Ch 4. Moments/Equivalent Systems Multimedia Engineering Statics Moment2-D Scalar Moment3-D Scalar Moment3-D Vector Couples and Equiv. System Distributed Loads, Intro
 Chapter 1. Basics 2. Vectors 3. Forces 4. Moments 5. Rigid Bodies 6. Structures 7. Centroids/Inertia 8. Internal Loads 9. Friction 10. Work & Energy Appendix Basic Math Units Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll

 STATICS - EXAMPLE Example New Tri-Axle Dump Truck Sand Load A new 10 kN tri-axle truck is designed to carry sand but the design engineer notes that if it is loaded incorrectly the truck can tip. Thus, it is necessary for the truck to be stable even with both of the rear wheels on 3rd axle go flat. The center of gravity (cg) of just the truck is located at 0.5 m from the front edge of the sand. The distributed load of the sand can be modeled as a third order equation,       -0.17 x3 + 0.38 x2 + 0.77 x + h  kN/m. The center of gravity (cg) of just the sand is not known. What will be the maximum allowable h before the load will be unsafe when the back rear wheels are flat? Solution When both rear wheels become flat, there will be no vertical force acting on them and the truck may become unstable. Additionally, when the truck starts tipping, the force on the front wheel will go to zero because at that time the moment about the middle wheels wants to rotate the truck. The model can be simplified as a beam with a pinned support at the location of the 2nd axle and two forces (weight of the truck and the weight of the sand pile). Notice the back rear wheels and front wheels do not exert a reaction force. Previously, it was determined that the resultant force FS for a distributed load over a line is,       It was also shown that the location of the line of action of FS is given by       The sand pile is distributed from 0 to 3 meters. The resulting point force is,      Also the location of this force can be found by      The moment of this force about the rear wheel should cancel with the moment of the weight      ΣM = 0       The equation can be simplified to      1.5 h = 2.079 This gives a maximum value of h as 1.386 m.

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