring was to have an initial preload, the minimum stress would be negative thus increasing the stress range greatly and decreasing life span.

One step that spring manufacturers perform to limit the magnitude of stress is to stress relieve.  The stress relieving process heats the spring to high temperatures so that permanent or residual stresses from the forming process are eliminated.  The temperature here is not so high that the spring would lose its initial temper.

When designing an extenstion spring, a fairly conservative design factor is used. THis is mostly because of the heightened risk for catestrophic failure if the spring fails. The stresses are generally less than 60% of the materials tensile strength.  Sixty percent is chosen because the endurance limit of most materials is around 60% of tensile strength. 

6. Spring Index

The spring index is a simple ratio of the mean diameter to the wire diameter.  It is a variable used in many spring calculations and give us an idea of what type of spring it is.  The index should be kept between 4 and 10. 

If it is lower than 4 and your spring is coiled too tight, this will require special tooling to form.  When forming a tight radius, you might also have problems with the material cracking internally and that will lead to premature failure. 

Spring indexes greater than 10 means the spring will be real flimsy and lead to problems with packaging and tangling.  Think of a slinky.  You also won’t be able to grind or perform other operations like plating. 

This should be a giant red flag that something is not right with your design.  Don’t design yourself into a hole!

7. Spring Calculations

So this is a little bit of a backwards approach.  There are five main ways of selecting a spring, but they only make sense once you understand the equations behind spring design.  For this we will go through the calculations and then come back to the methods of selection.

Most spring design is done based on dimensions so it is likely that you will have the outside diameter (OD) and the wire diameter.  From these two pieces of information, we can find the inside diameter (ID) and the mean diameter, Dm using the following equations:

From here we can calculate the shear stress on the spring coils (assuming the wire is round).

Where C is the spring index, P is the applied load and K is the stress concentration factor.  Note K is shown for extension and compression springs.  Other types of springs have different values.

Because of the radii needed to form the hook or loop, we also need to calculate the stress in both areas indicated below. Points A and B will be calculated with the formulas below.

To calculate the normal (bending) stress at point A, use the following formulas.

To calculate the shear stress at point B, use the following formula.

Once all these are know,n, the coil shear stress and the shear stress at point B will need to be summed using the square root of the sum of the squares. If you want to convert the shear stress to normal stress, use the maximum distortional energy theorum as a conversion. You will divide the shear stress by 0.577. (See the Mohr’s Circle Video for more information on this.)

The final step is to calculate the spring rate, R

Where G is the shear modulus of elasticity and δ is the deflection.  Many times you will be given the spring rate, so we can rearrange this equation to calculate the number of active coils in the spring.

8. Spring Selection Approaches

There are five basic methods for selecting a compression spring.  You will need to select one of the five that best fits your application. 

  1. Design based on physical dimensions
  2. Design based on spring rate
  3. Design based on two loads
  4. Design based on one load and spring rate
  5. Design based on one load and free length

Case 1: Design based on physical dimensions

This is probably the most common method of designing a spring.  When approaching the design you probably already have a certain envelope that the spring must fit into or around.  When selecting a spring by physical dimensions, you need to specify two of these three things:

  • Outside Diameter – OD
  • Inside Diameter – ID
  • Wire Diameter – d

Case 2: Design based on spring rate

When you know the spring rate you desire, you can start your selection there.  You will need other information such as free length, load capacity or some dimension of the spring to complete the selection.  Once you have selected this information you will probably be able to calculate the number of active coils and the free length using the following equations.

Case 3: Design based on two loads

For this design criteria, you will specify the spring rate based on two loads that are applied to the spring.  At each load, the spring will deflect differently.   Many spring manufacturers will have this option of choosing a spring on their website.  The variable, L, is the total height of the spring and not the deflection.

Once you have the rate, you are basically back in the previous design case (Case 2) and need to make other decisions to select your spring.

Case 4: Design based on one load and spring rate

This is really another subset of load Case 2.  The only difference is we will be able to calculate the free length without making any other selections.

Case 5: Design based on one load and free length

This is really just a subset of Case 3 where we assume that P2 is 0 lb and L2 is the free length.  Then use the equations in Case 2.

So let’s run an example

I need a spring that will sustain a load of 40 lb and a maximum load of 100 at 0.50 more deflection. We can start by using Case 3 above to determine our spring rate.

Since my maximum load is 100 lb, I want a spring that has 10% to 20% more capacity than needed.

Looking through the list, I immediately notice that there is only one option of spring. My design criteria is too tight! Nevertheless, we will run with the calculations to prove they work.

At this time, I will perform my hand calculations (left) and determine the results.  I have also displayed the additional information from the manufacturer (right).  Notice how my calculations match with the manufacturer. 

At this point, we can move on to the stress calculations for this spring.  Since we are using music wire, we will assume that for this diameter, the tensile strength is 250 ksi.

As you can see from the stress calculations, the intended load is only 53.6% of the tensile strength.  If you are curious as to where the 0.577 comes from, please watch the video on Mohr’s Circle.

As we calculate the stress aat points A and B, we can see that these are also in line with our design limit of 60% of the tensile strength. The stress at Point A is 5% higher and that is why extension springs tend to break in this location.

Stress at Point A

Stress at Point B

Some Verification

Just to prove that these calculations are correct, I ran an FEA of this spring. Let’s compare the difference.



It is pretty amazing how accurate these calculations were to the FEA results. Yes, you can trust this process.

Conclusion

Spring design seems complicated because there are so many different ways to approach it.  Just remember that there is no perfect solution for your application.  With all the off-the-shelf sizes, shapes and materials to choose from, you will select a very good spring for your application.  Just remember to watch for the red flags of improper spring index and always have plenty of alternative springs ready.

You got this!

Glossary of Terms


Good Advice for Using a Hydraulic Motor as a Pump

Hydraulic motors are wonderful hydraulic components. They can change fluid power into mechanical rotational power where direct driving cannot take place.

Yes, gear, gerotor, geroler and some vane hydraulic motors can be used as a pump. Bent axis or swashplate piston motors cannot. However, the designer needs consider the following:

  • Cavitation
  • Efficiency
  • Internal Motor Leakage
  • Accurate control of the motor

First of all, you bought a hydraulic motor so it should be used as a motor. Pumps and motors are designed differently so they are not completely interchangable. In fact, only gear pumps can be used as hydraulic motors. Piston and vane pumps will not work as a hydraulic motor at all. Motors are motors and pumps are pumps. If we compared this to an automobile, we would never design the starter to start the car and then remain engaged to act as an alternator.

So there are really two instances where you would want to use a hydraulic motor as a pump. The first is when you have a motor turning larger flywheel and then shut off the flow to the motor. The flywheel has an incredible amount of inertia and blocking flow to and from the motor would cause a pressure spike of infinite proportions and destroy your components. This is the same as flyback voltage in electronics where a circuit with an inductive component (motor or solenoid) is shut off. The energy stored creates very high voltages that can fry sensitive electronic components if not properly dealt with.

Another example of this is a hydrostatic drive system, like a lawnmower. Once the pump stops supplying flow, you will want the tractor to slow down to a stop, but not jerk to a stop. We want to turn the motor into pump so that we can gently brake the system to a stop.

To minimize the possibility of high pressure you will want to place a valve in the system to limit the pressure that can be produced. This pressure can be stored in an accumulator or other device (perhaps extending a cylinder under constant load) but most likely it will be converted to heat through a relief valve.

The second instance is more rare. It is when a hydraulic motor is used to perform one action and then another process takes over and changes the position of the motor. I recently designed a system where a cart was driven via a roller chain by three different processes. In the first process, cylinders would move the cart. The second process engaged a precision electric servo motor to position the cart for machining operations. Finally, a hydraulic motor was used to reset the cart to the original position so the entire process could be repeated. The hydraulic motor was constantly engaged the entire time even though it was only used for a small percentage of the time (<10%). As a result, the motor was being used as a pump for 90% of the time.

Cavitation

Cavitation generally occurs when a pump is suctioning oil from the reservoir, but the oil cannot come fast enough. As a result, the pump will try to vacuum the oil through the pump. Oil does not work this way and the pump instead will create large pockets of heat which locally boil the oil creating pockets of gas. As this moves through the pump, the increased pressure of the oil causes the oil gas to change state again back to a liquid by implosion. This process happens quickly and violently and usually damages the pump very quickly.

Cavitation is caused by:

  • Insufficient oil in reservoir
  • Clogged suction line filter
  • Clogged suction strainer
  • Clogged or non-existent breather
  • Too long of a suction line
  • Too small diameter of a suction line
  • Reservoir installed below the pump. (Pump is not naturally receiving oil)

Proper maintenance of the system will remedy most of the causes for cavitation. The preventative maintenance schedule should check for clogged filters and breathers and reservoir oil level.

The remaining items need to be addressed in the design phase. As hydraulic designers we need to consider the ‘head’ pressure at the pump inlet and we desire for it to be a positive number. Generally speaking, this is the vertical distance between the suction port of the pump and the oil level in the hydraulic tank. You want the hydraulic oil level to be higher than the pump.

You also must subtract the pressure it takes to move the oil through the suction line to the pump. A basic understanding of fluid mechanics tells us that shorter runs and larger diameter hoses will minimize this factor. Use our hose calculator to determine the hose size for the length you need to run. A well designed hose will have no more than 5 ft/s (1.5 m/s) fluid velocity and less than 1 psi (6.9 kPa) pressure drop. (After all a perfect vacuum is 14.7 psi (101.3 kPa)).

Even if you are only temporarily running your pump as a motor, you will need to address cavitation. Chances are that you chose a motor because you could not get a direct drive in the space necessary. It is likely that your hose run is relatively long and you will need larger hoses than desired to ward off cavitation.

Motor Efficiency

Because a motor is not designed as a pump, it will be very inefficient. Roughly 10% to 20% less than the manufacturer’s rated efficiency and that will vary greatly based on your specific pressure and flow. As a result, your oil will be creating more heat that needs to be rejected from the system. If you need to run your motor for an extended period of time as a pump, a high pressure heat exchanger may be a requirement. An alternate fluid path that draws fresh oil from the reservoir may also work.

Internal Motor Leakage

All hydraulic motors have internal leakage. This is any oil that passes from the high pressure section to the low pressure without producing motion. In addition to that, there are some cavities in the motor that can hold oil (for lubrication etc) that need to be maintained at low pressures. If high pressure oil gets into these cavities, the life of the motor can be shortened greatly.

If you intend to create large pressures with your motor being used as a pump, you will need to consult the manufacturers specifications on whether or not a case drain is needed at specific pressures and flows. Case drains prevent damage from internal leakage to the low pressure cavities The most common form of damage is a leaking shaft seal which will change your internal leakage to external leakage. This means there will be oil to clean up.

Motor Control

If accurate motor control is needed when in motor mode, you may have restrictions in the work lines to control the flow. These flow controls could be as simple as an orifice fitting or as complex as flow control valves in and out of the motor with flow controls and pressure compensators. However, a proportional control valve will accomplish flow control nicely by metering in and metering out the flow from the motor.

While the proportional control valve is the best for controlling the motor, it poses a problem when we switch to pump mode. You see, as we meter out the flow from the motor, the valve spool creates another restriction between the return work port and the tank. As the valve spool nears the center position, it will be closing this path off more and more. As a result, the center position of the valve can only be blocked or have a very small opening (orifice) called a motor spool.

If your schematic looks like the figure below, you will want to redesign. Any inertia on the motor when the valve is in the center position will create very large pressure spikes. The remainder of this article will unpack how to deal with this situation.

It is not possible to have a good valve for metering flow but yet an open center for use when it is a pump. It should be evident that something will have to give if we want to control the motor accurately and still allow for use as a pump.

There are two main ways to do this. If free flow of the motor is desired as in the case of the motor being driven by another apparatus, you can add a flow path between the work ports. This is usually done with a two position two way valve that is solenoid operated. When the motor is in motor mode, the valve is closed. When the motor needs to free flow to slow down, the valve will open. Your control system will need to handle the transition and may need to open before the directional control valve has closed.

Putting the two position, two way valve closer to the motor will minimize the hose size needed to prevent cavitation.

If you want to control the slow down, you can change the two position two way valve to be a proportional valve. This is called hydraulic braking. This way, the valve can be opened any amount depending on the pressure. With this setup, you can accurately control the speed of the motor at any given point.

Another way to do hydraulic braking is by adding cross port relief valves between the work ports so that the flow from A relieves to B and vice versa. This option is attractive because it costs less than the proportional valve and there is no control system involved. One downside is that the relief valves will eventually snap shut and the motor will come to an abrupt stop. The higher the pressure setting, the more abrupt the stop.

Conversely, the smoother the stop, the longer it takes to stop. Once again as engineers we are battling incompatible things.

Best Guide to Determining Deflection in Variable Cross Section Beams

Beam tables give information on and assume that the deflection calculation is based on a constant cross section.  So, what do we do if our beam has a cross section that changes over the length of the beam?

To determine the amount of deflection in a variable cross section beam, you must integrate the beam deflection formula with the moment of inertial being a variable with respect to the length and apply boundary conditions.  The beam deflection formula is v’’ = M(x)/[E*I(x)].

Continuous or Discrete – There are two types of beam sections, continuous and discrete.  Most beams are continuous beams and have either a constant section or a section that changes gradually over the length of the beam.  Roof beams in large steel buildings are a great example of a continuous variable beam.  The beam is relatively short in height on the ends and very tall in the middle.

Discrete beams are beams that have sudden discontinuities in the section.  Believe it or not, these are sometimes easier to calculate because the discrete sections are usually constant which leads to easier calculus.

The beam deflection formula is a universal formula that allows for the customization of multiple loadings and beam sections.  I will warn you that the more exact your calculation needs to be, the harder the math will be to do.  Simplification here will save a lot of time and effort.  As mentioned before the formula is:

v’’ = M(x)/[E*I(x)]

Where v’’ is the second derivative of deflection (the acceleration of the deflection), M is the moment which is usually a function of the position along the length of the beam, x.   E is the modulus of elasticity and I is the area moment of inertia of the beam.  All tabulated beams will consider this to be a constant and therefore none of the deflection formulas can be used.

Now when we integrate the equation above, we will be doing an indefinite integral which means that we have to add a constant, Cn, to the polynomial each time we integrate.  Since we will be integrating the equation two times, we will end up with two constants.  If we have a discrete case, we will have two or more equations. 

Boundary conditions are requirements that the beam deflection formula will need to abide by when it is in the final form.  The final form only comes when we use the boundary conditions to solve for the constants formed by the indefinite integral.  Common cases are the ends of a simply supported beam need to be 0 (in, mm etc.) or the slope of a cantilever beam needs to be 0 radians.

In this article, we are going to walk through three examples of common variable cross section beams.

  1. A two-section cantilever beam with point load on the end.
  2. A two section simply supported beam under its own weight.
  3. A constantly changing continuous simply supported beam with a constant distributed load.
https://mentoredengineer.com/the-best-guide-to-solving-statically-indeterminate-beams/

Example 1: A two-section cantilever beam with point load on the end.

This problem with consist of a 100 in. long cantilevered steel beam with a load of 500 lb. on the end.  The first 50 inches of the beam will have an area moment of inertia of 10 in^4 and the remaining beam will be 1 in^4.

Now we will determine the moment and integrate the beam deflection equation twice each time adding a variable for the indefinite integral.  I have selected to make my coordinate system (x variable) start from the base.  This makes the integration slightly harder, but the variables C1 and C2 will cancel out because of boundary conditions 1 and 2.  You’ll see in a second.

I only need to do the integration for one of the sections and then change I1 to I2 in the equations.  I have also kept the variable ‘v’ as the deflection of the beam, but changed the first derivative of deflection to the variable ‘s’, to indicate slope.  I also specified the variables.

Now that the problem is defined, let’s setup the boundary conditions.  We will want the position and slope at the fixed end of the beam to be 0 in and 0 radians.  We will also need two more boundary conditions at the joint between the segments.  The slope and position at this position will need to be the same.

Let’s solve for Boundary Conditions 1 and 2

As mentioned above, I foresaw that variables C1 and C2 would be equal to 0 when I chose to have the coordinate system start at the base. 

Next, we will look at boundary conditions 3 and 4.  These are slightly more complex.

Please note the check that I put in the Find block so that we could verify that the v1 = v2 and s1 = s2 at 50in.  This verifies that the position and slope will be continuous at this point. 

The next step is to verify the results.  This is done in two steps.  The first is to plot each segment over the entire length.  We looking for the four boundary conditions to be met.  As you can see, the lines intersect and are tangent at 50 in.  Also, v1 has no deflection or slope at the base. 

Finally, we will merge the two plots together forming the final equation for the deflection of our cantilevered beam.

As you can see, the deflection rapidly increases once past 50 inches from the base.  This is clearly indicated in both graphs.

The Best 4 Ways to Improve Torsional Beam Performance

Example 2: A two section simply-supported steel beam under its own weight.

This problem with consist of a 300 in. long simply-supported steel beam with a distributed load of 30 lb./in on the left end.  The right end has a distributed load of 50 lb./in.  The first 200 inches of the beam from the left will have an area moment of inertia of 10 in^4 and the remaining beam will be 1 in^4.

Now we will determine the moment and integrate the beam deflection equation twice each time adding a variable.  I have selected two coordinate systems.  The x coordinate goes from left to right and the y coordinate goes from right to left.  They are related by:

y = L-x

I have chosen this coordinate system so that C2 and C4 will cancel out when we solve for Boundary Conditions 1 and 2.  It also simplifies the math tremendously.  You’ll see in a second.

I only need to do the integration for one of the sections and then change I1 to I2 and w1 to w2 in the equations.  For the right-hand section equations, I will also substitute ‘y’ for ‘x’.  I have also kept the variable ‘v’ as the deflection of the beam, but changed the first derivative of deflection to the variable ‘s’, to indicate slope.  I also specified the variables.

Now that the problem is defined, let’s setup the boundary conditions.  We will want the ends of the beam to be 0 inches of deflection (BC 1 and 2).  We will also need two more boundary conditions at the joint between the segments.  The slope and position at this position will need to be the same where the segments join.

Let’s solve for Boundary Conditions 1 and 2

As mentioned above, I foresaw that variables C2 and C4 would be equal to 0 when I chose to have the coordinate system start at the base. 

Next, we will look at boundary conditions 3 and 4.  These are slightly more complex.

Please note the check that I put in the Find block so that we could verify that the v1 = v2 and s1 = s2 at 200in.  This verifies that the position and slope will be continuous at this point. 

The next step is to verify the results.  This is done in two steps.  The first is to plot each segment over the entire length.  We looking for the four boundary conditions to be met.

Uh-oh, what happened!?  The lines definitely intersect at 200 in and each end has 0 inches of deflection, but they are not tangent at the intersection.  Not only I am illustrating the power of graphing the solution for accuracy, but also demonstrating that using the two different coordinate systems posed a problem.  According to the equations, the slopes approach the location of the junction on a downward slope equal in magnitude.  However, to make this work one of the slopes actually needs to be coming up.  We can fix this issue by making one small change.

s1 = -s2

Let’s make this change and proceed with the solution.

Yes, much better!  Finally, we will merge the two plots together forming the final equation for the deflection of our cantilevered beam.

As expected, the longer stiffer section deflects less.

How to Calculate Beam Data When Your Case Isn’t in a Table

Example 3: A constantly changing, continuous, simply-supported beam with a constant distributed load.

This problem with consist of a 300 in. long simply-supported steel beam with a distributed load of 1000 lb./in across the beam.  The section starts off at a height of 10 inches increases linearly to the center where it reaches a height of 24 inches.  It then tapers back to 10 inches. 

To determine how the moment of inertia changes with respect to x, we will model in Solidworks and take sections every 30 inches.  We will tabulate this data and fit a line to it.

Now, you probably noticed that I only made the table for values of 0 in. to 150 in.  This is because I am going to use symmetry to simplify this complex problem.   We can use symmetry because both the load and beam section are symmetric from the midpoint of the beam.  Because of symmetry we will need to have the end point have a deflection of 0 in and the slope at the middle of the beam be 0 deg.  We can then mirror this to get the continuous deflection of the beam.  For this case, we will have the x coordinate go from left to right.

You can see here that the calculated values of I(x) closely match what is listed in the table above.  I have named the second derivative of position ‘a1’ (acceleration).  As you can see, with the top and bottom having the variable ‘x’, it will be super fun to integrate this.  So, there is one thing you need to know about me.  I have limits as to things I won’t do.  Integrating this is one of those things.  That’s why we have MathCAD!

As you can see, the very tedious work of integration was glossed over and we were able to directly solve for our boundary conditions.  In the equations of s(x) and v(x), there were actually natural logs and somehow an inverse tangent appeared (not shown).  I’m still not regretting letting MathCAD do the work.

The next step is to verify the results.  This is done in two steps.  The first is to plot each segment over the entire length.  We looking for our boundary conditions to be met.  As you can see, the deflection at x = 0 inches is 0 inches and the slope appears to be flat at x = 150 inches. 

Finally, we will mirror the plots together forming the final equation for the deflection of our cantilevered beam.

As you can see, the deflection is 0 inches at the end points and has the maximum deflection at the center.

The Best Guide to Solving Statically Indeterminate Beams

Conclusion

This article covers three popular load cases where a beam has variable cross sections.  While this does involve calculus, it is often very easy to do by hand because it is polynomials.  If not, be thankful for robust programs like MathCAD to perform this for you.  This article should give you a good handle on the procedure used to analyze beams like this.  If your beam isn’t loaded exactly like this, you can always find the moment calculation in a table and integrate your heart out.