Authored by:
Edited by Stephen J. Mraz
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Companies in the highly automated metal-fabricating industries are going through tough times, making efficient production and fast assembly lines crucial. And plants that can’t keep up will quickly disappear.
Metal fabrication often involves station-to-station handling of slippery, oily metal sheets or blanks. Suction cups are the most common tools used to handle and move such sheets. But as production speeds accelerate, suction cups must also become more capable of handling these speeds and loads. Fortunately, friction cups, suction cups with specialized surfaces, gives companies a viable way to handle today’s and tomorrow’s higher speeds and feeds.
Suction-cup challenges
Suction cups are easily adaptable to the changes going on in fabrication industries but they do present some challenges, however. One is to prevent cups from slipping out of position or even dropping parts due to oil or die lubrication used in forming operations. If cups slip out of position, the sheets will be placed incorrectly in the next station, necessitating a production stop to manually reset the sheets. Dropped parts can lead to scratched parts, line stoppages, and delays as technicians reset the part. Another challenge is keeping up with the technical developments and higher speeds of fabricating machines, a challenge that puts enormous demands on suction cups. (Ten years ago, press lines operated at 8 to 10 strokes/min. Today, that figure is 15 strokes/min. And it is estimated speeds will reach 20 to 25 strokes/min in the near future.)
Coping with higher speeds is critical in that it lets companies take full advantage of their machines. A synchronized cross-bar press handling large-body side apertures, for example, can be accelerated up to 30 m/sec^{2}, which is greater than 3 g. These same apertures must than be decelerated nearly as quickly to be placed in the next station. Emergency stops can put even higher stresses on suction cups.
One way to cope with higher speeds and quicker stops is to switch from suction cups to friction cups. They have high coefficients of friction and can exert more force. Most friction cups have their internal surface covered with patterns of knobs and oil channels. Many such cups have a good friction grip initially. However, wear on the cups from normal usage soon degrades the pattern and the cups have more or less the same performance as conventional cups. Users must frequently change cups to maintain performance.
Some friction cups, such as Duraflex cups from Piab, Hingham, Mass., carry a friction pattern that resembles those on winter tires. The cups and patterns are polyurethane, but the polyurethane used for the patterns has a harder durometer. They work best with vacuums from –40 kPa (–12-in. Hg) to –95 kPa (–28-in. Hg). And the friction performance will last the entire life of the cup. In fact, the initial wear slightly improves the friction grip.
Theoretical analysis
Suction cups will shift or move if the holding force or grip is less than the sum of all counteracting forces. In slow (up to 0.5-m/sec) handling applications, the force that determines the size, number, and performance of the suction cups is the weight of the object handled. In faster applications, especially those with rapid acceleration or decelerations, there are other forces that must be take into consideration when selecting suction cups, including the mass of the object being moved and accelerations and decelerations.
Suction cups generate a normal (vertical) lifting force (F_{n}), as well as a parallel friction force (F_{fr}). These frictional forces are related by their coefficient of friction ( μ). So F_{fr} = (F_{n}) × μ
Walking through an example
An automobile hood is being formed in a press line under the following conditions:
The 12-kg blank measures 1,000 × 1,500 × 1 mm and 2 gm^{2} of oil covers it. One suction cup measures 75- mm in diameter and exerts 150 N of vertical force with –60 kPa (or –18-in. Hg) of air pressure. This system uses four cups. Maximum speed and parallel acceleration are 12 m/sec and 20 m/sec^{2}, respectively, emergency stop decelerate parts at –30 m/sec^{2}, and the press operates at 12 strokes/min or 720 parts/hr.
Can the suction cups safely move this piece?
Step 1. Let’s start by just determining the number of suction cups needed. Typically, safety factors of 2 to 5 are used and, in this case, a minimum safety factor of 5 is needed. So the number of cups × F_{n} × 5 (safety factor) must exceed the force of gravity on the hood.
The blank is 12 kg, so force needed to support it is: 12 × 9,82 m/sec^{2} = 117,84 N. The combined lifting force of four cups is 600 N. Therefore the lifting force of all cups divided by the force needed to lift the hood or 600/117,84, gives us a safety factor of 5.09. But can the cups handle acceleration and braking forces?
Step 2. In this step we consider the maximum acceleration and braking forces that the combined cups’ friction force must withstand.
The maximum acceleration of the machine creates a force of 12 kg × 20 m/sec^{2} or 240 N. However, emergency stops generate 12 kg × 30 m/sec^{2}, or 360 N. To manage the worst case, an emergency stop at maximum speed, the cups must have a total friction force of 360 N. The friction coefficient of the cups is approximately 0.2 between rubber and oily metal.
The calculation becomes 360 N/(150 N × 0.2), so 12 cups are needed to handle emergency stops. And for a good safety factor, 16 cups would be recommended.
The four cups calculated in step 1 were far too few. The sheets would have slid even under acceleration. Instead of going to 16 cups, which costs a lot in rebuilding the end-of-arm tool, technicians could slow the press line.
The maximum speed, 12 m/sec, generates emergency-stop braking of –30 m/sec^{2}. To brake, the hood will create 12 kg × 360 N in braking force. Assuming we can add four more cups to the existing end-of-arm tool will give us eight conventional cups with a total F_{fr} of 150 × 0.2 × 8 = 240 N.
The emergency stop acceleration (A) can be determined by:
A = (V_{f}– V_{i})/t
where V_{f}= final speed, V_{i} = initial speed, and t = time. A = –30 m/sec^{2}, V_{f}= 0 (full braking), V_{i} = 12 m/sec, and t = 12/30 or 0.4 sec, the time needed at maximum speed to come to a complete stop. How fast can we run the machine if the eight cups can handle 240 N? This is actually determined by seeing how powerful a braking force it can withstand, which is calculated using:
F_{fr/m} + acceleration, or 240 N/12 kg + 20 m/sec^{2}.
If we assume the time (t) to completely stop braking is 0.4 sec, the maximum allowed speed (Vi) will be: 0.4 × 20 = 8 m/sec.
So the machine can run at a maximum speed of 8 m/sec, not the specified 12 m/sec, if it is supposed to manage a complete emergency stop without sliding cups. It would take 16 cups and rebuilding the end-of-arm tool to do that.
A 33% slower speed while the cups are used would reduce total cycle time by about 20 to –25%. So the machine capacity, 12 strokes/min, will not be reached. Instead, we will get only nine to 10 strokes/min. Production would go from a desired 720 to 600 parts/hr. A press line usually runs 20 to 24 hr/day. That means we make approximately 2,400 fewer parts/day.
Step 3. After discovering friction suction cups for oily sheets and parts, we decide to see if they will improve the situation. A friction cup can increase the coefficient of friction (μ_{rest}) from approximately 0.2 to nearly 0.5, or almost the same as on a dry metal sheet. Using the same calculations (with μ_{rest} = 0.45) as in step 2, the results show only six to eight friction cups are needed, including a safety margin to manage emergency stops.
The end-of-arm tooling can remain the same (no retooling or building), production can run at the maximum speed of the machine, 12 strokes/min, and emergency stops will not make the blanks slide.
In this simplified example we have just looked at a parallel motion with acceleration and braking. Vertical or angled motion with acceleration/braking will add even more to the needed friction force of the cups due to gravity. We have also assumed a decentralized vacuum system. How the vacuum is generated can also affect speed. A decentralized system is faster in most cases.