THE ART OF HEAVY TRANSPORT Marco J. van Daal
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F Nomenclature
The pendulum axle The pendulum axle is quite different from the fusee axle even though their function is similar. While assessing the pendulum axle, where the fusee axle would show a shaft running across the full width of the transporter, one will quickly see that such a shaft is not present on a pendulum axle. Instead there are two independent axle assemblies, one on the left and one on the right side of the transporter. Each assembly is mounted on a turn table against the underside of the transporter deck (see C in the picture). Therefore every pendulum axle can rotate in the horizontal plane. The turn table holds the upper leg of the axle which is fixed. The turn table and the upper leg are bound to make the same motion. The upper leg ends in a knee joint (see A in the picture) that joins the upper leg to the lower leg, the lower leg in turn connects to the wheel assembly. The knee joint allows the lower leg to pivot in respect to the upper leg. This pivoting motion is initiated by the hydraulic axle cylinder (see B in the picture). It is these hydraulic axle cylinders that can be plumbed into groups that form the hydraulic suspension for which transporters are so well known.
Figure F-18, Pendulum axle assembly. A -knee joint. B -hydraulic axle cylinder. C -turn table. Transporter brand: Goldhofer.
In summary, the difference between a fusee axle and pendulum axle is that the pendulum axle assembly includes a turntable that mounts the axle to the underside of the transporter. The fusee axle includes a shaft that mounts to both wheel sets. This difference in construction or design brings with it a possible clash in terminology. When talking about fusee axles, an axle really means one axle. However, when talking about pendulum axles, one axle in reality means two axles. For this reason the term “axle lines� is introduced. One axle line includes all axles between the left and the rights side of a transporter. THE ART OF HEAVY TRANSPORT
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Figure 1-8, One of a combination of two 12-axle lines of self propelled transporters rolling off a barge. The “knees” are facing forward direction. Transporters are carrying a 450 ton (991,189 LBS) conveyor crane. Location: Houston, Texas, USA. Transporter brand: Scheuerle self propelled.
Figure 1-9, It is difficult to see in this picture but the second and third axle (see red arrows) have their “knees’ facing in forward direction just as all the rest of the 12-axle lines. This transporter is carrying a 280 ton (616,740 LBS) transformer. Location: Al Taweelah, United Arab Emirates, Middle East. Transporter brand: pull type Cometto. THE ART OF HEAVY TRANSPORT
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1 Principle working of the Hydraulic Platform Transporter
Figure 1-10, three sets of 12-axle lines self propelled transporters carrying a 900 ton deck. Even though this is a side move there is a distinct forward and reverse direction. The “knees” of each set of transporters are facing the same forward direction. Location: Morgan City, Louisiana, USA. Transporter brand: Scheuerle self propelled.
1.6.3.2 Suspension groups This section explains how the operator and his “signal men” can communicate, avoiding confusion, by agreeing on a simple yet important naming convention. This naming convention is based on the principle of assigning a name to each of the four corners of the transporter. Each corner represents a suspension group. When the level of the transporter needs to be corrected, (“go up” or “go down”) the corner or suspension group in question is called as opposed to “go down on the front right corner”. In the noisy environment the names assigned to each corner should be unique and it should not be possible to mistake one name for another. The naming convention that I personally have had the most success (read: the least problems) with is the following: ALPHA BRAVO CHARLY DELTA THE ART OF HEAVY TRANSPORT
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1 Principle working of the Hydraulic Platform Transporter
Picture 3, 80 mtr (262'-5.6") long vessel weighing 370 ton (814,978 LBS). Location: Nanjing, People's Republic of China Transporter Brand: Nicolas self propelled and pull type
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8.1
8 Limiting stability angle
Longitudinal limiting stability angle for 4-point suspension
According to preceding sections, the longitudinal structural (S) and longitudinal hydraulic (H) stability angles are determined by the following formulas; tanSα S1 =
D Rcog −S1 D CoG − Rcog
and
tanHα P1 =
D Rcog −P1 D CoG − Rcog
The below two figures will aid in the understanding of this theory.
Figure 8-1, Visualization longitudinal limiting stability area, 4-point suspension
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∆y =
D DELTA −Y * WALPHA/BRAV O + D CHARLY −DELTA * WCHARLY − D DELTA −Y Wtot where D DELTA −Y = 1/2 * D CHARLY −DELTA where WCHARLY = 1/3 * Max. Cap
Substitution gives the following equation; D CHARLY −DELTA Max.Cap * WALPHA/BRAV O + D CHARLY −DELTA * D 2 3 ∆y = − CHARLY −DELTA Wtot 2
equation (1)
Moment calculation for DELTA in x-direction;
∑M
DELTA − X
= (D DELTA −X + ∆x) * Wtot − D DELTA− ALPHA * WALPHA/BRAV O = 0
(D DELTA −X + ∆x) * Wtot = D DELTA −ALPHA * WALPHA/BRAV O WALPHA/BRAV O =
(D DELTA −X + ∆x) * Wtot D DELTA −ALPHA
where D DELTA −X = 1/3 * D DELTA −ALPHA where ∆x = 0 (This is the assumption) Substitution gives the following equation;
WALPHA/BRAV O
D DELTA −ALPHA * Wtot W 3 = = tot equation (2) D DELTA −ALPHA 3
Substitute equation (2) in equation (1); D CHARLY −DELTA Wtot Max.Cap * + D CHARLY −DELTA * D 2 3 3 ∆y = − CHARLY −DELTA Wtot 2 Substitution in the stability angle formula gives; tanSβ S2 =
D Rcog −S2 D CoG −Rcog
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8 Limiting stability angle
tanSβ S2 =
∆y D CoG − Rcog
D CHARLY −DELTA Wtot Max.Cap * + D CHARLY −DELTA * D 2 3 3 − CHARLY −DELTA Wtot 2 = D CoG −Rcog
For the hydraulic stability formula; tanHβ P2 =
D Rcog −P2 D CoG −Rcog
where D Rcog −P2 = 2/3 * D CHARLY −Y = 1/2 * 2/3 * D CHARLY −DELTA Substitution gives the following equation;
tanHβ P2
2 * D CHARLY −DELTA D CHARLY −DELTA 2*3 3 = = D CoG −Rcog D CoG −Rcog
Set the structural (S) and hydraulic (H) stability angles equal to each other; tanSβ S2 = tanHβ P2 D CHARLY −DELTA Wtot Max.Cap * + D CHARLY − DELTA * D 2 3 3 D CHARLY −DELTA − CHARLY −DELTA Wtot 2 3 = D CoG −Rcog D CoG − Rcog D CHARLY −DELTA Wtot Max.Cap * + D CHARLY −DELTA * D D 2 3 3 − CHARLY − DELTA = CHARLY −DELTA Wtot 2 3 D CHARLY −DELTA Wtot Max.Cap * + D CHARLY −DELTA * D D 2 3 3 = CHARLY −DELTA + CHARLY −DELTA Wtot 2 3 1 1 5 * D CHARLY −DELTA * Wtot + * D CHARLY −DELTA * Max.Cap = * D CHARLY −DELTA * Wtot 6 3 6 1 4 * D CHARLY − DELTA * Max.Cap = * D CHARLY −DELTA * Wtot 3 6
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9
Engineering case studies
This section provides the case studies of real life example heavy transports. All the aspects covered so far in this book will be addressed in these case studies.
9.1
Case study 1
The cargo comprises a diesel engine, 175 ton (385,463 LBS) in weight (including the transport beams). Its CoG is located 2,000 mm (6’-6.7”) above the transporter deck (including the transport beams). The transport utilizes a 12-axle line configuration with 3-point suspension, the CoG is located in the centroid of the stability area. The stability area is located at half the axle height; this is at 300 mm (1’-0”). These are the transporter particulars; ALPHA/BRAVO 8 axles CHARLY 8 axles DELTA 8 axles Ad = axle distance Tl = transporter length Th = transporter height Tw = transporter width Dab = axle base distance Wa = axle line weight Ac = axle capacity
1,500 mm (4’-11.1”) 18,000 mm (59’-0.7”) 1,200 mm (3’-11.2”) 3,000 mm (9’-10.1”) 1,800 mm (5’-10.9”) 3.5 ton/axle line (7,709 LBS) 15 ton/axle (33,040 LBS)
Determine the hydraulic and structural stability.
Figure 9-1, A 175 ton (385,463 LBS) diesel engine, case study 1 THE ART OF HEAVY TRANSPORT
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11 External forces
11.3.1 Wind forces, example 1
Figure 11-10, Wind forces, example 1
As this figure is taken from a preceding section, a number of properties have already been calculated, below a summary. Coordinates Rcog Rcog (4,500; 1,500; 300) mm or Rcog (14’-9.2”; 4’-11.1”; 1’-0”) Coordinates CoG CoG (4,500; 1,500; 3,800) mm or CoG (14’-9.2”; 4’-11.1”; 12’-5.6”) DCoG-Rcog = 3,500 mm (11’-5.8”) DALPHA-BRAVO DBRAVO-CHARLY DCHARLY-DELTA DDELTA-ALPHA Initial loads; WALPHA WBRAVO WCHARLY WDELTA Wtot
= 1,800 mm = 4,500 mm = 1,800 mm = 4,500 mm
= 50.0 ton = 50.0 ton = 50.0 ton = 50.0 ton = 200 ton
(5’-10.9”) (14’-9.2”) (5’-10.9”) (14’-9.2”)
(110,132 LBS) (110,132 LBS) (110,132 LBS) (110,132 LBS) (440,529 LBS)
16.667 ton/axle 16.667 ton/axle 16.667 ton/axle 16.667 ton/axle
(36,784 LBS) (36,784 LBS) (36,784 LBS) (36,784 LBS)
For this example it is assumed that a Beaufort 4 wind is blowing with a velocity of 7.9 mtr/sec (17.67 MPH). The wind angle φ is 30 degrees and the shape factor (Sf) is 120% or 1.2. Furthermore AFRONT=AREAR=150 mtr2 (1,615 ft2) and ASIDE=400 mtr2 (4,306 ft2). The areas include the transporter areas. DCoW-Rcog = 3,000 mm (9’-10.1”)
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The Beaufort scale shows that for a wind velocity of 7.9 mtr/sec (17.67 MPH) the wind pressure (WP) is 3.9 kg/mtr2 (0.36 LBS/ft2). Resolving the wind pressure (WP) in longitudinal and transverse direction leads to;
WP - L = cos(ϕ ) * WP WP - L = cos(30) * 3.9 = 3.38 kg/mtr 2
WP -T = sin (ϕ ) * WP
and
WP -T = sin (30) * 3.9 = 1.95 kg/mtr 2
and
The wind force can be determined. FWIND-L = WP -L * A FRONT/REAR * Sf
and
FWIND-T = WP -T * A SIDE * Sf
3.38 *150 *1.2 = 0.608 ton (1,339 LBS) 1,000 1.95 = * 400 *1.2 = 0.936 ton (2,062 LBS) 1,000
FWIND-L = FWIND-T
The wind forces are now to be converted into an additional force on the suspension groups. The longitudinal wind force has an influence on the suspension groups ALPHA and BRAVO or the suspension groups CHARLY and DELTA. The base distance (DBASE) is DBRAVO-CHARLY. The transverse wind force has an influence on the suspension groups BRAVO and CHARLY or the suspension groups DELTA and ALPHA. The base distance (DBASE) is DALPHA-BRAVO. The figure below visualizes this.
Figure 11-11, Wind force visualization, plan view and schematics
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Figure 13-6, Lashing forces in the various quadrants
Lashing in quadrant Q1 Q1 Q1 Q2 Q2 Q2 Q3 Q3 Q3 Q4 Q4 Q4
Resolved component XQ1 YQ1 ZQ1 XQ2 YQ2 ZQ2 XQ3 YQ3 ZQ3 XQ4 YQ4 ZQ4
sin ϕ pos pos pos pos pos pos pos pos -
cos ϕ pos pos pos pos
sin Ω pos pos neg neg -
cos Ω pos neg neg pos -
Direction of resulting force X(pos) Y(pos) Z(pos) X(neg) Y(pos) Z(pos) X(neg) Y(neg) Z(pos) X(pos) Y(neg) Z(pos)
Table 13-5, Direction of each of the lashing components
THE ART OF HEAVY TRANSPORT