Fifth Wheel Load Calculation

Bodywork calculations 3Table 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 ................................ 11 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 ............19 © Scania CV AB 2003 1 Bodywork calculations 3 BODYWORK CALCULATIONS 2 © Scania CV AB 2003 . 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. the supports will each carry half of the weight. In this case. There is now equilibrium. the entire load will be carried by the second support. 50 kg each. The sum of the downward forces must always equal the sum of the upward forces. a weight of at least 20 kg must be placed over the first support.Bodywork calculations 3 PRINCIPLES OF CALCULATION The procedure for calculating a suitable platform length. © Scania CV AB 2003 3 . the plank will lift from the other support. If a plank (assumed to be weightless) is placed on two supports and a 100 kg weight is positioned at the mid-point. To prevent the plank from tipping. If the weight is placed on the outside of one of the supports. payload and axle weight is based on a fewsimple rules. The closer the weight is moved towards the wheel. Replace one of the supports by a wheel and the other by a man lifting. if the weight is moved beyond the axis of the wheel. How does the weight supported by the man vary in relation to the position of the weight? The weight (load) is designated L (kg). 4 © Scania CV AB 2003 . The distance between the loading points (wheel axis and man) is designated as A (mm).Bodywork calculations 3 This is popularly known as the lever principle. However. he must exert a downward force on the plank to prevent it from tilting upwards. the less of the 100 kg weight must be lifted by the man. If a weight is placed close to the man. he must lift a large part of the weight. The distance from the wheel axis to the centre of gravity of the weight (load) is designated as lever H (mm). The load (reaction force exerted on the man) is denoted F (kg). i. Axle weight and body calculations for trucks are based on this simple lever principle using the formula: Lx H = FxA 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). forces. distance (A). For the sake of simplicity we also use the unit (kg) for loads. The centre of gravity of the body and load are assumed to be at the mid-point of the platform. the load (L) multiplied by its lever (H) must be equal to the force of reaction (F) multiplied by its lever.Bodywork calculations 3 In order to achieve equilibrium. Load (L) x Lever (H) = Reaction (F) x Distance (A) Load (L) = Reaction (F) x Distance (A) Lever (H) Reaction (F) = Load (L) x Lever (H) Distance (A) Lever (H) = Reaction (F) x Distance (A) Load (L) © Scania CV AB 2003 5 . 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. The weight may be replaced by the truck body and load.e. This booklet deals with the main principles for bodywork calculations. taking note of legal and technical limitations. chassis weights are also available on the distributor’s website. Scania distributors and dealers have a PC-based calculation program for load/weight optimization and are able to offer assistance in bodywork calculation. The distributor has access to chassis weights. Optimizing the load requires data on the weight and dimensions of the chassis. 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. Example of PC calculation Front Chassis weight Additional weight Bodywork weight Weight 1-4 Bodywork equipment Unladen weight Load 0 Load 1-4 Weight of payload Unladen weight Weight of payload Gross train weight Maximum permissible weight Weight reserve 6445 0 1146 0 2135 9726 Rear Total 2585 9030 0 0 3404 4550 0 0 -135 2000 5854 15580 3885 11535 15420 0 0 0 3885 11535 15420 9726 5854 15580 3885 11535 15420 13611 17389 31000 14200 19000 32000 589 1611 1000 Weight on steered axles On steered front axles Slip limit Asphalt Slip limit Gravel roads 66% 43% 31% 18% 6 © Scania CV AB 2003 .Bodywork calculations 3 OPTIMIZING LOAD All types of transport work using a truck require that the truck chassis be equipped with some type of bodywork. In many countries. PB = FB + TB © Scania CV AB 2003 7 . FB = 10000 x 3600 = 8571 kg 4200 Rear axle weight (PB) is the sum of load distribution on the rear axle (FB) and the truck’s chassis weight on the rear axle (TB).Bodywork calculations 3 EXAMPLE OF CALCULATION Example 1 4X2 Tractor with two axles Calculation of front and rear axle weights (PA and PB) on a two-axle tractor with a kingpin load (L). Applying the lever principle gives the following equation: FB = L x H A FB A L H =Kingpin load carried by the rear axle = Wheelbase = Kingpin load =Distance between front axle and fifth wheel if: A = 4200 mm L = 10000 kg H = 3600 mm This gives the following distribution of load on the rear axle (FB). TA = 4500 kg this gives a front axle weight (PA) as follows: PA = 1429 + 4500 = 5929 kg 8 © Scania CV AB 2003 . the sum of load distribution on the front axle (FA) and the truck’s chassis weight at the front (TA). FA = L . i.FB Load distribution on the front axle (FA) in this example is as follows: FA = 10000 .8571 = 1429 kg Front axle weight (PA) is obtained in the same manner as the rear axle weight.Bodywork calculations 3 If chassis weight at the rear.e. PA = FA + TA If chassis weight at the front. TB = 4000 kg this gives a rear axle weight (PB) as follows: PB = 8571 + 4000 = 12571 kg Load distribution on the front axle (FA) is calculated by subtracting the load distribution on the rear axle from the total load (L). 5000 = 15000 kg The distance to the centre of gravity of the bogie (B) for the various chassis types is given in the main dimension drawings.Bodywork calculations 3 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: H = FB x (A + B) L By subtracting the truck’s chassis weight at the rear (TB) from the maximum permitted bogie weight. permitted load (kingpin) This gives the following distribution of the load on the bogie (FB) FB = 20000 . (PB) maximum permitted load on the bogie (FB) can be calculated. 20000 kg 5000 kg where: H FB A B L = Distance between front axle and kingpin = Maximum permitted load (kingpin) on bogie = Wheelbase = Distance to centre of gravity of bogie = Max. © Scania CV AB 2003 9 .TB if: PB TB = = max. FB = PB . the fifth wheel must be located 4300 mm from the front axle (100 mm behind the first driven axle). 10 © Scania CV AB 2003 .Bodywork calculations 3 Maximum permitted load (L) is calculated by adding the max. in order to optimize the utilization of axle weights.TA if: PA = 7000 kg TA = 5000 kg this gives the following max. permitted load on the front axle (FA) L = FB + FA The maximum permitted load on the front axle (FA) is calculated in the same manner as the max. permitted load on the bogie (FB) as follows: F A = P A .5000 = 2000 kg L = 15000 + 2000 = 17000 kg if: A = 4200 mm B = 675 mm (6x4) this gives the following optimum placing of the fifth wheel. permitted load (L): FA = 7000 . H = 15000 x (4200 + 675) = 4300 mm 17000 In other words. permitted load on the bogie (FB) and the max. See example 5. 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: KB = K x C A KB = 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 Crane weight on front axle (KA) will then be: KA = K . KB = 1950 x 802 = 364 kg 4300 The weight of the crane on the front axle (KA) and rear axle (KB) are then added to the chassis weight of the truck at the front (TA) or rear (TB) in order to carry out further bodywork calculations. If the truck is equipped with heavy optional equipment such as a crane behind the cab.KB KA = 1950 .Bodywork calculations 3 Example 3 4X2 Crane behind cab Equipment inside the wheelbase such as a crane behind the cab.364 = 1586 kg the rear axle (KB) will bear the following proportion of the total weight of the crane (K). © Scania CV AB 2003 11 . 3550 = -1050 kg Note that KA is negative meaning that the front axle will be loaded with 1050 kg. the weight distribution of the crane on the front and rear axle must be calculated before carrying out the bodywork calculations above. Crane weight on the rear axle (KB) is added to the chassis weight of the truck at the rear (TB) and the reduced crane weight on the front axle (KA) subtracted from the chassis weight of the truck at the front (TA) in order to carry out further bodywork calculations. If the truck has heavy optional equipment such as a rear-mounted crane.KB KA = 2500 .Bodywork calculations 3 Example 4 6X2 Rear-mounted crane Equipment outside the wheelbase such as a rearmounted crane. Applying the lever principle gives the following equation: KB = K x C (A+B) KB = 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 C A B = 2500 kg = 7400 mm = 4600 mm = 612 mm (6x2) The weight of the crane on the front axle (KA) will then be: KA = K . the following proportion of the total weight (K) of the crane is supported by the rear axle (KB) KB = 2500 x 7400 = 3550 kg (4600+612) 12 © Scania CV AB 2003 . KA if: PA = 6500 kg TA = 5000 kg KA = 1130 kg (as example 3) this gives the following maximum permitted load on the rear axle. Applying the lever principle gives the following equation: H = FB x A L The maximum permitted load on the rear axle (FB) is calculated by subtracting the truck’s chassis weight rear (TB) and crane weight rear (KB) from the maximum permitted rear axle weight (PB).: FA = PA .Bodywork calculations 3 Example 5 4X2 Calculating length Calculation of body length.KB if: PB = 10000 kg TB = 1780 kg KB = 364 kg (as example 3).4260 . L = FA + FB The maximum permitted load on the front axle (FA) is calculated in the same manner as the maximum permitted load on the rear axle (FB) i.TA .e.1780 .364 =7856 kg this gives the following maximum permitted load (L): FA = 7500 .TB .1586 = 1654 kg L = 1654 + 7856 = 9510 kg © Scania CV AB 2003 13 . Same truck and equipment as in example 3. The maximum permitted load (L) is calculated by adding the maximum permitted load on the front axle (FA) to that on the rear axle (FB). FB = 10000 . FB = PB . 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. i. The distance between the front axle and the centre of gravity of the platform + load is thus as follows: H = 7856 x 4300 = 3552 mm 9510 i.e. if such a change is more optimal from a weight distribution perspective. X/2 cannot be less than: X/2 = H . as in this example. The calculation program also makes it possible to gain some load capacity by choosing a front or rear axle with lower permissible weight capacity. X/2 is: X/2 = 3552 .3552 = 748 mm (Y) in front of the rear axle to optimally utilise the maximum permissible axle weight.D Rear overhang (J) may be calculated as follows: J=D+X-A J = 1352 + 4400 . If the centre of gravity of the platform + load is assumed to be located at the mid-point of the platform. wheelbase (A) = 4300 mm. The maximum length of the platform from the centre of gravity and forward is limited by the crane and its base. if D = 1352 mm. the platform + the truck’s centre of gravity must be located 3552 mm (H) behind the font axle or 4300 . 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.e.Bodywork calculations 3 In example 3. check the national regulations.4300 = 1452 mm Comments: In this example we have calculated backwards by establishing D through the use of SCANIA’s calculation program.1352 = 2200 mm Platform length is then: X = X/2 + X/2 X = 4400 mm 14 © Scania CV AB 2003 . However. In practice this has little or no significance. platform length may then be calculated as below. distance (D). Applying the lever principle gives the following equation: H = FB x (A + B) L H = Distance between front axle and centre of gravity of load for maximum axle weight utilization. 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 = 12000 x (5000 + 553) = 4595 mm 14500 If the body in this example is 8000 mm and the distance between the front axle and body is 650 mm. FB = Maximum permitted load on bogie A = Wheelbase B = Distance to centre of gravity of bogie L = Maximum permitted load including bodywork if: FB A B L = = = = 12000 kg 5000 mm 553 mm (6 x 2) 14500 kg this gives the following position of centre of gravity for maximum axle weight.Bodywork calculations 3 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.H E = 650 + 4000 . For calculation of (L) and (FB). © Scania CV AB 2003 15 .4595 = 55 mm Check with national regulations that this distance (E) is within the limits given. see earlier example. Applying the lever principle gives the following equation: H = FB x (A . permitted load (kingpin) on bogie A = Wheelbase B = Distance to centre of gravity of bogie L = Max. in order to achieve optimum use of front axle weight and bogie weight. with the tag axle in front of the driven axle. By subtracting truck chassis weight at the rear (TB) from the maximum permitted bogie weight (PB).Bodywork calculations 3 Example 7 6x2/4 Tractor Calculation of the position of the fifth wheel (H) on a threeaxle truck. maximum permitted load on the bogie (FB) can be calculated.B) L if: H = Distance between front axle and fifth wheel FB = 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.TB if: PB = max 20000 kg TB = 5000 kg This gives the following distribution of load on the bogie (FB) FB = 20000 . FB = PB .5000 = 15000 kg 16 © Scania CV AB 2003 . 675) = 3022 mm 17000 This means that in order to optimize the use of axle weight. © Scania CV AB 2003 17 .5000 = 2000 kg L = 15000 + 2000 = 17000 kg if: A = 4100 mm B = 675 mm this gives the following optimum location of the fifth wheel H = 15000 x (4100 .TA if: PA = 7000 kg TA = 5000 kg this gives the following maximum permitted load (L): FA = 7000 . permitted load on the bogie (FB) and the maximum permitted load on the front axle (FA) L = FB + FA The maximum permitted load on the front axle (FA) is calculated in the same manner as the max.Bodywork calculations 3 Maximum permitted load (L) is calculated by adding the max. permitted load on the bogie (FB) as follows: FA = PA . the fifth wheel should be located 3022 mm from the front axle. Target.Bodywork calculations 3 Example 8 8x4 Centre of gravity calculation To calculate: Dimension (E). · The chosen length of the bodywork (X) may be a dimension that the bodywork builder has chosen as standard. 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. FA = 14 000 FB = 18 000 F tot = 32 000 .5 mm = 970 mm = 650 mm = 14 000 kg = 18 000 kg = 23 305 kg = 7 000 mm Weight front Weight rear Weight tot. the distance between the centre of gravity of the platform and the optimum centre of gravity of the platform/load (H). From a stability perspective a shorter wheelbase would be preferable.970 = 4 707. Deviating from this standard may mean paying a much higher price. Also check national regulations to make sure that the distance (E) is within permissible limits. 18 © Scania CV AB 2003 .8 695 L = 23 305 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 PB = Load + bodywork load on Remarks: · The distance (D) may be a minimum dimension. 5000 mm.5 mm H= AT x PB L = 4707.5 .5 x 15170 23305 = 3064 mm E = X/2 + D – C – H = 3500 + 650 – 970 – 3064 = 116 mm Dimension (E). · The chosen wheelbase (A). if a front cylinder is to fitted between the cab and bodywork.g. But some countries require an even longer wheelbase to make it possible to load the vehicle to the maximum.5 865 PB = 8 135 . laden truck Chassis weight Load + bodywork A B C D FA FB L X = 5000 mm = 677. e. is very long for a tipper truck but for the sake of the calculation has no significance.2 830 PB = 15 170 . Calculation AT = A + B – C = 5000 + 677. D is as follows: D = H – X/2 = 4 131-3 350 = 1031 mm The distance between the front axle and bodywork is: D = 1031 mm och E = 0. Target. H should thus be equal to D + X/2 and E should be equal to 0. FA = 7 100 FB = 24 000 F tot = 31 100 .4 870 . © Scania CV AB 2003 19 .Bodywork calculations 3 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.9 455 PA = 2 230 PB = 19 415 L = 21 645 A = 3350 mm B = 1256 mm FA = 7100 kg FB = 24000 kg L = 21645 kg X = 6200 mm AT = 4606 mm (as on main dimen sion drawing) PB = 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: H= AT x PB = L 4606 x 19415 21645 = 4131 mm 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.4 585 . laden truck Chassis weight Load + bodywork Weight front Weight rear Weight tot.