Make a schematic diagram (as Fig. 3.5.1) and estimate the weight of the boom and it’s centre of gravity.
Fig. 3.5.1
For an example we use the following:
– Centre of rotation Gthe hinge point of the boom GO
– Weight of the boom: G G130 t
– Distance from O to the centre of
gravity: B G34 m
– (With boom down): L1G50 m
– (With boom up): L2G17 m
– Distance from hinge point to centre
of the line PQ: A G22 m
– The asked for time for
boom-hoisting: t1G5,5 min
– Creeping time for the boom when starting and ending the hoist
movement, including latching: t2G1 min
– Time for really hoisting the boom: t3Gt1At2G5,5A1 G4,5 min
– Average speed: ûGPQAPR – Total efficiency of the wirerope
sheaves, the drum and the gearbox: ηG0,86 – Force in the boom-hoist-tackle, when
starting to hoist the boom: F1GG · B
(During the hoisting of the boom the force F1 becomes lower as the distance A becomes greater.)
(The influence of the wind from the rear becomes greater when the boom is hoisted.)
Take the motor power: N G280 kW n G1500 rev兾min
25 percent; eventual 40 percent rating faG160 percent.
The tackle of the boom hoist mechanism is 2 times Z. 2 · Z G2 · 9 G18.
This means that the force F1 is taken by a bundle of 2 · 9 G18 wire ropes.
The efficiency of the sheaves before the drum is:ηsG0,91.
The force in each of the 2 wire ropes, which are connected to the boom hoist drum is then:
F2G F1
2 · Z ·ηs
F2G 201 000
2 · 9 · 0,91G12 271 kg
Note: When a computer calculation of the wire rope forces during hoisting is made, and the weight of the underpart of the boom-forestays is taken into account, it will be found that these underparts give some assistance in lifting the first few degrees from the horizontal.
This means that the force in the boom-hoist tackle is reduced over the first 6 degrees by some 7 percent, thus using less motor power. This phenomenon can help in difficult cases.
Normally the wire ropes of the boom-hoist mechanisms will offer a safety factor of about six, against rupture. However, a boom-hoist mechanism is a system that can give rise to difficulties if one of the two wire ropes should fail. Cases of a boom or even a whole crane collapsing when a wire rope breakdown occurs are well known in the annals of crane history. If one wire rope should fail, the safety factor against rupture of the remaining wire rope would only be three which is not high enough for safe operation.
The safest manner for the reeving-in of the boom-hoist mechanism is shown in Fig. 3.5.2.
Fig. 3.5.2 Ideal reeving system for boom-hoist mechanism
3.6 Calculating the needed power of the crane-travelling motors. Wheelslip control –how to calculate the forces through skewing of the crane and trolley
Factors to be considered are:
1. The resistance due to nominal travelling.
2. The resistance due to the influence of the wind.
3. The resistance due to the acceleration of the rotating masses.
4. The resistance due to the acceleration of the linear moving masses.
Crane travelling speed m兾min 45 m兾min m兾sec ûG0,75 m兾sec Total efficiency of the gearings ηt ηtG0,9 Wheel resistance of the crane
travelling wheels kN兾t f G5 kg兾tG0,05 kN兾t Influence of the (side) wind:
FwGΣ(A · C ·η2) · q kN
Example
Reduction between motor and wheel:
Jbrake sheaves兾couplings B
Jgearbox,reduced B
JtotalGJt JtG2 kg m2
Calculation kN kW
1. Resistance due to nominal travelling: 2. Resistance due to
wind:
Calculation kN kW 3. Resistance due to
the acceleration of 4. Resistance due to
the acceleration of Lateral forces on the
Addition travelling wheels Motor power
1. Nominal travelling F1G67 kN N1G55,8 kW 2. Wind q G275 N兾m2 F2G510 kN N2G425 kW
Total, for nominal
travellingCwind ΣF G577 kN ΣN G480,8 kW
Lateral forces on the
Addition travelling wheels Motor power
During acceleration:
1. Nominal travelling F1G67 kN N1G55,8 kW 2. Wind, q G275 N兾m2 F2G510 kN N2G425 kW 3. Acceler. rot. masses,
taG6 sec F3G−kN N3G11,8 kW
4. Acceler. lin. masses,
taG6 sec F4G167,5 kN N4G139,5 kW Total during acceleration ΣFaG744,5 kN ΣN4G632,1 kW Overload factor of motors
during acceleration
faGMmax: Mnom faG160 percent ΣN G632,1 1,6 G395 kW Take motors, each 20 kW; so 24 motors.
Total available power N G24 · 20 G480 kW.
The influence of a slope
If a crane has to drive up a slope, an additional resistance has to be overcome. Assume that a rubber tyred gantry (RTG) has to run against the slope of αG1 degree. The total weight of the loaded RTG is QtG 165 tons; the crane travelling speed is
û G135 m兾minG2,25 m兾sec
– Check the minimum wheel load on a driven crane wheel: P1G . . . . kN
– Check the maximum driving force that the crane travelling motor can exert on the circumference of the driven wheel (traction force):
P2Gfa· N ·η
û kN
– Friction coefficient between rail and wheel: mu GP2兾P1
– Allowed is mu G0,12.
Skewing of the crane and trolley
Cranes, and trolleys, can skew. This can cause severe wear and tear of the rails and the travelling wheels. The FEM standards mention the following (in booklet 2) about skew:
2.2.3.3 Transverse reactions due to rolling action
When two wheels (or two bogies) roll along a rail, the couple formed by the horizontal forces normal to the rail shall be taken into consideration. The components of this couple are obtained by multiplying the vertical load exerted on the wheels (or bogies) by a coefficient λ which depends upon the ratio of the spanρto the wheelbase a.(1)
(1)By ‘wheelbase’ is understood the centre distance between the outermost pairs of wheels, or, in the case of bogies, the centre distance between the fulcrum pins on the crane structure of the two bogies or bogie systems. Where horizontal guiding wheels are provided, the wheelbase shall be the distance between the rail contact points of two horizontal wheels.
As shown in the graph [Fig. 3.6.2], this coefficient lies between 0,05 and 0,2 for ratios ofρ兾a between 2 and 8.
However DIN and other standards give a more complex calculation about the horizontal forces through skewing. This calculation leads to greater forces than those mentioned in Fig. 3.6.2.
Advice
– Take the skew forces on crane- and trolley wheels as a minimum as 10 percent of the maximum wheel load.
– Take the skew forces on crane- and trolley wheels as 20 percent of the maximum wheel load for cranes and trolleys running û G 150 m兾m or more.
– Also check the calculations according to Fig. 3.6.2.
In order to keep the skewing forces on a crane travelling mechanism under reasonable control, the length兾width ratio, being the relation between the railspan or railgauge, and the centre distance between the fulcrum pins of the crane travelling mechanism under each corner ρ: a or L : B, should be at least 6 : 1.
Fig. 3.6.1
Fig. 3.6.2