Construction Crane Safety

Crane Accidents 04/12/2009 By : William A. Greco [email protected]

Page 1 of 7 Warrington,Pa.

Typical egg link

Abstract Slewing (rotating) a crane too quickly can induce forces which can cause major crane components to fail. This essay is being written to emphasis the importance of proper crane operation. Example: Most crane lifts use an elliptical egg link. Assume an egg link rating at 15,000 lbs tensile strength in this essay. . This essay will attempt to explore the possibility of cause by rotational torque, centrifugal force and load pendulum action which can decelerate too quickly, causing a dynamically induced inertial force far greater than the link working load limit. The essay will also identify the condition of load free fall and rapidly applied sheave braking. Lastly the essay will explore the forces involved in tag line movement. Lifting components are not static structural members, they are dynamic when loaded and are subjected to severe stresses when in use. To assume that a member of a lifting system would work for 100’s of lifts and then suddenly fail due to a manufacturing flaw or metallurgical instability has limited plausibility. Cracks and failures from stress do not develop over a long period of time after the stress is applied unless the component material is subjected to what is known as stress corrosion which is intergranular occurring at the metal grain boundaries.

Crane Accidents 04/12/2009

Page 2 of 7

By : William A. Greco [email protected]

Warrington,Pa.

Most of the dynamic component failures that we witness in our daily lives are caused by component wear causing tolerance shift, operator misuse, heat, friction, rapid temperature swing, poor lubrication, component control failures to name a few. This essay will not be able to find all the answer’s to tragic crane accidents but it should serve as a guide and warning against crane misuse. Crane Rotational Movement The dynamic effects accompanying rotational motion include. inertial torque, centrifugal force, and load pendulum action. For many cranes, centrifugal force produces a tensile force on a boom or jib that relieves the axial compressive forces to a minor extent. On the other hand, this same force acts to create an inertial force on the load. In the overall scheme of things, centrifugal force on the dead-load masses of cranes can usually be neglected, since swing speeds rarely exceed 1 radian/min 57.3 degrees, however cranes have the capability to operate at a slewing speed of 0-2 rpms or 12.5 radians per minute. If the boom of a crane is accelerated too quickly the load will lag the boom causing what is known in the heavy lift industry as side loading, if the boom and load get into this dynamic displacement and the boom rotation is suddenly stopped the load will begin to pendulum, the load component will become a damped harmonic which over time stabilizes to an undetectable movement. Centrifugal force throws the load out to an increased radius. If n is the number of slewing revolutions per minute, we have . 2

Fc

WR .

n

g

30

(equation-1)

where Fc = centrifugal force W = weight of load R = operating radius g - acceleration due to gravity Fc will act horizontally at the upper load sheave shaft away from the axis of rotation and parallel to the horizontal projection of the boom centerline Load pendulum action is another matter. This dynamic motion is induced by the inertia of the load as the crane powers up or brakes during swinging. The inertial force is horizontal, but it is tangential to the rotational arc (i.e., perpendicular to the centrifugal force). Consider a slewing (rotating) boom; when braking occurs and the boom slows, the load will continue and pull forward on the boom. After the boom stops, the load will continue to swing, pendulum fashion.

Crane Accidents 04/12/2009 By : William A. Greco [email protected] .

Page 3 of 7 Warrington,Pa.

The period of swing will vary with hoist line length L, the distance from load CG to the suspension point on the boom, with the relationship 1

Tp 2 . ( 3.14 ) .

L g

2 (equation-1A)

where Tp is the "swing period. Line length, is a random variable, (3.14 = pi), which makes Tp a random variable as well, but Tp,will increase with L. Should L be such that Tp corresponds to the natural period of vibration of the boom, resonance will occur. In a pure theoretical undamped system this will lead to a steady increase in vibration amplitude and eventual failure. In a real-life system with damping always present, amplitude will reach a peak value of some 5 to 10 times the effect of the same force statically applied. Fortunately, the structure period of vibration will always be shorter than load pendulum period and resonance will not take place. The crane industry therefore suggests that the problem be simply resolved by applying load and dead-weight inertial forces at twice the values obtained when using mean acceleration. The factor 2 is taken to account for the elasticity in the system. When the boom is rotated, acceleration takes place and the load will not be in the same vertical plane as the boom. During this side loading condition or load lag, the initial angle that the load makes with the vertical wire rope is given by Loadangle = Loadangle =

atan

a . 57.3 32.2 (equation-2)

where: atan = arc tangent = tan-1 a = the tangential swing acceleration. (Feet per second 2) 57.3 = degrees in one radian Following this representation, if W is the load weight, the inertial force F causing the lag can be expressed as angle Wd .

F W tan Load

g

ma

(equation-3)

In terms of rotational motion the basic inertial equation can be stated as Inertial Torque = polar moment of inertia (times) angular acceleration

Crane Accidents 04/12/2009 By : William A. Greco [email protected] .

Page 4 of 7 Warrington,Pa.

For any point on the rotating structure, the tangential acceleration (TA) = Radius of swing x angular acceleration or TA = (r) x (angular acceleration) where r is the radius to the point in question. The customary units for TA are radians per second squared. In calculating rotational inertial forces a convenient concept is that of equivalent mass, the single concentrated mass located at a particular radius and having the same rotational characteristics as the distributed masses it represents. Crane Rotational Force Example Calculate the centrifugal force when a 9,500 lb load and a 1,500 lb spreader bar, block and cable (11,000 lb total load) swings at 6.5 radians/minute (6.5 x 57.3 degrees) or 372 degrees per minute (angular slewing velocity) at a radius of 100 feet. From equation-1: 2

11000 . 100 . 6.5 . 3.14 32.2 30

= 15811.8 lbs of centrifugal lag on the system

Using the same lift parameters :

If the swing brake is applied forcing the boom to come to a stop in 1.4 seconds: A. What inertial force will be applied to the load ? B. What vertical angle will the cable be at between the load and the supporting boom block at the instant that the boom stops ? A.

Inertial force:

6.5. 2 3.14 = 0.68 60 radians per second initial angular velocity .

0.68 = 0.486 1.4 radians second squared angular deceleration 0.486. 100 = 48.6 ft/ second squared tangential deceleration from equation-3:

11000. 48.6 = 16602.5 32.2 lbs of inertial (stopping) force

Crane Accidents 04/12/2009

Page 5 of 7

By : William A. Greco [email protected] .

Warrington,Pa.

Crane Rotational Force Example (continued) B. What vertical angle will the cable be at between the load and the supporting boom block at the instant that the boom stops ? from equation-2: tan-1(48.6 / 32.2) = 0.986 x 57.3 = 56.5 degrees (load lag angle) Equation 1A deals with pendulum action, which is of no concern in this essay since we are only interested in maximum loading on the lift system. From this example it can be seen that with high boom acceleration and quick braking an egg link with a 15,000 pound rating could be subjected to forces which are above the working strength of it’s design if lifting only 9,500 pound loads. Load Drop Force When a load is dropped in free fall the acceleration is retarded by friction at the sheaves, the load continues to drop after the sheave brake is applied because the hoist cables stretch and kinetic energy vs spring rate of the cable interact such that a ratio of kinetic force to potential force must be calculated. The ratio of cable to lifted load is calculated thus: Where k = number of cables x spring rate (lbs/in.)x 12 (in/ft) And 1

F 1 W

2 1 k. v W.g

2

where : W = load weight, g= 32.2 v = final velocity of load Load linear drop example: If a 9,500 lb load went into upper block free fall for 6 feet and then the sheave brake was fully suddenly applied on a three part line crane where the friction loss could be taken at 2% and the cable spring constant were 2300 lb/in, what peak force will develop in the spreader bar connections ? (3) part cable sheave with 2% friction =

1 0.98

1 0.98

1 2

0.98

= 3.124

3 load friction effect factor

Crane Accidents 04/12/2009 By : William A. Greco [email protected]

Page 6 of 7 Warrington,Pa.

Load linear drop example: (Continued) The nominal mechanical advantage is 3 for a three part line. Acceleration (a) due to gravity accounting for friction = a = 32.2 ft/sec2 x (3 / 3.124) = 30.96 ft / sec accounting for friction And final velocity is v = (2ah)1/2 where h = height of drop ½

Or [2 (30.96) (6) ] = 19.275 ft sec The ratio of cable to lifted load is = 1

F/W=1

1

2 3. ( 2300) . ( 12). 19.275 . 950032.2

2

= 11.078

F = 11.078 x 9,500 = 105,241 lbs of force on the egg link Obviously a short free fall of only a few feet and a quick braking action could damage components and exceed working, yield and ultimate strengths. Tag Line Loads To move a 9,500 lb load 6 inches horizontally which is suspended beneath a boom 30 feet above the loads center line with a tag line will require a tag line crewman to exert 161.5 lbs of horizontal force (if the tag line were considered to be level horizontally with the center of gravity). Tag line movement can be a very difficult and exhausting task. Angled Lifts The load hoist line must remain vertical to maintain the load radius and the load chart rating. Another condition that can cause the load line to swing out, thereby increasing the load radius and resulting in lost capacity or extra force development is that the crane is initially hooked up beyond or inside of the the boom tip radius which creates a swinging load condition as soon as the lift begins. Crawler Cranes (on treads) Have the capability of moving with the load, rapid tread acceleration with the load in place can cause load lag. Wheeled vehicles normally have their outriggers spread and do not move with the load in place, crawlers on the other hand have adjustable tread assemblies that are spread out after site delivery. Conclusion: The majority of crane fatalities occur due to poor boom field assembly, the next major cause for crane accidents is booms contacting live electrical conductors. Poor operating procedures are always a problem, high accelerations in either slewing action or hoist movement can destroy major components of the lift system. We have surveyed some of the forces that can occur when cranes are mishandled. A structural failure does not necessarily mean an immediate fracture. Cranes which were overloaded on previous lifts, have suffered from sudden structural failure during normal safe lifts.

Crane Accidents 04/12/2009 By : William A. Greco [email protected]

Page 7 of 7 Warrington,Pa.

Sidenote : Hydrogen Embrittlement and Stress Corrosion Cracking Hydrogen is a byproduct of corrosion and electrochemical processes and can become trapped inside of steel at the time of manufacture or by permeation, solubility, and diffusion. Hydrogen trap interactions cause cracks to occur which grow over time. Cyclic loading and load pulsing interplay with corrosion cracking from Hydrogen embrittlement and often cause failures. Although the phenomenon has been the subject of many engineering and research papers, books, conferences and symposiums, the full understanding of these characteristics are still a mystery. Delayed failures in high strength steels often occur under circumstances where tensile or bend tests show no evidence of britTleness. Long fracture delay times are not uncommon.