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

2. Concurrent and Non-Concurrent Force Systems living with the lab FBFB FAFA x y W 45° x y BB AA FBFB FAFA W = 100 lb A B C D A D AyAy AxAx B C concurrent force systems non-concurrent force systems lines of action of forces intersect at single point lines of action of forces do not intersect at single point 2

3. Free Body Diagrams living with the lab A B CD x y 2.Draw the body of interest 3.Show loads exerted by interacting bodies; name the loads 4.Define a coordinate system 5.Label distances and angles 12 ft 8 ft 20 ft 3000 lbs B B=1500 lbs C C=1500 lbs AyAy D D A AxAx pinned joint resists motion in x and y directions roller support resists motion in y direction 1.Choose bodies to include on FBD STEPS: Assume center of mass of car is half way between the front and rear wheels 3

4. living with the lab Solve for Unknown Forces Strategically choosing the order in which the three equilibrium equations are applied can make the problem easier to solve. x y 12 ft 8 ft 20 ft B B=1500 lbs C C=1500 lbs AyAy D D A AxAx + Now we can sum forces in x and y. The order doesn’t matter in this case. 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 Should A y be larger than D? Why? Think critically to evaluate the solution. 4

5. 5 living with the lab A D AyAy AxAx B C 20° x y A B C D Free Body Diagram Tip 12 ft 8 ft 20 ft 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.

6. 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. a.Draw a FBD of the beam. b.Sum forces in the x‐direction to find Ax c.Sum moments about point A to find the reaction at B (By). d.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. 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). a.Draw a FBD of the beam for x = 2 m. b.Determine the reactions at A and C for x = 2 m. c.Derive an equation for the reactions at A and B as a function of x. d.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,... )