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Folding Trailer Assignment 1

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Folding Trailer Assignment 1
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    6. Design calculations: 6.1 Load Capacity: The maximum load  for which the trailer is designed to carry will mostly determine the strength of the trailer. It will also determine how strong the trailer should be. If a trailer is designed for 800Kg it will obviously have much lighter construction than a trailer designed for 1000Kg, but they should perform equally well at their rated capacity. The maximum load capacity which is applied to the EX2000  is 500Kg . For obtaining maximum strength all the side rails and cross members are welded  together rather than using bolt and nuts. 6.2 Axle location: The axle should be located behind the load center for stability . Often trailers have the axle at or just barely behind the center of the bed. The load will be heaviest at the front. But it is also an indication of weakness. For most applications, the axle should be noticeably behind the center of the bed, and for maximum versatility, allow the axle position to be adjusted 32 . In design of EX2000  the axle is located just behind the centre  of the trailer.    6.3 Load distribution: It is the most common cause of stability concerns. The trailer load should be reasonably distributed from side to side with at least 10%  of the trailer weight on the hitch. With typical trailers, more weight on the hitch is better for stability. 15%  is usually a good number  33 . CORRECT WAY TO LOAD THE TRAILER 6.4 Calculations: The folding trailer chassis is constructed of 2”H x 3”H galvanized steel with C channel shape with thickness of 3mm. The cross section of the rail channel is shown following picture:    6.4.1 Calculation of second moment of area (I): I = 12 33 bd12BD   = 120.0702x 0.0478120.0762x 0.0508 33   = -6-6 10x 1.378010x 1.8730   = 7 10x.95 4 mm  The diagram of the load distribution, acting and reacting forces on folding trailer and tires are shown in following picture: The diagram shows a uniformly distributed load 4910 N/m across the whole span. The total load carried = 4910 N/m x 1m = 4910 N      The bending moment M  is found by treating the distributed load acting at its centre of gravity. M = (2455 x 0.5)  –  [(4910 x 0.5) x 0.25] = 613.75 Nm 6.4.2 Calculation of bending stress 34 : = I My    =   7- 10x 4.950.0381x 613.75   = 47.2 MN/m 2   6.4.3 Calculation of Factors of Safety 35 : Ultimate Strength  for Steel is 1860 x 10   N/m 2 . Factors of Safety = )) 22 mN(StressWorkingmN(StrengthMaterial  = 66 10x 47.210x 1860  = 39.4   Result:  As factor of safety is bigger than 1  it will not break and will withstand the load. 6.4.4 Calculation of Double Shear Stress:   F = 24910   = 2455 N D  is the diameter of suspension unit connecting end to hub which is 25.4mm. = 2AF   = 2 0.0127x x 2 2455     = 2423731 2 mN   Factors of safety = 242373110x 1860 6  = 767.41 which is bigger than 1 therefore it will not break.
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