Answer to Quiz No. 1When the chock is removed will the Spool roll down the ramp?

Answer:

No, the spool will not roll as shown, because the radius from it's centerline to the outer layer of cable is less than the outside radius of the flanges. Imagine the spool rolling through 90 degrees from its present position. If 90 degree arcs for the cable and flanges are laid out parallel to the ramp, it can be seen that the resulting centerline positions of the spool at the end of each arc are not in the same place along the ramp.

If the spool was full of cable, then the radii would be the same and the spool would start to roll when the chock was removed. As the radius of the cable decreased due to cable being payed out, the spool would come to a stop on the ramp. This is assuming that the safe working load of the cable and the back stop were strong enough to resist the resulting momentum of the spool.

Note that neither the weight of the spool and cable nor the angle of the ramp have anything to do with whether the spool will roll or not.

To prove to your self that the spool will not roll, place a reel of cable, a spool of electrical wire, or a yo-yo on a flat surface and start to slowly unwind it. Note that as the spool is unwinding, the end of the cable/string is slowly following along behind the spool. If someone stands on the end of the cable, the flanges on the spool have to be slipped in order to continue unwinding the cable.

Answer to Quiz No. 2

The tensioning system shown is one of three being used to plumb and support a flare stack. The other two tensioning systems are not shown for clarity.

How many parts of hoist line are there in the tensioning system shown below?

Two or Three?

Extra credit: If T = 3,000 lb,

what is: R1______________

what is: R2______________

Answer:

There are three (3) parts of hoist line for the tensioning system shown.R1 = 2,000 lbsR2 = 1,000 lbs.

In order to solve for the number of parts of hoist line in a pulley system, the traveling load block must first be identified. Then an imaginary cut is made just above the traveling block. The "cut" hoist lines that are supporting the traveling load block are counted. The number of "cut" hoist lines is the number of parts of line.

In this case the traveling block is the top block, because as the winch is tightened up, the top block moves or travels toward the bottom or stationary block. If the winch is let out, the top block moves away from the stationary block. The traveling block in this example is similar to the load block on a crane, except that the imaginary cut here is made below the traveling block instead of above it.

The stationary block here is similar to the point sheave on a crane.

Answer to Quiz No. 3

ERECTION PROCEDURE:

1. Hook up cranes to tower boom at position 1.2. Hoist the tower boom to position 2.3. Pin tower boom to mast.4. Release load on Crane B.5. Hook up boom pendants.6. Tighten up on the boom suspension.7. Unhook Crane A.8. End of procedure.

From a crane capacity stand point, is this a safe lift? YES or NO

No, this is not a safe lift.

At POSITION 1, crane A has a 13.33 kip load and crane B has a 26.67 kip load. A kip is 1,000 lbs. 100% of capacity chart for crane A is 13.33/.9 = 14.81 kips and crane B is 33.34 kips. At POSITION 2, after the tower is pinned to the mast and the load to crane B is released, the load to crane A increases to 30.00 kips. This is an increase in load of 125%, well out side of the capacity chart and into the accident range.

At POSITION 2, if the load to crane A is released first, the load to crane B increases to 48.00 kips. This is an increase in load of 80%, well out side of the capacity chart and again into the accident range.

Many times, the transfer of a load in the air from one crane to another is a subtle thing and usually happens when using two lift cranes. All lifts should be carefully looked at from a load transfer point of view.

This rigging quiz is the result of a lesson learned experience where the load to crane B was release first and crane A started to turn over. To save the crane, the operator dropped his end of the tower boom and it was severely damaged when the tip hit the ground.

Answer to Quiz No. 4

Given the sketch below:

For each sling, find the tension in pounds.

T1=____________ T2=____________

The tension in the left sling T1 = 35,619 lbs.The tension in the right sling T2 = 79,919 lbs.

The main purpose of the monthly quizzes is not to test the readers skill in algebra or trigonometry, although a good knowledge of both is required in today's construction world, but to see if the reader recognizes one or more concepts that are fundamental in solving the problem. The reader can then use various ways to solve it, including mathematics, laying it out to scale, etc.

In the number 4 quiz, one way to solve it would be to solve for all the angles at the hook and then sum the forces vertically for T1 & T2.

A second and recommended way, because the sketch is easy to lay out to scale, is to choose a horizontal reference line (HRL) that runs through the center of gravity (CG) and the attachment point of the right sling to the load. Extend the left sling down until it intersects the HRL. Scale off or solve the horizontal distances from the CG to the intersections of the HRL and the right sling and the extended line of the left sling. These distances are approximately 3.25' and 10.07' respectively.

Now solve for the vertical reactions at the intersection points. The vertical reaction at the intersection of the left sling R1 = 100,000 lbs*3.25'/ (10.07' + 3.25') = 24,400 lbs. The vertical reaction at the intersection of the right sling R2 = 100,000* 10.07'/13.32 = 100,000 - 24,400 = 75,600 lbs.

Solve for the vertical distance h from the CG to bearing on the hook mathematically or by scaling it off the sketch, h = 9.46'.. Solve for the total length of the left sling plus the extended length down to the point of intersection with the HRL = 13.81'. Therefore T1 = R1*13.81'/h = 35,619 lbs. T2 = R2*10'/h = 79,916 lbs.

If the sketch is laid out carefully and to as large a scale as possible, the accuracy of the results should be acceptable for this kind of problem where the safety factor of the slings is at a 5:1.

Lastly, it should be remembered that a HRL can be chosen at any vertical point on the sketch and the one chosen in this example was for convenience only.

Answer to Quiz No. 5

SETTING A STACK SECTION

Traditionally, stack sections up to 30 tons are set using a method where the hoist line runs over the jib point sheave, then down and around the sheaves on the spreader bar and finally dead ended up on the boom tip.

Is it possible to set a 60 ton stack section rigged as shown if there is adequate crane capacity, jib capacity, crane stability and head room?

The answer is YES and NO..

YES, because the stack section can be lifted safely as shown. And NO because the problem is that when it lifts off the ground, it will hang at approximately a 75 degree angle and would be hard to set on top of an existing stack. Also the lugs would have to be designed to take the resulting bending..

If the stack section lifted off the ground plumb, the vertical reaction at each lift lug would be 30 ton and the

jib would be over loaded..

But due to the equalization of the sheaves, the tension in each part will be the same and as the section is being lifted, the bottom of the stack on the inside will raise first and the CG will start to move toward a equalized loading position between the hoist lines on the boom and the hoist line on the jib..

Before the bottom of the stack on the out side clears the ground, the CG will reach the equalized loading position where it is approximately 2.5' from the vertical extension of the boom's hoist lines and 7.5' from the jib's line. This position of the CG will produce a vertical reaction at the left lug of 45 tons and a vertical reaction at the right lug of 15 tons and the system is then in equilibrium with 15 tons per part..

Note that until the CG is in the equalized position, the support will carry any excess load on the outer lug over 15 tons. When the CG is in the equalized position, then the stack will lift off the support without overloading the jib, but will be hanging at approximately a 75 degree angle with the bottom tilted toward the crane..

If the crane could physically walk forward until the low edge of the top stack section could be lowered down on the far side of the existing stack, dogged off to prevent horizontal movement toward the crane, stabilized with chain hoists against horizontal rotation, and then slowly lowered as the crane was walked backwards until the top stack section was resting solidly in a plumb position, the above rigging hook up would work. But, as they say on TV, don't do this at home as it is a very complicated procedure, one for a Rigging Engineer to design and execute.

Answer to Quiz No. 6If there is adequate crane capacity, crane stability, boom clearance & rigging capacity, is this a safe lift? YES or NO

The answer is no, this is not a safe lift. It is unstable against tipping in the transverse direction and borderline stable against tipping in the longitudinal direction.

This is because the angle Alpha below the center of gravity (CG) in the transverse direction is 45 degrees, and the angle Beta at the spreader bar is approximately 24 degrees. In order to be stable, Beta must be greater than Alpha.

For the longitudinal direction, angle Alpha is 32.75 degrees and angle Beta at the spreader bar is approximately 35 degrees. As Beta is greater than Alpha, theoretically, it is stable.

To better understand why the angle Beta at the spreader bar (or slings if they are being used above the spreader bar) must be larger than the corresponding angle Alpha at the CG in both the transverse and longitudinal direction, consider the following. Assume that the pick point "B1" in the transverse direction is actually located at "C" and centered between the shackle points. If the CG was located directly under the

hook and could be kept there, then the load would be stable. But the location of the CG for a load is hard to calculate and is usually off by at least several inches to a foot. During lifting, dynamics of the lift, wind, different lengths of rigging, etc., tend to also shift the location of the CG. Therefore, if the CG is not under the hook as the load is lifted, the offset CG will cause the load to rotate away from the hook, and in so doing, the spreader bar and the skid platform will form a parallelogram as they rotate from the horizontal. As there is no resisting force against overturning or tipping, the spreader bar and skid platform will continue rotating into a more acute parallelogram until the slings bear up against the sides of the turbine. If the trunnions were located outside of the turbine frame, then the skid would turn up side down. 

Consider one more situation where the skid is being lifted in the transverse direction without spreader bars, but with slings attached at the trunnions on the "D" line and connected to the pick points or hooks at a point below the CG. Also assume that there is no interference between the inclined slings and the turbine housing. As in the example above, there is no resisting force against overturning as the CG is located above the pick points or hooks & the skid is therefore unstable. Now, if the slings are lengthened until the pick points are above the CG, they provide a resisting force against overturning and the load will be stable.

So whether the slings are connected to the tunnnions and run directly to the hook, or they start at the spreader bar and go to the pick points or hooks, they must form an angle Beta that is greater than the angle Alpha at the CG. If the CG is not centered between the skid pick points, then both Alpha angles in the transverse and both Alpha angles in the longitudinal must be computed and compared with the corresponding Beta angles.

Now the most often asked question is how much greater must angle Beta be than angle Alpha. Theoretically and on paper, if Beta is as great as say one degree larger than Alpha, the lift should be stable. But due to the dynamics of lifting, the CG location not always being know accurately, the slings not always matched for length, etc, it is recommended that Beta be at least 20 degrees larger than Alpha. In most cases this will be conservative but safe.     

Answer to Quiz No. 7

The answer to question "A" is that case I & case II both have the same maximum tire loading.

The loading to the tractor's rear tandem axles is 13.33 kips and the loading to the trailer's tridem axles is 26.67 kips, and it is the same loading for both case I & case II. The important thing to remember here is that once a load (and the center of gravity (C.G.)) is positioned on the trailer, the loading to the tires will remain constant, no matter how the load is supported on the trailer.

The answer to question "B" is that case II has the worst trailer loading.

In case I, the front support has a load of 18.18 kips and the rear support has a load of 21.82 kips. In case II, the front support has a load of 28.57 kips and the rear support has a load of 11.43 kips. The load to the front support for case II causes maximum bending in the trailer frame. Note that the load to the rear support and trailer frame for case II is 11.43 kips, but the load to the trailer's tridem axles due to the C.G. placement is 26.67 kips. Therefore, there are two things to consider when placing a load on a trailer. The first is placing the load to achieve optimum loading to all axles, especially if it has to be a highway legal load. The second is placing the supports under the load to minimize bending in the trailer frame.   

Answer to Quiz No. 8

 

Yes, the 1000 Te reactor can be safely jacked up using eight 150 Te hydraulic jacks. In order to be able to jack up the reactor, the eight jacks must be plumbed into hydraulic groups such that no individual jack can be over loaded. The hydraulic pump is equipped with a manifold that has 8 outlets so one way to plumb the jacks would be to run individual hoses from the manifold to each of the eight jacks. The only problem with an eight point support is that without a computerized console, it is almost impossible to keep from overloading any one jack during the jacking process. As most jobs would not have a computerized pumping console available, this method will not be considered as feasible in this quiz.

Another way would be to plumb the jacks into a four point hydraulic support system with two jacks making up each hydraulic group. The problem with this method is that unless the jacks are supported on a very firm footing (say on a deep concrete foundation or on large load spreaders), settlement of one or two jacks in a hydraulic group will cause other jack groups to become overloaded, due to the table leg effect where one leg shorter than the other three will overload the two diagonal legs. Also the table leg effect can occur when jacking a load up (even without differential settlement between jacks) if the pressure in each hydraulic group is not monitored very closely. The bottom figure on page three shows a graphic schematic for a four point support system. Note that the hydraulic points (centroids) for hydraulic groups (circuits) 1 & 2 are located 2 + 3 = 5 units left of the center of gravity (CG) and the hydraulic point for circuits 3 & 4 are located 4 + 3 = 7 units right of the CG. Therefore, the individual load per jack for saddles 1 & 2 is 1000 Te * 7 units / (12 units * 4 jacks) = 145.83 Te. The individual load per jack for saddles 3 & 4 is 1000 Te * 5 units / (12 units * 4 jacks) = 104.17 Te. 

The method most often used to jack up a heavy load is the three point hydraulic support system. This method gives the user the ability to raise or lower either end of a load such as a reactor or to roll it from side to side if desired, providing that in doing so none of the jack rams bottom out while they are extending or retracting. This means that differential settlement of the jacks is not a problem with this method. Refer to the top figure in the sketch below which shows a schematic for a three point hydraulic support system. Note that jacks L1+ L2 make up circuit 1, jacks R1 + R2 make up circuit 2 and jacks L3 +L4 + R3 + R4 make up circuit 3. The centroids or hydraulic points of circuits 1 & 2 are both located 5 units left of the CG and the centroid of circuit 3 is located 7 units right of the CG. This makes the reactor very stable during jacking because the CG is located closer to the two hydraulic points in circuits 1 & 2 that control the roll than it is to the third hydraulic point that controls elevation. This eliminates the wheel borrow effect where a load that is located over the front tire rather than back closer to the two handle bars causes the wheel borrow is be very unstable and wants to tip over. Based on the above analogy, the user should always try to keep the CG closer to the base of the stability triangle than to it's top. The stability triangle is formed by connecting the hydraulic points of the three hydraulic groups with lines and the CG must always stay inside of these lines for the load to be stable. The three point hydraulic support system is similar to the four point system except that circuits 3 & 4 have been connected together to form circuit 3. Therefore, the individual jack loads for each saddle are the same, namely 145.83 Te for saddles 1 & 2, and 104.17 Te for saddles 3 & 4. 

Note that another way to make a three point hydraulic support system for this example is to plumb jacks L1 + L2 + L3 into circuit 1, plumb jacks R1 + R2 + R3 into circuit 2, and plumb jacks L4 + R4 into circuit 3. This will lower the individual jack loads to 138.89 Te for the jacks at saddles 1, 2 & 3, and 83.33 Te for the jacks at saddle 4. The down side is that the CG of the reactor will then be located 2 units from the base of the stability triangle. Depending on the jacking requirements, the CG being that close to the base of the stability triangle may not provide enough stability for tipping in the longitudinal direction of the reactor. It is left to the reader to draw the schematic for the above three point support system and to verify the individual jack loads.

In general, it is a good rigging practice to design jack loads for a maximum of 50 to 60% of the jack's capacity. The high jack loads in this quiz were only used to make the quiz more interesting.     

Answer to Quiz No. 9

The answers are:

1a. The left wing tip navigation light is RED

1b. The right wing tip navigation light is GREEN 

2a. The left side of a ship is called the PORT side

2b. The right side of a ship is called the STARBOARD side

3. The common factor is the number of letters in each word, i.e., notice that each word in the top line below has fewer letters than the corresponding value on the bottom line.

LEFT     RED     PORTRIGHT GREEN STARBOARD

You are probably wondering what airplanes and ships have to do with rigging.Well, airplane navigation lights don't have anything to do with rigging and the information is included as interesting trivia, but it might come in handy the next time you are looking up in the night sky at a commercial jet liner and you want to know which way it is heading (this works only if you can see the lights). In rigging, we deal with the ocean shipment of heavy plant equipment quite frequently, so it shows some maritime knowledge to be able to properly name the sides of a ship when discussing a stowage plan, etc. 

COMMENTS ON THE QUIZ:It has been brought to my attention by some "old salts" that:1. There is a red navigation light on the left or port side of a ship and a green light on the right or starboard side, similar to an airplane. When two ships "A" and "B" meet at sea in any orientation other than on a parallel course, ship "A" to the right has the right-of-way. As the ships meet, ship "B" to the left sees a red (stop) light, while ship "A" to the right sees a green (go) light. 

2. When exiting port, the buoy lights on the left of the channel are red and the ones on the right are green. This is true for US waters only. The rest of the world have the red buoys on the left when returning to port.

Answer to Quiz No. 10

FIND:1. If soil stability is not a problem and the crane has been set up properly (blocked firm and level), what point on the crane do you watch in order to tell when tipping is starting to occur:

FIG. 1 A___ B___ C___ D___

FIG. 2 A___ B___ C___

2. When tipping starts to occur, what percentage of the crane's lifting capacity chart is the crane working at?

FIG. 1 _____________%

FIG. 2 _____________%    

The answer to question 1, figure 1 is "C" and question 1, figure 2 is "C". When the truck crane starts to tip, the outrigger housing at point "C" will lift off the heavy outrigger(s) and float(s) and you can see daylight or space between them. When the crawler starts to tip, the track rollers at point "C" will lift off the heavy track pads and you can see daylight or space between them.

The answer to question 2, figure 1 is approximately 117 % of chart and question 2, figure 2 is approximately 133 %.The answer to figure 1 is found by solving the ratio of 100 % of chart * 100 % of tipping/ 85 % of tipping = 117.65 % of chart. Figure 2 is found the same way except 75 % of tipping is used in the ratio instead of 85 %. The answers are approximate as the exact % of chart required to actually tip the crane over depends on the dynamics of the lift, wind, soil stability, etc.

The main purpose of this quiz is to show what percent of chart the cranes are actually working at if tipping is allowed to happen and the danger in doing so. Cranes should never be worked at capacities greater than 100 % of chart. At 100 % of the capacity chart, the cranes should still be down hard on the outriggers or track rollers with no signs of tipping. A lot of operators and riggers continue to load cranes until tipping occurs, rationalizing that they have (in the case of a crawler) "25 % in their pocket". In loading a crane to tipping, the operator is eating into the safety factor provided by the Manufacture and in some cases, working as few as 1 to 2 % outside the chart will change the rating from tipping to structural. Therefore, working a crane outside the lifting capacity chart is not only a bad rigging practice, it is a very dangerous practice. The crane manufactures, OSHA, ASME B30.5 forbids anyone working a crane outside of the capacity chart, to do so is an illegal practice.

Follow the link below to an article by Ron Kohner that explains the ramifications and dangers of working outside of the lifting capacity chart from a Crane Manufactures view point.

The secondary purpose of this quiz is to show the location on a crane where the signs of tipping first occur. Even though the lift must be planned to be within the lifting capacity chart, it is a good rigging practice to have someone watching the points of tipping on the counterweight side as well as watching the soil and load spreaders for settlement under the outrigger pads or crawler tracks on the load side. The lift should be stopped immediately if either signs of tipping or settlement are observed. Constructing a firm and level lift pad is one of the most important parts of a lift, but settlement can occur from unknowns such as poor soil compaction, underground water, underground voids, crushing of underground pipes, etc.

     Ron Kohner's Article

Operating in the Twilight Zone of Crane Capacitiesby Ron Kohner

The rating chart for a mobile crane spells out operating parameters which are critical to the machine's safe use. While some charts can be very complex, most crane users understand that a chart divides rated capacities into two general categories: (l) those limited by tipping (stability) and (2) those limited by structural factors. This information is clearly indicated in some fashion on all rating charts. Beyond this point, however, a frightening number of users mistakenly read dangerous misinformation into the chart.

How many times have you heard a conversation like this? "We had a heavy piece to lift. The weight was over the chart, but we were in a tipping range, so I knew she'd get light before anything would break." 

Sound familiar? -- Believe it's true? -- It absolutely is not!

By relying on tipping to signal a problem, an operator can overload a crane's structural components by a tremendous amount. In over 25 years in the crane and rigging industry, I have heard the above "logic" too many times to believe it constitutes an isolated misunderstanding. Rather, it would seem that many users are guilty of trying to second guess the crane manufacturer. 

US-OSHA (Occupational Safety and Health Administration) requires that, "When ratings are limited by structural competence, such ratings shall be clearly shown and emphasized on the rating charts." There can be no doubt that the intent here is to alert crane operators to the structural limitations for their cranes. OSHA clearly recognizes the importance of structural limits. Their further requirements also state, "When loads which are limited by structural competence rather than by stability are to be handled, it shall be ascertained that the weight of the load has been determined within 10 percent before it is lifted." 

That's clear enough. Who cannot imagine the potential disaster of a structural failure? Most operators agree -- you don't cheat on structural limitations. However, the rating chart does not tell us that it is safe for the crane to approach tipping without suffering structural consequences. It only indicates that no structural limitations are exceeded by the stability ratings shown on the chart. These are not ratings which will tip the crane! They are only a certain percentage of the load at which the crane will tip. If we load the crane beyond the manufacturer's ratings -- even in a stability-limited area of the rating chart -- we have no idea which structural components may be overloaded before the crane actually shows signs of tipping. 

A chart may list a rating of 5,000 pounds limited by stability. In fact, the strength of the boom may only be 5,100 pounds. Under current federal stability regulations for a crawler crane, it would take 6,700 pounds to actually tip this crane. If the operator lifts "until she gets light," the boom is being overloaded by 1,600 pounds or 32 percent. There is a "twilight zone" between the stability ratings shown on a chart and the load it actually takes to tip a crane. A rating chart does not tell us what structural limits are violated when the crane is taken beyond the chart and approaches tipping. 

A simple story helps to illustrate this problem. Imagine you and I have just built a crane. We have scratched our heads bald and gone cross-eyed, working at a computer screen to determine just how strong our crane is and what it can lift. It's a crawler-mounted crane, so OSHA regulations tell us that we can only rate the crane at "75 percent of the load, which produces tipping." We do some calculations to develop a rating chart. We also do some testing to verify that chart. Sure enough, the computer was right. Under carefully controlled test conditions, our crane begins to tip at a 20-foot radius with a 100,000 pound load. 

Some simple math quickly tells us that if the crane tips with a 100,000 pound load, our rating for the crane (remember, it's a crawler) can be no more than 75 percent x 100,000 pounds or 75,000 pounds -- if tipping is the limiting factor. We dig through piles of computer printouts that predict stresses in our crane, and find that only two components are near their strength limits for this particular boom length and radius. Our boom can handle 76,000 pounds at a 20-foot radius. Rope strength limits the crane's capacity to 85,000 pounds. Our tipping rating of 75,000 pounds is, therefore, the lowest of all the factors. We publish our rating chart showing 75,000 pounds at a 20-foot radius, and we don't asterisk the rating because it is stability limited -- not structural.

We sell our first crane. A few weeks later, we are at a jobsite proudly watching our brainchild at work. The rigger hooks onto a large bridge girder to off-load it from a truck. Painted on the side of the girder is the

weight: 97,000 pounds. We pull out our trusty tape measure and check the radius: 20 feet. We think back to our rating chart -- 75,000 pounds at 20 feet. Certainly, the operator will try to get closer before lifting. But no -- the engine revs and the crane begins to take a strain. We run screaming toward the operator. He's going to collapse our new crane! 

Looking more than a little disgusted, the operator calmly explains to us that, yes, he knows the load is beyond the rating chart. He smugly points out that he's working in a tipping range on the chart, so logically, it is safe to overload the crane "until she gets light." He goes on to assure us that it should tip before anything will break. How wrong can he be!

We begin to sputter, "I designed this crane. The boom is only good for 76,000 pounds and the rope for 85,000 pounds." Both will be grossly overloaded if he tries to pick the 97,000 pound girder at 20 feet, relying only on the feeling of tipping to stop him. 

The key here, is that the rating he sees on the chart is not the tipping capacity. It is only a certain percentage of the tipping load (in this case 75 percent) as the law requires. There is a "gray" area between the rated capacity listed on the chart and the load that will actually tip the crane. Many structural limits exist in this "gray" area. If he relies only on tipping to warn him of a problem, he can be asking for big trouble. He will substantially overload structural components of the crane. When you are relying on "the bubble in your butt" to warn you of tipping, only the Good Lord and the crane designer know what is being damaged. 

Some will argue that there is a safety margin built into the boom and other structural components which should protect us during overloading. Yes, there is a margin. However, that margin does not belong to the user. It is there to protect against unforeseen problems beyond the user's control -- material variations, wind, soft ground, etc. If that margin is deliberately usurped at the start, nothing remains to prevent catastrophe when Mr. Murphy and his Law visit our work site.

A crane will certainly not collapse under every overload, but make no mistake, the overloading does damage. Project managers consider the operator a hero when he makes a lift beyond the chart. Six months later, when the boom on that crane collapses with a light load, no one can figure out why. Think about it. Did the manager and operator help cause that accident by pushing the crane near tipping in the past? Each overload should be viewed as cutting one strand of a rope. The damage is cumulative, and it is only a matter of time until the rope breaks. 

OSHA mandates, "No crane shall be loaded beyond its rated load." That's pretty hard to misinterpret, but too many users think they know more, and are tempted to bend the rules. The next time you are tempted to do it, or even to ask someone else to do it, remember our little story. When you appear to be in a stability-limited range of a rating chart, a structural limitation may be only a few pounds away -- and you don't know it. Stick to the charts.

 

Ron Kohner is the president of Landmark Engineering Services, Ltd. in Roseville, Minnesota. He has worked as a crane and heavy lifting engineer since 1970. 

Maximum Reach Quiz No. 11

FIND:As the reactor is up ended, does the initial tail load gradually decrease to zero?

Case 1 - Yes______ No_____ 

Case 2 - Yes______ No_____ 

Case 3 - Yes______ No_____ 

Answer to Quiz No. 11Case 1 is YES

Refer to page two for reactor R-1 and note that the Tail Load at initial pick in the horizontal is 240 kips . This value is found by the following formula: 60' * 200 ton/ 100' = 120 ton = 240 kips. As the reactor is up ended, the Tail Load gradually decreases in a predictable manner down to 0.0 kips until the lift crane has the full load. As a general rule of thumb for this case, the Tail Load at 85 degrees is still 50 to 70 % of the original Tail Load at initial pick, depending on the location of the center of gravity and the off set of the tail lug from the reactor centerline. Final Up ending between 85 and 90 degrees should be done very slowly so the full load of the reactor is gradually transferred to the lift

crane.

Case 2 is NO

Refer to page three, reactor R-2. The reactor in case II acts like an equalizer beam as the lift point, the tail point and the center of gravity are all on the same line. The result is that the Lift Load and the Tail Load for initial pick remain constant through reactor up ending to the point just before it reaches vertical, a tenth or two degrees less than 90 degrees. This is so because the horizontal distances between the center of gravity and the lift and tail points remain in proportion to each other through the up ending. In that last tenth or two of a degree, the full Tail Load is transferred to the lift crane. Therefore, final up ending with Case II should also be done very slowly and smoothly so

the full load of the reactor is gradually transferred to the lift crane.

Case 3 is NO

Refer to page four, reactor R-3. During up ending, the Tail Load will gradually increase in a predictable manner until the center of gravity is centered over the tail lug. At this point, the tail device will have the full weight of the reactor. At this point in the up ending, a horizontal force must be applied at the top of the reactor to keep it from rotating out of control over the tail point as up ending continues. Most heavy reactors are tailed by a tailing device like the one shown in the quiz, but are designed so that as the reactor is up ended to about 45 degrees, the tail point is shifted to a

location that is above the centerline of the vessel and then the up ending becomes like Case I.

Maximum Reach Quiz No. 12

Answer to Quiz No. 12

Most loads that are lifted are handled by pulley systems of some kind. For example, block and tackle or the reeving of the point sheaves and load block on a lift crane. This quiz is given to help the rigger get a better feel for "parts of line" and "mechanical advantage". 

Parts of line are the number of lines supporting the load block and the load. For example, make an imaginary horizontal cut just above the load block in system 2 and count only the wires below the cut. There are two "parts" of wire in this example.

Mechanical advantage is a ratio of the output force to the input force. In the above quiz, it is assumed that the systems are frictionless and in equilibrium. Therefore, the M.A. is a ratio of the load and a force required to keep the system in equilibrium and = W/P.

Some comments on the quiz:

1. The left pulley in system 2 is just a diverter sheave for changing the direction of the wire rope and contributes nothing to the mechanical advantage. The down side is that the use of diverter sheaves does increase the lead line pull required to raise or move the load.

2. If the load "W" in system 2 must be raised, a force greater than "P" would be required due to the additional force required to bend the wire rope around both the sheave hooked to the support at the top, the sheave in the traveling block, the diverter sheave and the additional force required to overcome friction between all sheaves and their axles.

3. In a single pulley system ( one comprised of one wire rope) the M.A. and the number of parts of line are always the same. Therefore, P = W/No. of Parts

4. System 5 shows are double pulley system. Multiply the number of parts in each system to get the M.A. Note that there are two traveling blocks in this system. Therefore, the M.A. of system 5 = 2*3 = 6.

5. The force "F" shown in the quiz is the force required to support only those pulleys and their loads that are hooked to the support structure.

Note that if the load "W" in system 1 was a deer about to be strung up and dressed out, that a tree with a branch strong enough to support 200 lbs would have to be located. Also note that if a pulley arrangement similar to that shown for systems 2, 3, 4 or 5 were used that only a branch strong enough to support 150, 150, 125 and 116.8 lbs respectively would be required.     

Maximum Reach Quiz No. 13Given: Four rigging hook up's as shown below:Find: A. Will the load lift level? B. Which sling will have the most load?

Answer to Quiz No. 13

The main points to remember concerning the center of gravity (CG) are: 

1. The CG always hangs under the hook when the load is suspended. 

2. Equal angles between the slings and the horizontal means equal tensions. 

3. Draw the load to scale and size the slings so that the hook is forced over the CG at initial pick. This will ensure that the load will pick level. Compare rigging hook up no. 2 to no. 4. 

4. When figuring the vertical reaction for use in determining sling tensions, use ANY HORIZONTAL REFERENCE LINE. See rigging hook up no. 1 

Maximum Reach Quiz No. 14During a critical lift over operating equipment, an attempt is made to lift a load at a site with poor soil conditions. At initial pick, the crane operator, noticing that both of his tracks are starting to sink, lowers the load back to the ground and calls for crane mats to be brought out before he will proceed.

The rigging superintendent, hurrying to meet schedule, brings out one-inch thick trench plates and tells the operator to hurry up and walk his crane onto the plates. The operator refuses and insists on timber mats. The superintendent says he has enough plates to stack four-high under each track and "that's got to be better than one layer of timber mats!" The operator still refuses.

1. Is the operator being unreasonable?

2. Is the rigging superintendent correct in saying that (4) 1" thick steel plates are better load spreaders than (1) layer of 12" thick timber mats? 

Answer to Quiz No. 14

The answer to question 1: The operator was not being unreasonable.The answer to question 2: The rigging superintendent was completely wrong.

All lifts and especially critical lifts should have a firm base for a crane to work safely. One of the best ways to provide that is to use timber mats that are in good condition and free of rot or pests. One-inch thick steel plates, even stacked four-high are not comparable substitutes for timber mats. In fact, it would take more than (13) layers of one-inch thick steel plates to provide the same load spreading capacity as one layer of 12" thick timber mats. If the plates could be welded together on four sides, four layers of one-inch thick steel plates would be comparable, or a single layer of a solid four-inch thick plate, such as a SuperLift counterweight slab, would also be comparable.

Steel plates are useful when a crane is working on areas paved with asphalt or concrete mainly to avoid damage to the paved surface and for increased shear resistance to punching through thin paving. It should be noted, however that mats should still be used on paved surfaces if there are sewer lines, storm drains, or other underground structures that must be protected.

To illustrate, if we assume that the timber mat or steel plate acts like a cantilever beam fixed by the

tracks and extending out on both sides of each track (refer to the rigging quiz no. 14 to the dimension Leff), and the crane is working on an area that has an allowable soil bearing capacity of 2,000 lb/sq ft, we have:

*Maximum bending moment supported by the "beam", Mmax = SFb , where Fb = 2,050 psi for a timber mat and 21,600 psi for the steel plate, and S = section modulus = (width)(thk2) / 6 = 288 in3 for a 12" wide x 12" thk timber and 2 in3 for 12" wide x 1" thk steel plate. Then Mmax = 590,400 in-lb for the timber and 43,200 in-lb for the steel.

*The effective length of the "beam", Leff = Sqr(2Mmax /w), where "w" is the load distributed by the "beam" and is 166.7 lb/in in this example. Then Leff = 84.2" for the timber and only 22.8", or only about one-fourth as long for the steel. Adding another steel plate doubles the value of "S" as S = (2plates)(12"wide)(1"thk2) / 6 = 4 in3, but not the value of "Leff". The total thickness of the plates can be squared in the equation only if the plates are welded together to act as a unit. If you go through the math, it will take (14) layers of plates for "Leff" to equal 84.2".

*To illustrate the above, try supporting something on a cantilevered beam made up of (20) sheets of paper laying flat. Then try it again after stapling all of the edges together. Big difference! 

Maximum Reach Quiz No. 151. Given:

2. For long boom erection or for capacity lifts, most crane manufacturers recommend lifting over the front of the crawlers with the idler tumblers blocked.

3. Crane mats are not shown under the tracks for clarity.

FIND:

1. What are the three advantages of blocking the front idler tumblers. 2. To correctly block the idler tumblers, should blocking be placed as shown at location:a______, b_______, or both_________

3. To prevent the crawlers from rolling backward during erection, should the following type of blocking be used:c_______, d________, or neither_______ 

Answer to Quiz No. 15

Question 1:The three advantages of blocking and lifting over the front idler tumblers are: 

a. When the front idler tumblers are blocked, the fulcrum or tipping point is moved from the centerline of the front rollers to the centerline of the front idler tumblers. This increases the stability of the crane because the lever arm from the composite center of gravity of the crane to the new fulcrum point is now longer, as much as 2' to 3' for some cranes. This is similar to a teeter totter where a small child sitting near the end of the board, balances a young man sitting on the other side, but nearer to the fulcrum point. If the small child moves to the very end of the board, he can now balance a larger man sitting at the same distance from the fulcrum point as the young man was sitting. Lifts should be made over the front of the crawlers with the idler tumblers blocked; not to get increased lifting capacity, but to increase the safety factor of the lift. 

b. The longer lever arm also reduces the soil bearing pressure under the toes of the tracks, because there is now a longer portion of the track bearing on the soil. This is true, only if the crane is set up level and the front idler tumblers are blocked, using the manufacturer's recommended thickness of steel plate or hard wood. If a thicker plate is used than recommended or if the crane is set "up hill" with the track toes high, there is a danger that the front idler tumblers will become point loaded and, therefore, will have extremely high soil bearing pressures. 

c. Increased stability is also attained by lifting over the front tumblers because the weight of the drive sprocket and chains or drive motors that connect to the drive tumblers provide additional counterweight. Since strength factors are generally closely related to a crane's stability, blocking the idler tumblers should never be used to attempt to increase a cranes capacity without written approval of the crane manufacturer. 

Question 2: To correctly block the idler tumblers, steel plate or hard wood blocking should be placed at location "b", and in a thickness recommended by the crane manufacturer. 

The thickness will range from about 1/2" to 1-3/8", depending on the crane manufacturer. The blocking width should be a maximum of 24" so it will not extend back under the front rollers. The length of the blocking should be equal to the width of each tread + 12" (for a 6" extension either side of the track). 

The following procedure should be used to block the front tumblers: 

a. Lay crane mats transverse to the tracks. See Quiz No. 14 for crane mat layout.b. Walk the crane upon the mats until it is within the set radius for the lift.c. Set the travel locks on the crane.d. Walk the crane back against the travel locks.e. Check the radius to make sure the crane is still within the set radius.f. Mark the location of the centerline of the front idler tumblers on the crane mats.g. Walk the crane slightly forward until the travel locks can be disengaged.h. Walk the crane backward about 5 feet.i. Place the steel plate or hard wood blocking on the crane mats. Center the blocking so that there is 12" either side of the idler tumbler centerline mark on the crane mats.j. Walk the crane forward until the centerline of the front idler tumblers is a few inches past the centerline of the blocking on the mats.k. Engage the travel locks.l. Walk the crane back against the travel locks.m. At this point, the centerline of the front idler tumblers and the centerline of the blocking should coincide.n. Recheck the radius.o. Make the lift.

Blocking the tracks at position "a" is not required nor recommended. 

Question 3: Neither type "c" or "d" blocking is required to prevent the crane from rolling backwards during erection. 

The travel locks are designed to provide ample resistance to rolling. 

There are some engineered lifts, when a crane is equipped with a heavy lift attachment, where sliding of the tracks must be considered. This is when a capacity load is being lifted with a long boom at a low boom angle. For this case, the resisting force generated by the friction between the tracks and the crane mats must be greater than the sliding force generated by the horizontal component of the axial boom force. If this ratio is below 2.5 : 1, chocks similar to "d" must be used to bring it up to a 2.5 : 1 or greater. As stated above, this type of lift would be an engineered lift and is presented here for information only. 

Maximum Reach Quiz No. 16

Answer to Quiz No. 16

Refer to this Graphic      

Question 1: The pivot point is at point “e”

Question 2: The pivot point is at point “k”

The pivot point is at point “k” because it is the point of trailer rotation where there is the least amount of scrubbing of the tires during turning as tire scrubbing takes the path of least resistance. In the case shown, none of the trailer axles are steerable.

The complete tractor-trailer pivots around point “k”. This is why so much maneuvering room is required to move a trailer sideways a foot or two, i.e., if a rig is backed up to a loading dock but must be moved over 1 foot for better alignment, the driver needs a large clear area to pull out into in order to move the pivot point of the trailer over 1 foot and get it backed up square with the dock.

The next time you are driving down the freeway next to an 18 wheeler and are concerned that it is going to swerve over into your lane, just keep your good eye on the tractor’s front steering tire on your side. If the driver is going to move over in your lane, the front tire will move over quite a bit before the tractor or trailer tires will show signs of moving sideways.

Question 3: The minimum turning radius for the tractor is: 27’- 6”The min. turning radius for the tractor-trailer combination is: 60’- 0”

Question 4: The maximum turning radius for the tractor is: 42’- 8”The max. turning radius for the tractor-trailer combination is: 79’- 5”

Question 5: The answer is: NO

The above turning radii are for one particular set of turning parameters and will remain constant only if the tractor is kept at an angle of 25 degrees with the trailer as shown.In fact, if the road has been laid out with the maximum and minimum turning radii for the tractor-trailer combination as shown, the driver can keep the complete rig on the road but to do so he will have to watch the left trailer tires in his rearview mirror to see that they are kept right on the inside edge of the road and he will have to watch his right front fender to see that it stays on the outside edge of the road. To accomplish the above, the driver will have to adjust the steering angle of the front tires of the tractor somewhat from 25 degrees to maintain the 25 degrees between the tractor and the trailer.If the driver keeps the front tires turned at a constant 25 degrees for a full 360-degree turn, the turning radii will become progressively smaller. If he continues the turn, the tractor will eventually end up approximately perpendicular to the trailer as shown on the small sketch A. 

On the other hand, when considering the tractor by itself, if the driver keeps the front tires turned at a constant 25 degrees, the tractor will turn around at a constant radius and in a near perfect circle. 

Maximum Reach Quiz No. 17GIVEN:A vertical vessel 15’ O.D. X 114’ and that weights 360,000 lbs is to be lifted by two Manitowoc 4100W S2 cranes.

FIND:First, Considering only crane capacity, rate the lifts in order of safety, with “a” being the safest.

a_____ b_____ c_____ d_____

Second, Considering only crane capacity and potential hook height differences between the two cranes during lifting, rate the lifts in order of safety, with “a” being the safest.

a_____ b_____ c_____ d_____

 

Answer to Quiz No. 17

Maximum Reach Quiz No. 18

Answer to Quiz No. 18

Maximum Reach Quiz No. 19

Given the Following three Figures:

FIND:

1.The civil department at a construction site needs to cover a crane lift pad with a 25mm X 7m X 10m long steel plate.a. From the information in Figure 1, show how the two pieces of steel plate can be made into one 7m x 10m plate by making one continuous cut on the 8m x 8m plate and then fitting & welding the resulting three pieces together. The cut can be straight, stepped or curved.

2. Figure 2 shows a typical floor slab for either a home or a building. Surveyors usually lay it out with appropriate instruments.a. What is one easy way for a carpenter to layout and/or check a corner of say an exterior wall for square by using just a measuring tape?

3.The load to each front tire on the caterpillar 966 wheel loader shown in Figure 3 when the bucket is empty is 10,000 lbs. and is 18,000 lbs. when the bucket is full. The bucket is not resting on the ground in either case.a. What is the approximate soil loading in psi under the front tires when the bucket is empty?b. What is the approximate soil loading in psi under the front tires when the bucket is full?Answer to Quiz No. 19

Question 1:

Cut the 8 m x 8 m plate in two pieces as shown in Figure-a. Move the crosshatched piece (the one on the right side) up and lap it over the left piece so that a total width of both pieces is 7 m and a total height is 10 m. See Figure-b.

Place the 1 m x 6 m piece of plate in the void between the two big pieces of plate. Weld the three pieces of plate together.

Question 2:

One easy way to check the corner of two walls for square is to use a 3 --4 --5 triangle relationship. Using a measuring tape, the carpenter starts at the inside corner of the wall and measures a distance of 3’ along one wall and makes a point A. He then measures from the corner a distance of 4’ along the other wall and makes a point B. He then measures diagonally from point A to point B. If the measurement is 5’, then the two walls have a square corner. The carpenter can use any multiples of the 3--4---5 triangle, for example, he can use 30’ to point A and 40’ to point B. The diagonal distance between points A & B would then be 50’ for the two walls to have a square corner.

Question 3:

a. When the bucket is empty, the soil loading under the front tires is approximately 55 psi.

b. When the bucket is full, the soil loading under the front tires is approximately 55 psi.

The approximate soil loading under the front loader tires is 55 psi in both cases above because:

When the bucket is empty, the width of the foot print under each front tire bearing on the ground is somewhat less than the width of the tire and the length of the foot print is very small, making the foot print a narrow band. By dividing the load exerted on the ground from a front tire by the area of the footprint, the soil bearing pressure will be approximately 55 psi.

When the bucket is full, the additional weight flattens out the front tires, so that now the width of the foot print under each front tire bearing on the ground is a little wider but quite a bit longer, making the foot print a rectangle. Again, by dividing the load to the ground from the front tire by the area of the footprint, the soil bearing pressure will still be approximately 55 psi.

The 55 psi under the front loader tires is equal to 7,900 psf. On most of our construction projects, an average allowable soil bearing pressure is about 4,000 psf. If you followed this loader around on 4,000 psf soil, you would notice it leaving tire tracks as it is over loading the soil and the tires sink into the soil until they pick up a bearing area that equals 8,000 psf. This is usually not a problem, unless the loader is running on asphalt, because we don’t care if some settlements of the tires occur.

A loaded 10 wheeler dump truck with a tire pressure of 75 psi has a soil loading of about 10,800 psf, which is generally much higher than the loading to the soil from most heavy lifts. Running it along the haul route or over the crane pad is a good way to do some compaction and find any soft spots.

Maximum Reach Quiz No. 20

GIVEN:

1. Sketch of a horizontal vessel being lifted using double basket hitches (Two individual slings are required, one for each basket hitch with all four of the eyes over the hook).2. The vessel diameter is 10 feet. This diameter is from the centerline to centerline of the slings. See section A-A on the sketch.3. The total weight of the vessel including rigging is 200 kips.4. The center of gravity (CG) is at the center of the vessel in all directions.5. The longitudinal angle of the slings with the horizontal is 65 degrees.6. The span is 20 feet.7. The total length of each sling is 54.08 feet, i.e., (the sling length from the hook to the point of tangency on the shell plus the sling contact length along the shell above the centerline of the vessel plus the sling contact length along the shell below the vessel centerline down to the bottom)*2 parts = 54.08 feet.

FIND:

1. The compound or true angle of the legs of the slings with the horizontal. The horizontal base of this angle is from the point of sling tangency to the vertical centerline of the vessel.2. The vertical height from bearing on the hook down to the bottom of the vessel.3. The tension in kips for the inclined portion of the legs of the slings that are not in contact with the shell. A kip equals 1,000 lbs.4. Select a sling that has the required SWL for the above tension.

Answer to Quiz No. 20

Question 1: The true angle is 60.55 degrees.

Question 2: The vertical height from hook to bottom of vessel is 21.44 feet.

Question 3: The tension in each leg of the sling is 57.42kips.

Question 4:

A 2” dia. IPS, IWRC, sling good for 60 kips could be usedORA 1 7/8” dia EIPS, IWRC, sling good for 64 kips could be used

COMMENTS:

As the section of each sling that is in contact with the shell follows an ellipsis, it is pretty complex to calculate the above answers. By making some assumptions, approximate answers can be obtained. It is left to the user to make a layout for the following steps.

1. The approximate distance H from the hook to the bottom of the vessel can be found by the following formula:

H = .5((20 feet Span/Tan (90-65))-10 feet Vessel OD) + 10 feet Vessel OD/2 = 21.45 feet.

2. Assume that the slings are tangent to the vessel at the horizontal centerline and then solve for the true angle using the H obtained above.

The approximate true angle is 60.99 degrees.

3. From the true angle, the approximate tension in each leg is 57.17 kips.

These answers are pretty close to the ones listed for the questions to the quiz because the slings are actually tangent to the shell 1.21 feet above the centerline of the vessel. If shorter slings or a larger diameter vessel had been used, the difference between the answers would be much greater as the assumption that the slings are tangent to the vessel at the centerline would be farther off. Of course, an original assumption could have been made that tangency was above the centerline some distance.

The answers listed for the questions in the quiz were obtained by a basket hitch program that can be found in the RIGGING DESIGN PROGRAMS section of this website. Just use the basket hitch program for a vessel without saddles.

    Maximum Reach Quiz No. 21

FIND:

1. Assuming that the crane has adequate lifting capacity and that the trailers are stable for tipping, can the vertical vessel shown in sketch 1 be safely offloaded and placed on cribbing using one crane lifting at either the trunnions or the tail lug? (a.) Yes (b.) No

2. What is the smallest size shackle that can be used to connect the two 2.5" diameter slings shown in the sketch 2?(a.) 35 ton (b.) 55 ton (c.) 85 ton (d.) 110 ton

3. Assume there is no physical interface when mating Crosby shackles in a straight pull as shown in sketch 3, can they be connected to each other:(a.) pin to pin (b.) pin to bail (c.) bail to bail 

Answer to Quiz No. 21

 ANSWERS TO RIGGING QUIZ No.  23  1.       When using a      Choker     hitch, 20 to 25 percent of single leg capacity is lost.

2.       The most basic hitch is the single leg or      vertical      hitch.

3.       A      basket       hitch can have twice the capacity of a single leg.

4.       When a shackle is used to help form a choker hitch; the pin of the shackle must be placed into the      eye if the sling     .

 5.       A double wrap choker or basket hitch has excellent grip on a load because

the contact is   greater than 360 degrees     .  6.       Never use a double basket hitch (single wrap) at a horizontal sling angle of

less than      60 degrees_.  7.       Never use a double choker  (double wrap) at a horizontal sling angle of less

than      45 degrees     .  8.       A single leg sling will often provide poor      load control     .  9.       A three legged sling will often provide improved load control, but the

capacity of the sling depends on how each leg      shares the load     .  10.   It is possible for only     two     legs of a four-legged bridle to carry the full load

at times.

11.   A single basket hitch (single wrap) has poor grip on the load because it has    less than 360 degrees      contact with the load.

12.   Place     block of wood      under the eye of a choker hitch to avoid “beating down the eye” and to increase the grip on the load.

13.       Three      legs of a three-legged bridle will carry the load if the lengths are close to the same length.

14.   When using synthetic slings, snug the choker before lifting and slowly take up slack to avoid    friction melting     .

15.   A wire rope sling used in an inverted basket hitch over a hook can lose 50 %     of its capacity.

Answers A.     Choker

B.     Basket

C.     vertical

D.     load control

E.      greater than 360 degrees

F.      two

G.     synthetic sling

H.     three

I.        eye of sling

J.       shares the load

K.     four

L.      60 degrees

M.    45 degrees

N.     50 percent

O.     block of wood

P.      center of gravity 

16.   The sling connected      closest      to the center of gravity will carry the greatest share of the load.

 

Q.     closest

R.     90 degrees

S.      friction melting

T.      less than 360 degrees

U.     100 percent    From: “The Crosby group product application seminar workbook on hitches”    

Maximum Reach Quiz No. 24

ANSWER TO RIGGING QUIZ No. 24 Question 1:           TWO tag lines should be used for this load.                                 There are no written rules that tell how many tag lines should be used to tag lifted loads. As a

rule of  thumb, use as many as necessary to adequately control the load in a desired manner.  For example,  loads similar to a convection section need two tag lines to keep it square with the boom during hoisting and swinging.  Also, if the hook is locked to the load block, tag lines will keep the load from rotating which will keep the hoist lines from spinning up.  

 Question 2:       One tag line should be hooked to corner A of the convection section  and laid out to the West along

the North side of the lift crane.  The other tag line should be hooked to corner B of the convection section and laid out to the West along the South side of the lift crane.  See the Tag Line Plot Plan below. 

                             With this tag line configuration, the riggers on each tag line will not have to walk as far (244’ Vs

420’) and only about  half as fast as they would if the tag lines were hooked to corners C & D. 

                            The major benefit in using this tag line configuration is that both teams on the tag lines can see each other and will be able to maintain about the same tension on each tag line.  Someone monitoring  the convection section can keep it square with the boom by directing  one tag team to pull a little harder and the other team to let off a little.  This way, the tag teams are not trying to out pull each other.  In most cases, all the two tag teams have to do is just hold their tag lines snug and walk slowly to keep their tag lines square with the convection section.

 

Question 3:         In this case the tag line at corner A should be hooked to the top of the convection section in order for the tag line to clear the top of the heater during swinging.  The tag line at corner B should be hooked to the bottom of the convection section to keep the tag line as flat as possible.

 Question 4:         200’ minimum ((80’ + 5’ + 15’)*2) for tag line A and 170’ minimum  ((80’ + 5’)*2) for tag line

B.   To be effective, tag lines should be at a 2:1 slope or flatter.  

Question 5:         ½” minimum.  This was found by first solving for the friction force F between the thrust bearing on the hook shank and support plate on the load block.  F =  65,000 lbs of convection section weight * .11 coefficient of friction = 7,150 lbs.  The torque T = 7,150 lbs * 3” radius of the thrust bearing = 21,450 lb-in.  The horizontal force H required to rotate the convection section  = 21,450 lb-in / 240” half length of the convection section  = 90 lbs.  The SWL of a ½” diameter manila rope = 550 lbs., and is therefore plenty safe even when considering a mild wind. 

                               A 1” diameter manila rope is recommended for a lift like this, as it is easier to hold on to.  Also if

the load control is real critical, is recommended that two tag lines be hooked at corner A and two hooked to corner B just incase one would come loose or break.

 Question 6:         Two minimum.  A 150 lb. rigger can pull a force equal to his own weight if he is leaning

backwards, has his feet firmly planted and the tag line is quite flat.  The problem is that the rigger can’t always firmly plant his feet, so the friction force between his feet and the soil is equal to .5 coefficient of friction * 150 lbs. = 75 lbs.  Therefore two riggers can pull approximately 150 lbs. > 90 lbs. == Okay

 COMMENT ON TAG LINE USE: In general, the use of tag lines is one of the most important and least understood operations in construction.  In lifting operations where the load is high above the ground with very little clearance between it and the boom, there is usually mass confusion in keeping the load square with the boom. Everyone is hollering and running back and forth to pull first in one direction and then in another as the load sways back and forth.   It doesn’t have to be this way.  Using a planned tag line procedure makes the lift go very smoothly without anybody getting nervous or upset.  One person should be able to control the complete tag line operation by just using hand signals.

RIGGING QUIZ  No.  25 

GIVEN:

1.      Outrigger Float Diameter = 24”.2.      Boom Is 81’ Long.3.      Total Load (Load Block + Rigging + Pipe) = 14,000 lbs.4.      Soil Is A Sandy Loam With A Bearing Strength Of 3000 psf (pounds per sq. ft.). 

 FIND:1.       Is the crane positioned far enough away from the shoulder of the excavation?

      Case A:                        Yes_____          No_____  if no, what should “D” be?_______ft.      Case B:                        Yes_____          No_____  if no, what should “D” be?_______ft.

2.       List five things to check for in regards to ground work prior to setting up a crane near a slope.

3.       Is the outrigger blocking area “B” large enough?      Case A:            Yes_____          No_____      Case B:            Yes_____          No_____

4.       In general, does the maximum outrigger loading occur at:      A. Minimum radius with minimum boom & maximum load, or _____      B. Extended radius with long boom & capacity load _____

 ANSWER TO RIGGING QUIZ No. 25

There are really two parts to this quiz.  First is determining where to safely set up the crane relative to the shoulder of the slope and second is making sure that the outrigger float loads are distributed down through the blocking or load spreaders such that they don’t exceed the allowable soil bearing pressure. QUESTION No.  1: Safe distance from crane to excavation Case A                   The crane is not positioned far enough away from the shoulder of the slope. 

 According to CIRIA (Construction Industry Research and Information Association, Westminster, London), the horizontal distance “D” should be greater than 4 times the width of the outrigger float blocking  “B”, without having to consider soil type. 

Therefore “D” = 4 * 2.5 feet = 10 feet Using cut slope criteria from OSHA (Occupational Safety & Health Administration, USA), it is  recommended that for this type of soil, the horizontal distance from the edge of the blocking under the outrigger float to the toe of the cut slope be a minimum of a 1.5 Horizontal:1Vertical.  This would be applicable to fill slopes as well.  The criteria is also applicable to crawler cranes with or without crane mats. 

Therefore  “D” = 15 feet – 10 feet = 5.0 feet  (with a minimum “D” distance of H/2).  See the supplemental information on page 3 for a graphic layout of recommended minimum “D” distances based on different types of soil defined by OSHA.

 Case B                   The crane is not positioned far enough away from the retaining wall. Per CIRIA, the criteria for this situation is that “D” should be the larger of 4*B or 1.5*H                 Therefore “D” = 10 feet Again, using OSHA criteria, it is recommended that for this type of soil the horizontal distance from the edge of the blocking under the outrigger float to the edge of the retaining wall be a minimum of a 1.5 H: 1V.  This criteria is also applicable to crawler cranes. 

Therefore  “D” = 6 feet  (with a H/2 minimum “D” distance).  QUESTION No.  2:  Ground condition checklist Five things that should be checked for in regards to ground work prior to setting up a crane near a slope are:

1.        Are there cracks in the asphalt/concrete paving?  Have there been any new patches?  This could be an indication of settlement below the pavement.  Also watch the pavement during the lift to see if the cracks enlarge.

 2.        If the soil has been recently placed, check for compaction. 3.        Are there underground utilities or hazards such as culverts, electrical duct banks, water lines, sewer

lines, steam lines,  gas lines or underground cavities (such as a slow leak in a underground steam line).4.        The presence of a water table.  The higher the water table, the less stable the ground is. 5.        Loosely compacted soil close to new buildings or foundations is susceptible to settlement.

 QUESTION No 3:  Outrigger blocking size Case A   and Case B are both                       No From a past article entitled “Outrigger Expertise” printed in International Cranes (IC), a guideline was presented for determining the area of blocking or load spreaders required under the outrigger floats for compacted sand clay aggregate soil:  The area of blocking in square feet that is required under each outrigger float is equal to the maximum capacity of the crane in US tons divided by five, equivalent to a ground pressure of 70 pounds/square inch (10,080 pounds /square feet (psf)).  The area of blocking per the above is 45 tons/5  = 9 square feet.                               Therefore, the size of the blocking required under the outrigger floats = 36 inches square                                                                                                                                                = 40 inches in diameter

 It is recommended that for cranes up to 100 tons in capacity to size the blocking or load spreader so that it is 4 times larger than the area of the outrigger float. In this case 4 times the area of a 24 inch diameter float is 12.6 square feet.                 Therefore, the size of the blocking required under the outrigger floats = 42 inches square                                                                                                                                                = 48 inches in diameter Using outrigger blocking of 48 inches in diameter and the OSHA criteria for locating the crane next to a slope, the operating radius for Case A = 33.25 feet.  For a 81 foot boom, 35 foot radius and a 14,000 lbs. load, Grove calculates that each front outrigger will have a vertical force of 35,000 lbs.  This yields a soil bearing pressure under each outrigger blocking of 35,000 lbs/12.6 square feet = 2,780 psf < 3,000 psf === good. Using outrigger blocking of 40 inches in diameter per the criteria in the IC article yields a soil bearing pressure of 35,000/9 = 3,890 psf > 3,000 psf  === no good. If the load was 18,650 lbs., the load to each front outrigger would be 40,000 lbs. and the soil bearing pressure under the outrigger blocking would be 40,000 /12.6 = 3,175 psf > 3,000 psf === not good.  Some settlement of the blocking could be expected. In this quiz, the load of 14,000 lbs. was chosen because as a non-engineered lift, The author would restrict the lift to 75% of the load capacity chart.  Note that the actual lifting capacity for a 35 foot radius is 18,650 lbs.     QUESTION No. 4:  Maximum outrigger loadings The answer is “A” With a 34 foot boom over the front of the crane at a radius of 10 feet and with a 90,000 lb. load, the  loading to each of the front outriggers is 60,000 lbs.With a 81 foot boom over the front of the crane at a radius of 35 feet and with a 18,650 lb. load, the  loading to each of the front outriggers is 40,000 lbs. SUMMARY: Safe distance from crane to excavationThe guidelines presented for placing cranes next to a slope are somewhat conservative, with CIRIA being more conservative.  If the user needs to set the crane closer to the shoulder of the slope than is indicated in the guidelines, then an engineering assessment should be made to determine the crane location relative to the shoulder of the slope. Outrigger blocking sizeThe criteria used by the author for sizing the blocking area under the outrigger floats is applicable to structural fill compacted properly within a plant site where the allowable soil bearing capacity is between 3,000 and 4,000 psf and the load being lifted is not more than 50% of the load capacity chart. If the allowable soil bearing capacity is less than 3,000 psf or the load being lifted is greater than 50% of the load capacity chart or the lifting capacity of the crane is greater than 100 tons, then an engineering assessment should be made.  The study should determine the outrigger loadings, the allowable soil bearing capacity of the soil and then the sizing of the blocking under the outrigger floats so that the actual soil bearing pressure is less than the allowable. SUPPLIMENTAL INFORMATION ON STABILITY SLOPES  

Type A Soils - Cohesive soils with an unconfined compressive strength greater than 3,000 lb./square foot (psf). Examples are clay, silty clay, sandy clay and clay loam. Cemented soils such as caliche and hardpan are also considered type A.  

Type B Soils - Cohesive soils similar to above but with an unconfined compressive strength greater than 1,000 psf but less than 3,000 psf. Also granular noncohesive soils including angular gravel, crushed rock, dry rock that is unstable such as shale, previously disturbed soils, and soils that are fissured. 

Type C Soils - Cohesive soils with an unconfined compressive strength of 1,000 psf or less. Also granular soils including gravel, sand and loamy sand, submerged soils, soils from which water is freely seeping, and submerged rock that is not stable. NOTE:    These categories are to be selected by a competent person based on an analysis of the soil properties, the

excavation performance and the environmental exposure.    

RIGGING QUIZ  No.  26 

GIVEN: 1.         The sketch of a pipe spreader bar

2.         Design information not shown on the sketch            a.         Tension in the left vertical sling below the spreader bar   = 304.63 kips            b.         Tension in the right vertical sling below the spreader bar = 345.37 kips

c.         Fy                    = 36 ksi            d.         Impact factor    = 1.8            e.         Each prong of the duplex hook is 8” in diameter

f.          The vertical height from the centerline of the spreader bar down to the center of the lifting lug holes on the vessel is 180.00 in. 

3.         The rigging hook up for each side above the spreader is per the sketch, ie an inclined sling over the hook and connected to a top lug with a shackle

4.         The rigging hook up for each side below the spreader is a vertical sling connected to a bottom lug on the spreader bar with a shackle and connected to the lug on the vessel with a shackle

5.         The information on the sketch was obtained from the spreader bar program with lugs top and bottom (English version) in the design section of this website.  To see the complete input/output information, open the program, click on the sample values and then input the information off the sketch.  Leave the properties for a 18” x 0.375 pipe as is.  Calculate and note that with this configuration there is very little bending due to the influence of the slings.  Also note that the combined stress check is 0.72, a very nice safety factor 

6.         The 19.25 in. for the vertical distance from the centerline of the top lug holes down to the centerline of the spreader bar was used to allow ample room for pads, pad welds and the lug welds

 

FIND:           1.         Design lifting lugs for the above spreader bar to carry a 650 kip load

a.         Feel free to use the pad eye lug design program in the design section of this websiteb.         Use the same design criteria as in the pad eye lug program, ie, Fy = 36 ksi, impact factor = 1.8,

allowable force on the weld = 14.85 kips/in, etc.2.         Make a detail drawing of the lifting lugs showing their shape and how they will be attached to the pipe

spreader bar, ie butt welds, fillet welds, etc.  Include weld sizes3.         Use the same shackle size for all four lugs on the spreader bar and for both lifting lugs on the vessel4.         Select a sling for the right side above the spreader bar.  Use this same size of sling for the other three

slings.  Feel free to use the program that calculates the safe working load of a sling when bent over a hook, the bail of a shackle, etc, located in the design section of this website

5.         Calculate the total length of the four slings required

ANSWER TO RIGGING QUIZ 26QUESTION: 1.         LUG DESIGN:     In designing a lug, the first step is deciding what shackle you are going to use, as the

diameter of the pin and bail are important in the lug design/rigging hookup.  The maximum sling tension of 366.18 kips occurs in the right sling above the spreader bar.  A Crosby G2140 x 200 shackle was chosen because its SWL is 200 metric tons (Te) or 440.92 kips, which is greater than 366.18 kips. 

             When the Crosby G2140 x 200 shackle was used in the Pad Eye Lug Program found on this website, it

resulted in the following lug design. This is not the only lug design that would work for this lift, but it is felt that it is the safest and most economical.  This same lug design will be used for all four lugs:

             a.         5.00” lug pin hole diameter    (0.25” greater than the shackle pin diameter)            b.         8.00” lug radius                        (Maximum effective lug radius = 7.83”)            c.         3.50” lug plate thickness            d.         16.00” lug plate width                          e.         1.25” pad thickness            f.          7.50” pad radius          

g.         366.18 kips force on the lug    (Worst case force)            h.         90.00 degree angle                   (Parallel to the centerline of the lug plate)            i.          0.54                                         (Combined stress factor for the lug plate)

j.          0.30” fillet weld                        (Between the pads and the lug plate)k.         1.39” fillet weld                        (Between the lug plate and the pipe if a

pad eye type lug will be used) 2.         LUG DRAWING:     See the lug drawing below for details.    

With the lug design complete, the next step is deciding how the lug plates will be connected to the spreader bar. 

 One way would be to use four pad eye type lugs and butt weld the lug plates to the pipe with 1.39” fillet welds.  As the 0.38” wall pipe is to thin for this size of fillet weld, this method would require using a 36” long heavy wall pipe section at each end of the spreader bar that was at least 1.39” thick or using doubler pads under the lug plates.  Another consideration is that butt welds require strict QC to insure their structural integrity.  Also, the pipe sections at the lugs would have to be checked for over stressing or deformation.  The lugs would tend to make the pipe sections take on an oval shape.  It might be necessary

to weld a stiffener plate inside the pipe between each set of pad eye lugs or weld a stiffener ring around the outside of the pipe at each set of lugs.  The stiffener ring could be welded to the pipe and the outboard edge of the pay eye lugs.   Considering the above, this is not the best way to go. The recommended way would be to fabricate one curved lug plate for each end of the spreader bar.  See the lug drawing below for details of this type of lug plate with identical lugs at each end.  The spreader bar would have to be slotted top and bottom at each end with a saw so that the lug plates could slide right in to place with a very close fit.  They would then be welded in place with a 0.38” fillet weld.  The strips from the slots would then be welded back into place.  The capacity of the 0.38” welds on each lug plate equals 4 welds * 16 in. x 14.85 kips/in * 0.38  = 361.15 kips.  This is greater than the horizontal compressive force induced by the inclined slings of 123.10 kips * 1.8 I.F. = 221.58 kips.  Very little weld would be required to keep the lug plates stable relative to the spreader bar in the vertical direction.  The 0.38” welds would provide much more than enough capacity in both directions. 

 

 

 The lug drawing shows the details and final shape of the lug plates.   Note that the top part of the lug plate is 16.00” wide down to where it first intersects the centerline of the pipe.  In order to maintain this minimum width, the lug plate at the centerline of the pipe is 2 * 8.48” = 16.96” wide.  The lug plate for the left end of the spreader bar is identical to the one shown above except that the angle is 68 degrees and the lug plate width at the centerline of the pipe is 2 * 8.63” = 17.26” wide. 

3.         SHACKLE SIZE:     The shackle size selected is a Crosby G2140 x 200. Six shackles would be required for connecting the slings to the four lugs on the spreader bar and the two lifting lugs on the vessel.  As stated above, the capacity of these shackles is 200 Te.

 4.         SLING SIZE:     The size selected is a 3.5” diameter, EIPS sling.  

In selecting a sling for the right side above the spreader bar, the program for determining the SWL of a sling bent around pins, etc was used.  It was found that for a hook diameter of 8”and a shackle bail diameter of 4.5”, that a 3.5” diameter EIPS sling in a three-part configuration would provide a SWL of 372.32 kips with 6.14 kips reserve capacity.  All four slings would require the 3.5” diameter sling in a three-part hook up.  Each sling above the spreader bar would be hooked up as follows:  One eye would be placed over the hook.  The body of the sling would then run down around the bail on the shackle, back up and over the hook and then back down where the other eye would be placed over the bail of the shackle.  Each sling below the spreader bar would be rigged the same way except there would be a shackle at both ends.

 5.         SLING LENGTHS:     All four slings are in a three part configuration.             Top right sling length:                101’ – 8.52”            Top left sling length:                  103’ – 5.76”            Both bottom vertical slings:          33’ – 2.97”             These lengths were calculated using the following information:

a.         The distance from the center of the 200 Te shackle pin to bearing on the bail = 18.07”b.         The sling length in contact with the 4.5” dia. bail of the shackle = 12.57”c.         The sling length in contact with the 8.0” dia. hook = 18.06”

             For example, the length of the top right sling was calculated by the following formula:   3 * (414.70” –

18.07”) + 12.57” +18.06” = 1220.52” = 101’ – 8.52”                       Where 414.70” is the distance from bearing on the hook to the centerline of the top right lug and the 3 is for

three parts of sling.                       In order to maintain the vertical distance from bearing on the duplex hooks down to the centerline of the top

lugs on the spreader bar of 391.06”, and from the centerline of the spreader bar down to the centerline of the lifting lugs on the vessel of 180”, slings should be used that have been measured for length under a SWL during load testing.  Most sling manufacturers are able to fabricate matched slings so their length, measured under a SWL, is within 1.0” of each other and within +/- 6.0” of the length specified on the purchase order.  Therefore each pair of slings should be purchased with the specification that their lengths should be within the plus 6.0” tolerance or all slings should be within the minus 6.0” tolerance, not one 6.0” minus and the other 6.0” plus.  If both the top slings ended up being 6.0” longer than the specified length

on the P.O., then the net increase in the vertical distance 391.06” would be about 2.0” and the effect on the strength of the spreader bar would be less than 0.01%.

 SUMMARY: Even though this was a quiz, the information from it can be used to make a spreader bar drawing and a rigging hook up drawing.  They can then be used to make an actual 650 kip lift.

RIGGING QUIZ NO. 27 

GIVEN:1.         The two layouts below.  Note that the geometric shapes in both layouts are the same size and/or

dimensions.  They have just been arranged differently in each layout.

FIND:

1.         Determine where the hole came from when the geometric shapes are re-arranged as in the lower layout. Note:  This quiz was submitted by Marco Van Daal

ANSWER TO RIGGING QUIZ 27 QUESTION:  

Determine where the hole came from when the geometric shapes are re-arranged as in the lower layout. ANSWER:                  Make a large scale drawing of the two layouts and calculate the following areas:             TOP LAYOUT:             The area of the four geometric shapes = 32.0 units            The area of the right triangle inclosing the four geometric shapes = 32.5 units             Therefore the slope of the line forming the top of the geometric shapes is an under slope when compared to

the cord or hypotenuse of the right triangle, where the triangle area is bigger by 0.5 units.             BOTTOM LAYOUT:             The total area of the four geometric shapes plus the 1.0 unit hole = 33.0 units             Therefore the slope of the line forming the top of the geometric shapes is an over slope when compared to

the cord or hypotenuse of the right triangle, where the area of the triangle is smaller by 0.5 units.             SUMMARY:             The 0.5 unit difference between the area of the triangle and the area of the geometric shapes in the top layout

plus the 0.5 unit difference between the area of the geometric shapes and the area of the triangle in the lower layout = 1.0 unit = the 1.0 unit hole. 

             This quiz was included in the monthly quizzes because it shows that some times the answer to a problem is

not obvious at first glance and a more detailed study is required to see the difference.