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Equilibrium of Non-Concurrent Force Systems in Living with the Lab

This text discusses the equilibrium of non-concurrent force systems, specifically the weight of a bridge deck on a beam and the support reactions involved. It includes step-by-step instructions and examples of solving for unknown forces.

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Equilibrium of Non-Concurrent Force Systems in Living with the Lab

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  1. living with the lab Equilibrium of Non-Concurrent Force Systems weight of bridge deck on beam BEAM supportreactions supportreactions

  2. living with the lab Concurrent and Non-Concurrent Force Systems non-concurrent force systems concurrent force systems lines of action of forces do not intersect at single point lines of action of forces intersect at single point y FA A C D B x W FB 45° C B A y Ax FB FA D qA qB Ay x W = 100 lb

  3. living with the lab STEPS: Choose bodies to include on FBD Draw the body of interest Show loads exerted by interacting bodies; name the loads pinned joint Define a coordinate system resists motion in x and y directions Free Body Diagrams Label distances and angles 3000 lbs A C D B resists motion in y direction Assume center of mass of car is half way between the front and rear wheels C=1500 lbs B=1500 lbs roller support y C B Ax x A D 20 ft 8 ft 12 ft Ay D

  4. living with the lab Solve for Unknown Forces B=1500 lbs C=1500 lbs y B C Ax x A D 20 ft 8 ft 12 ft Ay D Strategically choosing the order in which the three equilibrium equations are applied can make the problem easier to solve. When solving for the “reaction forces” on a beam like this, it is usually easiest to sum moments about the point where there are the most unknowns. Point A has two unknowns, so let’s begin by summing moments about that point. one equation one unknown Now we can sum forces in x and y. The order doesn’t matter in this case. + Should Ay be larger than D? Why? Think critically to evaluate the solution.

  5. living with the lab Free Body Diagram Tip Sometimes it’s helpful to strategically align your coordinate system; here, the coordinate system is aligned with the beam. In this case, it really doesn’t make that much difference in solution difficulty (compared with a horizontal / vertical alignment), but it may be a little easier since the distances used in moment calculations are clearly labeled on the beam. A C D B 20° B C 20° 20° A Ax x D y Ay 20 ft 8 ft 12 ft

  6. living with the lab Class Problem • A man who weighs 890 N stands on the end of a diving board as he plans his dive. • Draw a FBD of the beam. • Sum forces in the x‐direction to find Ax • Sum moments about point A to find the reaction at B (By). • Sum forces in the y‐direction to find Ay. • Assumptions: • • Ignore dynamic effects. • • Ignore deflection (bending) of the diving board. • • Assume the weight of the man can be lumped exactly 3m horizontally from point B.

  7. living with the lab Class Problem • A stunt motorcycle driver rides a wheelie across a bridge. The combined weight of the • rider and the motorcycle is 2.45 kN (about 550 lbs). • Draw a FBD of the beam for x = 2 m. • Determine the reactions at A and C for x = 2 m. • Derive an equation for the reactions at A and B as a function of x. • Enter the equation from (c) into Mathcad, and plot Ay and Cy versus x on the same plot. • Assumptions: • The tire will have frictional forces with the road that could lead to a non‐zero value for Ax. • Ignore these forces when computing reactions. • Ignore dynamic effects (bumps, bouncing, change in motorcycle angle, . . . )

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