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**Problem 4.1 Applied Loads & Reactions**

Loading up that pickup truck! Use spacebar or mouse to advance slides.

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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.**

B 1.8 m 1.2 m 0.75 m

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**This weight is distributed over the two axles, at RA and RB.**

1.8 m 1.2 m 0.75 m RB

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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**

Analyze as though a simple beam. F = 0 = kg + RA + RB MB = 1400(1.2 m) - RA(3.0 m) RA = 560 kg x 9.81m/sec2 and RB = 840 kg x 9.81m/sec2 A B RA 1.8 m 1.2 m 0.75 m RB

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**The reactions at the wheels are: RA = 5.49 kN RB = 8.24 kN**

1.8 m 1.2 m 0.75 m RB

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**Now let’s add some crates to the bed of the truck and determine the reactions at the wheels.**

RB

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**Each crate has a mass of 350 kg and is positioned as shown.**

B RA 1.8 m 1.2 m 0.75 m RB

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**Repeat the same calculations with the added loads as shown.**

B RA 1.8 m 1.2 m 0.75 m RB

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**F= 0 = -350 -350 -1400 + RA + RB RA RB 1.7 m 2.8 m C D 350 kg 350 kg**

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**MB = 1400(1.2 m) - RA(3.0 m) + 350(2.05 m) + 350(3.75 m)**

C D 350 kg 350 kg 1400 kg A B RA 1.8 m 1.2 m 0.75 m RB

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**MB = 1400(1.2 m) - RA(3.0 m) + 350(2.05 m) + 350(3.75 m)**

1, , = RA(3.0 m) 3,7100 kg-m = RA(3.0 m) 3,7100 kg-m = RA (3.0 m) RA = 1, kg x 9.81 m/sec2 RA = kN

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**F= 0 = -350 -350 -1400 + RA + RB. RB = 350 + 350 + 1400 - RA**

F= 0 = RA + RB RB = RA RB = 2100 kg kg RB = kg x 9.81 m/sec2 RB = 8.47 kN 350 kg 350 kg 1400 kg A B RA = kN RB = kN

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**Final results are shown. Each rear wheel = 6**

Final results are shown. Each rear wheel = 6.07 kN Each front wheel = 4.24 kN 350 kg 350 kg 1400 kg A B RA = kN RB = kN

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