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Published byTerrence Daisley Modified about 1 year ago

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Use spacebar or mouse to advance slides. Problem 4.1 Applied Loads & Reactions Loading up that pickup truck!

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A 1400-kg pickup truck is loaded with two crates, each having a mass of 350 kg. Determine the reactions at each of the two rear wheels and front wheels.

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Consider just the pickup truck.

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Its weight is concentrated at its center of gravity. 1.8 m1.2 m AB 0.75 m

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This weight is distributed over the two axles, at R A and R B. 1.8 m1.2 m AB RARA RBRB

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SUGGESTION: Since the loads are given in kilograms with even numbers, it might be best to calculate with kg and convert to Newtons at the end.

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Analyze as though a simple beam. F = 0 = kg + RA RA + RBRB MB MB = 1400(1.2 m) - R A (3.0 m) RA RA = 560 kg x 9.81m/sec 2 and RB RB = 840 kg x 9.81m/sec m1.2 m AB 0.75 m RARA RBRB

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The reactions at the wheels are: R A = 5.49 kN R B = 8.24 kN 1.8 m1.2 m AB 0.75 m RARA RBRB

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Now let’s add some crates to the bed of the truck and determine the reactions at the wheels. 1.8 m1.2 m AB RARA RBRB

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Each crate has a mass of 350 kg and is positioned as shown. 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD

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Repeat the same calculations with the added loads as shown. 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD

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F= 0 = RA RA + RBRB 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD 350 kg 1400 kg 350 kg

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M B = 1400(1.2 m) - R A (3.0 m) + 350(2.05 m) + 350(3.75 m) 1.8 m1.2 m AB 0.75 m RARA RBRB 1.7 m2.8 m CD 350 kg 1400 kg 350 kg 2.05 m

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M B = 1400(1.2 m) - R A (3.0 m) + 350(2.05 m) + 350(3.75 m) 1, ,312.5 = R A (3.0 m) 3,7100 kg-m = R A (3.0 m) 3,7100 kg-m = R A (3.0 m) R A = 1, kg x 9.81 m/sec 2 R A = kN

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F= 0 = RA RA + RBRB R B = RARA RB RB = 2100 kg kg RB RB = kg x 9.81 m/sec 2 RB RB = 8.47 kN AB R A = kNR B = 8.47 kN 350 kg 1400 kg 350 kg

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Final results are shown. Each rear wheel = 6.07 kN Each front wheel = 4.24 kN AB R A = kNR B = 8.47 kN 350 kg 1400 kg

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