BODYWORK CALCULATIONS 2

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1 Table of content BODYWORK CALCULATIONS...2 PRINCIPLES OF CALCULATION...3 OPTIMIZING LOAD...6 EXAMPLE OF CALCULATION...7 Example 1 4X2 Tractor with two axles...7 Example 2 6X4 Tractor with three axles...9 Example 3 4X2 Crane behind cab Example 4 6X2 Rear-mounted crane...12 Example 5 4X2 Calculating length...13 Example 6 6X2 Calculating centre of gravity...15 Example 7 6x2/4 Tractor...16 Example 8 8x4 Centre of gravity calculation...18 Example 9 8x4*4 Centre of gravity calculation

2 BODYWORK CALCULATIONS 2

3 PRINCIPLES OF CALCULATION The procedure for calculating a suitable platform length, payload and axle weight is based on a fewsimple rules. The sum of the downward forces must always equal the sum of the upward forces. If a plank (assumed to be weightless) is placed on two supports and a 100 kg weight is positioned at the mid-point, the supports will each carry half of the weight, 50 kg each. If the weight is positioned above one of the supports, this will be loaded with 100 kg and the other support will bear no load. If the weight is placed on the outside of one of the supports, the plank will lift from the other support. To prevent the plank from tipping, a weight of at least 20 kg must be placed over the first support. There is now equilibrium. In this case, the entire load will be carried by the second support. 3

4 This is popularly known as the lever principle. Replace one of the supports by a wheel and the other by a man lifting. If a weight is placed close to the man, he must lift a large part of the weight. The closer the weight is moved towards the wheel, the less of the 100 kg weight must be lifted by the man. However, if the weight is moved beyond the axis of the wheel, he must exert a downward force on the plank to prevent it from tilting upwards. How does the weight supported by the man vary in relation to the position of the weight? The weight (load) is designated L (kg). The load (reaction force exerted on the man) is denoted F (kg). The distance from the wheel axis to the centre of gravity of the weight (load) is designated as lever H (mm). The distance between the loading points (wheel axis and man) is designated as A (mm). 4

5 In order to achieve equilibrium, the load (L) multiplied by its lever (H) must be equal to the force of reaction (F) multiplied by its lever, distance (A). Axle weight and body calculations for trucks are based on this simple lever principle using the formula: L x H = F x A Load (L) x Lever (H) = Reaction (F) x Distance (A) This equation may be rearranged in order to calculate load (L), reaction (F) or lever (H). The wheel in the above example can be replaced by the front wheel of the truck and the man can be replaced by the rear wheel of the truck. Load (L) x Lever (H) = Reaction (F) x Distance (A) Load (L) = Reaction (F) = Reaction (F) x Distance (A) Lever (H) Load (L) x Lever (H) Distance (A) The weight may be replaced by the truck body and load. The centre of gravity of the body and load are assumed to be at the mid-point of the platform. For the sake of simplicity we also use the unit (kg) for loads, i.e. forces. Lever (H) = Reaction (F) x Distance (A) Load (L) 5

6 OPTIMIZING LOAD All types of transport work using a truck require that the truck chassis be equipped with some type of bodywork. The intention of bodywork calculations is to optimize the chassis and the location of the bodywork in order to achieve maximum payload without exceeding maximum permissible axle weight and bogie weight, taking note of legal and technical limitations. This booklet deals with the main principles for bodywork calculations. Scania distributors and dealers have a PC-based calculation program for load/weight optimization and are able to offer assistance in bodywork calculation. Optimizing the load requires data on the weight and dimensions of the chassis. The distributor has access to chassis weights. In many countries, chassis weights are also available on the distributor s website. Example of PC calculation Front Rear Total Chassis weight Additional weight Bodywork weight Weight Bodywork equipment Unladen weight Load Load Weight of payload Unladen weight Weight of payload Gross train weight Maximum permissible weight Weight reserve Weight on steered axles 66% On steered front axles 43% Slip limit Asphalt 31% Slip limit Gravel roads 18% 6

7 EXAMPLE OF CALCULATION Example 1 4X2 Tractor with two axles Calculation of front and rear axle weights (P A and P B ) on a two-axle tractor with a kingpin load (L). Applying the lever principle gives the following equation: if: A = 4200 mm L = kg H = 3600 mm F = L x H B A This gives the following distribution of load on the rear axle (F B ). FB =Kingpin load carried by the rear axle A = Wheelbase L = Kingpin load H =Distance between front axle and fifth wheel F = x 3600 = 8571 kg B 4200 Rear axle weight (P B ) is the sum of load distribution on the rear axle (F B ) and the truck s chassis weight on the rear axle (T B ). P B = F B + T B 7

8 If chassis weight at the rear, T B = 4000 kg this gives a rear axle weight (P B ) as follows: P B = = kg Load distribution on the front axle (F A ) is calculated by subtracting the load distribution on the rear axle from the total load (L). F A = L - F B Load distribution on the front axle (F A ) in this example is as follows: F A = = 1429 kg Front axle weight (P A ) is obtained in the same manner as the rear axle weight, i.e. the sum of load distribution on the front axle (F A ) and the truck s chassis weight at the front (T A ). P A = F A + T A If chassis weight at the front, T A = 4500 kg this gives a front axle weight (P A ) as follows: P A = = 5929 kg 8

9 Example 2 6X4 Tractor with three axles Calculation of the position of the fifth wheel (H) on a three-axle truck in order to achieve optimum use of front axle weight and bogie weight. Applying the lever principle gives the following equation: where: H F B A B L H = F B x (A + B) L = Distance between front axle and kingpin = Maximum permitted load (kingpin) on bogie = Wheelbase = Distance to centre of gravity of bogie = Max. permitted load (kingpin) The distance to the centre of gravity of the bogie (B) for the various chassis types is given in the main dimension drawings. By subtracting the truck s chassis weight at the rear (T B ) from the maximum permitted bogie weight, (P B) maximum permitted load on the bogie (F B ) can be calculated. F B = P B - T B if: P B = max kg T B = 5000 kg This gives the following distribution of the load on the bogie (F B ) F B = = kg 9

10 Maximum permitted load (L) is calculated by adding the max. permitted load on the bogie (F B ) and the max. permitted load on the front axle (F A ) L = F B + F A The maximum permitted load on the front axle (F A ) is calculated in the same manner as the max. permitted load on the bogie (F B ) as follows: F A = P A - T A if: P A = 7000 kg T A = 5000 kg this gives the following max. permitted load (L): F A = = 2000 kg L = = kg if: A = 4200 mm B = 675 mm (6x4) this gives the following optimum placing of the fifth wheel. H = x ( ) = 4300 mm In other words, in order to optimize the utilization of axle weights, the fifth wheel must be located 4300 mm from the front axle (100 mm behind the first driven axle). 10

11 Example 3 4X2 Crane behind cab Equipment inside the wheelbase such as a crane behind the cab. If the truck is equipped with heavy optional equipment such as a crane behind the cab, the weight distribution of the crane on the front axle and rear axle must be calculated before the body calculations above can be carried out. Applying the lever principle gives the following equation: K = K x C B A K B = Weight of crane carried by rear axle K = Total weight of crane C = Distance between front axle and centre of gravity of crane A = Wheelbase if: K = 1950 kg C = 802 mm A = 4300 mm the rear axle (K B ) will bear the following proportion of the total weight of the crane (K). Crane weight on front axle (K A ) will then be: K A = K - K B K A = = 1586 kg The weight of the crane on the front axle (K A ) and rear axle (K B ) are then added to the chassis weight of the truck at the front (T A ) or rear (T B ) in order to carry out further bodywork calculations. See example 5. K = 1950 x 802 = 364 kg B

12 Example 4 6X2 Rear-mounted crane Equipment outside the wheelbase such as a rearmounted crane. If the truck has heavy optional equipment such as a rear-mounted crane, the weight distribution of the crane on the front and rear axle must be calculated before carrying out the bodywork calculations above. Applying the lever principle gives the following equation: K = K x C B (A+B) K B = Weight of crane carried by rear axle K = Total weight of crane C = Distance between front axle and centre of gravity of crane A = Wheelbase B = Distance to centre of gravity of bogie if: K = 2500 kg C = 7400 mm A = 4600 mm B = 612 mm (6x2) the following proportion of the total weight (K) of the crane is supported by the rear axle (K B ) The weight of the crane on the front axle (K A ) will then be: K A = K - K B K A = = kg Note that K A is negative meaning that the front axle will be loaded with 1050 kg. Crane weight on the rear axle (K B ) is added to the chassis weight of the truck at the rear (T B ) and the reduced crane weight on the front axle (K A ) subtracted from the chassis weight of the truck at the front (T A ) in order to carry out further bodywork calculations. K = 2500 x 7400 = 3550 kg B ( ) 12

13 Example 5 4X2 Calculating length Calculation of body length. Same truck and equipment as in example 3. Applying the lever principle gives the following equation: H = F B x A L The maximum permitted load on the rear axle (F B ) is calculated by subtracting the truck s chassis weight rear (T B ) and crane weight rear (K B ) from the maximum permitted rear axle weight (P B ). The maximum permitted load (L) is calculated by adding the maximum permitted load on the front axle (F A ) to that on the rear axle (F B ). L = F A + F B The maximum permitted load on the front axle (F A ) is calculated in the same manner as the maximum permitted load on the rear axle (F B ) i.e.: F B = P B - T B - K B F A = P A - T A - K A if: P B = kg T B = 1780 kg K B = 364 kg (as example 3). this gives the following maximum permitted load on the rear axle. F B = =7856 kg if: P A = 6500 kg T A = 5000 kg K A = 1130 kg (as example 3) this gives the following maximum permitted load (L): F A = = 1654 kg L = = 9510 kg 13

14 In example 3, wheelbase (A) = 4300 mm. The distance between the front axle and the centre of gravity of the platform + load is thus as follows: Rear overhang (J) may be calculated as follows: H = 7856 x 4300 = 3552 mm 9510 J = D + X - A i.e. the platform + the truck s centre of gravity must be located 3552 mm (H) behind the font axle or = 748 mm (Y) in front of the rear axle to optimally utilise the maximum permissible axle weight. If the centre of gravity of the platform + load is assumed to be located at the mid-point of the platform, as in this example, platform length may then be calculated as below. The maximum length of the platform from the centre of gravity and forward is limited by the crane and its base, i.e, distance (D). X/2 cannot be less than: X/2 = H - D if D = 1352 mm, X/2 is: X/2 = = 2200 mm Platform length is then: J = = 1452 mm Comments: In this example we have calculated backwards by establishing D through the use of SCANIA s calculation program. This naturally makes it simpler and quicker to find a suitable vehicle. The result is a vehicle with weights and locations of crane and platform entirely optimized. The calculation program also makes it possible to gain some load capacity by choosing a front or rear axle with lower permissible weight capacity, if such a change is more optimal from a weight distribution perspective. The authorities in most countries will approve the vehicle even if the centre of gravity of the load is not at precisely the same location as that of the platform. In practice this has little or no significance. However, check the national regulations. X = X/2 + X/2 X = 4400 mm 14

15 Example 6 6X2 Calculating centre of gravity Calculation of the distance (E) between the mid-point of a given body (theoretical centre of gravity) and the centre of gravity in order to reach maximum axle weight. Applying the lever principle gives the following equation: H = F B x (A + B) L H = Distance between front axle and centre of gravity of load for maximum axle weight utilization. FB = Maximum permitted load on bogie A = Wheelbase B = Distance to centre of gravity of bogie L = Maximum permitted load including bodywork this gives the following position of centre of gravity for maximum axle weight. H = x ( ) = 4595 mm If the body in this example is 8000 mm and the distance between the front axle and body is 650 mm, the distance (E) between the centre of gravity for maximum axle weight and the mid-point of the body (theoretical centre of gravity) is as follows: E = D + X/2 - H if: F B = kg A = 5000 mm B = 553 mm (6 x 2) L = kg For calculation of (L) and (F B ), see earlier example. E = = 55 mm Check with national regulations that this distance (E) is within the limits given. 15

16 Example 7 6x2/4 Tractor Calculation of the position of the fifth wheel (H) on a threeaxle truck, with the tag axle in front of the driven axle, in order to achieve optimum use of front axle weight and bogie weight. Applying the lever principle gives the following equation: H = F B x (A - B) L if: H = Distance between front axle and fifth wheel F B = Max. permitted load (kingpin) on bogie A = Wheelbase B = Distance to centre of gravity of bogie L = Max. permitted load (kingpin) The distance to the centre of gravity (B) of the bogie for the different types of chassis is given in the main dimension drawings. By subtracting truck chassis weight at the rear (T B ) from the maximum permitted bogie weight (P B ), maximum permitted load on the bogie (F B ) can be calculated. if: F B = P B - T B P B = max kg T B = 5000 kg This gives the following distribution of load on the bogie (F B ) F B = = kg 16

17 Maximum permitted load (L) is calculated by adding the max. permitted load on the bogie (F B ) and the maximum permitted load on the front axle (F A ) L = F B + F A The maximum permitted load on the front axle (F A ) is calculated in the same manner as the max. permitted load on the bogie (F B ) as follows: F A = P A - T A if: P A = 7000 kg T A = 5000 kg this gives the following maximum permitted load (L): F A = = 2000 kg L = = kg if: A = 4100 mm B = 675 mm this gives the following optimum location of the fifth wheel H = x ( ) = 3022 mm This means that in order to optimize the use of axle weight, the fifth wheel should be located 3022 mm from the front axle. 17

18 Example 8 8x4 Centre of gravity calculation To calculate: Dimension (E), the distance between the centre of gravity of the platform and the optimum centre of gravity of the platform/load (H). Weight front Weight rear Weight tot. Target, laden FA = FB = F tot = truck Chassis weight Load + bodywork PB = PB = L = A = 5000 mm B = 677,5 mm C = 970 mm D = 650 mm F A = kg F B = kg L = kg X = mm Calculation H = 18 AT = Theoretical wh eelbase H = Optimal centre of gravity load/platform L = Maximum weight load + bodywork E = Distance between H and the centre of the bodywork = Load + bodywork load on P B AT = A + B C = ,5-970 = 4 707,5 mm AT x PB 4707,5 x = L = 3064 mm E = X/2 + D C H = = 116 mm Dimension (E), the distance between the practical and optimal values of H is = 116 mm. The bodywork should be 116 mm further forward towards the cab to achieve optimal load distribution. Remarks: The distance (D) may be a minimum dimension, e.g. if a front cylinder is to fitted between the cab and bodywork. The chosen length of the bodywork (X) may be a dimension that the bodywork builder has chosen as standard. Deviating from this standard may mean paying a much higher price. The chosen wheelbase (A), 5000 mm, is very long for a tipper truck but for the sake of the calculation has no significance. From a stability perspective a shorter wheelbase would be preferable. But some countries require an even longer wheelbase to make it possible to load the vehicle to the maximum. Also check national regulations to make sure that the distance (E) is within permissible limits.

19 Example 9 8x4*4 Centre of gravity calculation To calculate: The optimal centre of gravity of the bodywork/ load should coincide with the mid-point of the bodywork. H should thus be equal to D + X/2 and E should be equal to 0. Weight front Weight rear Weight tot. Target, laden truck FA = FB = F tot = Chassis weight Load + bodywork PA = PB = L = A = 3350 mm B = 1256 mm F A = 7100 kg F B = kg L = kg X = 6200 mm AT = 4606 mm (as on main dimen sion drawing) P B = Load + bodywork load on rear axles AT = Theoretical wheelbase H = Optimal centre of gravity load/platform L = Maximum weight for load + bodywork E = Distance between H and the centre of the bodywork X = Length of bodywork D = Dimension between front axle and front edge of bodywork Calculation: AT x PB 4606 x H = = = 4131 mm L Since the requirement is that the centre of gravity of the bodywork should be precisely above the centre of gravity of the load and bodywork, D is as follows: D = H X/2 = = 1031 mm The distance between the front axle and bodywork is: D = 1031 mm och E = 0. 19

Axle load calculations

All types of transportation work using trucks require the truck chassis to be supplemented by some form of bodywork. The aim of axle load calculations is to optimise the chassis and bodywork location.